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Why light shows its wave-like properties only when it interacts with objects with dimensions close to the wavelength of light? In Young's Double Slit Experiment, we were taught that light behaves as a wave here because the width of the slits are very close to the wavelength of light itself. But why does light behave like a wave only when it interacts with objects that have dimensions close to the wavelength of light? Even in my book no explanation is given as to why this is true. Can someone please explain as to why this is true? I am so confused.

Consider an optical lens, say a magnifying glass. The light interacts with it, its path bending as it passes through. Yet the lens is far larger than the wavelength, while the individual atoms are much smaller. It is simply that interference effects are easiest to see when the dimensions are similar, as evidenced in the Young's Slits experiment.

Purely resistive AC circuits Is it possible to have a purely resistive RC circuit? Recently came across a question: A 130Ω resistor and a 40μF capacitor are connected in series to an AC source of frequency ω. For the combination to be purely resistive what must be the value of frequency ω? Is this question right or are they missing the info about inductance?

The only way the circuit can look "purely resistive" is if there is an inductor in series with the capacitor and resistor and where the frequency is such that the inductive reactance equals the capacitive reactance, or $$\omega L=\frac{1}{\omega C}$$ $$\omega=\frac{1}{\sqrt {LC}}$$ where $\omega$ is now the resonant frequency. Hope this helps.

How do Standing Waves satisfy the criteria in order to be considered as Waves? Why are standing waves even categorised as waves if they don't transfer energy from one point to another? Waves are generally defined as disturbances that transfer energy but standing waves don't fulfil that criterion. Even though they are formed by the superposition of two travelling waves (of the same frequency) but if we just consider the resultant standing wave, I am not sure that how a standing wave can be considered as a wave in the first place.

Energy is being transported. A standing wave can be written as a superposition of a left-moving wave, which transports energy to the left, and a right-moving wave, which transports energy to the right. For a standing wave, it happens that there is no net transfer of energy to the right or to the left. But there are still processes transferring energy around the string. A standing wave is clearly different from a string at rest. Let's try an economic analogy, although as a warning analogies are prone to breaking down. My analogy is that we could say that an economy is the transfer of goods and services. If the exports and imports into a country balance out, there is still trade occurring and therefore an active economy, even though the net trade is zero. The economy is different from the state where no trade of any kind is occuring.

If we free a metal plate of free electrons will it become weaker to external mechanic forces? If we free a metal plate of free electrons by means of a strong electric field will it become then weaker in case of applied external forces? To make it simpler, do free electrons have any role in connections between lattice atoms?

When one puts a capacitor at the ends of a battery, the metal plates on the negative side will have a lot more charge and the other plate will display a positive charge, which means a lack of electrons. So in this sense you do free electrons from one side and add them on the other. do free electrons have any role in connections between lattice atoms? As long as the currents induced in the circuit while "freeing electrons" are within values that will not start melting the circuit, the plates will be unaffected too. High currents start melts, i.e. destroying lattices. A good electrical conductivity is the same as a small electrical resistance. An electric current will flow through all metals, however they still have some resistance, meaning the current needs to be pushed (by a battery) in order to keep flowing. The bigger the resistance, the harder we have to push (and the smaller the current is). Current flows easily through copper thanks to its small electrical resistance, without much loss of energy. This is why copper wires are used in mains cables in houses and underground (although overhead cables tend be aluminium because it is less dense). However, where size rather than weight is important, copper is the best choice. Thick copper strip is used for lightning conductors on tall buildings like church spires. The copper strip has to be thick so that it can carry a large current without melting. So it will be the process of removing electrons that generates melting currents destroying the solid state lattice.

Low-energy excited states of $^{13}C$ While doing some studying for an exam in introductory nuclear physics, I stumbled upon a question I can't answer. I'm supposed to explain the ground state and the three lowest energy excited states of $^{13}C$ using the shell model. The ground state has spin and parity $1/2^-$. The excited state with $E = 3.09 \text{ MeV}$ corresponds to $1/2^+$, the $3.68 \text{ MeV}$ state to $3/2^-$ and $3.85 \text{ MeV}$ to $5/2^+$. I know that $^{13}C$ has one unpaired neutron, which will determine the spin and parity of the nucleus. $1/2^-$ must then correspond to a $p_{1/2}$-level, $1/2^+$ to an $s_{1/2}$-level, $3/2^-$ to $p_{3/2}$ and $5/2^+$ to $d_{5/2}$. (See table below.) The unpaired nucleon must reside in the level corresponding to each state. \begin{align} 0 \text{ MeV} && 1/2^- && p_{1/2} \\ 3.09 \text{ MeV} && 1/2^+ && s_{1/2} \\ 3.68 \text{ MeV} && 3/2^- && p_{3/2} \\ 3.85 \text{ MeV} && 5/2^+ && d_{5/2} \\ \end{align} In order to transition from $1/2^-$ to $1/2^+$, we can either let the unpaired neutron in $1p_{1/2}$ jump to $2s_{1/2}$ or let a paired neutron jump from $1s_{1/2}$ to $1p_{1/2}$. Other options seem out of the question, since the $3s_{1/2}$-level lies much higher and would therefore require too much energy to achieve. How would I determine which of the two would happen?

