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Asymmetry between space and time in special relativity Consider 2D spacetime with two inertial reference frame S and $S'$, where $S'$ is moving in the $S$ positive spatial direction at velocity $vt$, along with the usual graphical representation with $t$ on the vertical axis and $x$ on the horizontal axis. Use units such that $c = 1$. Consider the space-time displacement vector which points forward in time in the coordinates of $S'$ at $(0,0)$. I.e., it is displacement vector that connects the clock at point $(0,0)$ in $S'$ to the clock at point $(1,0)$ in $S'$. Is it true that it follows that the clock at $(1,0)$ in $S'$ is at $(t,vt)$ in $S$, for some value of $t$? If so, does it follow that viewing the time-axis of $S'$ in the $S$ coordinates, it is rotated to the right from vertical by an angle $\theta$ such that $v = \tan(\theta)$? Is there any analogous statement that can be made about the direction in $S$ of the $S'$ space-displacement vector, i.e. the vector with components $(0,1)$ in $S'$? I can't come up with one and it seems that the answer has to be "no" because I haven't yet used anything about Lorentz invariance of the interval and the rotation of the space and time axes is equivalent to Lorentz invariance. But I would not have expected an asymmetry here between time and space.

If you choose to draw the frames so that the $S$ axes are horizontal and vertical on your paper, then they look like this (with the $S$ frame in black and the $S'$ frame in blue): The marked points are $(1,0)$ and $(0,1)$ in the $S'$ frame.

Newtonian vs Lagrangian symmetry Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as: $$ ma = -mg $$ From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian: $$ L = \frac{mv^2}{2} - mgx $$ because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see). Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?

* *The (infinitesimal) translation $$\delta x~=~\epsilon$$ changes OP's Lagrangian with a total time-derivative $$\delta L~=~mg \epsilon~=~ \frac{d}{dt}(mg \epsilon t).$$ This is known as a quasi-symmetry. Noether's theorem does also hold for quasi-symmetries. *Concerning symmetries of action vs. EOM, see also e.g. this related Phys.SE post.

Contradiction of Units in Polytropic Process A polytropic process is a process that obeys the relation $pV^n=C.$ However, when I try to solve a problem involving this relationship with, for example, $n = 1.5$ my use of units in my equations breaks down. At some point, I find myself having to solve for the constant $C$ to evaluate the integral which in the case of $n = 1.5$ gives the constant $C$ with the units $\mathrm{\frac{kg\cdot m^{3.5}}{s^2}}$ and does not match up with the expected units for energy. (Assuming units of $\mathrm{m^3}$ for $V$ and units of $\mathrm{\frac{kg}{m\cdot s^2}}$ for $p$.) Usually I drop the units entirely at this point and arrive at the correct answer regardless, but this does not feel like a rigorous solution. How would I avoid this contradiction in my dimensional analysis, and why does it exist?

The product $pV$ (where $p$ is pressure and $V$ is volume, each with their standard units in your favorite system) has units of energy. As noted in the comments, the product $pV^n$ when $n\neq 1$ does not. Nor should you expect it to. We're no longer dealing with energy but with an invariant in certain adiabatic processes, for example. I don't recommend ever dropping units entirely, as this will catch you up someday. Instead, consider replacing the constant $C$ with $p_0V_0^n$, where $p_0$ and $V_0$ describe a reference state (which could be a final state) measured in the same units.

Why is $J^P=0$ for all even-even nuclei? I have the following question, for which I have answered fully, but I am questioning the logic behind finding the value of $J^P$ for ${}^{20}_{10}\mathrm{Ne}$: N.B, when I wrote "Number per level" in the picture above this is really just the degeneracy, $2j+1$. Now to answer the question at the bottom for ${}^{7}_{3}\mathrm{Li}$, ($Z=3\,$ & $N=4$) since there is one unpaired proton, it will go in the $1p_{3/2}$ level and since the subscript is the value of $J$, then $$J^{P}=\frac32^{-}$$ where I used a minus sign in the superscript since the parity $P=(-1)^{\ell}$ and $\ell=1$ for a $p$ state. Onto ${}^{29}_{14}\mathrm{Si}$, ($Z=14\,$ & $N=15$) there is only one unpaired neutron, using the table I added this will go in the $2s_{1/2}$ level and therefore $$J^P={\frac12}^{+}$$ now, with a positive sign since $\ell=0$ for a $s$ state ($P=(-1)^{0}=1$). I left the ${}^{20}_{10}\mathrm{Ne}$, ($Z=10\,$ & $N=10$) to the end on purpose as although I note that all nucleons are 'paired off', looking at the table the $10$th proton (or neutron), ought to go in the $1d_{5/2}$ level and hence $$J^P={\frac52}^{+}$$ parity is $+1$ here as $\ell=2$. But this is incorrect and the correct answer is actually $$J^P=0$$ But, from the logic I just showed this does not make sense. ${}^{16}_{\,8}\mathrm{O}$ is another example when $N=Z$ (an even-even nucleus), with $J^P=0$. Am I just to accept that for some (unknown) reason $J^P=0$ for all even-even nuclei? So in summary; why can't I just use the table (I inserted in the picture above) to determine the value of $J^P$?

I'll answer by asking you some questions: For $^7_3$Li, why did you ignore the 4 neutrons? Why did you ignore the two $1p_{3/2}$ neutrons? You mentioned that the third proton was "unpaired." For $^{29}_{14}$Si, why did you ignore the $14$ protons and their contribution to spin and parity? Again, you mentioned there was an "unpaired" neutron. Maybe you should consider the idea of pairing, in the ground state, to be balancing the angular momentum contribution of nucleons, so that paired nucleons contribute zero, and the parity is even, regardless of the particle state. The basic rule for ground state nuclei is that pairs of neutrons or pairs of protons occupy the same $\ell,j$ state with opposite $m_j$ values and contribute 0 to the total J. You should have said "there are no unpaired nucleons for $^{20}_{10}$Ne, so the ground state angular momentum will be zero."

Superinsulator infinite voltage Everybody knows about superconductivity, at cryogenic temperatures conductor resistance drops to zero for direct current. There is lesser known related phenomena called superinsulator where material gains infinite resistance. Does it really block infinite high voltage? That seems to me too extreme to be true. Even the strongest insulators like telfon or fused quartz eventually breakdown under high voltage, so can these superinsulators really block infinite voltage?

To quote Wikipedia article on superinsulator: The superinsulating state is the exact dual to the superconducting state and can be destroyed by increasing the temperature and applying an external magnetic field and voltage. The reason for calling them superinsulators is then not because it cannot conduct at infinitely high voltage. Rather the definition of insulator here is as the material in which the excitations are localized, as opposed to a metal - a material where the excitations are extended. The two are dual, in the sense that they represent two thermodynamic phases separated by the metal-insulator transition. While in normal state the conducting excitations are electrons (more precisely the Landau quasiparticles), in the superconducting state these are Cooper pairs. It is thus natural to add prefix super- to an insulating material whose excitations are Cooper pairs.

For capacitors, why does the dielectric need to be inserted at a small constant speed? Likely a very silly question. I am aware that there is an attractive force from a charged parallel-plate capacitor in an open circuit without battery pulling the dielectric in, and that a decrease in energy will result upon insertion of the dielectric. I am told that this difference in energy stems from the person's retarding force against the attractive force, but my question is why must this retarding force be applied? Why can't one just let the capacitor pull it in on its own? What would the energy transfers be in that case? Thanks in advance!

If it pulls it on his own, and you disregard friction it will come out on the other side, and so on and oscillate.

What does the absolute value of an operator mean? In the text, we are given a Hamiltonian for two bodies with a potential energy of interaction that depends on the magnitude of the distance between them as: $$\hat{H}=\frac{\hat{\mathbf{p}}_{1}^{2}}{2 m_{1}}+\frac{\hat{\mathbf{p}}_{2}^{2}}{2 m_{2}}+V\left(\left|\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2}\right|\right).$$ What does the absolute value of an operator mean and how does it operate and why?

Consider any quantum mechanical observable $\hat O$ so that $$\hat O|\psi⟩=o|\psi⟩$$ with eigenvalue $o$. Then for any function $f$ of the observable $\hat O$ we have, $$f(\hat O) \mid\psi\rangle = f(o) \mid\psi\rangle$$ The same applies for a function of the absolute value operator. So if we consider $V(\mid\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2}|)$ and the two-particle state $\mid \psi(\mathbf{r_1}, \mathbf{r_2})\rangle$ then $$V(\mid\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2}|)\mid \psi(\mathbf{r_1}, \mathbf{r_2})\rangle=V(\mid{\mathbf{r}}_{1}-{\mathbf{r}}_{2}\mid)\mid \psi(\mathbf{r_1}, \mathbf{r_2})\rangle$$ where the potential on the RHS is now a function of the eigenvalue $(\mid{\mathbf{r}}_{1}-{\mathbf{r}}_{2}\mid)$.

How long ago was the Universe small enough for interstellar travel? Currently, even the nearest stars are lightyears away, and impossible to reach in our lifetimes. If space is always expanding, and was once infinitely smaller, then at what point in the past was space so much smaller that the average distance between stars was less than light days? Was there ever such a time?

As the universe expands each individual galaxy stays roughly the same size, with stars on orbits of roughly constant diameter, so the stars within any given galaxy were no closer together a long time ago than they are now (at least as far as cosmic expansion effects are concerned). The distances between galaxy clusters were smaller in the past, and a good way to get a sense of this is to note that the ratio of distance between them now to distance between them a long time ago is equal to the ratio of wavelengths in the light received and emitted. If we receive light from a galaxy and the light arriving has a wavelength twice as large as when it set out, then the universe was half as small when the light set out (that is, distances between galaxy clusters were then on average half what they now are). To find a time when galaxies were not many lightyears apart you have to go so far back that you arrive at times before the formation of galaxies, so there never was such a time. [Added remark in answer to a question in the comments concerning galaxy clusters. One galaxy cluster drifts away from another because the initial conditions gave them velocities of this form. This general condition is called the "Hubble flow" and it leads to the cosmic expansion. It is what things would do if they only experienced the average cosmic gravitation, without any local bumps owing to a non-homogeneous matter distribution such as a galaxy. Meanwhile everything attracts stuff near to it and this can lead to bound groups such as solar systems, galaxies and galaxy clusters. This binding is sufficient to turn the relative velocities around so that each bound group does not drift apart, nor does it expand (unless some other process intervenes).]

Can an electron near thar the positive terminal of a battery make it all the way to the positive terminal? Following Drude's model of conductivity, can an electron near thar the positive terminal of a battery make it all the way to the positive terminal? The electron is in the wire but near the positive terminal. At least, is it probable? Assuming that the circuit size is commonly small. The reason I asked this is that I do now know how to interpret the average velocity of an electron in this model. Because I know it bounces off impurities until reaching a stationary average velocity. Can I use this average velocity to estimate a time to go from one side to the other using the formula of uniform rectilinear motion?

This question has no answer as electrons are indistinguishable particles.

When a person pulls or pushes a cart, why is it advantageous for their body be tilted forward? This is not a homework question. I attempted to draw a free body diagram for a person pulling or pushing a cart. Based on Newton's third law, the following forces act on the body of the person: * *forward reaction force done by the ground because of friction between the person and the ground. *downward force (the person's weight) done by the earth. *backward reaction force done by the cart. I am wondering why the body of the person must be tilted forward. I have not seen any relationship between this posture with the magnitude of the forces acting on the person. Could you tell me why the person's body must be tilted forward? How does this posture provide mechanical advantages?

It adds a horizontal component to the vertical force exerted by your legs. When you're standing straight up, the forces exerted by your legs are straight up and down. Your legs are designed to exert a force to counter your bodyweight and allow you to stand like this, so they're quite strong at countering this force. When you're leaning over like this, each step you take restores you a leaning version of standing straight position, and it allows you to exert this standing-up force your legs exert in a horizontal direction - in each of these pictures, you can see the person with one leg straight back, and the other bent in front of it; at the completion of each step, you could consider them to be standing with both legs straight back for a brief moment before they move the next leg.

When a car accelerates relative to earth, why can't we say earth accelerates relative to car? When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not? Please explain in simple terms.

To be pedantically correct, one should perhaps say that the friction of the wheels on the ground serves to both accelerate the car by a macroscopic amount in one direction, and accelerate the combined center of mass of everything else on Earth by a microscopic (or maybe "femtoscopic") amount in the opposite direction. On the other hand, many factors are causing all locations on the surface of Earth to be accelerated in unpredictable directions by unpredictable directions, and these are so many orders of magnitude larger than the acceleration resulting from the friction of the car's tires on the road as to render the latter meaningless. If a 1,000kg car were to accelerate at 10m/s², the acceleration of the Earth would be about 0.0000000000000016 μm/s². If instead of looking at a car on the Earth one were to instead consider a 1,000kg car driving around the deck of a 10,000,000kg ferry floating in calm water with no wind, then a 10m/s² acceleration around the deck would cause a roughly 1mm/s² acceleration of the ferry. While it would probably be unusual for the ferry and the water to be sufficiently still that acceleration of the ferry due to friction with the car's tires on the deck would be more significant than acceleration due to waves or friction with the water, the acceleration would be within bounds of what commonplace equipment could measure. On the other hand, the acceleration of the ferry in that scenario is about 592,000,000,000,000,000 times as large as the acceleration of the planet.

Calculating uncertainty from significant figures of a value A question in Giancoli's Physics for Scientists and Engineers (2. ed) has me confused. Here it is (Ch 1, Problem 3): What is the area, and its approximate uncertainty, of a circle of radius $2.7 \times 10^4$ cm? I got the correct answer of $2.3 \times 10^9 \text{ cm}^2$, but the uncertainty provided in the answer was $0.2 \times 10^9 \text{ cm}^2$. How was this uncertainty calculated? As far as I can tell, it is not possible to determine the uncertainty of a measure just from its value, because I have no idea with what device/technology the radius was measured. All we can know is that the doubtful figure is in the $10^3$ position, and therefore the uncertainty in the area will be $x \times 10^9$, but $x$ could be anything? Why is $x = 0.2$?

Most likely, the authors assume Gauss error propagation in which the error on a function $f(x)$ of a variable $x$ is calculated as $$ \Delta f = \frac{\partial f}{\partial x} \Delta x~.$$ In your case, $f(x) = \pi x^2$ and $\frac{\partial f}{\partial x} = 2\pi x$, and $\Delta x = 0.1\times 10^4$cm (this is the worst case - realistically, we could also assume $\Delta x = 0.05\times 10^4$cm). Taking $\Delta x = 0.1\times 10^4$cm leads to $$\Delta f = 2\pi x \Delta x = 2\pi \times 2.7\times 10^4 \text{cm} \times 0.1\times 10^4\text{cm} = 0.1696 \times 10^9\text{cm}\simeq 0.2 \times 10^9\text{cm}~.$$

Relation between potential energy and conservative force Does potential energy only happen when the work done is by a conservative force? Or does work done by non-conservative forces also create potential energy?

Forces can be conservative or non-conservative. But conservative forces do work where this work is equal to the change in potential energy. Conservative forces are also characterized by the fact that the work done by the force that moves an object from one point to another is independent of the path taken between these points (and the total work done will be zero when the path forms a closed loop). However, a non-conservative force is one where the work done will indeed depend on the path. A good example of a non-conservative force is friction. The work done against a frictional force will depend on the length of the path between the two points, and due to this path dependence, there will be no potential energy we can associate with this force, and indeed the same is true for all non-conservative forces. Non-conservative forces will either add or remove mechanical energy from a system. Friction, energy dissipation in the form of heat, removes energy from a system which cannot be fully converted back to work.

Are galaxies and clusters of galaxies much denser when we observe the 'past' far universe than the 'present' closer part of the universe? Are galaxies and clusters of galaxies much denser when we observe the 'past' far universe?If the universe is expanding, the galaxies should have been much closer in the past so should we see a larger distribution of images of galaxies from the past than these young closer galaxies and clusters of galaxies? If it were no space expansion should the density of galaxies on the picture be just proportional with the square of the distance from our position which should not be if the universe is expanding...

Clusters of galaxies generally become denser over time, because they accrete more nearby galaxies as time goes by. In addition, galaxies orbiting within a cluster will tend to lose energy to dynamical friction caused by their interaction with the dark matter in the cluster, so they will gradually sink towards the center of the cluster. Galaxies are a bit more complicated. They can become denser by accreting gas, which can concentrate in the center and form new stars, making the galaxy denser. They can also merge with other galaxies; if there is little or no gas in the two galaxies, then the energy of their orbital motion is converted into the kinetic energy of the stellar motions within the merged galaxy; the result is that the outer parts of the galaxy are more extended, and so the galaxy is, on large scales, less dense. It is important to understand that galaxies and galaxy clusters do not participate in the general expansion of the universe -- that applies to the space outside galaxy groups or clusters, or to the space outside isolated galaxies that are not in groups or clusters.

Why do we consider coherent waves interfere as if there is no distribution of energy as the wave progresses? In my textbooks, youtube videos, different websites, I have seen that during calculations regarding interference of waves we derive the equations in such a way as if there is no distribution of energy by waves., I mean if it is a circular wave then the diameter of locus wavefronts will increase witth the passage of time by indulging more and more particles and involving new particles has to be done by distributing energy(as far as I know). So shouldn't this cause change in amplitude as time passes? Moreover aren't each wavefronts subject to countless interferences causing it to distort largely from its initial form? An intuitive approach will be much more helpful. Thank you.