Unfortunately there is no easy rule to unambiguously determine whether the valence neutron is in 2s1/2 or 1s1/2 orbital. The situation is even more complex if you consider the interacting shell model, which include 2-body interaction between those single-particle states. Usually if you consider some specific model space in shell model (for this case is spsdpf) you can roughly estimate based on the single-particle energy but still it is a nontrivial task for many cases. A proper shell model calculation will give you the (fractional) occupation number for each orbital or the spectroscopic factor for specific configuration, which are the information you need. For the specific case you ask, from the particle-vibration coupling (PVC) model by N. Vinh Mau, Nuclear Physics A 592, 33 (1995), the 1/2+ is mainly (91.5%) from a valence neutron 2s1/2 + 12C core in ground state. It is interesting to note that it also include a small 8.5% fraction from the configuration of 1d5/2 + 12C core in 2+ excited state.

Is Landau theory for phase transitions valid only for "order to disorder" phase transition? In the Landau theory we assume order parameter that is equal to zero at $T>T_c$ and none zero at $T<T_c$ which is valid only for order to disorder phase transition according to my understanding. So that is mean that I can't use Landau theory on Liquid - Gas phase transition?

The Landau paradigm is getting a bit outdated in my opinion (not only) . There are plenty of different phase transitions which cannot really be categorised as first or second order. Enough to look at lattice models of quantum gravity or the standard model, but in analytic models also the study of Kibir-Żurek mechanism shows that. Many phases such as transition to spin glass, cannot be categorised according to order/disorder... Other aspects like long range entanglement can come into the picture. The Landau description is a model related to the 20th century , when with 20th century technology we were able to create 20th century materials. Most of them could be analysed fairly easily in the labs. Nowadays we need to trap ions with lasers, excite composite systems, create a 2d lattice to give rise to anyons, create well designed lattices. One should look at phase transitions from a different perspective, because the old categorisation can easily break in many cases. The Landau theory is not only valid to order-disorder but this is the most typical type of distinction, when your order parameter is zero or nonzero on the two sides.

Where is energy in energy density? I was learning about energy density and it seemed to be defined as the potential energy per unit volume in an electric field $\frac{dU}{dV} = \frac{1}{2}\epsilon E^2$ But how can just the electric field have a potential energy on its own without presence of any charge? What is causing this energy to be present in an electric field?

There are charges present. They are the cause for the electric field. Defining something based on the electric field is indirectly defining it based on the charges and charge configuration.

Why is the reflection coefficient 1 for step potentials where energy is less than the potential? Consider a potential $V(x)$ which is zero when $x<0$ and $V_0>0$ when $x>0$. Suppose there is an incident particle with momentum $p=\hbar k$ and energy $E = \hbar^2 k^2 / 2m < V_0$ coming from $x= - \infty$. Now then after scattering ($t \to \infty$) the wavefunction should be $$\Psi = \psi_{reflected} + \psi_{tunneling}$$ where $$\psi_{reflected} = A e^{-ikx} \;\; ,x<0$$ $$\psi_{tunneling} = B e^{- \kappa x}\;\; ,x>0$$ Because the wavefunction at $x>0$ is real, there is no probability flow there. So, using the definition of reflection coefficient where it is the ratio of the reflected to incident probability current, the reflection coefficient comes out to be 1. But the reflection coefficient is also defined as $$R = \int |\psi_{reflected}|^2 dx \;\;\;, t \to \infty$$ which doesn't make sense because there should be some non-zero probability that the particle is in the region where $x$ is positive.

There are two ways of thinking about it, one is strictly mathematical (shut up and calculate, in a way), and another is maybe not very precise but rather intuitive. The first one you did correctly, $R=1$. In a potential $V=V_0$ for all $x>0$ (continuing infinitely "long" to the right) there is no way for the particle with $E<V$ be "transmitted", so reflectivity is indeed 1. This means that all particles will be reflected sooner or later (simply because they cannot be transmitted). It does not mean that you cannot register a particle at some point under the potential; even it gets there, it will still be reflected. The apparent contradiction originates in that we tend to think that the act of reflection happens at the point $x=0$ only; this is not true.

Does mass have an effect on Centripetal Acceleration? I am using an online simulation for a lab concerning Centripetal Acceleration. When I change the mass the graph indicates that the magnitude of the acceleration is constant. According to the Centripetal Acceleration formula: $a=v^2/r$, this is true because no mass is present in the relationship. However, when I use Newton's Second Law of Motion, $a=f/m$, I can see that the mass and the acceleration are inversely proportional. Both of these ideas are found when I look them up online, now I am a bit confused on which one might be more valid.

Here's a more math-y restatement of the same general answer :) Just from your question, we see that $v^2/r = a = f/m$. If the simulation keeps $v^2/r$ constant ($a_1$ = $a_2$), then when $m$ changes, $f$ should change. Depending how sophisticated the simulation is, it may or may not report the value of $f$. If you solve for $m$, the result is something like $m=\frac{fr}{v^2}$. From this you can see that when $m$ changes, at least one of the other terms will change. Maybe they all do; maybe it's just $f$.

Occlusion angle subtended by a black hole Suppose we have a Schwarzschild black hole of mass $M$ at a distance $R$ to an observer. Due to the curvature of space, the apparent angle (or half-angle) of occlusion $\theta_a$ of the silhouette of the blackhole will be larger than the angle described by its Schwarzschild radius, $\theta_s$. But how much larger? Taking no relativistic effects into account, the angle subtended by the center of a sphere to the edge of a sphere with a radius the same size as the Schwarzschild radius $r_s=\frac{2GM}{c^2}$ can be deduced to be $\theta_s=\arcsin\frac{r_s}{R}=\arcsin\frac{2GM}{Rc^2}$. However, a light ray cast from the observer at an even larger angle (maximally $\theta_a$) may find it's way falling into orbit or directly into the blackhole, causing the apparent size of the blackhole to be larger than its Schwarzschild radius. How is $\theta_a$ described by $R$ and $M$? Is it the angle (or half-angle) subtended by the photon sphere?