Normally we use coherent lasers or collimated beams of coherent atoms in interferometers. The spreading of the waves in the transverse directions relative to the propagation is taken to be negligible over the extent of the wave propagation. Even if this isn't so, the phase of the wave is independent of the amplitude, the subsequent interference caused by a phase difference only diminishing in amplitude with any spreading of the wave over space. It is not necessary to discuss the additional complexity of how the amplitude of a particular wave evolves to explain the cause of interference, a phase shift. This is probably the most likely reason why explanations of interference leave this out.

Clarification of labeling of Lorentz transformation I read a labeling of Lorentz matrix labeling $\Lambda^\mu_\nu$. To be more specific, it's used as $x^\mu\rightarrow \Lambda^\mu_\nu x^\nu$. I want to ask: * *What does the indices $\mu$, $\nu$ mean in $\Lambda^\mu_\nu$? *What does $x^\mu\rightarrow \Lambda^\mu_\nu x^\nu$ mean? Are $x^\mu$ and $x^\nu$ in the same frame?

* *Τhe indices $\,\mu,\nu\,$ take the values $\,0,1,2,3$. The value $\,0\,$ corresponds to the time coordinate $x^0=c\,t\,$ while the values $\,1,2,3\,$ correspond to the space coordinates $\:\:x^1=\mathrm x,\:\:\:x^2=\mathrm y\:$ and $x^3=\mathrm z$. *Equation $\,x^\mu\rightarrow \Lambda^\mu_\nu x^\nu\,$ is better to expressed as $\,x'^\mu=\Lambda^\mu_\nu x^\nu\,$ and is the Lorentz transformation of the space-time position 4-vector from an inertial frame $\,\rm S\,$ to another inertial frame $\,\rm S'\,$ moving with constant velocity 3-vector with respect to the unprimed one.

Normal force shift for ball kept in cavity Adapted from JEE advanced paper-1 of 2020 If you see the left side of the Planck which the ball touches, it seems so that as we vary $\theta$, the contribution of the force from that point of contact drops to zero. I have marked the point of interest in paint: As we reduce $\theta$ , we see both the edges touching the sphere contribute to supporting it's weight. But, as we increase $\theta$ it drops off. It is intuitive to understand for me, but I can't give a precise reason why as to this happens. Hence, the question: For what exact reason does the normal force shift to the right edge as we vary $\theta$?

To simplify first you can visualize what happens with the cavity when incline has an angle zero. The normal reaction on both the edges is same. Direction of normal reaction is vertical upwards opposite to gravity. When there is incline the sphere is actually also falling towards the right edge. More the θ, more the sphere falls towards the right edge. There will be two components of weight of sphere. One is mgcosθ, other is mgsinθ. mgcosθ is supported by normal components at both the edges. mgsinθ is supported just by the right edge. So on the right edge you can think of two normal forces - One stopping sphere from falling into cavity and other is just the plank surface stopping forward motion on incline. When θ is 90, sphere is actually not falling into cavity its just hitting the right edge plank

Does Earth's rotation affect the orbital velocity of its satellites? We all know about the spin of black holes. That spin and of course it's gravity determines the minimum distance a star can orbit the black hole with a min orbital velocity. Do we see similar result? Does spin of earth also effect the orbital velocity of its satellites?

The effect of black holes on prograde and retrograde orbits is due to frame dragging, which is a general relativistic effect. Hence the effect is microscopic on Earth satellites. The LAGEOS satellites tried to detect frame dragging around Earth, but did not succeeded directly. NASA has claimed success with GRACE, though. A more recent paper used LAGEOS and other satellites to find the effect. The drift due to the effect is about 30.68 milli-second of arc per year, so it is totally minuscule. The big source of noise in these measurements is tidal effects due to the lumpiness of Earth, which is also changing over time due to other bodies. That brings up a bigger, classical effect on satellites that does matter in practice. A rotating planet with a tidal bulge will pull and push on the orbiting satellite depending on how the orbital period fits with the planetary rotation. There is also a difference in how easy it is to capture natural satellites in prograde and retrograde orbits (classic paper). Retrograde orbits average the bulginess more, and are somewhat more stable, while prograde orbits tend to change more. For big enough satellites (or rather moons), the bulges they induce themselves matter. If they are above prograde geosynchronous orbit the bulge will tend to be ahead of them (Earth rotates faster than the moon orbits) and pull them ahead, moving them into a higher orbit and slowing Earth's rotation. The opposite would happen in a retrograde orbit. But this is all classical physics.

How do we know that forces "add" like vectors? As we all know the basic principle of vector addition. construct a parallelogram as shown, and the diagonal would be the resultant vector. As a mathematical method or formula, it is perfect to define the sum of two vectors in this way, because it is nothing but a binary operation. But when it comes to practical life, say an object is at $P$, and two force vector $A$ and $B$ is applied on the object. How would you know that the resultant force would act on the object in the direction of vector R? even if so, how do you even know its magnitude is going to be equal to the diagonal length of this parallelogram?

As far as I see, it is an experimental fact. Applying the forces on an object through springs, it is possible to know the modulus of the forces by the spring displacements. Knowing also the direction of them, the forces can be modelled as vectors. It happens that adding them using the parallelogram rule works either for a body at rest (where the vectorial sum of all forces must be zero), or for an accelerated one, where $\mathbf F = m\mathbf a$ holds.

Confusion on Normal force and its resolution on inclined plane I had the following doubt regarding normal forces and its resolution on inclined plane. (Also I am quite new to SE, so I have no idea how do we upload diagrams to better represent my arguement). Consider two cases, in both cases there is a inclined plane with some angle of inclination ($\theta$), on which there is a block of mass m that is kept. In Case I, we fix our coordinate axes such that the x-axis lies on the base of the inclined plane, and the y-axis perpendicular to x-axis, and in the next case the x-axis is placed along the inclined plane and the y-axis is perpendicular to the x-axis. The problem I was facing was that if we consider the normal force in case I and say that the normal force is $mg cos(\theta)$ then this is usually accompanied by the reasoning that there is no acceleration in the direction of normal force, therefore net force in that direction is 0, and hence acceleration is 0 as well. But in this coordinate plane,saying that acceleration normal to the inclined plane is 0, is just like saying that its rectangular components is also 0, because if we have a vector whose magnitude is 0, then its rectangular components must also have a 0 magnitude. So in one sense we will be saying that the horizontal component of the block's acceleration is also 0, which is not true. But in the second case, where we have a inclined coordinate system, we can easily make such a assumption that normal force is mgcos($\theta$) because in the inclined coordinate system, due to the normal force being perpendicular to the x-axis (which we placed on the inclined plane), it doesn't have any rectangular x-component and thus we can easily conclude the normal force to be mgcos($\theta$). I don't understand how different coordinate system can predict different normal forces? Where exactly am I going wrong?

The net force normal to the plane is 0 not $mg cos(\theta)$. The block moving on a one-dimensional plane has only one degree of freedom - along the plane. And the force along the plane is $mg sin(\theta)$

Are neutrinos' velocities related to their energies? As they are with other particles? I have read extensively about neutrino energies, including in the popular press, and yet velocities of the neutrinos themselves (not the detectors they 'run into' or the leptons they 'create' after interaction) are rarely mentioned.... How can a neutrino's speed be measured to begin with? How sure are experimenters and theorists about their calculations of neutrino energies, anyway?

Yes, the speed of neutrinos is related to their energy via the usual formula for relativistic energy: $E_{\mathrm{kinetic}}=E-E_0=\gamma m_0 c^2 - m_0 c^2$ Now, neutrinos have masses of electronvolts ($eV$) or less - we don't quite know yet. But their (kinetic) energies usually are many orders of magnitude more than that: radioactive decay or processes in stars gives energies in the 100keV-MeV ballpark, and at accelerators even GeV are possible. So the restmass of the neutrinos is utterly negligible, and one does not need to distinguish between relativistic, total, or kinetic energy. Their speed is thus simply the speed of light. Or, more precisely, for a 1MeV neutrino e.g. from the Sun, the $\gamma$-factor is $10^6$: $E_{\mathrm{kinetic}} / (m_0 c^2) = \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ Your pocket calculator can't even easily calculate that as this speed is so close to the speed of light, so use the relation $\frac{v}{c}=\beta=\sqrt{1-\frac{1}{\gamma^2}}\approx 1-\frac{1}{2\gamma^2}=1-5\times10^{13}$ to see that this neutrino travels at the speed of light to excellent precision.

Where the reaction force from Newton's third law is acting on this body? In the diagram below I have a fire extinguisher sitting upright on a skateboard. Gas is being expelled out of the fire extinguisher and causing the skateboard to move forward (which is to the right in this picture). This would work in a vacuum as well I am told. My understanding is that in Newton's third law when a body Y is pushing on a body Z then body Z is pushing back on body Y in the opposite direction but equal in magnitude. You often see big arrows like the white ones in my diagram that illustrate Newton's third law. In my example though, where is the force B actually occurring and what causes it? I guess I'm still confused. There is gas which pushes the gas in front of it on the way out of the fire extinguisher and in return the gas in front pushes back on the gas behind it which causes a net force like force B? That's the best explanation I can think of. So again I guess my question is: where is the force B actually occurring and what causes it?

When compressed air is forced to flow through a narrow tube like nozzle it is obvious that it will exert force on the nozzle and that force is marked here as B. Well the force will be torque hence will cast force at the upper part of the extinguisher, now if the extinguisher is attached to the skate board to move forward. Well now how actually is force B is created , the force is created by the sudden decrease in the pressure of the gas when released. When the gas releases sudden drop of pressure causes the gas to expand and that expansion of gas forces the extinguisher backward.

Direction of shear along an oblique plane under a compressive force I think this is a low hanging fruit mechanics question. Here is a picture depicting what I am interested in: Say I have a compressive force (green arrow) acting perpendicular to the blue block's front surface. As the blue block's oblique surface comes into contact with the orange triangular block, a force will be imparted. Below I have depicted two possible force diagrams of the forces applied by the blue block (note that the normal force applied by the rear rigid wall has been omitted in these diagrams). What I am uncertain of is the direction along which the shear arrow (light blue arrow) should be oriented. If anyone could offer some intuition as to why one direction is more logical than the other, I would appreciate it. My impression is that Case 2 is the correct direction because if the rigid wall were lower (as depicted below), then the blue block would rise up and to the right. As it traverses this path, the blue block is scraping against the orange oblique surface in the direction of its motion.

Sorry if I overlook the difficulty of your question, but you simply decompose the original force applied (green arrow) into a normal and tangential component. The sum of these components should therefore give you the original force. This doesn't work in your case 1. Case 2 is the correct decomposition.

How seriously can we take the success of the Standard Model when it has so many input parameters? The Standard Model of particle physics is immensely successful. However, it has many experimentally fitted input parameters (e.g. the fermion masses, mixing angles, etc). How seriously can we take the success of the Standard Model when it has so many input parameters? On face value, if a model has many input parameters, it can fit a large chunk of data. Are there qualitative and more importantly, quantitative, predictions of the Standard Model that are independent of these experimentally fitted parameters? Again, I do not doubt the success of the SM but this is a concern I would like to be addressed and be demystified.

The Standard Model may have many parameters, but it also talks about many things, each typically only involving a very limited number of parameters. For example, the muon lifetime$^\dagger$ $$\tau_\mu=\frac{6144\pi^3M_W^4}{g^4m_\mu^5}$$depends on only $M_W,\,g,\,m_\mu$ ($g$ is the weak isospin coupling), and the tauon lifetime $\tau_\tau$ satisfies$$\frac{\tau_\tau}{\tau_\mu}=\frac{1}{3(|V_{ud}|^2+|V_{us}|^2)+2}\frac{m_\mu^5}{m_\tau^5},$$which only depends on $|V_{ud}|,\,|V_{us}|,\,m_\mu,\,m_\tau$. So these two predictions need six parameters, which doesn't in itself sound that impressive. But that misses the point here. The full slate of SM predictions uses all the parameters in a variety of subsets, creating a system of far, far more simultaneous equations than we have parameters. If (to take one estimate discussed herein) there are $37$ parameters, it's not like we're fitting a degree-$36$ polynomial $y=p(x)$ by OLS. It's more like requiring the same few coefficients to simultaneously fit many different regression problems. $^\dagger$ I'm sure someone will object my formula for $\tau_\mu$ has a time on the LHS and an inverse mass on the RHS, is using natural units, and should read$$\tau_\mu=\frac{\hbar}{c^2}\frac{6144\pi^3M_W^4}{g^4m_\mu^5}$$in SI units to expose the use of two more parameters $c,\,\hbar$. Bear in mind, however, these both have exact SI values by definition. Anyway, it doesn't really affect the argument.

Effective Lorentz Factor in Cold Plasma, Razin Effect I was studying about the synchrotron radiation in plasma medium and got stuck at a point. Usually the Lorentz factor is defined in the following way $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $\gamma$ is the Lorentz factor, $v$ is the particle's velocity and $c$ is the speed of light in the vacuum. But when the particle moves in a medium (for example, plasma medium) with some refractive index it's $\gamma$ gets changed. For cold plasma the refractive index is written as, $$n(\omega)^2\approx 1-\frac{\omega_{pe}^2}{\omega^2}$$ where, $n(\omega)$ is the refractive index that depends on the frequency $\omega$ and $\omega_p$ is the plasma frequency. Now in a medium with some refractive index the speed of light gets modified in the following way $$c\rightarrow\frac{c}{n}$$ following this the Lorentz factor also changes as $$\gamma_{*}=\frac{1}{\sqrt{1-\frac{n^2v^2}{c^2}}}\\ \implies \gamma_{*}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}(1-\frac{\omega_{pe}^2}{\omega^2})}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2} + \frac{v^2\omega_{pe}^2}{c^2\omega^2}}}=\frac{1}{\sqrt{\frac{1}{\gamma^2}+ \frac{v^2\omega_{pe}^2}{c^2\omega^2}}}$$ But in this paper just below equation (2) the Lorentz factor in cold plasma is defined as $\gamma^{-2}_{*}=(\gamma^{-2}+ {\omega_{pe}^2}/{\omega^2})$. I do not understand how the $v^2/c^2$ term is getting vanished from the denominator in the definition of $\gamma_{*}$ ?

Let me start by defining $\beta = \tfrac{ v }{ c }$, then we can show that: $$ \beta^{2} = \frac{ \gamma^{2} - 1 }{ \gamma^{2} } \tag{0} $$ where $\gamma$ is the Lorentz factor. We can then show that: $$ \begin{align} \gamma^{-2} + \beta^{2} \tilde{\omega}^{-2} & = \gamma^{-2} + \left( \frac{ \gamma^{2} - 1 }{ \gamma^{2} } \right) \tilde{\omega}^{-2} \tag{1a} \\ & = \gamma^{-2} + \tilde{\omega}^{-2} - \gamma^{-2} \tilde{\omega}^{-2} \tag{1b} \end{align} $$ where $\tilde{\omega} = \frac{ \omega }{ \omega_{pe} }$ and $\omega_{pe}$ is the plasma frequency. In the limit of large $\gamma$, the last term can be dropped leaving one with the expression in that paper.

How to generalize this hydrostatics problem? Preamble Consider an ideal, incompressible fluid of density $\rho$ in a uniform gravitational field, in a rigid container and in contact with two massless pistons of differing area $A_1$ and $A_2$. At the same level in the fluid as the pistons is a pressure gauge reading zero absolute pressure. There is no pressure outside the container. See the below drawing: Two masses $m_1$ and $m_2$ are placed on their respective pistons. We may find the new pressure reading at the gauge by considering the new rest positions of the pistons and masses. Let $y_1$ and $y_2$ denote both positive (upward) and negative (downward) displacement of the pistons. Then to conserve volume one requires: $$ y_1 A_1 + y_2 A_2 = 0 $$ Using Pascal's law, one can also relate the difference in pressure between the two piston's positions as: $$ \frac{ m_1 g }{ A_1 } = \frac{ m_2 g }{A_2} + (y_2 - y_1 )\rho g $$ These equations can be solved for the positions of the pistons, and ​then choose either piston to sum up the pressure change to the $y=0$ gauge. Choosing piston 2: $$ P = \frac{m_2 g}{A_2} + y_2 \rho g $$ When one perform the subsitution the values for $y$ and $\rho$ cancel out, leaving: $$P = \frac{ m_1 + m_2 }{ A_1 + A_2 } g $$ Conjecture For N number of pistons and masses arranged as above, the final equilibrium pressure at the $y=0$ pressure gauge will be: $$ P = \frac{ \Sigma m_k }{ \Sigma A_k }g $$ Question Can the above conjecture be shown from a physical argument, without resorting to solving the system of linear equations? I.e., why would an arbitrary number of pistons act like a single piston of the same total area and mass?

The force in any given pipe is $PA_{i} = gm_{i} + y_{i}\rho g A_{i}$, intuitively it is carrying the mass and the water difference. The total force is then $P\sum{A_{i}} = g\sum{m_{i}} + \rho g \sum{y_{i}A_{i}}$. Argue from conservation that $\sum{y_{i}A_{i}} = 0$. Then $P = \frac{\sum{m_{i}}}{\sum{A_{i}}}g$ In words, the total force (pressure at y=0 * total area) must balance the total mass (water + masses) against gravity, and the total effect of the water is 0 by definition of the water level, leaving only the weight of the masses.

Magnetic scalar potential of a straight wire If we have a finite straight wire carrying a current $I$, we can integrate the Biot-Savart Law for the magnetic field along the wire and find the field at every point of the space as: $$ \mathbf{B} = [0,B_\phi,0]; \quad B_\phi = \frac{\mu_0 I}{4\pi a}(\cos\alpha_1-\cos\alpha_2); $$ where $a$ is the distance from the wire and $\alpha_1, \alpha_2$ are the angles between the evaluation point and the end-points of the wire. Similarly, by integration of the Biot-Savart Law for the magnetic vector potential, we find: $$ \mathbf{A} = [0,0,A_z]; \quad A_z = \frac{\mu_0 I}{4\pi}\ln\left(\frac{z_1+\sqrt{z_1^2+a^2}}{z_2+\sqrt{z_2^2+a^2}} \right); $$ where $z_1, z_2$ are the segments between the projection of the evaluation point on the straight wire and the end-points of the wire. It can be verified that $\nabla\times\mathbf{A}=\mathbf{B} $. I'm now struggling to find a formulation for the magnetic scalar potential in a region of the space that does not contain the straight filament, so that: $$ \mathbf{B} = -\nabla\Psi. $$ Any idea on this? I was not able to find any reference online or on Electromagnetics books. Many thanks in advance for any suggestion.

The two crucial theorems surrounding vector and scalar potentials are Irrotational Fields Have Scalar Potentials A vector field $\mathbf{V}$ may be written as $\mathbf{V} = \nabla \Psi$ if and only if $\nabla \times \mathbf{V} = 0$. Divergence-free Fields Have Vector Potential A vector field $\mathbf{V}$ may be written as $\mathbf{V} = \nabla \times \mathbf{A}$ if and only if $\nabla \cdot \mathbf{V} = 0$. There are no magnetic monopoles as far as we know, meaning that magnetic fields are always divergence free and therefore expresssible as the curl of a vector potential $\mathbf{A}$. Maxwell's equations tell us that $\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$, where $\mathbf{J}$ is the current density. If the RHS is zero, the magnetic field is zero. This means that if there is a nonvanishing magnetic field, $\mathbf{B}$ has nonzero curl and so cannot be expressed in terms of a scalar potential.

If atoms both absorb and emit photons, then why are there still gaps in an absorption line spectrum? In my physics notes there is a section on line spectra, and describing how absorbance line spectra can by used by astronomers to find out what gases are present in a star, should it be emitting white light from its core. I understand the concept of the absorbance by the atoms of photons of only specific energies, associated with the different energy levels of the electrons within it. In that sense, I can see how the absorbance line spectrum would form. The notes then speak about how the excited electrons return to the ground state and release a photon. My question is this: surely that light would “counteract” the absorption? If the absorbed photons are just re-emitted at a later stage so wouldn’t that just “fill in” the spectrum again? While editing this question I read my notes again and checked the wording closely. It doesn’t specify that the emitted photon has to be of equal energy as the absorbed one - is that the answer? Can an electron drop down a few energy levels (emitting a photon) without losing all of its energy? Many thanks, Hugo

Yes, materials can absorb high energy photons and release it as multiple lower energy photons. But also important is that the direction of the release is not the same as the absorption. If we have a beam of particular power on an absorptive material that can come to equilibrium, it must radiate that same power. But if the emission is spherically symmetric, the power in any one direction will be much lower than that of the initial beam. It may only be compared the background of the brighter source that the material appears to be an absorptive spectrum. If viewed from another angle with a dim background it may show an emission spectrum.

Conditions for a planet to become spherical I have a question, from which size/mass will a body in space adopt a spherical shape? Over 500 kilometers wide and/or 1/4 the mass of Pluto? Something like that, I always had this doubt.

For a body to become a sphere, it must have sufficient self-gravity to pull itself into the shape of one. However, because the self-gravity of an object depends on its mass rather than size, it means that a body made of a denser material would become spherical at smaller radii than that of less dense material. Additionally, there are other factors such as how easy the materials are to mould into a sphere which gives further deviations. (The second reason tends to win). Hence, for bodies made of rock, the minimum size to become a self-gravitating sphere is about 600km in diameter, whereas, for ice, the minimum size is only 400km.

How much is the inherent quantum-mechanical uncertainty in the definition of the second? Inspired by this other question. The second is defined such that the electromagnetic radiation whose energy equals the hyperfine splitting of the ground-state of the Cs-133 atom has a frequency of exactly 9,192,631,770 Hz. However, any such transition has a "natural width" due to the fact that excited states that are subject to spontaneous emission are not eigenstates of the full interaction Hamiltonian; this implies that they don't have an exact energy, and even the most perfect measurements of the frequency of the radiation would therefore exhibit random variation. How much is this inherent uncertainty in the case of the definition of the second?

This NIST publication is a great source for understanding these errors. In practice, the clock emits radiation of some frequency to try to excite as many Caesium atoms as possible. The maximum number are excited when the clock's emitted radiation matches the peak resonance frequency of the transition. So, even though there is a quantum mechanical linewidth, its effect just introduces an uncertainty which can be reduced by observing a larger number of atoms (millions) for a longer time (large compared to the period). This is the benefit of spatially longer clocks and the fountain design. From the NIST pub., the Q factor of NIST-F1 is $Q = \frac{f_{0}}{\Delta{}f_{a}}\sim Tf_{0}$ where $f_{0}$ is the transition frequency, $T$ is the observation time, and $\Delta{}f_{a}$ is the resonance width. Observation times of ~1 second provide a Q of $10^{10}$ (where larger is better). The effective line width in this case is ~1 Hz. Other large effects come about from biased errors (as opposed to symmetric increases in width). These include a blackbody shift and a density shift. The uncertainty usually comes from the uncertainty in trying to correct for these biases. You can see here to find a laundry list of other effects they correct for, such as gravitational redshift, second-order Zeeman, and microwave amplitude shift. There are several other effects with either small bias or small added uncertainty in the correction. To summarize, quantum mechanical line width does not directly place an uncertainty on clocks, as its effect on uncertainty can be reduced with longer times and more atoms. Other relevant effects currently limit the precision of Caesium clocks.

Why do rain drops fall with a constant velocity? While reading my physics book. I came across a line that says that: Rain drop falls with a constant velocity because the weight(which is the force of gravity acting on body) of the drop is balanced by the sum of the buoyant force and force due to friction(or viscosity )of air. Thus the net force on the drop is zero so it falls down with a constant velocity. I was not satisfied by the explanation So I searched the internet which too had similar explanations: The falling drop increases speed until the resistance of the air equals the pull of gravity, at which point the drop begins to fall at a constant speed, its terminal velocity. My confusion regarding the matter is that if the net force acting on a body (here the rain drop) is zero then it should remain suspended in air rather than falling towards the earth. So how come the rain drop keeps falling when net force acting on it becomes zero? How the air resistance and other forces stops the rain drop from acquiring accelerated downward motion?

Here is a slightly different way to think of this. If the net force is zero, the acceleration of the droplet is zero- even though its velocity is not zero. With the acceleration zero, the velocity remains constant as it falls.

Confusion with Impulse and Work I get Momentum and Impulse as well as Work and energy but struggle when it comes to connecting the two ideas. I understand that an objects KE can change without its momentum changing, Like in a inelastic collision were the KE of the system decreases but the amount of matter moving to left with some speed - the amount of matter moving to the right with some speed will be equal before and after the collision. Hence Momentum is Conserved. My confusion starts with the question is an impulse of a force on an object always accompanied by some work done on the object by that force?

The standard example of a force doing no work is when an object is moving in a circle at constant speed - the force points towards the centre of the circle. Over 180 degrees there is a non-zero impulse; the direction of movement has changed so the momentum vector has changed. However the speed has not changed, so no work has been done. It is actually impossible to change the kinetic energy of a single object without changing its momentum, because if you change the kinetic energy you must change the velocity. In an interaction between more than one object the total momentum stays fixed though, even if the total kinetic energy changes.

Finding Average Acceleration with only given angle A car enters a curve in the road with a speed of 32 m/s and emerges from this curve 4 s later with the same speed. However, the direction of the velocity changes by 150 degrees during this time. What is the magnitude of the average acceleration during this time interval? My question is, that how are we supposed to calculate the radius when we only have angle. We don't even know the length of curve

We don’t need to know the length of the curve or its radius. We have $$\displaystyle \text{average acceleration} = \frac{|\Delta \vec v|}{t}$$ We know the change in velocity $\Delta \vec v$ and the time period $t$ so we have everything we need to find the average acceleration.

Doubt regarding anamolous expansion of water and ice lakes I had a question regarding anomalous expansion of water. So if we have water and we start cooling it up to 4 degrees, water contracts, but after that in the range of 0-4 degree water starts expanding due to hydrogen bonding. The problem was not in this statement but what the book said after it: This is the primary reason as to why when during winters, when the temperature drops, not all water in the water body freezes, only the surface layer freezes, because as water is being cooled, after 4 degrees, any ice that is formed comes to the top due to it being less dense and ice being a bad conductor of heat insulates the water below it from the outside region hence preventing it from freezing altogether. But if that is the case, shouldn't we observe a similar thing whenever we freeze water in our refrigerators , we should not have had ice cubes at all. since a similar thing happens here, and the ice tray also being a good insulator, insulating the bottom most water from freezing?

The volume of water in ice trays is very small. Hence it has no problem freezing completely as the icy layer doesn't get thick enough to prevent heat loss to the freezer. In a lake or sea, the thickness of the icy layer is a lot more and it provides a lot of thermal resistance. So the lakes do not freeze completely.

How can the value of charge on an atom be less than that of a proton? The dipole moment of an $\text{HCl}$ molecule is about $3.6×10^{-30} \ \mathrm{Cm}$ and the separation between the atoms of it is about $1.27×10^{-10}\ \mathrm m$. We know that dipole moment, $p=qd$. So substituting the values of $p$ and $d$, we can calculate the value of charge on the atoms of the $\text{HCl}$ molecule and it will be approximately $q=p/d=2.8×10^{-20}\ \mathrm C$. This value is definitely smaller than the charge of a proton. How is it possible?

Whether you believe that this is some sort of time-averaging of point-like electrons moving very quickly, or that the electrons are actually smeared out by the wave function, this is because the electrons in orbit around the atoms are smeared out in orbitals. So, even though we describe the Cl atom as "stealing" the electron from the H, it doesn't, completely. The reality is more akin to a water molecule - where outer-most filled electron orbital is more concentrated around the more electro-negative atom than the lesser one. It just so happens that for water this imbalance is relatively small, for HCl it's big, and for NaCl it's a massive $8.971$ Debye ($2.992 \times 10^{-29}\,\mathrm{C\, m}$). Dividing that by the spacing, from the same web page, of $2.361\,$Å gives $1.267\times 10^{-19}\,\mathrm{C}$ (about 79% of the charge of an electron).

The relationship between the pressure at the base of a container and the normal reaction on the container This is NOT a homework question. The question here is just an example. In this question, after both the liquids are mixed and a homogenous mixture is created, I found the effective density of the mixture and found that the pressure at the base increases by 1/2(rho)gh. This was calculated by taking a smaller cylinder with area a inside the system and as the fluid inside this imaginary cylinder is in equilibrium, I took the pressure at the bottom into the area of that smaller cylinder as the weight of the fluid. The normal reaction on the container by the ground remains the same obviously as there is no change in the mass of the system. Now what I think is that the pressure on the base × base area = normal force As the base area hasn't changed but the pressure has, the normal force also changes? Which is clearly wrong. I have clearly misunderstood something here and would like to know if the pressure at the bottom of a container and normal force are related and how.

To get the pressure at the base increases by 1/2(rho)gh presumably you've found that the new density is $\frac{3\rho}{2}$ and done this: Original pressure is $$2H\rho g + H (2\rho) g = 4H\rho g\tag 1$$ and the new pressure is $$2H\frac{3\rho}{2} g + H\frac{3\rho}{2} g = \frac{9}{2}H\rho g\tag 2$$ But you should divide the contribution of the top cylinder, to the pressure at the base, by 2 due to the increase of area, as the force due to the weight of the top cylinder spreads evenly throughout, (first term) and change equations 1) and 2) to 3) and 4) $$H\rho g + H (2\rho) g = 3H\rho g\tag 3$$ $$H\frac{3\rho}{2} g + H\frac{3\rho}{2} g = 3H\rho g\tag 4$$ so A) and B) seem ok.

Clausius inequality leading to absurd result Background: After deriving Clausius inequality, the author of this book derives the following relation: Consider the cycle shown in the figure in which leg $A \rightarrow B$ is irreversible. In the equation $$ 0>\oint\frac{\mathrm{d}Q}{T}=\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d}Q}{T}+ \int_{B \operatorname{rev}}^{A} \frac{\mathrm{d}Q}{T} $$ the second term on the right-hand side of this equation is given by $S(A)-S(B)$ because it is taken over a reversible path. When we move this quantity to the left-hand side, we find that $$ S(B)-S(A)>\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}. $$ Thus the difference in entropy between the points is greater than the integral of $\mathrm{d} Q / T$ over an irreversible change. Problem: Entropy is a state function so $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}= {\Delta} S. $ By the inequality derived we have ${\Delta} S>{\Delta} S$ which is absurd.

For the irreversible path between the same two end states, dQ is different than dQ for the reversible path, and in the integral of dQ/T for the irreversible path, you are supposed to use the temperature at the boundary interface between the system and surroundings $T_B$. So for the irreversible path, you should be using $$\int{\frac{dQ_{irrev}}{T_B}}$$So the two integrals are nothing like one-another. The correct form of the inequality should read: $$\Delta S\geq \int{\frac{dQ_{irrev}}{T_B}}$$

How much longer could Titanic have stayed afloat if it gotten rid of its anchor and chain right after hitting the iceberg? I am wondering how much longer the RMS Titanic could have stayed afloat if the crew had allowed the ship's anchor and anchor chain to fall to the bottom of the ocean immediately after the ship had hit the iceberg. (I am not even sure if a ship's anchor chain can be unfastened from a ship, but let's just say for the sake of this question that it can be unfastened.) The combined weight of Titanic's anchor and anchor chain was approximately 116 tons according to this Wikipedia article: "...In 1911, the company manufactured the anchors and chain for the ocean liner RMS Titanic. The largest of the anchors weighed 15.5 tons and on completion was drawn through the streets of Netherton on a wagon drawn by 20 shire horses.[15] The chain and fittings for the anchors weighed around 100 tons..." https://en.wikipedia.org/wiki/N._Hingley_%26_Sons_Ltd Since the Titanic went down bow first, and the anchor and chain was located in the bow section, immediately getting rid of 116 tons in the bow section would have increased the time it had stayed afloat before it sank.

Titanic weighed 46,000 tons, so removing say 120 tons would have made almost no difference. She had several long tears well below the water line. Flooding was such that there was no means that could have prevented her sinking. It is quoted as about 7 tons per second. It would have taken just 17 seconds to replace all the weight of anchors and chains. It's worth noting that her chief (?) architect and captain would both have considered counter-flooding in order to keep her afloat. This would be an automatic thought to them. They did not do these things because they knew it would make no practical difference to her. She was mortally wounded. Such measures would also have made launching boat difficult and greatly increased panic.

From momemtum Hamiltonian to real space Hamiltonian I know how to calculate a bulk momentum space Hamiltonian from a real space one. For example, given a SSH model $$H=v\sum_ic^\dagger_{iB}c_{iA}+w\sum_ic^\dagger_{i+1,A}c_{iB}+h.c.,$$ its bulk momentum-space Hamiltonian is $$H(k)=\left(\begin{matrix}0&v+we^{-ik}\\v+we^{ik}&0\end{matrix}\right).$$ How do I implement inverse Fourier transform to gain the real-space Hamiltonian?

The momentum space Hamiltonian you have written down is the full Hamiltonian in the sector of the Hilbert space indexed by the quantum number called the (crystal) momentum. It is a 2x2 matrix because, once we specify the momentum $k$, we still need to specify which sublattice $\sigma$ we are in. Therefore, our full Hamiltonian can be expressed as: $$H = \sum_k \sum_{\sigma \sigma'} H_{\sigma, \sigma'}(k) c_{k, \sigma}^{\dagger}c_{k, \sigma'}$$ where $\sigma, \sigma' \in \{ A, B \}$ is the sublattice index. Now, to go back to real space, we simply need to write our $c_k$'s in terms of $c_i$'s as: $$c_{k, \sigma}^{\dagger} = \frac{1}{\sqrt{N}} \sum_k e^{ik\sigma} e^{ikx}c_x^{\dagger}$$ which, when substituted back into the full Hamiltonian yields the SSH Hamiltonian in real space!

Why $E$ is neglected at large and small $r$ of quantum harmonic oscillator? In obtaining radial solution of quantum oscillator why E is neglected? Radial equation: Resource: nouredine zettili.

$E$ is neglected because in both cases (where $r \rightarrow 0$ and when $r \rightarrow \infty$) there is a term multiplying $U(r)$ that gets very large when compared to the constant $E$. When $r \rightarrow 0$, this is the term which depends on $\frac{1}{r^2}$; when $r \rightarrow \infty$ this is the term which depends on $r^2$.

What is the physical meaning of "correlation length"? I am studying phase transitions right now and trying to understand the physical meaning of the concept correlation length. I saw the equations but I still couldn't quite wrap my head around the physical meaning of it. Like is it the length of the correlation between neighbouring atoms or what? And what does it mean for correlation length to be large or small? System is considered more organized when correlation length is large, right? And what is the meaning of correlation length being infinite, like when it is infinite, can we say that all atoms in the system are correlated with each other?