Any light ray that crosses the photon sphere will hit the horizon, and any light ray that doesn't, doesn't. So, your guess that it would be the angle subtended by the photonsphere is pretty good. However, in reality the angle is slightly bigger, since two light rays that start parallel at infinity will have bend further inwards by the time they reach the photon sphere. The quantity you want to look at as the critical impact parameter $b_{crit}$. This is the distance between two light rays that start parallel at infinity at are tangent to the photon sphere. For a Schwarzschild black hole with mass $M$: $$b_{crit} = 6\sqrt{3}\frac{G M}{c^2} $$ So for a sufficiently distant observer at distance $R$ (ignoring any cosmological effects due to expansion) the angle $\theta$ subtended by the black hole is given by: $$ \theta \sim 6\sqrt{3}\frac{G M}{ R c^2}.$$

Lepton Universality Does universality of lepton also mean that two lepton families have the same renormalized coupling if they interact with a particular particle?

Wikipedia: “The coupling of leptons to all types of gauge boson are flavour-independent: The interaction between leptons and a gauge boson measures the same for each lepton.” [Note: Recent experiments are raising questions about this.] Lepton universality does not apply to their Higgs couplings; if it did, they would have the same mass. As ACuriousMind put it so well here, “Lepton universality means all leptons behave the same, except when they don't.”

Change in the mass of a star due to nuclear fussion of Hydrogen Let's suppose a star is initially composed only of hydrogen. After a certain time $t$, it has consumed the $5\%$ of its initial hydrogen, fusing it to produce helium, according to the following reaction: $$4 H^1 \rightarrow He^4 +\Delta E$$ I'm trying to find out the ratio of the initial mass $M$ and the mass after the fussion of the 5% of the initial Hydrogen, $M/M'$. My attempt Assuming that this nuclear reaction is the only one that occurs, if initially there were $N_H$ attoms of Hydrogen and no attoms of Helium, then after the $5\%$ of the H is fussed, there will be $N'_H=0.95N_H$ and $N'_{He}=\frac{0.05}{4}N_H$. However, since the mass of H is 1 a.m.u. and the mass of He is 4 a.m.u., $M'=(0.95N_H·1 + \frac{0.05}{4}N_H·4)\text{ u.m.a.}=N_H\text{ u.m.a.}=M$ Therefore, does the mass remain constant during the process of nuclear fussion of Hydrogen?

If the atomic mass of $^4\text{He}$ were exactly $4$ times the atomic mass of $^1\text H$ then $\Delta E$ would be zero and the mass of the star would be constant (and we would not be here to think about this problem). However, the atomic mass of $^4\text{He}$ is slightly less than $4$ times the atomic mass of $^1\text H$, so ...

EM wave power dependency on frequency Does a (classical) radio wave with a given amplitude carry more power if that wave is at a higher frequency than at a lower frequency?

Perhaps you need to ask a similar question on Amateur Radio StackExchange. There they could explain to you the observable details of how a given antenna rod and a given power from a wave generator change the output of the antenna. I will give you an idea of how the achievable output power of a radio wave changes at constant input power. The power of a radio wave depends from the number of involved surface electrons on the antenna rod, the rate of emission from these electrons per time and the wavelength of the emitted photons. Depending from the length of the rod there is an optimal frequency of the wave generator. For this frequency the radiated energy has its maximum. For lower frequency the electrons get accelerated forth and back less often and the number of emitted photons decreases. For higher frequency among other things the heat losses increases and the radiated power goes down.

Are the physical structures in our sun of comparable complexity to those in the human brain? The writings of Rupert Sheldrake tend to provoke strong emotions, be they ridicule, curiosity, outrage, sympathy, disgust, or otherwise. While Physics SE is not an appropriate forum in which either to debunk or to promote his general worldview, it does strike me as an appropriate place in which to examine some of the more specific claims he has made. One recent claim which struck me as interesting - I lack the expertise either to support or to denigrate it any further than that - was that various celestial bodies could be conscious - note the conditional - on the grounds that the physical structures within, for example, our sun are at least as complex as those found within the human brain. (The paper in which Sheldrake sets out these ideas can be read here.) Is there any evidence in support of the specific claim about the complexity of structures within our sun? Is there any evidence to the contrary? Is it even meaningful, within the language of physics, to talk about Structure A being more complex than Structure B?

We cannot compare the complexity of objects; we can only compare the complexity of our models for these objects. And when we talk about models, the complexity of a model increases when our understanding of the object increases. Given enough resources and motivation, we can research an object forever. Observations and research of the human brain predates the written history. So does the knowledge of the human nature and behavior. Everyone needs and learns at least some of this knowledge. On the other hand, the research of the Sun internals is rather recent - as much as few centuries and limited to a small group of physicists. We have quite a clear understanding that we are not in a position to control anything that happens in the Sun, so our motivation is somewhat limited to almost pure scientific interest. To sum up, our models of human brain are much more developed and much more complex than anything we know about the Sun.