In a magnet, atoms with positive spin and atoms with negative spin will cluster together and the correlation length $\xi$ measures the typical size of these clusters. When $\xi$ first reaches infinity, you do not have a single infinitely big cluster yet because even though exponentially decaying correlations are gone, you still have correlations which decay as a power law. This indicates that there are clusters of all sizes. To truly make all spins the same, you need the coefficient of this power law to disappear as well which only happens at zero temperature. I discussed this a bit in a previous answer where the context was renormalization.

Why proper time is a measure of space? Recently I've been trying to learn General and Special Relativity by myself. There is an specific thing I do not understand perfectly, proper time in the metric of the space-time. Take the case of an empty space-time: $$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2$$ where $\tau$ is the proper time of an object. I don't understand when $c^2 \mathrm d \tau^2$ is used as $ds^2$. Why is this possible? It is related to the worldline traced by an object? Could you calculate the line integral in order to find the length of the path the object? This line integral would give the path length of a geodesic? Could you get the speed of an object by doing this (correct me if I'm wrong): $$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2 \rightarrow \\ -c^2 \left( \frac{\mathrm d \tau}{\mathrm d t} \right)^2 = -c^2 + \left( \frac{\mathrm d x}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d y}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d z}{\mathrm d t} \right)^2 $$ If you assume that: $\dot{x}^2 + \dot{y}^2 + \dot{z}^2 = ||v||^2$, then by rearranging a little bit the equation, you could get the velocity of an object.

$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$ If the result is positive, the equation corresponds to a spacelike worldline segment. It is always possible to find a frame, so that $dt = 0$ and two events linked by the segment are simultaneous for this frame, happening at different spatial locations. If the result is negative, it is a timelike segment. It is always possible to find a frame, so that $dx^2 + dy^2 + dz^2 = 0$ and two events linked by the segment happen in the same place for this frame but at different times. In this last situation, the quantity $dt^2 = d\tau^2$, is the proper time. The time of a clock at rest in the frame.

How does Upthrust/buoyant force act on an object? I came across this question where, In a container there's water at the bottom , kerosene on top and an ice cube floating between them and I was asked to calculate the ratio of height of cube in ice to that in water: Now till now my intuition for up thrust was that there needs to be some fluid below the object to give an upward perpendicular force.As simple as this: Therefore for this question I thought there should be no upthrust from kerosene but I was surprised when I saw the free body diagram in the solution: And therefore now I wish to understand: * *What's wrong with my understanding of upthrust? *Where is the upthrust from kerosene actually coming from and How does upthrust acts on a body in general?

A buoyant force results when the pressure pushing up from the bottom of an object is greater than the pressure pushing down from above. For your situation, there is an increase in the pressure as you go down from the top of the cube through the kerosene and another increase as you go down to the bottom through the water.

Would somebody feel a magnetic field if they are travelling at the same velocity as a charge? I am little bit curious about how magnetic fields are being generated when a charge moves. I want to check if somebody travelling along with a charged particle, would that person experience a magnetic field? How are magnetic fields really generated?

Magnetic fields are really created by moving charges. "somebody " would need to describe how to measure or "feel" a magnetic field, anyway she would "feel" the electric field of the charge.

A question regarding commutators in quantum mechanics I propose the following thought experiment: Suppose we have a beam of identically prepared electrons that is splits into two. The first goes through detector A that detects the $x+y$ where $x$ is the coordinate along x direction and $y$ is the coordinate along the $y$ direction. Then, we measure the difference of the momenta of the electrons in the $x$ and $y$ directions i.e. $p_{x}-p_{y}$. Then, according to the postulates of quantum mechanics, we can measure the both quantities to arbitrary precision since $$ [x+y, p_{x}-p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$ The second beam of electrons is subjected to a similar measurement by a detector B but this time we measure $x-y$ and then measure the sum of momenta i.e. $p_{x}+p_{y}$. Then, again we can measure $x-y$ and $p_{x}+p_{y}$ to arbitrary precision because $$ [x-y, p_{x}+p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$ Then, adding the results of the measurements we have $(x+y)+(x-y)=2x$ and then $(p_{x}-p_{y})+(p_{x}-p_{y})=2p_{x}$. Both of which, $x$ and $p_{x}$ can be measured to arbitrary precision thus violating the uncertainty principle. If on the other hand, we carry out this experiment and find that we are not able to measure the above quantities to arbitrary precision then it follows that the postulates of quantum mechanics do not correctly predict the outcome of the experiment in the sense that the commutator vanishes but we can't measure the quantities to arbitrary precision. Does this mean that the postulates of quantum mechanics are inconsistent? (I certainly don't hope so!)

You are assuming that the two beams of electrons are two different systems in identical quantum states. The uncertainty principle limits measurement of two non-commuting observables on one system, but says nothing about measurements on separate systems. If I had two identical systems in identical states, I could just measure $x$ in one system and $p_x$ in the other system which would give me an accurate measurement of $x$ and $p_x$ at the same time. There is no need to go through the complicated process you have described.

Does deceleration require energy? Consider an apple falling from a tree and striking the ground. The ground decelerates the apple once it hits it, but the force is not applied over any "distance" - it is experiencing the force when it is in contact with the ground - so no work is done, yet there is a change in momentum, what is going on with the energy here?

The ground decelerates the apple once it hits it, but the force is not applied over any "distance" - it is experiencing the force when it is in contact with the ground - so no work is done, yet there is a change in momentum, what is going on with the energy here? Assuming that the ground is rigid and the earth is much more massive than the apple and is at rest, then you are correct the force exchanges momentum but no energy is exchanged. This implies that the energy of the apple is constant. All of the KE must go to some form of internal energy. If we replace the apple with a spring, then we can see that the spring will compress which will increase the internal energy of the spring. Similarly, an apple will store energy in internal deformation. An apple is less elastic than a spring, so there will probably be some plastic deformation resulting in the production of thermal energy. So no work will be done on the apple, no energy transferred to it from the ground, but the apple will exchange its KE for elastic potential energy and heat.

How to interpret Einstein's Field equations for 1+0 dimension $x$ in General Relativity? The Einstein field equation for our $3+1$ spacetime dimensions is $$ R_{\mu \nu} - {1 \over 2} g_{\mu \nu} R + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$ I am learning to navigate through this equation and would appreciate some help in interpreting the Einstein field equation in the trivial 1+0 dimension case, please. Let us consider only the $x$ space dimension, and no time dimension in this question. I have some questions regarding the form of the equation in this 1+0 dimensional case, as follows: * *What is the form of the Ricci curvature $$R_{\mu \nu}$$ in one dimension. Is it a 1x1 matrix, ie. it is a scalar? Is its value always zero? *Is the value of the Ricci scalar $$R$$ always equal to zero in one dimension? *What is the form of the metric tensor $$g_{\mu \nu}$$ for one dimension? Is it just a scalar? *What is the form of the stress-energy-momentum tensor $$T_{\mu \nu}$$ in one dimension? Is it a scalar? What can generally be interpreted from this equation in the 1 dimensional case? What information can be obtain from this equation?

In a $1$-dimensional manifold the Riemann tensor's only component is $R_{0000}=0$ by the tensor's symmetries, so $R_{\mu\nu}=0,\,R=0$. The EFE simplifies to $\Lambda g_{00}=\kappa T_{00}$. The two sides are still not scalar-valued, as they depend on the chosen coordinate system viz. $ds^2=g_{00}(dx^0)^2$.

Does energy conversion always requires Kinetic energy in some manner to be present? Does energy always needs to be converted into Kinetic form in order to convert into another form? What I mean, is Kinetic energy meta form for all energy conversion? As we can see for electricity to light up the bulb the electrons has to be set into motion. Also in Nuclear power plants we basically covert the nuclear tension among the atoms into kinetic form which heats off the turbine in turn produces the energy

To put the point another way, energy transfer always requires motion. In an entirely motionless environment, there can be no energy transfer. Given that motion equates to KE, KE is always involved in the transfer of energy.

Angular Velocity via Extrinsic Euler Angles I am wondering if the angular velocity of a rotating coordinate system, if expressed through extrinsic Euler angles, is $(\dot{\alpha},\dot{\beta}, \dot{\gamma})$ since extrinsic Euler angles are rotations about fixed axes so the rates should be orthogonal to each other.

The Rotation matrix is created with those 3 matrices Rotation about x-axes with the angle $~\alpha~$ $$\mathbf R_x= \left[ \begin {array}{ccc} 1&0&0\\0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0& \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right] $$ Rotation about y-axes with the angle $~\beta~$ $$\mathbf R_y= \left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left( \beta \right) \\ 0&1&0\\ -\sin \left( \beta \right) &0&\cos \left( \beta \right) \end {array} \right] $$ Rotation about z-axes with the angle $~\gamma~$ $$\mathbf R_z=\left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left( \gamma \right) &0\\ \sin \left( \gamma \right) &\cos \left( \gamma \right) &0\\ 0&0&1\end {array} \right] $$ Example * *first rotation about the z-axes $~\mathbf R_z(\gamma)$ *second rotation about the new axes y' $~\mathbf R_{y'}(\beta)$ *third rotation about the new axes z' $~\mathbf R_{z'}(\alpha)$ hence the rotation matrix $\mathbf R~$ is $$\mathbf R=\mathbf R_z(\gamma)\,~\mathbf R_{y'}(\beta)\,\mathbf R_{z'}(\alpha)$$ from here you obtain that angular velocity $~\mathbf\omega$ $$\mathbf\omega=\mathbf A(\alpha~,\beta~,\gamma)\,\begin{bmatrix} \dot\alpha \\ \dot\beta\\ \dot\gamma\\ \end{bmatrix} \Rightarrow\quad \begin{bmatrix} \alpha \\ \beta\\ \gamma\\ \end{bmatrix}=\int\,\mathbf A^{-1}(\alpha~,\beta~,\gamma)\,\mathbf\omega\,dt $$ your question. for a "small" rotation angle $~\varphi~,$ $~\cos(\varphi)=1~,\sin(\varphi)=\varphi~$ the rotation matrix is now: $$\mathbf R=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}+ \left[ \begin {array}{ccc} 0&-\gamma&\beta\\ \gamma &0&-\alpha\\ -\beta&\alpha&0\end {array} \right] $$ and the angular velocity $$\mathbf\omega=\begin{bmatrix} \dot\alpha \\ \dot\beta\\ \dot\gamma\\ \end{bmatrix} \Rightarrow\quad \underbrace{\begin{bmatrix} \alpha \\ \beta\\ \gamma\\ \end{bmatrix}}_{\mathbf \phi}=\int\mathbf\omega\,dt $$ $\mathbf \phi~$ is now a pseudo vector. hence: only for a small angles $~\alpha~,\beta~,\gamma~$ the angles are rotation about the axes $~x~,y'~,z'~$ Edit how to obtain the angular velocity from the rotation matrix $~\mathbf R$ with $$\mathbf{\dot{R}}=\mathbf R\, \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}} \\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \quad \Rightarrow\quad \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}} \\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]=\mathbf R^T\,\mathbf{\dot{R}} $$ I use MAPLE program to obtain the result , for the example above ,you obtain $$ \left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y} }\\\omega_{{z}}\end {array} \right] = \underbrace{\left[ \begin {array}{ccc} 1&0&-\sin \left( \beta \right) \\ 0&\cos \left( \alpha \right) &\sin \left( \alpha \right) \cos \left( \beta \right) \\ 0&-\sin \left( \alpha \right) &\cos \left( \alpha \right) \cos \left( \beta \right) \end {array} \right]}_{\mathbf A(\alpha,\beta)} \,\underbrace{\begin{bmatrix} \dot\alpha \\ \dot\beta \\ \dot\gamma \\ \end{bmatrix}}_{\mathbf{\dot{\phi}}}\\ \begin{bmatrix} \dot\alpha \\ \dot\beta \\ \dot\gamma \\ \end{bmatrix}=\left[ \begin {array}{ccc} 1&{\frac {\sin \left( \alpha \right) \sin \left( \beta \right) }{\cos \left( \beta \right) }}&{\frac {\cos \left( \alpha \right) \sin \left( \beta \right) }{\cos \left( \beta \right) }}\\ 0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0&{\frac {\sin \left( \alpha \right) }{\cos \left( \beta \right) }}&{\frac {\cos \left( \alpha \right) }{\cos \left( \beta \right) }}\end {array} \right] \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} $$ the components of the angular velocity are given in the rotating system, not in inertial system. the components of the angular velocity in inertial system are $$\mathbf\omega_I=\mathbf R\,\mathbf\omega$$ notice the singularity at $~\beta=\pi/2~$ . each rotation matrix has singularity at some rotation angle.

How to calculate the density of states near a point for the given dispersion relation? The problem's statement is as follows: Given that $ E(\vec{k}) = ak_x^2 + bk_y^2 + c|k_z|$, Calculate the density of states near $(0,0,0)$. It's easy to do the integration if $E$ is quadratic in all of the components, but in this case there's a linear dependence for the $z$ component. I've tried to calculate it from the definition: $$ D(E) = \frac{2V}{(2\pi)^3}\int \delta (E(\vec{k}) - E)\, \mathrm{d}\vec{k}$$ But I have no idea how to do the integration through change of coordinates. Can anyone help?

To give you the main ideas of the resolution without giving it away, you can work in several steps: First, you can do a change of variable for $k_x$, $k_y$ and $k_z$ in order to absorb the constants a,b and c so that the spectrum in the new coordinates is $E=q_x^2+q_y^2+|q_z|$ then, you can do -a polar change of variable in the plane qx-qy to reduce the integration since the energy depend on the radius and not the angle -a change of variable with $q_z$ since $\int_{-a}^a dk f(|k|)=2\int_{0}^a dk f(k)$ Then you can integrate on the radius, reminding that for a function f with zeros at positions $x_i$ we have $\Delta(f(x))=\sum_i \frac{\delta(x-x_i)}{|f'(x_i)|}$ separating the $q_z$ axis into region where this has no solution and some where it has (depending on the value of E). Finally the integration on $q_z$ should be doable allowing you to conclude.

Fourier Transform in wavefunction of Free particle I was reading Introduction to Quantum Mechanics by David Griffiths and I am at Chapter 2, page 45. He says that The general solution to the time-dependent Schrodinger equation is still a linear combination of separable solutions (only this time it's an integral over the continuous variable $k$, instead of a sum over the discrete index). This means that we can write free particle with general wavefunction with $\Psi(x,t)$ as Fourier transform of eigenfunction of Schrodinger equation (I think). i.e., \begin{equation} \Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk \end{equation} where $\phi(k)$ is eigenfunction of Schrodinger Equation for free particle. (I think) But then he writes \begin{equation} \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0)e^{-ikx}dx \end{equation} from Plancherel's theorem. This clearly means that eigenfunction of the Schrodinger Equation will depend on the initial wave function of the particle. Is my interpretation correct? I am not sure because this is not very intuitive to me. Also, does this mean that Fourier Transform of any general free particle wavefunction $\Psi(x,t)$ is eigenfunction of Schrodinger equation?

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk$$ where $\phi(k)$ is eigenfunction of Schrodinger Equation for free particle.(I think) No, you misunderstood your text book. $\phi(k)$ is a completely arbitrary function. For any function $\phi(k)$ this $\Psi(x,t)$ will be a solution of Schrödinger's equation $$i\hbar\frac{\partial}{\partial t} \Psi(x,t)= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t)$$ just because $$e^{i(kx-\frac{\hbar k^2}{2m}t)}$$ for any $k$ is a solution of Schrödinger's equation. When you take the above solution $\Psi(x,t)$ at time $t=0$, then you have $$\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{ikx}dk$$ This means that $\phi(k)$ is the Fourier transform of $\Psi(x,0)$. You can invert this transformation and get $$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0) e^{-ikx}dx$$

What do we mean by saying light is a form of energy? "Energy" means "Capability to do work" We can now write "light is a form of energy" into "light is a form of capability to do work" Capability can be thought as a Capacity Whenever we talk about Capability of doing work we have an object that contains the ability to do work. So in this case who is having the ability to do work? The answer could be electromagnetic wave or something but what it is and how it is having the capability? Is it necessary to have mass to have Capability to do work?

Energy is a physical property of a physical system. There is nothing like pure energy but maybe in Dragon Ball cartoons. Not in our universe. Therefore, the question if the light is a form of energy is not consistent with known physics. What can be safely said is that light, either described as electromagnetic waves or as photons, is a physical system that has some energy. Then, it may do work. Having mass is not necessary for the ability to do work. Systems with or without mass do have energy.

Bull in a china shop: Is there a measure of level of organization of energy (flow) to do work? The discussion around the second law of thermodynamics and creation of orderly structures (e.g. life, evolution, etc.) is an old one. Now that creationists learned the difference between open vs. closed/isolated systems, one argument that I commonly see is a "bull in a china shop" analogy. They say "Fine, energy flows in open systems allowing their entropy to decrease locally, but this energy is completely 'raw' and 'disorderly'. Actual work in man-made systems are highly fine-tuned designs that allow energy to flow in certain ways. Energy from the Sun is just randomly flowing energy and it cannot do meaningful work towards complexity, just like a bull has raw energy in a china shop but it will never create meaningful designs." I understand how they are trying to deflect their misunderstanding of the second law, but it got me thinking: Is there a measure for such an "level of organization of energy flow"? And how would you approach such an argument?

Yes, it is called exergy, which is the work that can be extracted from a thermodynamic energy flow. It is equal to the amount of energy input times the Carnot efficiency. For photosynthesis the relevant amount of energy is about $1 \ \text{ kW m}^{-2}$ and the relevant Carnot efficiency is about 0.95 (the Carnot efficiency between the sun at 5778 K and the earth at an average 288 K). So we are left with about $950 \text{ W m}^{-2}$ of useable work. The reality is that what the sun provides is a very low entropy more than a high energy. The bull in a China shop analogy is a rather poor analogy.

A Question About the Surface of a Black Hole Singularity In Kip Thorne's book, Black Holes and Time Warps, he states that the mass of the core of a star shrinks until quantum gravity takes over. And then discusses that at this distance, the singularity cannot be distinguished from quantum foam. I am confused. Is he stating that the singularity is covered in quantum foam or does the mass shrink so small it becomes part of the quantum foam. If it is the latter, how can the quantum gravity, affect the macroscopic world? Is he discussing the surface is smooth, except for the fluctuations required by quantum gravity? Sorry - I am not a scientist or student...

In Thorne's book, on page 477, it says "Because all conceivable curvatures and topologies are permitted in side the singularity, no matter how wild, one says that the singularity is made from a probabilistic foam. John Wheeler, who first argued that this must be the nature of space when the laws of quantum gravity hold sway, has called it quantum foam." So what you are thinking of as a foam is really only a probabilistic foam. This is just a way of saying that every possible state in the singularity is only a probability. (Takes you back to the probabilistic nature of any particle which is a discussion far beyond this answer). But remember Thorne's central point of the chapter, that time does not exist at the singularity. Space and time have separated from each other. Time stops at the event horizon, but space continues to dilate all the way down to the singularity. This quantum foam is Wheeler's way of describing the situation of an unknowable, but only probabilistic, nature of the singularity.

Why are photons being identical particles? Recently, I study quantum optics and deal with quantization of EM field in a cavity. We know we can express/quantize vector potential in terms of $\hat{a},\hat{a}^{\dagger}$ to get a quantized EM field in a cavity. $$ \vec{A}(\vec{r},t)=\sum_{n,\sigma}\sqrt{\frac{\hbar}{2\epsilon_0\omega_n V}}\vec{e}_{n,\sigma}\Big[\hat{a}_{n,\sigma}e^{i(\vec{k}_n\cdot\vec{r}-\omega_nt)}+\hat{a}_{n,\sigma}^{\dagger}e^{-i(\vec{k}_n\cdot\vec{r}-\omega_nt)}\Big] $$ The quantized Hamiltonian is: $$ \hat{H}=\sum_{k}\hbar\omega_k(\hat{n}_k+\frac{1}{2}) $$ The eigenstate of quantized Hamiltonian is: $\left| n_1,n_2,n_3,... \right>=\left|n_1\right>\otimes\left|n_2\right>\otimes\left|n_3\right>...$ The state means there are $n_1$ photons in the first mode and $n_2$ photons in second mode and so on... So every mode has it own number of photons and photons in the different modes are not at the same frequencies. But why do we take photons as indistinguishable particles?

Calling photons indistinguishable particles is an artefact of thinking of them as fermion-like particles, rather than excitation quanta of electromagnetic field - the phrase is used to make the point to those (yet) unfamiliar with second quantization (see also this answer). If you take a specific mode of the field, with occupation number $n_k$, there is no way of distinguishing its $m_k$-th quantum from $m_k+1$-th or any other. But, since your already know how to quantize the field, this even does not come into question. The photons in different modes (with different k-vectors, frequencies and polarization) are of course distinguishable.

How do stars produce energy if fusion reactions are not viable for us? From what I've learned, fusion reactions are not currently economically viable as of right now because the energy required to start the reaction is more than the energy actually released. However, in stars they have immense pressures and temperatures which are able to allow these reactions to take place. However, if these reactions are considered endothermic for us, how are they exothermic in stars? i.e. how are stars able to release energy? Moreover, why are such fusion reactions for us endothermic in the first place? Given we are fusing elements smaller than iron, wouldn't the binding energy per nucleons products be higher and hence shouldn't energy be released?

The problem is not with “creating” fusion: we have H-bombs that will do this. The technical difficulty is with controlling the very exothermic fusion process, so the whole thing doesn’t blow up violently.

Will Helium any time in the future make up the most part of the Univers? as Stars are using Nuclear Fusion to produce energy, will there theoretically be a time in the future of the universe where helium will take over hydrogens first place as the most common element in the universe?

It is actually probable that this will happen. It's not easy to work out when in the universes' history this will happen, but some scientists (by looking at the ratio of hydrogen/helium in the universe after 13.7 billion years of universal evolution) estimate that helium, by the fusion of hydrogen, will become more abundant than hydrogen in about 1,000 to 1,000,000 times the current age of the universe (in about 10,000,000,000,000 to 10,000,000,000,000,000 years). Then following similar periods of time, the next most abundant element will be carbon by the fusion of helium, and then similarly oxygen (maybe) by the fusion of carbon and helium and so on.

Testing Lorentz force on a moving macroscopic charged conductor? Did anyone ever made an experiment with a moving macroscopic charged conductor in a magnetic field testing its deviation due to Lorentz force? I mean: what moves here is the massive charged conductor itself, not only the charges inside it. Would it be feasible with our technology? If it is then why does this force only appears in late 1800's in the scientific literature?

A macroscopic object with charge $q$ will not experience a magnetic force of the form $${\bf F}=q(\bf v\times B)$$ if it were to pass through a magnetic field $\bf B$ And of course if there is an electric field, then we have the full form of the Lorentz (electromagnetic) force $${\bf F}=q(\bf E+ v\times B)$$ and charged particles experience a local electromagnetic force according to this equation. That is, this equation does not work for macroscopic objects. If we wanted to extend the Lorentz force law to macroscopic objects, we would need to take into account additional quantities like the polarization (density) $\bf P$ and magnetization (density) $\bf M$ to be sources of the electromagnetic field and this can severely complicate the situation. In such a case, the Lorentz force$^1$ (density) takes on a much different form $${\bf F} = (\rho_f −\nabla\cdot {\bf P}){\bf E} + ({\bf J_f}+ \frac{\partial {\bf P}}{\partial t})\times \mu_0{\bf H} − (\nabla\cdot {\bf M}){\bf H} − (\frac{\partial {\bf M}}{\partial t})\times \epsilon_0 {\bf E}$$ where $\rho_f$ and $\bf J_f$ are free charge and current densities respectively, where $${\bf B}= \mu_0 {\bf H} + \bf M$$ Note that the equation for the Lorentz force density requires many assumptions to be valid, and an explanation of this and a derivation can be found here. This equation has been tested, but for very simple systems and the experiment appears to be consistent with the equations. $^1$ Note that many authors have many other forms to this equation depending on the setup, assumptions, boundary conditions etc., and the equation above for the electromagnetic force density is in the most general form.

Why is there a discrepancy between $m_n-m_p$ and $m_d-m_u$? The difference in mass between a neutron and a proton is $1.3\space {\rm MeV}/c^2$, but the mass difference between an up quark and a down quark is $2.5\space{\rm MeV}/c^2$. How come the mass differences aren't the same?

Some comments mention that calculating the mass of a nucleon is extremely complicated, which is true: see lattice QCD. It does require a supercomputer. But the strong force is insensitive to quark flavor, so I'm not sure that any of that complexity is related to this question. The quark flavors differ not only in mass but also in electric charge, and a naive back-of-the-envelope calculation suggests that the rest of the mass difference could be electromagnetic. The energy needed to bring three charges $q_1,q_2,q_3$ from infinity to a triangle of side $r$ is $k(q_1q_2{+}q_2q_3{+}q_3q_1)/r$. For two down quarks and one up quark, that's $-\frac13 ke^2/r$, and for two up quarks and one down quark, it's zero. That would account for the discrepancy if $\frac13 k e^2/r \approx 1.2\text{ MeV}$, or $r\approx 0.4\text{ fm}$, which is close to the size of a nucleon. The paper "Electromagnetic proton-neutron mass difference" by Oleksandr Tomalak (DOI, arXiv) quotes the results of much more sophisticated calculations of the electromagnetic contribution, which hover around 1 MeV with large error margins.

How do we measure time? I'm having a little trouble trying to put to words my problem and I apologize in advance for any causation of trouble in trying to interpret it. We define periodic events as those events that occur over equal intervals of time. But, don't we use periodic events themselves to measure time (like a pendulum or the SI unit definition of transition frequency of Cesium)? Then how is it we know we have equal intervals of time? Another way to put my problem would be: We metaphorically describe time in terms of the physical idea of motion, i.e., 'time moves from a to b', but how do we deal with how fast it moves because to know how fast it moves, we must know its rate and to know its rate is like taking the ratio of time with time? This is all very confusing. I apologize again for any problem in trying to understand.

As Marco Ocram describes in the answer above, you will not notice any change in the rate of time inside a closed system. You can only measure a changing rate of time with relation to some other place where time runs at a different rate. So, for example if we were to take your house, family and all your clocks and put it all on Jupiter (where time runs slower due to greater gravity), then you would not notice any difference the the rate of time. You would continue to live your life and age and cook a 3 minute egg in exactly the same way. But if, after a while, you called your cousins back on Earth, then you would notice that your cousins have aged more, and their clocks would have ticked more. It is only by comparing between two systems that a difference can be measured. It was this discovery, that time runs at a different rate in different frames of reference, that is a major source of the study of physics today.

Euclidean propagator expression for massless particle Let $\Delta_F(\tilde{x})$ denote the Feynman propagator in the Euclidean variable $\tilde{x}$, in $D$ dimensions, $$\Delta_F(\tilde{x}) = \int \frac{\text{d}^D\tilde{p}}{(2\pi)^D}\frac{e^{i\,\tilde{p}\cdot\tilde{x}}}{\tilde{p}^2+m^2}.\tag{1}$$ Since this expression is $\mathrm{O}(D)$ invariant, one can change variables to spherical coordinates and simplify the expression, yielding $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}\tilde{p}\text{d}\theta\,\frac{\tilde{p}^{D-1}}{\tilde{p}^2+m^2}e^{i\,\tilde{p}\cdot\tilde{x}}\, \left(\sin(\theta)\right)^{D-2}.\tag{2}$$ For $m = 0$, $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}\tilde{p}\text{d}\theta\,\tilde{p}^{D-3}\, \left(\sin(\theta)\right)^{D-2}e^{i\,|\tilde{p}||\tilde{x}|\cos(\theta)}.\tag{3}$$ However, I am supposed to get $$\Delta_F(\tilde{x}) = \frac{1}{(D-2)S_{D-1}}\frac{1}{r^{D-2}}.\tag{4}$$ Any ideas on how one can proceed further? Edit: Had forgotten to add some extra steps. Edit 2: Using $u = i\,|\tilde{p}||\tilde{x}|\text{cos}(\theta)$ as suggested, $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}u \text{d}\theta\,(\tan(\theta))^{D-2}\,\frac{u^{D-3}e^{u}}{(ir)^{D-2}}.\tag{5}$$ Am I missing some identity that involves gamma functions?

$$ e^{ip\cdot x}= e^{i|p|r \cos \theta} $$

How do conservative forces do work on a mechanical system if they conserve mechanical energy? The question arises from my confusion over the two definitions of work (relevant to classical mechanics) I've encountered: * *$W= \int \vec F\cdot\mathrm{d}\vec x$ *$W=$ net change in energy of a system We define the mechanical energy for a state to be the sum of the kinetic and potential energies, and then define a conservative force as one that doesn't change/conserves a system's mechanical energy. By the second definition, this implies that conservative forces like gravitational force do no work on a mechanical system, but this conflicts with the statements in my reference book and the ones I have read on the internet. They reason, with the (aforementioned) first definition: if a non-zero conservative force $\vec F$ acts along and throughout a body's non-zero displacement, the work ($W=\int\vec F\cdot\mathrm{d}\vec x$) should be non-zero, and therefore a conservative force does work. Which of the aforesaid is actually true, and why?

Your second definition is incorrect. The total work done on a system by all forces is equal to the change in kinetic energy, not total energy. So, gravity pulling a mass down does work that increases its kinetic energy. Non-conservative forces change total mechanical energy, conservative forces do not. They can both do work by changing the kinetic energy of a mass or system of masses.

How can the electron in a hydrogen atom have energy without angular momentum ($n>0$, $\ell=0$)? Been struggling with this concept. Or is this just one of those things in quantum mechanics which attempting to understand is futile? Guess we can see it as the electron just linearly oscillating back and forth but it doesn't feel right.

Classically a particle without angular momentum will have a radial orbit. This is the closest classical picture to an s-orbital. The orbital is spherically symmetric because the orientation of the orbital is undetermined

Do sunrises and sunsets look the same in a still image? A question that popped into my head: if I see a picture of the sun close to the horizon, in an unknown place, can I know if it was taken at sunset or sunrise? Do sunrises and sunsets look the same in a still image? Can one tell them apart?

Yes, the temperature of the air that the sunlight goes through to reach our eyes would be different. At sunset the air would be warm, at sunrise it's colder and that causes the light to refract differently. This website shows more about it

How can we make the energy levels of an one-dimensional infinite square well potential equispaced? We know that the energy values for an one-dimensional infinite square well potential is given by $$E_n = \frac{n^2{\pi}^2{\hbar}^2}{2ma^2}$$ where $a$ is the width of the well. Now, as we can see that the difference in energy of two consecutive levels is given by $$E_{n+1} - E_n = (2n+1)E_1.$$ In my PhD interview, it was asked that how can we make these energy levels equispaced? I know that the quantum harmonic oscillator energy levels are equispaced, but how can we make the levels equispaced for one-dimensional infinite square well potential? Is this related to Bohr Correspondence Principle somehow?

the question is not very well defined because it is not clear what we are allowed to change. One thing I can think of is if we make the particle massless, then the energy disperssion becomes linear. We don't have $E_p \propto p^2$ but rather $E_p \propto p$. So if we have $H=\sqrt{c^2p^2 + m^2c^4} + V(x)$ and set $m=0$ then we get the same eigenstates of the standard version (sine and cosine) but the energy is $E \propto n$, making it linear and equally spaced.

If space is a vacuum, how do stars form? According to what I have read, stars are formed due to the accumulation of gas and dust, which collapses due to gravity and starts to form stars. But then, if space is a vacuum, what is that gas that gets accumulated?

Basically speaking, the space used to be a gigantic nebula, but most of it collapsed, forming the primordial stars. These stars went supernova, spreading it's material throughout space and making nearby nebula collapse. This process goes on, with the difference that the new stars have heavier elements. This is an oversimplification, though.

Why are time-ordered Greens functions equal to retarded Greens functions at zero temperature? When I calculate a photon polarization diagram: I get the same answer: * *If I calculate it in equilibrium (retarded Greens functions) with finite chemical potential, in the limit of zero temperature, or *If I calculate it as a scattering amplitude (time-ordered Greens functions) in a particle bath $|\psi\rangle = \prod_{|\vec{p}|<p_F}c^\dagger_{\vec{p}}|0\rangle$. The equality follows from a weird cancellation from the Fermi-Dirac distribution at zero temperature. Is there a theorem or some argument that this should happen in general (i.e. for more general correlation functions)?

assuming you work in perturbation theory, at zero temperature the (unperturbed) state is the vacuum, meaning that $\psi_k(t)|0\rangle = 0$, so the only part of the retarded GF $$g^r(k, t) = \theta(t)\langle 0 | \psi_k^{\dagger}(0)\psi_k(t) + \psi_k(t) \psi_k^{\dagger}(0) | 0 \rangle$$ that survives will give the same result as the time-ordered one $$g(k, t) = \theta(t)\langle 0 | \psi_k(t)\psi_k^{\dagger}(0)|0\rangle + \theta(-t)\langle 0 | \psi^{\dagger}(0)\psi(t)|0\rangle$$

Rotating Rod and conservation of energy The question says "A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Figure is attached below). The rod is released from rest in the horizontal direction" (a) What is the angular speed of the block when it reaches its lowest position I have tried the problem assuming the total mass to be concentrated at the center of mass so and the reduce in potential energy would give the increase in rotational energy $\frac{MgL}{2}$ must be equal to $\frac{Iω^2}{2}$ , but as I assumed the total mass is to be at center of mass which makes I as $\frac{ML^2}{4}$ but am ending with an incorrect solution. The textbook solution has taken moment of inertia of the straight rod as have assumed to reduce in potential energy of Centre of mass. When taking potential energy of COM then why is moment of inertia of rod considered than considering moment of inertia of COM with respect to the axis? Please suggest me why am I wrong?

I assumed the total mass is to be at center of mass which makes I as M(L/2)^2 but am ending with an incorrect solution. Its very incorrect way of finding moment of inertia. Moment of inertia is found by: $$I=\int r^2dm$$ So $I$ depends on the distribution of mass from the axis, and not the center of mass. The standard value of $I$ for a rod (treated as 1 dimensional object), about center of mass is: $I=ml^2/12$. I am leaving you a link, where you can see the derivations of $I$ for standard objects:Link (Make sure you know basic calculus for derivations) Here, the rod is hinged from one end, so by Parallel axis theorem, you can find $I$, about one end of rod ($I=ml^2/3$). Note that always moment of inertia is calculated about the axis about which rotation is taking place Then you can proceed with your question. Hope it helps!