A clarification on acceleration and velocity This is one of those questions which require an answer that does not take practical limitations into account. It is a theoretical physics question, perhaps. If there are any loopholes used, please explicitly state them. If the position is known as $x(t)$ from t=0 to t=1 second, how do I get the velocity at the initial and end points, since velocity at the end point will require $x(1-(\Delta t)/2)$ and $x(1+ (\Delta t)/2)$, which are added and divided by $\Delta t$ ? It gets worse if I want to know the acceleration at the end point, which requires the $v(1+(\Delta t)/2)$ which in turn requires $x(1+(\Delta t))$, which is simply not available. Is this an order thing or is it just neglected in calculus?

This question is more interesting if you refer to numerical calculations. For analytical tratement, it is simple: velocity is the derivative of position. If you say you know $x(t)$ for $t\in[0,1]$, then velocity is $v(t)=dx/dt$. But yes, the endpoints can be more problematic. However, real numbers are so dense that you can easily extrapolate the velocity function. In other words, if you know $v(t)$ for $t\in (0,1)$, then you can extend it to close the interval. $$v(t=0)=\lim_{t\rightarrow 0} v(t)$$ On the other hand, if you work with discrete time for numerical calculations, then this extension is more problematic if the interval size is big enough. The initial value is also given sometimes.

What does an $x$ Watt bulb actually means? Let's say I have a 11W bulb in my home, connected to a 220V power supply. What exactly does it mean that this bulb is 11W? As far as I know, the Wattage is determined by the formula $W = V * I$, so it really depends on the voltage and the current. The current is determined by the resistance: $V = R * I$. So I don't understand what is the meaning of a 11W bulb?

Yes, the current depends on the internal resistance. And that resistance has been adjusted to give a current that corresponds to 11 W. (By the way, note that we typically don't symbolise the wattage, the power, with $W$ but rather with $P$.)

Are Lie algebras of groups unique? Take for example $GL(2,\mathbb R)$, the group of $2\times2$ invertible matrices with real entries. By considering small variations from the identity, it is clear that one needs four parameters to parametrize this group, and hence we will need four "infinitesimal" generators. If I am thinking about this correctly, we could take as generators the matrices $$ J_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad J_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad J_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \quad J_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. \tag{1}$$ But we could equally well take some inspiration from the Pauli matrices and use $$ L_1 = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad L_2 = \frac{1}{2} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad L_3 = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad L_4 = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \tag{2}$$ Both of these sets are linearly independent, so they seem like they should be fine, but they produce different Lie bracket (commutation) relations. Specifically, for the first set we have the (somewhat messy) relations $$[J_1, J_2] = J_2, \quad [J_1, J_3] = -J_3, \quad [J_1, J_4] = 0, \quad [J_2, J_3] = J_1 - J_4, \quad [J_2, J_4] = J_2 \quad [J_3, J_4] = -J_3,\tag{3}$$ while for the second set we have $$[L_1, L_2] = L_3, \quad [L_1, L_3] = L_2, \quad [L_2, L_3] = L_1, \quad [L_i, L_4] = 0.\tag{4}$$ My question When speaking about certain Lie algebras, I often read/hear people refer to the Lie bracket structure, say $[S_\alpha, S_\beta] = i\hbar\epsilon_{\alpha\beta\gamma}S_\gamma$, as the Lie algebra of the group (representation) in question. But, if my argument above is correct, it seems that this structure is not unique, but indeed depends on the generators chosen. So I can see two possible resolutions: * *One (or both) of my suggested sets of generators is wrong. If this is the case, could you tell me why? *The Lie bracket structure for a given group is not unique, and those that say so are being sloppy with language somehow. Some of the more mathematically oriented sources seem to imply that the Lie algebra is actually a sort of tangent space about the identity transformation, with the generators as basis vectors. Then maybe it is this space that is unique, while the bracket structure is basis dependent?

Like most mathematical objects, when we ask whether two Lie algebras are "the same" we should really be asking whether these is an isomorphism of Lie algebras between them. Since a Lie algebra is a vector space with extra structure (the Lie bracket), such an isomorphism is an isomorphism of vector spaces $f : L_1 \to L_2$ that also has the property that $$ [f(v), f(w)]_2 = f([v,w]_1)$$ for all $v,w\in L_1$. So when someone writes down a definition of a Lie algebra on $n$ generators $e_i$ with some bracket $[e_i,e_j] = f_{ijk}e_k$ and someone else a definition on $n$ generators $e'_i$ with $[e'_i, e'_j] = f'_{ijk}e'_k$ these two definitions describe the "same" Lie algebra if there is an isomorphism of Lie algebras between them. As you have explicitly demonstrated by choosing different bases for the same Lie algebra, two such descriptions can be isomorphic even if their structure constants $f$ and $f'$ look different. This should be familiar - $f$ and $f'$ are basis-dependent expressions of a structure on the vector space like matrices for inner products: If you have an orthonormal basis, the inner product matrix is just the identity matrix, but if your basis isn't orthonormal, it's something else. That doesn't mean the inner product is somehow not uniquely given by taking one basis and saying it's a certain matrix in that basis, nor does it mean any other choice of basis is somehow less "valid" than another.

Solution to differential equation If I have a differential equations of the form $$\frac {d^2y}{dt^2}=\alpha^2y$$ Assuming the roots of the characteristic equation is complex the solution to the differential equation is: $$y=C_1e^{j\alpha t}+ C_2e^{-j\alpha t}$$ and after that we take only the real part of the solution. Why do we take only the real part of the solution? I'm solving the wave equation and this confusion stems from the solution of the Helmholtz wave equation.