Intuition for Spin operator in arbitrary direction I understand why the Spin operators in $x$, $y$ and $z$ direction are given by : $\begin{align*} S_x = \begin{pmatrix} 0 &\hbar/2\\ \hbar/2 & 0 \end{pmatrix} S_y = \begin{pmatrix} 0 & -i\hbar/2\\ i\hbar/2 & 0 \end{pmatrix} S_z = \begin{pmatrix} \hbar/2 & 0\\ 0 & -\hbar/2 \end{pmatrix} \end{align*}$ But why is the spin operator along an arbitrary direction $\vec{n}$ given by : $S_{\vec{n}} = n_x \cdot{S_x} + n_y \cdot{S_y} +n_z \cdot{S_z}$ ? I can see that it works along the $x$, $y$ and $z$ axis, and that is look like a scalar product between $\vec{n}$ and $ \textbf{S} = (S_x,S_y,S_z)$. I don't need a rigorous proof, a more physical explanation would be ok. I saw this post related, but no satisfactory answer. EDIT: Little precision, what is not clear for me is why I can do stuff with $\textbf{S}$ like if it was a vector. Also I would not be satisfied if you just say "it transforms like a vector". It is also not really clear what it would mean to take a scalar product with $\textbf{S}$.

I'm not sure what reasoning took you to accept that, $$S_{x}=\boldsymbol{e}_{x}\cdot\boldsymbol{S}$$ is OK, but let's take it from there. There's nothing special about “$x$”, as opposed to “$y$”, or “$z$”. You could have said $\boldsymbol{n}$, instead of $\boldsymbol{e}_{x}$, and you would have, $$S_{\boldsymbol{n}}=\boldsymbol{n}\cdot\boldsymbol{S}$$ IOW, $x$ is a dummy parameter there. As a simpler example that hopefully will clarify the question (Pauli matrices carry "internal space" representations), consider this: Suppose you have a physical law that tells you that a certain scalar $\sigma$ that depends on a given direction, x –that's relevant to your problem– is: $$\sigma=x$$ This is unsatisfactory as a physical law, because it does not have any definite transformation law under rotations. $\sigma$ is a scalar, but $x$ is not. If you want to make it right, you have to upgrade it to a physical law but referring it to an arbitrary direction and make the transformation law transparent: $$\sigma\left({\boldsymbol{n}}\right)=\boldsymbol{x}\cdot\boldsymbol{n}$$

Decomposition of product of two antisymmetric Lorentz tensors Suppose I have a tensor $A_{\mu\nu}$ in the $(3,1)\oplus (1,3)$ representation of the Lorentz group where $(a,b) =(2s_a+1,2s_b+1)$. I was wondering on how to decompose explictly in terms of tensors the prouct $A_{\mu\nu}\otimes A_{\rho\sigma}$ (where it is the same antisymmetric $A$ in the two factors of the product) in a sum of irreducibile representations. If I am not wrong I have that: $$[(3,1)\oplus (1,3)]\otimes [(3,1)\oplus (1,3)] = (5,1)\oplus (1,5)\oplus 2\, (3,3)\oplus [(3,1)\oplus (1,3)]\oplus 2(1,1)$$ However it is not at all clear to me how to translate this into an explicit representation in terms of tensors. i.e. I would like to do the analogous of: the product of two vectors $V_\mu\otimes V_\nu$ is: $$(2,2)\otimes (2,2) = (3,1)\oplus (1,3)\oplus (3,3)\oplus (1,1)$$ which can be easily written as: $$V_\mu\otimes V_\nu = A_{\mu\nu}+S_{\mu\nu}+\frac{1}{4}g_{\mu\nu}T$$ where $A_{\mu\nu}$ is antisymmetric, $S_{\mu\nu}$ is symmetric and traceless, while $T$ is a scalar times $g_{\mu\nu}$ which is Lorentz invariant. In the case of vectors it is trivial to explictly build these tensors. However, in the case I am asking about, I don't find the procedure to build the analogous objects so obvious or straightforward.

In general dimension, the decomposition is well understood, it follows from basic Young manipulations. In terms of Dynkin labels $$ (0,1,0,\dots,0)^2=(0,\dots,0)+(2,0,\dots,0)+(0,1,0,\dots,0)+(0,0,0,1,0,\dots,0)+(1,0,1,0,\dots,0)+(0,2,0,\dots,0) $$ of dimension $$ \left(\frac12 d(d-1)\right)^2=1+\left(\frac12d(d+1)-1\right)+\frac12d(d-1)+\frac{1}{4!}d(d-1)(d-2)(d-3)+\frac{1}{8}d(d-3)(d-1) (d+2)+\frac{1}{12} d(d-3) (d+1) (d+2) $$ or, in words, a scalar, a rank2 traceless symmetric, a rank2 anti-symmetric, a rank4 anti-symmetric, a rank4 tensor with mixed symmetry (anti with respect to three indices, the fourth one is not), and a rank4 tensor with the symmetries of the Riemann tensor (and traceless). In $d=4$ things are a little bit different because some of these break into self-dual and anti-self-dual parts; this happens to the rank2 anti-symmetric and the Riemann tensor. Also, some rank4 tensors become isomorphic to rank2 tensors, in particular the rank4 anti-symmetric becomes a scalar and the rank4 tensor with mixed symmetry becomes a rank2 traceless symmetric. So, to summarize, \begin{align} 2\, (3,3)\oplus 2(1,1)&:\text{ two rank2 symmetric tensors}\\ (3,1)\oplus (1,3)&:\text{ dual and anti-self dual parts of a rank2 anti-symmetric tensor}\\ (5,1)\oplus (1,5)&:\text{ dual and anti-self dual parts of a rank4 traceless tensor with the symmetries of the Riemann tensor} \end{align} i.e., schematically $$ A^2=2S+A+R $$ where both $A$ and $R$ can be broken into chiral parts if so desired, and $S$ into trace and traceless parts.

RMS velocity of a gas vs RMS velocity of a gas molecule, which is a more appropriate term? I'm a bit confused by the terminology. Is it the RMS velocity of a gas, or the RMS velocity of the gas molecules or of a gas molecule? Similarly, is it the mean velocity of a gas, or the mean velocity of the gas molecules or of a gas molecule?

The RMS velocity of a gas is the square root of the arithmetic mean of the squares (a particular generalised mean) of its molecules' instantaneous velocities.

Sound proofing: mass-spring-mass or mass-mass-spring? I am trying to improve the sound proofing of a metal box. The box is made of steel of thickness $0.8 \;\mathrm{mm}$. I have additional sheets of steel with $3 \;\mathrm{mm}$ thickness for reinforcement. For vibration reduction, I have access to some bitumen mats (anti-vibration mats for damping cars) and some alubutyl (butyl rubber with an aluminum foil on top). What would be the best way to improve the insulation properties of the box? (A) Glue both steel plates together $(0.8 \;\mathrm{mm}+3.0 \;\mathrm{mm})$, then add bitumen/alubutyl to reduce vibrancy. (B) Put the anti-vibration mat right into the middle to achieve kind of a mass-spring system. The bitumen mats are pretty stiff and thus will probably be not sufficient to act like a damping spring. The alubutyl however is more like gooey grubber and might be better for attenuating the mechanical vibration. On the other hand: It is only $2 \;\mathrm{mm}$ thick. My take on that is that (A) might be a more rigid construction (if glued tightly) while (B) might be better for blocking mechanical vibration to travel from the inside to the outside. The noise source I want to isolate (computer hard drive) adds both structure-borne and air-borne resonance in/to the case. I am looking forward to your suggestions.

Try sheets of cardboard. They don't transmit sound well. Consider if you leave an opening for cooling air flow, you have an opening for sound. Open cell foam might help? Pointing the opening away from listeners? Computer fans are typically noisier than hard drives. But if it is really the drive, you can get a solid state drive.

There is a way of making a Balloon lighter than Helium/Hydrogen (or vacuum)? I don't know much about physics, but I wonder if there is a way of making something like a Balloon to float more than Hydrogen, Helium or Balloon with Vacuum. I say "Vacuum Balloon" because, supposedly, if you had a strong and light enough material to make a Balloon with vacuum to not be crushed by earth atmosphere, it would float. From what I heard from questions about Vacuum Balloons is that, supposedly, you would just be around 5% to 7% lighter than conventional Helium and Hydrogen Balloons. And of course, when someone asks about something beyond that, it always gets to a simple Jet Engine or Anti-gravity, which is... Well, not possible right now. Well, I know that if there were something as good as anti-gravity out there, there would be billions of investment being thrown into it. But I'm just Wondering if there is something out there that kinda looks promising for "floating purposes".

No, not in air. Helium is good. Hydrogen is better. Warm hydrogen has an even lower density and is thus even better. Vacuum is as low as it gets, but you need a massive hull to avoid crushing. All that matters for buoyancy is the difference in density between the inside of your balloon and the outside. So, take an air balloon and put it in water, and it'll push upwards much, /much/ more than a hydrogen balloon in air.

What allows this tornado to form in a frying pan? I have just seen this video here where a mini tornado has formed in a frying pan: https://www.youtube.com/watch?v=6CJ-8ze2FjE Does anyone know how this happens or where I can find an explanation?

The window on one side is open providing an asymmetric air supply, and you can hear what sounds like a loud overhead extractor fan running. With the hot pan providing a source of rising smokey hot air, these are great conditions for producing an indoor tornado. The window appears to be cracked open, forming a vertical slit (off-camera to the left) letting air into the kitchen very close to the pan. As Mike Dunleavy noted in his answer to a Whirlpools and Tornados question, "Tornadoes happen when angular momentum of air is concentrated sufficiently in a vortex vertical to the ground." The window slit here provides a slightly off-center air supply that has non-zero angular momentum relative to the hot air rising from the pan. Just like a figure skater spinning faster as they draw in their arms and leg, angular momentum conservation makes the air spin faster as it is drawn into the vortex core. For examples of how simply having an asymmetric vertical-slit air supply can turn hot rising air into a tornado, check out Best Fire Tornado - DIY - no moving parts! and How to build a DIY Dry ice tornado! - Fluid dynamics in a vortex!. An overhead fan also helps to increase and concentrate the vertical air flow, as shown in How To Make An Indoor Tornado In 3 Easy Steps!.

Polarization of natural light for double slit interference Last week I performed Young's double slit experiment using a laser. As expected, I obtained an interference pattern as predicted by Fraunhofer theory (enveloped by the 1 slit diffraction curve). Then, I added two polarizers with perpendicular axes in front of each slit. Since the light coming from the slits had now different polarizations, I no longer noticed the interference pattern (only the diffraction from each separate slit). It is generally known that in order to perform Young's double slit experiment using natural light, we should first make the natural light "more" coherent by placing a single slit before the double slit. However, from my understanding, natural light is unpolarized. So now my question is, why does natural light produce an interference pattern if the incoming light in unpolarized? Shouldn't the light from the 2 slits not interfere because the polarizations are different just like in the laser experiment? Does the first slit polarize the light? Or is it that the light exiting the 2 slits is synchronously unpolarized (the electric field vectors are always on the same line) hence creating a pattern?

Unpolarized light has a polarization vector which varies randomly with time, so at any given time, the polarization at the left and right slots will be the same (and thus they will interfere). However, natural light also has low spatial coherence (i.e. the phase varies across the wavefront over a relatively short distance). So this incoherence can manifest as a polarization discrepancy between slits (not likely a 90$^\circ$ discrepancy if the slits are close; generally something less).

What are the magnitudes of the acceleration of the falling balls B and C relative to A? (Need help with binomial approximation) I'm mostly having a problem incorporating the approximation. My trouble is towards the bottom of the post, but I wrote down all of my steps just in case I made an error somewhere. So in this question 3 balls start off with zero velocity and begin to fall towards the earth. They're vertically aligned, with ball A in the middle, ball B 22m directly above A, and ball C 22m directly below A. The question asks me to find the magnitude of the acceleration of ball B and C with respect to A. The question also says to use the binomial approximation because otherwise when you numerically compute the magnitude your calculator will not store enough sig figs to give you an accurate result. So far I did: $F = ma $ The acceleration for ball A is $ \vec{a}_A = \frac{F_A}{m_A} $, and ditto for balls B and C Force due to gravity is $ F_B = \frac{G M_e m_B}{r_B^2}$ where $M_e$ is the mass of the earth, and $r_B$ is the radius between the center of the earth and the center of ball B. Now the acceleration due to the earth on ball B is $ \vec{a}_B = \frac{G M_e}{r_B^2}$ We want the acceleration of B with respect to A so we can write $ \vec{a}_B - \vec{a}_A = \frac{G M_e}{r_B^2} - \frac{G M_e}{r_A^2} $ We can rewrite $r_B = r_A + 22m$ So $ \vec{a}_B - \vec{a}_A = \frac{G M_e}{(r_A + 22m)^2} - \frac{G M_e}{r_A^2} $ $ \vec{a}_B - \vec{a}_A = G M_e \left[\frac{1}{(r_A + 22m)^2} - \frac{1}{r_A^2}\right] $ Now this is where I'm lost. I don't know how to incorporate the approximation $ (1+x)^n \approx 1+nx $ $ (r_A + 22m)^{-2}$ doesn't follow the $(1+x)^n$ format I put the expression $\frac{1}{(r + x)^2}$ into wolfram to see if there are any alternate forms of the expression that I can apply to the approximation and I couldn't find any. I also tried to factor out $\frac{1}{r_A^2}$ to get $ \frac{G M_e}{r_A^2} \left[\frac{r_A^2}{(r_A + 22m)^2} - 1 \right] $ But this doesn't work with the approximation either. If anyone could point me in the right direction or give me a hint it would be very much appreciated Edit: The question is P1.5 from Moore's General Relativity Workbook

When you have an expression of the form $(a + b)^n$, where $|a| \gg |b|$, you can rewrite it as $$(a + b)^n = \left(a\left(1+\frac ba\right)\right)^n = a^n \left(1+\frac ba\right)^n$$ and then, given that $\left|\frac{nb}a\right| \ll 1$ (and $a + b > 0$), apply the binomial approximation to obtain $$a^n \left(1+\frac ba\right)^n ≈ a^n \left(1 + n\frac ba\right).$$ In your example $r_A ≈ 6371\,{\rm km} \gg 22\,{\rm m}$, so we can write $$\frac1{(r_A + 22\,{\rm m})^2} = (r_A + 22\,{\rm m})^{-2} = r_A^{-2}\left(1 + \frac{22\,{\rm m}}{r_A}\right)^{-2} ≈ r_A^{-2}\left(1 - 2 \cdot \frac{22\,{\rm m}}{r_A}\right)$$ and then subtract $r_A^{-2}$ to obtain $$\frac1{(r_A + 22\,{\rm m})^2} - \frac1{r_A^2} ≈ r_A^{-2}\left(1 - 2 \cdot \frac{22\,{\rm m}}{r_A}\right) - r_A^{-2} = - 2 \cdot 22\,{\rm m} \cdot r_A^{-3}.$$ Multiply this by $GM_e$, and you'll have your answer.

Time evolution of $\pi/2$ pulses This is the topic Ramsey interferometry. I want to do this without referencing the Bloch sphere, just with the Hamiltonian (given on Wikipedia and below) and Time-Dependent Schrodinger Equation. A $\pi/2$ pulse in a two-level system is defined as a pulse that sends the excited and ground states as follows, $$| g \rangle \rightarrow \frac{1}{\sqrt{2}}( | g \rangle+| e \rangle).$$ $$| e \rangle \rightarrow \frac{1}{\sqrt{2}}( -| g \rangle+| e \rangle).$$ And the Hamiltonian of the system is given as follows: $$\hat H =\frac{-\hbar}2 (\begin{matrix} -A & B \\ B & A\end{matrix})$$ So the eigenvalues/vectors are $$E_{1,2}= \pm \hbar C/2$$ $$|\psi_{1,2} \rangle = (A \mp C,-B)$$ Where $$C = \sqrt{A^2+B^2}$$ Two $\pi/2$ pulses are given at $t=0$ and $t=\tau$. How do I find the final wavefunction after both pulses have been applied $| \psi(t>\tau) \rangle = c_g(t)| g \rangle + c_e(t) | e \rangle $? My question is mainly: what time evolution operators do you apply and when, and when do you apply the unitary operator of the $\pi/2$ pulse?

My question is mainly: what time evolution operators do you apply and when, and when do you apply the unitary operator of the /2 π / 2 pulse? You apply the unitary time evolution operator in between the $\pi/2$-pulses. This is in the order that they physically happen during a Ramsey sequence. The evolution of the state can then be written as $$|\text{out}\rangle=U_{\pi/2}U(t)U_{\pi/2}|\text{in}\rangle,$$ where $U(t)$ is the free evolution of the atom in-between the $\pi/2$-pulses. One is then normally concerned with something such as the difference in population between the ground and excited state or the population in just one. For the difference, one would then measure $$|\langle\text{out}|\sigma_z|\text{out}\rangle|^2,$$ where $\sigma_z=|e\rangle\langle e|-|g\rangle\langle g|$. The same can be done for $|c_{g,e}(t)|^2$ the correct operator just needs to be used in the above, i.e. $|e\rangle\langle e|$ or $|g\rangle\langle g|$.

How fast can manned spacecraft slingshot from Black Holes? I've read of Black Holes launching particles at over 99% the speed of light. Could manned spacecraft use Black Holes to slingshot ourselves at this speed, or would the G forces kill us? Intuitively, I worry the inertia of turning the curve would kill the crew.