This is a common question in classical physics. The idea is that we are really treating a problem where everything is real number-valued, but complex numbers offer convenient mathematical short-cuts. In the present example you have a differential equation $$ \frac{d^2y}{dt^2} = \alpha y $$ and let's agree at the outset that all of $y$, $t$ and $\alpha$ are real. The idea is that we can introduce a complex variable, let's call it $z$, which also satisfies the differential equation we are interested in: $$ \frac{d^2z}{dt^2} = \alpha z . $$ Now bring in the fact that $z$ can always be written $$ z = f + j g $$ where $f$ and $g$ are real. Then we have $$ \frac{d^2}{dt^2}( f + j g) = \alpha (f + j g) . $$ This gives $$ \left(\frac{d^2 f}{dt^2} - \alpha f\right) + j \left( \frac{d^2 g}{dt^2} - \alpha g\right) = 0 $$ where everything in the equation is real apart from $j$. But the only way a complex number can be zero is if both its real and imaginary parts are zero, so we have $$ \frac{d^2 f}{dt^2} = \alpha f $$ and $$ \frac{d^2 g}{dt^2} = \alpha g . $$ The lesson is if a complex function satisfies a linear differential equation, then its real and imaginary parts also satisfy that same differential equation. That is the thought which you need to ponder, and which lies at the heart of a lot of applications of complex analysis to physics and engineering. The general idea is that we find a complex solution, because that is convenient to do, and then take the real part at the end, since we know that it too will be a solution. We can set this up at the start by announcing "the actual physical thing I want to find out about ($y(t)$ in your example) is the real part of this complex thing which I am introducing". We get so used to doing this that often we just carry on using the same symbol (such as $y$) rather than going via another name (such as $z$). People then begin to get into a muddle between what is the function they are really finding out (i.e. the real function) and what is the mathematical tool they have introduced (the complex function).

Why aren't all actions conformally invariant? I am very confused about coordinate invariance of actions in classical field theories on arbitrary background spacetime or even with dynamical metric. From this question, we see that if the integrated term, namely the Lagrangian density, of the action is well-defined, i.e. is a 0-tensor (a scalar), then the action is necessarily coordinate invariant. Now, as a conformal transformation is a particular type of coordinate transformation, shouldn't any action be conformally invariant? I know that is certainly false but I am really interested in a rigorous explanation.

Note the words "conformal transformation" can mean slightly different things in different places. But conformal transformations, in the sense that it's used in conformal field theories, are not just coordinate transformations. Instead they are a simultaneous coordinate and field transformation such that the metric is left invariant. This often feels rather odd as conformal transformations are usually only talked about as being coordinate transformations, but keeping careful track of what's going on in sources such as the Di Francesco et at text will reveal that while they perform coordinate transformations, they always employ the flat metric. Polchinski's string book is a little more explicit about this. But if you want the real hard proof that conformal transformations, as they appear in CFTs, are not just coordinate transformations, look no further than the transformation of the stress tensor under such a transformation. This transformation rule can be found in any source on CFTs and is usually one of the very first things written down/derived. You will note that the transformation is very simply not the transformation of a rank 2 tensor under a coordinate transformation. Instead there is an additional term (the Schwartzian) in the transformation rule which comes from the fact that we are transforming the metric directly as a field at the same time (it's simultaneous Weyl transformation). For a couple specific references, see eq (4.31) in David Tong's string theory notes or eq (2.4.26) in Polchinski's strings book.

Can the mass-shell equation be derived from the path integral formulation I've recently been trying to wrap my head around the notion of virtual particles, which as far as I understand live in quantum histories which can never be observed directly and which are not bound by certain laws of physics, but which may evolve into physically possible histories and thus produce a noticeable effect on the expectation value of quantum observables through interference. The equation I've seen which separates "real" states which we can potentially observe from "virtual" ones which we can't is the "mass shell equation" $$E^2=m_0^2c^4+{\bf p}^2c^2.$$ This reminds me of Richard Feynman's informal derivation of the action principle towards the end of this lecture https://www.feynmanlectures.caltech.edu/II_19.html , where by treating both physically possible and impossible histories on equal footing, he gives a reason why histories with non-stationary action should be vanishingly unlikely. Can a relativistic version of this approach give us the mass shell equation on top of the Euler-Lagrange equations? Can the mass shell equation be derived from the action principle somehow? Am I seeing a connection where there isn't one and the virtual particles of QFT are different from Feynman's non-stationary paths?

Since Feynman considers point particles let us discuss this case. Yes, OP's speculations can indeed be realized. The Hamiltonian action $$\begin{align}S_H[x,p,e]~=~&\int d\tau ~L_H, \cr L_H~=~&p_{\mu}\dot{x}^{\mu}-\frac{e}{2}(p_{\mu}p^{\mu}+m^2), \end{align} $$ for a relativistic point particle has the mass-shell constraint $$p_{\mu}p^{\mu}+m^2~\approx~ 0$$ as one of its Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post.

Amperian loop for different situations Why are amperian loops always taken as circles? What if we take it as a triangle or rectangle? Give an example by taking amperian loop as rectangle for current carrying straight wire and derive equation for magnetic field? How will the equation change?

An Amperian loop can be any closed path. Ampere's law is true for any closed path. $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\rm enc} $$ Whether the law is useful or not depends on whether you are able to easily calclulate the LHS (the closed line integral of the B-field around the loop), and the RHS (the total current passing through that loop). It is usually the LHS that determines what shape of loop you should choose. The ideal case would be where the B-field is either parallel or perpendicular to the loop line element, all the way around the loop, and/or of constant magnitude. That makes the LHS easy to calculate. In the case of a long wire, there is symmetry around the axis of the wire, such that the B-field curves around the the wire and thus is constant in magnitude at a given distance from the wire and will always be parallel to the line element on a circular path. Thus $$\oint \vec{B}\cdot d\vec{l} \rightarrow 2\pi r B_r\ .$$ There are examples where non-circular loops are used. The most well known would be the long solenoid. A rectangular loop placed with one of its straight sides along the axis of the solenoid and its opposite side outside the solenoid is most useful here, since the B-field in the solenoid is directed along the axis and the B-field outside the solenoid is $\sim 0$. Thus $$\oint \vec{B}\cdot d\vec{l} \rightarrow L B_z\ , $$ where $L$ is the length of the rectangle.

How to generate electric current without a permanent magnet? The question is pretty simple: Can we build a device that coverts mechanical work in electric current1 without employing a permanent magnet and without access to any external source of current? The restrictions in place seem to rule out the possibility of current generation via induction; and I cannot think of another practical method. I have heard that industrial alternators sometimes work with electromagnets, but we don't have access to any external source of current, so this path doesn't seem viable. Do we really need stupid magnetic rocks to produce current? Unacceptable. To be more specific and minimize to risk of misunderstandings: my question is more or less equivalent to the following one Can we build a device, powered by hand via some sort of rotating lever, that produces electric current, crucially without employing any external current and without any permanent magnet? [1]: Usable electric current, let's say sufficient to properly power up a lamp; doesn't matter if AC or DC.

To answer this question one needs to think of a form of energy and then a device that will convert that energy into electrical energy. Mechanical - Wimshurst machine, Van de Graaff generator, piezoelectric crystal, perhaps a self-exciting generator? Light - Solar cell Thermal - Thermocouple Nuclear - Atomic battery Sound - Contact microphone Chemical - battery Gravitational - via a mechanical device without a permanent magnet In terms of lighting a bulb then a store of electrical energy, eg a capacitor, might be needed.

How could Leptoquarks explain Lepton Flavour universality (LFU) violation? I recently read about the possibility of Leptoquarks and that this new particle could also explain a possible LFU violation. Why would introducing a new particle explain LFU violation?

What is a leptoquark? Leptoquarks (LQs) are hypothetical particles that would interact with quarks and leptons. Leptoquarks are color-triplet bosons that carry both lepton and baryon numbers. Their other quantum numbers, like spin, (fractional) electric charge and weak isospin vary among theories. Leptoquarks are encountered in various extensions of the Standard Model, In general extra elementary particles in a new model modify the effect of higher loop corrections in the calculations of an interaction. The fact that leptoquarks carry lepton number means that higher order loop corrections can be different for decays that end up in leptons depending on the type of lepton and the mass of the leptoquark, thus affecting universality.

Is pseudo force just an ad hoc number to explain motion in non-inertial frames? Consider an observer in a non-inertial frame $S$ who observes a particle's motion with a relative acceleration $\vec a_s$ and further calculates (or was told by his fellow observer in an inertial frame) the net real forces acting on it as $\vec F$. Now the observer adds a pseudo force $\vec f_s$ to the net real forces acting on the particle to explain relative acceleration $\vec a_s$. Question: Is what I described above the "proper" method to calculate pseudo forces? In other words, is a pseudo force just an arbitrary constant that arose out of our desperation to explain non-inertial motion?

Inertial forces aren't desperate attempts to explain motion in non-inertial frame, but rather correct explanation of motion in non-intertial frames. In physics, everything is about frames of reference. For observer at inertial frame looking at someone at merry go round, centrifugal force doesn't exist, however this force is real for someone on merry go round since from his perspective it is imposible to correctly explain its motion (equation of motion) without it. There is no absolute frame of reference in physics. Electromagnetism manifests itself as electric or magnetic force in different reference frames. Theory of relativity is fundamentally derived from differences which arise when the same phenomenon is observed in different frames of reference only here one observer moves relatively to another at speeds close to light speed. Sometimes physicists say that inertial forces are pseudo forces because they don't exist in all frames of reference while some other forces do (like centripetal force).

Problem with gravity Sorry if this question is dumb, but I don't seem to have a grasp on it. Suppose you are on a rock in space, with no external forces acting. The rock attracts you with a force given by $$F=G\frac{m_1m_2}{r^2},$$ and you also attract the rock with an equal and opposite force. The ground exerts the normal force which keeps you stationary, but what stops the rock from accelerating, however small that be?

Just like the ground exerts force on you stopping you from accelerating, your legs push the rock stopping it from accelerating.

I am moving right and rain is falling vertically down.Why should I hold umbrella at an angle? This is related to relative velocity.I get that, from my moving frame of reference rain is making an angle.But still... it doesn't make sense to hold umbrella at an angle when rain is falling vertically down.

Imagine yourself in a train car with no roof (to let the rain in) and no windows (so you don't know you're moving). In that car, all you know is that the rain is coming at you at an angle to the vertical. That the rain is falling vertically to someone standing by the train tracks is not relevant to your experience. To best prevent from getting wet, you will hold the umbrella at an angle to the vertical. There are other observers moving in different directions and at different speeds relative to the train tracks. In their frames of reference, they all experience the rain falling at different angles. When you say "the rain is falling vertically down" that is just referring to one particular frame of reference and no one frame has the "correct" or "actual" direction of the rain.