This won't work unless the black hole is orbiting something else. It doesn't make a lot of difference to this whether we use a black hole or any other type of body - it's just gravity and conservation fo momentum and energy at work. If you just had an isolated body and approached it, a slingshot won't happen. All that happens is that your trajectory will change (and so will that of the other body - conservation of momentum etc.) You don't gain any speed. If the body you approach is orbiting another (as planets in the solar system orbit the Sun), then you can do a slingshot. What is different is that you are, in essence, "dragged along" with the planet's orbital motion. By choosing direction of approach properly you can get that boost in velocity (a slingshot, or gravity assist) because of the "dragging along" effect. Now in your case you just have an isolated body (a black hole). That doesn't get you a gravty assist. You can use it to change course but not gain energy. Energy you gain in falling towards it will be lost climbing away.

Is there any sensor that creates a visual image of magnetic fields? I am working on a project where I am trying to create a magnetic tag that can be detected from $20 - 50$ cm away. Is there any sensor on the market capable of this?

You maybe interested in looking here: Magnetic field gel based viewing film or for more advanced applications the ferrolens (commercial product named Ferrocell). However these optical magnetic sensors need more likely to be in physical contact with the magnetized matter and have a sensitivity at about 400 Gauss (i.e. 40mT) in average magnetic field strength. I doubt you can find such an optical magnetic sensor at 20cm-50cm that works at that distance range from the magnetic source? Unless we are talking about a very strong magnetic field source as for example in MRI where you can have 1T (10,000 Gauss) strength and above fields. Note: MRI field strength value indicated above, refers to the Βr residual magnetism value thus field strength on contact, zero distance from the magnetic source which in a multi-poles MRI configuration can be translated into 0.3T (3,000 Gauss) and above field strength on air many cm away from the source. You may try this product here the mini-Ferrocell which has a sensitivity of 100 Gauss (10mT) which translates to a field strength on air a few cm away from a normal N42 grade Neodymium dipole magnet with Br value on contact about 5,000-6,000 Gauss (i.e. Br 500 to 600 mT). If you increase the strength of the magnetic source on contact, say for example at 1T (i.e. 10,000 Gauss) Br value, you should gain more distance separation and still get a visible response from this sensor but no more than 8 cm distance at maximum in my experience. Nevertheless, all these should not be applicable in your case since your magnetic tag as you described it should have no more than a few mT, Br residual magnetism strength on contact.

The effect of gravitational lensing during the lunar eclipse During the lunar eclipse, the Moon turns into bloody colour while the shadow of the Earth is casting over the lunar surface. The red hue can be explained by means of the refraction of light and Thomson scattering. Besides these two, will the gravitational lensing effect take a role in the eclipse? That is, the Sun light is bent due to the Earth and shone onto the lunar surface, and then the light is reflected towards the Earth. If so, how significant will this effect be?

Earth's gravity is far to weak to make much difference. But the Sun's gravity makes a solar lens that could be used as a telescope, if we devoted enough resources and ingenuity to it. We could literally map the surface of an exoplanet $100$ light years away at $25$ km/pixel. See The Solar Gravitational Lens will Map Exoplanets. Seriously.

If a person drops a briefcase in an elevator and it does not fall to the floor, what is the elevator's aceleration? I read this question on my Physics book and I'm still wondering whether my answer is right. My first thought is that the elevator is accelerating downwards. If it were accelerating upwards the briefcase would fall to the floor even quicker than if the elevator was not accelerating at all. Then, the question is, how much is it accelerating? My answer is $a \ge g$. If the acceleration of the elevator is the same as that due to gravity, the elevator and the briefcase will experience the same acceleration, and the briefcase will never touch the floor. If, on the other hand, the acceleration of the elevator is greater than the acceleration of the suitcase, then the elevator is moving faster than the suitcase and the suitcase will eventually touch the elevator's roof. Is my reasoning correct? I found a good explanation online, but they claim $a = g$ is the only right answer. Am I right to think that $a > g$ is plausible too?

If the briefcase remains in equilibrium with the elevator, then the elevator is accelerating downwards. This means that $a=-g$, where the minus sign means the motion is downwards. If the briefcase rises to the elevator's ceiling, then $a \geq -g$.

Question about indices and matrix This is essentially a trivial question, which can be answer probably immediately, but i have this doubt anyway. If, say, $$\Lambda^{a}_{b} = \begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}$$ $$\Lambda^{b}_{a} = \begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}^T or \begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}^{-1}$$ ?? For example, suppose $$T^{c'd'}=T^{cd}\Lambda^{c'}_{c}\Lambda^{d'}_{d} = = \begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}T\begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}^T or \begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}T\begin{pmatrix} f & -fc\\ -fc & f \end{pmatrix}^{-1} $$ ?

The tensor indices can represent matrix elements, but to do so you have to give them an order! Placing them one on top of each other is ambiguous. For example: $ \Lambda_a \,^b $ has as rows $\Lambda_0 \,^b$. Then there is another issue. Transposing a matrix means interchanging its indices only if they are both contravariant or covariant. So what you have done there is lowered (or raised) one index, transposing the tensor, and then raising (or lowering) againg. In general, it doesn't have to be either the transpose or the inverse, although it can happen (as with the lorentz transformation tensors). For example take this matric together with the minkowski space-like metric: $ \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix} = \Lambda^a \, _b $ And see that it is neither the transpose or the inverse. (The edit was for this final sentece xD)

Deriving Hartree from fundamental atomic units leads to error? I am trying to find how many newton meters are in a Hartree by using the following definition in terms of other physical constants: $$E_h = \frac{\hbar^2}{m_ea_0^2} $$ The values of the other physical constants in SI units (according to Wikipedia and the NIST reference on constants) are: \begin{align} m_e &= 9.109\ 383\ 7015(28) \times 10^{-31}\ \mathrm{kg}\\ \hbar = \frac{h}{2\pi}; h &= 6.626\ 070\ 15 \times 10^{-34}\ \mathrm{kg} \frac{\rm m^2}{\rm s} \\ a_0 &= 5.291\ 772\ 109\ 03(80)\times 10^{−11}\ \mathrm m \end{align} I cannot reproduce the answer from the NIST database, which says that $$E_h = 4.359\ 744\ 722\ 2071(85)\times10^{−18}\ \mathrm J $$ However, the energy I get is $4.359\ 744\ 722\ 223\ 2755 \times 10^{-18} ... $ Can anyone reproduce their answer from the above numbers? Or alternatively, is there an explanation as to why the energy seems slightly off from the calculated value? (I am using the mpmath library in python with high precision. The code used to calculate and print the answer is below) mpmath.mp.dps = 75 m_e = mpmath.mpf("9.1093837015e-31") a_0 = mpmath.mpf("5.29177210903e-11") h = mpmath.mpf("6.62607015e-34") hartrees = mpmath.power((h/(mpmath.mpf(2)*mpmath.mp.pi)),2)/(m_e*a_0**2) mpmath.nprint(hartrees,50)

You are comparing \begin{align} E_h &= 4.359\ 744\ 722\ 2071(85)\times10^{−18}\ \mathrm J \\ \\ \frac{\hbar^2}{m_e a_0^2} &= 4.359\ 744\ 722\ 2232\ 755 \times 10^{-18}\ \mathrm J \end{align} Your final digits $\cdots 755$ are superfluous, smaller than the uncertainty. Your disagreement is $$ \frac{(\cdots 2232) - (\cdots 2017)}{85} =1.9\sigma $$ which is a little high. However, you’re using a different expression for $E_h$ than the NIST website. If I use their expression, I get $$ \alpha^2 m_e c^2 = 4.359\ 744\ 722\ 1987 \times10^{-18}\ \mathrm J $$ which is off by $-1.0\sigma$. The NIST expression is probably preferable to yours, because $\alpha$ and $m_e$ are independent of each other, while the definition of $a_0$ includes the electron mass. So to the extent that the electron mass is uncertain, that uncertainty multiplies your expression three times (once from $m_e$ and twice from $a_0^2$), while it enters the NIST expression only once. If you really want to understand what’s happening here, you probably want to read the ugly details of CODATA’s enormous least-squares fit of all the data on physical constants; see the links in this other constants-related answer. Note that you can get these results using regular double-precision arithmetic, rather than using a multiple-precision library. (Multiple-precision libraries have their uses; they also have their quirks.) The problem from an arithmetic standpoint is that, in a world where the Hartree energy is a constant of nature, the measured values of $\alpha$ and $m_e$ are inconsistent with each other at the one-ish-sigma level. Dealing with those kinds of inconsistencies is the purpose of CODATA's least-squares fit to the entirety of the constants literature.

Why doesn't a parallel circuit violate conservation of energy? Let's imagine a hypothetical circuit where there are a large number of wires placed in parallel to each other, hooked up to a simple power source. We know that voltage at each wire would be equal $V_{total}=V_1=V_2=...=V_n$ where $n$ approaches a large number; and that each wire is of some arbitrary constant length. Next, assume that at the start of each wire there is a single charge of $+1C$, in each wire placed in parallel. Since work done on a charge is $W=VQ$; where $W=$ work done, thus we apply the same voltage to each charge in each wire placed in parallel. Since the voltage across each wire would be the same (say, $Resistance$ is ineligible, but $\neq0$) the work done would be same. Additionally, we know $W=\vec{F}.\vec{s}$; Since the charge is displaced to a significant length (i.e of the wire) Thus work is done even if we may not be able to easily quantify force. My questions is this - if the number of parallely-placed wires increases, $W\uparrow$. Thus, we can gain infinite joules by placing more and more parallel wires violating the conservation of energy: \begin{equation} \sum_{i=0}^{\infty}W_i = V_i \times1 \end{equation} by moving the $+1C$ charge in each parallel wire. How is that possible?

It does not violate the conservation of energy rule. If you add more parallel wires instead of a single wire, the charge will get distributed between multiple wires but the total charge $Q$, which in second case will be the sum of all the smaller charges pushed through multiple wires, will stay the same. Attempting to drink a beverage through 10 straws at once, instead of 1 straw, won't make you empty the cup 10 times faster.

Vector operator formulation I showed that the ladder operators: $ \hat{\overrightarrow{a}}=(a_x, a_y , a_z)$ and $\hat{\overrightarrow{a}}^{\dagger} = (a_x^{\dagger}, a_y^{\dagger} , a_z^{\dagger})$ can form a vector operator by proving: $$ [J_k, a_l] = i \hbar \varepsilon_{klm} a_m \hspace{1,5cm} [J_k, a_l^{\dagger}] = i \hbar \varepsilon_{klm} a_m^{\dagger}$$ I also know that to construct spherical components it should look something like this: $$ V_1 = - \frac{1}{\sqrt{2}} (V_x + i V_y) \hspace{0,8cm}V_0 = V_z \hspace{0,8cm}V_{-1} = \frac{1}{\sqrt{2}} (V_x - i V_y) \hspace{0,8cm}(1) $$ I would simply plug them in the expressions $V_1$ , $V_0$ and $V_{-1}$ in order to get the spherical components for $\hat{\overrightarrow{a}}$ and $\hat{\overrightarrow{a}}^{\dagger}$ but now I'm wondering, how do I arrive at these general expressions for the spherical components of a vector operator $(1)$?

You would first find a linear combination of your operators so that $$ [\hat J_+,\hat{T}^\ell_m]=0\, , $$ and once you have that you can ladder down using $$ [\hat J_-,\hat T^\ell_m]=\sqrt{(\ell+m)(\ell-m+1)}\,\hat{T}^{\ell}_{m-1}\, . $$ This does not fix the “norm” of the operator, i.e. $A\hat T^{\ell}_m$ also has the right transformation properties for any constant $A$. $A$ can be considered a normalization factor. In practice, one can often “guess” at the form of the operator $\hat T^{k}_k$ by comparing with the spherical harmonics in Cartesian coordinates: since $(x+iy)\sim \hat{T}^{1}_1\sim Y_1^1(\theta,\varphi)$, then $(x+iy)^k\sim \hat T^{k}_k$ and ladder down from there. Indeed if you compare your $V_k$ with the spherical harmonics $Y_{1}^m(\theta,\phi)$ in Cartesian form you can immediately see how the combination $V_x\pm i V_y$ occur.

Solve the stationary Schrödinger equation with the finite difference method I am trying to solve the stationary Schrödinger equation for a double-level well potential through the finite difference method. Here is the shape of the potential I would like to solve it for where the second walls are infinite. The Schrödinger equation for this problem is the following $$ -\frac{\hbar^2}{2m} \partial_{xx}\psi(x) = [E-V(x)]\psi(x)$$ which can be turned into $$ \partial_{xx}\psi(x) = \frac{2m}{\hbar^2}[V(x)-E]\psi(x) $$ we can discretise the wave function and approximate the second derivative of $\psi_i$ as $$\partial_{xx}\psi_i \sim \frac{\psi_{i-1} - 2\psi_i + \psi_{i+1}}{h^2}$$ Then the Schrödinger equation becomes $$ \frac{\psi_{i-1} - 2\psi_i + \psi_{i+1}}{h^2} = \frac{2m}{\hbar^2}[V(x_i)-E]\psi(x) $$ which can be put as an eigenvalue problem as this $$ \begin{pmatrix} -2 & 1 & 0 & ... & 0\\ 1 & -2 & 1 & ... & 0\\ 0 & .&.&. & 0\\ 0 & ... & 1 & -2 & 1 \end{pmatrix} \cdot \begin{pmatrix} \psi_1\\ \psi_2\\ .\\ .\\ .\\ \psi_N \end{pmatrix} = \frac{2m}{\hbar^2} \begin{pmatrix} V_1 - E & 0 & ... & 0\\ 0 & V_2 - E & ... & 0\\ 0 & 0 & ... & 0\\ 0 & 0 & ... & V_N - E \end{pmatrix} \cdot \begin{pmatrix} \psi_1\\ \psi_2\\ .\\ .\\ .\\ \psi_N \end{pmatrix} $$ Here is my question: If I didn't have $V_i$ I would be using scipy.linalg.eigh() to solve this problem for $E$, however $E$ is now inside the second matrix. Is there a way to solve this equation finding the values of $E$ and relative sets of eigenvectors with Python? Follow up I implemented it as suggested in the comments as $$ \begin{pmatrix} -2-\frac{2m}{\hbar^2}V_1 & 1 & 0 & ... & 0\\ 1 & -2-\frac{2m}{\hbar^2}V_2 & 1 & ... & 0\\ 0 & .&.&. & 0\\ 0 & 0 &... & 1 & -2-\frac{2m}{\hbar^2}V_N \end{pmatrix} \cdot \begin{pmatrix} \psi_1\\ \psi_2\\ .\\ .\\ .\\ \psi_N \end{pmatrix} = -\frac{2m}{\hbar^2}E \begin{pmatrix} \psi_1\\ \psi_2\\ .\\ .\\ .\\ \psi_N \end{pmatrix} $$ and used scipy.linalg.eigh(). Here is my code import numpy as np from matplotlib import pyplot as plt from scipy.linalg import eigh # number of discrete bins in which the domain is divided N = 200 # create discretised arrays for the x axis and the value of the wave funciton xaxis = np.linspace(-1, 1, N) Vpot = np.linspace(-1, 1, N) # factor with energy hbar and mass m = 1 hbar = 1 # define the potential def V(x): if np.abs(x)<0.5: val = 0. else: val = 4. return val # fill an array for the potential for i in range(len(xaxis)): Vpot[i] = V(xaxis[i]) # initialize NxN matrix Hmat = [[0 for x in range(N)] for y in range(N)] # fill Hmat like the following # -2-2m/h^2*V1 1 0 0 ...0 # 1 -2-2m/h^2*V2 1 0 ...0 # 0 1 -2-2m/h^2*V3 1 ...0 # .....................0 # .........1 -2-2m/h^2*VN for row in range(N): for elem in range(N): if row == elem: Hmat[row][elem] = -2 -(2*m)/(hbar**2)*Vpot[elem] if np.abs(row-elem) == 1: Hmat[row][elem] = 1 # now I get the eigenvalue with w[i] and corresponding eigenvector v[i] energies, psi = eigh(Hmat, b=None, eigvals_only=False, turbo=True) # plot potential #plt.plot(xaxis, Vpot) # plot ground state plt.scatter(xaxis, psi[0], s = 1) plt.show() Two questions 1 - If I just put the potential (Vpot) to zero I get the classical particle in a hole solution. The solution clearly resembles the analytical one, however I get the signs mixed somehow I am not sure why this is happening. 2 - When I introduce the double level in the potential, I start getting something wrong, even here I don't understand what's happening. Any ideas?

First of all, I suspect one can solve this problem analytically. As for scipy.linalg.eigh() (https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.eigh.html), you just include V in matrix a.

Interpretation of $\oint PdV\neq0$ I hope you are excellent. I'd like you to help me make sense of the integral $ \oint PdV \neq 0 $ for some thermodynamic process. What can it mean for the integral to be nonzero? I can only interpret it as if there is work, however my deep understanding is very limited. I appreciate your comments.

Given integral repersent the work done to a closed path . Which is non zero means think as a non conservative force you will get intuition. Since force is directly related to pressure and then work done work done. hope this answer might help you to why work done through a closed path is non zero.