Shift Vector in Warp Equation In the Alcubierre metric, why is there a beta with subscript multiplied by a beta with superscript? I know beta with subscript is the shift vector, but what is the difference between the two? $$\text ds^2 = -(\alpha^2 - \beta_i \beta^i) \text dt^2 + 2 \beta_i \text dx^i \text dt + \gamma_{ij} \text dx^i \text dx^j$$

This is how the dot product is defined for covariant and contravariant vectors (without explicitly inserting the metric) i.e, with the metric, the dot product would look like $$\beta \cdot \beta = \beta_i \beta^i = \gamma_{ij} \beta^j \beta^i$$ Note that $$\beta_i = \gamma_{ij} \beta^j $$ and $$\beta^i = \gamma^{ij}\beta_{j}$$ The metric $\gamma_{ij}$ raises and lowers indices in the equation you have mentioned.

Is temperature of 1 Kelvin equivalent to 1 eV in natural unit? We know that the Boltzmann's constant, $k_B$=8.617 $\times$ $10^{-5}$ eV/K. Now in the natural unit, $k_B=1$. So can I say, in the natural unit, 1 K temperature is equivalent to 1 eV in energy? 300 K is equivalent to 300 eV? Or am I missing something?

No. The situation is similar to the situation in special relativity where we use units in which $c = 1$ instead of $c = 3.00 \times 10^8 \text{ m/s}$. Using $c = 1$ does not mean that 1 meter is now equivalent to 1 second. It means that we are defining our unit for distance to be the distance traveled by light in 1 second. In other words, "1 second of distance" is the distance traveled by light in 1 second of time. In the case you're interested in, where we use "natural units" in which $k_B = 1$, temperatures are measured in units of energy. In other words, we could say that a system has a temperature of 1 eV, or 300 eV; but these are not equivalent to 1 K or 300 K any more than "1 second of distance" is equivalent to 1 meter. Rather, we are defining our temperature scale as something like, "if the average KE of an atom in a monatomic ideal gas is 3 eV, then the temperature of the gas is defined to be 2 eV." The change from 2 to 3 is because of the $\frac32$ factor in the equipartition theorem for the ideal gas: $$ \langle E \rangle = \frac{3}{2} T $$ if $k = 1$. (Finally, a nitpick on your question: in conventional units, $k_B = 8.617 \times 10^{-5} \text{ eV/K}$, not $8.617 \times 10^{-5} \text{ eV}$. This may have been a typo, but note that the correct units make it clearer that Boltzmann's constant is a "conversion factor" between temperature and energy.)

What is the definition of a magnet or a magnetic field? Electric forces are the forces which come about between two types of charges, positive and negative. Gravitational forces are the forces between matter. Nuclear forces are the forces which act on the atomic scale and are quantum mechanical forces, they act between nucleons. And, magnetic forces are the forces between magnets? I felt this definition wasn't specific enough so I searched up the definition of a magnet. "A magnet is a material or object that produces a magnetic field" Well, what is a magnetic field? "A magnetic field is a vector field that describes the magnetic influence on magnetic materials" I feel like this is a loop. I am sorry, but I am not satisfied and feel like there is a more fundamental definition of magnetic field or magnet that I am not aware of. So, what is a magnet?

There is a very good explanation of how electric charges in motion create the thing we call a magnetic field in this paper, "Magnetism, Radiation, and Relativity", Supplementary notes for a calculus-based introductory physics course by Daniel V. Schroeder, Weber State University (http://physics.weber.edu/schroeder/ dschroeder@cc.weber.edu). I struggled for years to understand this topic and I did not make any real progress until I read this paper. Schroeder's basic point is that moving charges exert special forces on charges that are not moving, and we call those special forces "magnetism" for convenience. I recommend you have a look at the paper and let us know if this helps.

What are the physical principles at play when a glass is stuck to a wet table? I've decided to write a relatively detailed paper on the following situation, but I'm finding the topic quite hard-to-google. Imagine a glass table with water spilled onto it. Once a drinking glass is placed on top, it becomes quite difficult to remove. It is more than the weight of the drinking glass that you have to overcome in order to lift it from the table. Some more specifics and details: * *The contact surface of the drinking glass is a circle. *Both the table and the glass surfaces are completely flat. *There is no air trapped between the table and the drinking glass when the latter is placed onto the puddle of water resting on the former. So, main question: how can we model this situation, taking all relevant things into consideration, in terms of the force required to lift the drinking glass? Secondary question bombardment: * *Does the thickness of the layer of water matter? (If so, in what way?) *How thick is a layer of water anyway? *Is atmospheric pressure at play here? *I assume the surface tension/viscosity of the water too?

Let's model the glass as a cylinder with radius 4cm and height 10cm. Air pressure is about 101,000Pa, there is a downward force due to this pressure on the top of the glass, usually balanced by a similar upward force due to the air inside the glass. The water makes an airtight seal if the surface is flat enough. When you try and lift the glass, the water stays in contact with it, due to surface tension, let's say the water can 'stretch' 1mm without breaking, when you try and lift the glass straight up - but any higher it breaks. then from the formula $$PV=nRT$$ the volume is increases by a factor 10.1/10 = 1.01 and the pressure inside the glass decreases by the same factor, to 100,000Pa, giving a pressure difference, between the inside and outside of the glass, of 1000Pa from $$P=\frac{F}{A}$$ with an area of $\pi\times 0.04^2$, about 0.005 square meters, we find the force needed is $$F = 5N$$ plus the weight of the glass. About the weight of 5 apples plus the weight of the glass.