If I lift a body with a force greater than its weight, what will happen to the excess energy provided to the body I will give an example to explain my question. Case 1: An elevator lifts body a with force equal to its weight for a distance $d$ * *Energy given to the body (work done)$=$ Weight $×$ $d$ *Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ Weight $× \ d$ Case 2: An elevator lifts the same body with force equal to twice weight it’s for a distance $d$ * *Energy given to the body (work done) $=$ $2 \ ×$ weight $×$ $d$ *Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ weight $×$ $d$ So doubled the amount of energy I gave to the body yet it’s capacity to do work by falling down has not changed. Where is the excess energy the lift provided the body? (I am in 11th grade so please make your explanation simple enough for me to understand.)

Elevators go through a phase of acceleration, constant speed and then deceleration. It is only during the acceleration phase that the force is greater than the weight of the elevator and contents. The 'excess' force is converted into kinetic energy. During constant speed (by definition) the force is EQUAL to the weight (if it were greater then the acceleration would continue) As you approach the destination level the force reduces to be less than the weight and hence the kinetic energy is reduced and is converted to the final PE required to reach the destination. It is only if the deceleration rate is greater than 1 g (9.8m/s^2) that you would actually lift off the floor. Lifts always decelerate at a slower rate - so you only feel a slight 'lifting' of your weight on the floor. You may also look at the meaning of 'jerk' in this context - which is a measure of the increase or decrease in acceleration, and gives you that feeling of a bump as the elevator starts or stops.

Why does a body accelerate when there is a force applied to it? Why does a body accelerate or changes velocity when a force is applied on it? How force acts upon things to make them accelerate?

Frame challenge: Your question doesn't make sense; nor can it be readily answered in words cause i am from another world and i don't know about these things One thing that we assume for physics is that the laws of physics (such as we understand them) apply everywhere. Time and space may behave differently around mass, sure, but that's still following the same laws on Earth as elsewhere. So wherever you are in the universe, you understand pull or push. All that you don't understand is the words. You say "explain pull or push". I physically grasp your body, and pull and push. Now you understand pull and push. Words are merely descriptive of a physical action. At some point, the physical action must be observed in a way we both agree on. Adding extra words does not help explain this.

Microstates of the canonical ensemble In the micro canonical ensemble the microstates of a system in an arbitrary macrostate, are also eigenstates of the Hamiltonian. Does the same apply to the microstates of the canonical ensemble? Are they eigenstates of the the Hamiltonian? I would expect them not to be, since here the energy is not constant. But I am not sure

When talking about the canonical ensemble, one has to distinguish * *the Hamiltonian of the system of interest, $H_S$ *the Hamiltonian of the system of interest + bath/thermostat/reservoir, $H_{tot} = H_S + H_B + V_{SB}$ $H_{tot}$ is treated in a microcanonical ensemble framework, and hence we are discussing its eigenstates. Generally it will not commute with $H_S$, since there is some interaction between the system and the bath. The derivation of the canonical ensemble is however based on solid reasoning that the interaction energy is smaller than the energy of the system, and can be neglected in tehrmodynamic limit (roughly speaking, the energy of the system is proportional to its volume, whereas the interaction energy is proportional to its surface, i.e., scales as volume to power $2/3$.) Hence, once this logic is accepted and we talk about microcanonical ensemble, the microstates of the system are its eigenstates.

What is the relationship between a boost transformation and a high energy limit? S-matrix element is calculated by a following formula, $$S_{\beta \alpha}=\langle \beta_{in}|U(+\infty,-\infty)|\alpha_{in}\rangle.$$ The time-evolution operator $U(+\infty,-\infty)$ is $$U(+\infty,-\infty)=T\exp\Big(i\int^\infty_{-\infty}d^4x\ \mathcal{L}_I( \phi_{in}(x))\Big).$$ On the other hand, according to these slides, the scattering amplitude under the eikonal limit is defined by $$S_{\beta \alpha}\equiv\lim_{\omega\rightarrow+\infty}\langle \beta_{in}|e^{-i\omega K^3}U(+\infty,-\infty) e^{i\omega K^3} |\alpha_{in}\rangle,$$ where $K^3$ is the generator of Lorentz boosts in the +z direction. My question is, why is the Eikonal limit, a kind of high energy limit, related to the Lorentz boost?

Start at the classical level. Say that in frame 1, a point particle is moving slowly in the $x$ direction. Frame 2, is related to frame 1 by a big boost in the $z$ direction. What is the particle doing in frame 2? It is still moving slowly in the $x$ direction, but it also has a huge momentum in the $z$ direction. So a physicist in frame 2 can safely take the approximation that the particle is a relativistic particle moving along the $z$ direction. That is more or less what is going on in the last formula of yours. One of the key things to keep in mind is that transition probabilities are Lorentz scalars, so you can compute them in any inertial frame. So the first formula and the last formula give exactly the same probabilities---for any finite value of $\omega$. But when you take the limit, you are making a certain approximation.

Physical reasons for why systems are chaotic? Are there any reasons why a system would exhibit chaotic behavior? Or is this something only found through numerical modelling or experimental testing? For example, the simple forced, damped pendulum or the duffing oscillator. Were these experimented on and it was found that they were sensitive to initial conditions, and then examined further to prove the 3 chaotic properties and finally deemed to be chaotic? Or is there something physical about them that gives away a possibility to chaos? If it is the former, how would we determine chaotic systems? Just trial and error until all 3 properties are proven?

In a chaotic system, if you start with two initial states that are nearly identical, they will diverge from each other exponentially. Soon they will be in completely different states. Note that this implies the system will have no stable repetitive paths through its state space. Consider the example of the frictionless billiard table. Two initial states have a ball strike another in slightly different spots. They reflect at slightly different angles. They hit the next ball in spots that are further separated, and their angle of reflection is increased. Soon one ball will miss the next ball entirely. I do not know why exponential separation is required, as opposed to polynomial. If the system evolves long enough, exponential separation will always be larger. But it seems that polynomial would be enough to ensure that the system does not repeat itself.

Will electric field cause electrons to move in disconnected wire? Imagine a straight piece of wire, not connected to anything. Parallel to one of it's ends is a section of circuit with direct current in it. They're placed in such a way, that only about 1/4 (or less) of disconnected wire is near a powered circuit. Question: will electric field in circuit cause at least some of electrons in disconnected wire to move to other end (upper end on picture)? Will this happen, if circuit is a toroidal solenoid, and end of disconnected wire in inserted in hole in the middle of said toroid?

If you had an electric field, charge separation would build up in the disconnected conductor such that the electric field inside the conductor was zero. This movement of charges would constitute a current that very rapidly decayed to zero. However, neither of the diagrams you show appear have a changing magnetic field because the current in the blue wire appears to be depicted as constant. Therefore no electric field is induced and no current flows in either orange wire.

Finding the resonance frequency for forced damped oscillations I have a problem regarding a forced, damped harmonic oscillator, where I'm trying to find the resonance frequency. I have calculated the frequency for free oscillations as $$\omega_{free}=\sqrt{\frac{\kappa}{I}-\left(\frac{b}{2I}\right)^2},$$ where $b$ is the damping coefficient. To the best of my knowledge, $\omega_{free}$ should be the same as the resonance frequency, but when I try to calculate the resonance frequency from the amplitude $$A=\frac{\tau_0}{I \sqrt{(\frac{\kappa}{I}-\omega^2)^2 + (\omega \frac{b}{I})^2 }}$$ by finding the maximum value, I get a slightly different equation: $$\omega_{max}=\sqrt{\frac{\kappa}{I}-\left(\frac{\sqrt{2}\cdot b}{2I}\right)^2}.$$ Which one is correct to use as the resonance frequency, and why is $b$ in $\omega_{max}$ scaled by a factor of $\sqrt{2}$ compared to $\omega_{free}$?

Resonant frequency which people define as $\omega_0$ is not the frequency with maximum osccilation, https://youtu.be/Y_DmzZcQR7A Walter lewin explains this at 20:00 $\omega_{max}$ is.

Why there is no reaction Deuterium + Deuterium $=\rm {}^{4}He$? Why there is no reaction like $D+D={}^{4}He$ specified here and in other places like this? Apparently $2\times2.0141-4.0026=0.0256$ is positive. What is the problem with this reaction?

According to Krane's Introduction to Nuclear Physics (Chapter 14), the reaction $$ D + D \to {}^4\mathrm{He} + \gamma $$ is possible but rare. The gamma ray is necessary because (as you note) there is energy released in the reaction, and it has to go somewhere. However, the $Q$-value for this reaction (23.8 MeV) is greater than either the neutron or proton separation energy for helium-4, and so the reactions $D + D \to {}^3 \mathrm{H} + p$ or $D + D \to {}^3 \mathrm{He} + n$ are more likely. The likelihood of this reaction actually occurring is discussed in Wilkinson & Cecil, "2H(d,$\gamma$)4He reaction at low energies." Phys. Rev. C31, 2036 (1985) and references therein. Figure 5 of that article shows branching ratios for the $D + D \to {}^4\mathrm{He} + \gamma$ reaction of between $10^{-4}$ to $10^{-7}$, depending on the energy. (These branching ratios are calculated relative to the $D + D \to {}^3 \mathrm{H} + p$ reaction.) In other words, this reaction is somewhere between ten thousand and ten million times less frequent than the reaction in which a proton is released.

What would happen if a teaspoon of neutronium crashed through Earth's atmosphere? Pretty self explanatory hypothetical. I realize this is probably an impossibility and maybe it'd be more likely to be hit by a small black hole or primordial black hole. I'm just curious, because we always talk about how much a teaspoon of neutronium would weigh on earth.

The teaspoon of neutronium would have a mass of about $10^{12}$ kg. Due to it's small size (and area), the pressure caused by the impact would be very high, so it's likely to plough straight through the earth and out the other side. It depends on the velocity, but things coming from space are usually moving fast. As the earth has a mass of $6\times 10^{24}$ kg the orbit of the earth wouldn't be changed much.

Simple oscillator displacement, speed, and acceleration diagram I'm currently studying the textbook Fundamentals of Acoustics (2000) by Kinsler et al. Chapter 1.2 The Simple Oscillator says the following: $$\dfrac{d^2 x}{dt^2} + \omega_0^2 x = 0 \tag{1.2.5}$$ This is an important linear differential equation whose general solution is well known and may be obtained by several methods. One method is to assume a trial solution of the form $$x = A_1 \cos(\gamma t) \tag{1.2.6}$$ Differentiation and substitution into (1.2.5) shows that this is a solution if $\gamma = \omega_0$. It may similarly be shown that $$x = A_2 \sin(\omega_0 t) \tag{1.2.7}$$ is also a solution. The complete general solution is the sum of these two, $$x = A_1 \cos(\omega_0 t) + A_2 \sin(\omega_0 t) \tag{1.2.8}$$ where $A_1$ and $A_2$ are arbitrary constants and the parameter $\omega_0$ is the natural angular frequency in radians per second (rad/s). Chapter 1.3 Initial Conditions says the following: If at time $t = 0$ the mass has an initial displacement $x_0$ and an initial speed $u_0$, then the arbitrary constants $A_1$ and $A_2$ are fixed by these initial conditions and the subsequent motion of the mass is completely determined. Direct substitution into (1.2.8) of $x = x_0$ at $t = 0$ will show that $A_1$ equals the initial displacement $x_0$. Differentiation of (1.2.8) and substitution of the initial speed at $t = 0$ gives $u_0 = \omega_0 A_2$, and (1.2.8) becomes $$x = x_0 \cos(\omega_0 t) + (u_0/\omega_0) \sin(\omega_0 t) \tag{1.3.1}$$ Another form of (1.2.8) may be obtained by letting $A_1 = A\cos(\phi)$ and $A_2 = -A\sin(\phi)$, where $A$ and $\phi$ are two new arbitrary constants. Substitution and simplification then gives $$x = A\cos(\omega_0 t + \phi) \tag{1.3.2}$$ where $A$ is the amplitude of the motion and $\phi$ is the initial phase angle of the motion. The values of $A$ and $\phi$ are determined by the initial conditions and are $$A = [x_0^2 + (u_0/\omega_0)^2]^{1/2} \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \phi = \tan^{-1}(-u_0/\omega_0 x_0) \tag{1.3.3}$$ Successive differentiation of (1.3.2) shows that the speed of the mass is $$u = -U \sin(\omega_0 t + \phi) \tag{1.3.4}$$ where $U = \omega_0 A$ is the speed amplitude, and the acceleration of the mass is $$a = - \omega_0 U \cos(\omega_0 t + \phi) \tag{1.3.5}$$ In these forms it is seen that the displacement lags $90^\circ$ ($\pi/2$ rad) behind the speed and that the acceleration is $180^\circ$ ($\pi$ rad) out of phase with the displacement, as shown in Fig. 1.3.1. (Arrows in figure 1.3.1 are mine.) We can see from figure 1.3.1 that the displacement is out of phase with the acceleration by $\pi$ radians (green arrow), as stated, but it seems to me that, according to figure 1.3.1, displacement is actually $3\pi/2$ radians out of phase with speed (blue arrow), rather than the stated $\pi/2$ radians (red arrow). Is this an error, or am I misunderstanding this?

Express displacement $x$, speed $u$, and acceleration $a$ via the same trigonometric function with positive amplitude: $$x(t) = A \cos (\omega_0 t + \phi) = A \sin (\omega_0 t + \phi + \frac{\pi}{2})$$ $$u(t) = - \omega_0 A \sin(\omega_0 t + \phi) = \omega_0 A \sin(\omega_0 t + \phi + \pi)$$ $$a(t) = -\omega_0^2 A \cos(\omega_0 t + \phi) = \omega_0^2 A \sin(\omega_0 t + \phi - \frac{\pi}{2})$$ Note that if we used $\cos$ instead of $\sin$, the final result would have been the same. From the three equations above, the phase difference of: * *acceleration to displacement is $-\frac{\pi}{2} - \frac{\pi}{2} = -\pi = \pi$ *acceleration to speed is $-\frac{\pi}{2} - \pi = -\frac{3\pi}{2} = \frac{\pi}{2}$ *displacement to speed is $\frac{\pi}{2} - \pi = -\frac{\pi}{2} = \frac{3\pi}{2}$ From the graph, by looking at positive to negative transition points, the phase difference of: * *acceleration to displacement is $\frac{3\pi}{2} - \frac{\pi}{2} = \pi = -\pi$ *acceleration to speed is $\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2} = \frac{3\pi}{2}$ *displacement to speed is $\frac{\pi}{2} - 2\pi = -\frac{3\pi}{2} = \frac{\pi}{2}$ You can see that acceleration-speed phase difference for the two methods above is off by a factor $\pi$. (The same applies to the displacement-speed phase difference.) Note that I did this on purpose! The catch is in the convention you use, i.e. which angle comes first in the angle difference. It does not matter which convention you use as long as you stick to one convention! This means acceleration-speed phase difference of $\frac{\pi}{2}$ is valid only if displacement-speed phase difference is $-\frac{\pi}{2}$, which is also clear from the acceleration-displacement phase difference. You can read the above results as follows: * *acceleration leads speed by $\frac{\pi}{2}$, or acceleration lags speed by $\frac{3\pi}{2} = -\frac{\pi}{2}$ *displacement leads speed by $\frac{3\pi}{2} = -\frac{\pi}{2}$, or displacement lags speed by $\frac{\pi}{2}$ To see this side-by-side, acceleration and displacement * *lead speed by $\frac{\pi}{2}$ and $-\frac{\pi}{2}$, respectively; *lag speed by $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, respectively.

Observing relativistic motion from afar If I were to look at a clock from a very far distance ( let's say 10 light-years away ), and the clock starts to move on a direction perpendicular to my line of sight, with a speed approaching c (e.g. 0.999c), will I see the clock ticking at a different rate (slower/faster) than it was before starting to move? The large distance and it being perpendicular to the direction of movement are important because it makes the distance between the observation point and the moving clock constant (it would take a lot of time before the angle changes significantly).

Yes, you would see the clock running slowly- the result of the transverse relativistic Doppler effect. See https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect

Mathematical background required for Lagrangian Field Theory? I want to start teaching myself Lagrangian Field Theory. I can do multivariable calc, tensor calc, Lagrangian mechanics, and some calculus of variations. Are there other math fields I should study before diving in to field theories? I’m hoping to work my way up to QFT.

I am no expert on the field, but when I was in college I was taught many concepts in Lagrangian and Hamiltonian mechanics that required group theory. It would be nice to have it on your maths background, apart from the ones you already have.

How does spontaneous symmetry breaking (SSB) happen? I've just finished studying for an exam on the Standard Model (so electroweak theory and symmetry breaking) and I can't figure out how this question never crossed my mind. I'm now studying the QCD chiral symmetry breaking, but I think my question applies to any (physical) SSB. I know what SSB is (symmetry of the Lagrangian but not of the states) and I also know how one implements it in a theory (scalar sector with a mexican-hat potential) and it's clear to me the implications of the two different phases, the broken one and the restored one (vev for the nonbroken scalars, Goldstone bosons eventually "eaten" by the gauge bosons). What I don't understand is, how does the phase transition work? How did the universe go from one phase to the other? Does the potential just "switch on"? Is it always on but at high energies the quantum/thermal fluctuations don't "see" its structure?

I'm just sketching the trail map of where your question might wish to go... it is a subject of limitless complexity. The universe cools down, and thermal QFT dictates mutation of the Higgs potential with temperature, section 3. This mutation of the effective Higgs potential from the form favoring the symmetric phase to one favoring SSB at lower temperature describes when the phase transition is likely to switch on, at $T_c$.