Does inverse square law indicate beam angle of radiation source? I was told that the radiation intensity from a particular X-ray machine falls in half every 4 feet. I think this should tell us something about the angle of the beam from the machine. I use trigonometry to restate the assertion as follows: Radiation is emitted at some angle θ producing a cone whose diameter grows by a factor of $\sqrt{2}$ every 4 feet. Therefore: I think $\theta = \tan^{-1}(\frac{\sqrt{2}}{4})$. The problem I see is that the term $\sqrt{2}$ is unitless, so this would be taking the arctan of inverse feet, not of a unitless ratio. (And since angles aren't in units of inverse-length the statement appears to be meaningless; "4" is not anchored to anything – it could just as well be units of meters or lightyears.) What am I missing here? The initial statement seems valid, and I don't see how it can be true of more than one beam angle.

the radiation intensity from a particular X-ray machine falls in half every $4$ feet This is the important part. After $4$ feet you have half the original intensity. After $4$ more feet, you have half of that, or $1/4$ of the original intensity. After $4$ more, $1/8$. And so on. This is exponential decay $$I = I_0 \cdot 2^{-x/(4 ft)}$$ Try it out with x = 0 ft, 4 ft, 8 ft, etc. Exponential decay is common when light is absorbed. The x-rays are not spreading out. They do not travel far in air or whatever is around the machine.

Is there any experiment to check discreteness of space? This article from 2015 seems to suggest that there will be experiments to check discreteness of space: If space-time is discrete, there should be imperfections. And even if rare, these imperfections will affect the passage of light through space. No one has looked for this yet, and I’m planning to start such a search in the coming months. I can't find any reference of the experiment being done, so I'm wondering if it was carried out and what did the experiment show.

In the article itself it mentions that there have been previous (failed) efforts to detect discreteness of spacetime: If true, this would distort images from far-away stellar objects, either by smearing out the image or by tearing apart the arrival times of particles that were emitted simultaneously and would otherwise arrive on Earth simultaneously. Astrophysicists have looked for both of these signals, but they haven’t found the slightest evidence for graininess. The author's proposed experiment was apparently to look for "imperfections" in a hypothetical lattice. However, since the author's name doesn't appear to be supplied in the article (?) it's difficult to find what experiments they have actually performed. It is safe to surmise that if the experiment has been performed then it did not detect discreteness: any evidence of discrete spacetime would be a major discovery and would have made a big splash.

Time evolution in quantum mechanics of states not contained in the Hilbert space Eigenstates of, for example, $\hat p$, are not elements of the standard quantum mechanical Hilbert space, i.e. $\psi(x)=e^{ipx}\notin\mathcal L^2(\Bbb R)$. This prompts the question of - given that after measurement the state of the system becomes one of these seemingly problematic states - how the time evolution can be defined such that we are able to "re-enter" the space $\mathcal L^2(\Bbb R)$ in such a way that the time-evolution is a continuous operation.

The issue doesn't come up in practice, because perfect momentum eigenstates are idealizations that don't occur in the real world. In order to measure a particle's momentum with infinite precision and end up with a perfect plane wave, your measurement apparatus would need to be infinitely spatially large. Any real-world measurement apparatus comes with a range of experimental uncertainty, so the post-measurement state will be some kind of wave packet (in $\mathcal{L}^2(\mathbb{R}^3)$) narrowly but not perfectly centered around some average momentum.

Why don't we add atmospheric pressure the same we do pressure from other liquids? If we have two fluids $1$ on top of $2$, I know that the absolute pressure of a fluid $2$ is $p_2 = p_1 + \rho gh$ where $h$ is the height of the second fluid, and $p_1$ is the absolute pressure at the bottom of fluid $1$. In other words, we add the pressures. Now, consider a thin closed off pipe filled with water as shown, such that the Rayleigh-Taylor instability does not apply: However, looking at the drawing, why would the absolute pressure at $P_1$ be $P_1=p_0 + \rho gh$ and not $2p_0 + 2 \rho gH$, and similarly, why is $P_2 = p_0+2\rho gh$ and not $2p_0 + 2\rho gh$. Why don't we add atmospheric pressure the same we do pressure from other liquids?

We do "add the pressures," but only when the effect is big enough to matter. The density of air at sea level is about $1.2\,\text{kg}/\text{m}^3$. So the pressure change in a column of air $1\,\text{m}$ high is about $1.2\times 9.8 \approx 12\,\text{Pa}$. Compare that with change the atmospheric pressure at sea level of about $100,000\,\text{Pa}$. In most situations a change of $0.012\%$ over a height of $1\,\text{m}$ can be ignored. However if the "column of air" is $1\,\text{km}$ or $10\,\text{km}$ high, the pressure change is not negligible, and this is the reason why atmospheric pressure changes with altitude!

Why doesn't the double-slit experiment violate the uncertainty principle? In the double-slit experiment, when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy. From this data, we can calculate what momentum the electron had when it was passing through the hole. So we can know the electron's position (as it was inside the hole) with arbitrary accuracy and also simultaneously know its momentum. Doesn't this violate the uncertainty principle? Where have I gone wrong?

Where have I gone wrong? Here, the noted with italics. when an electron reaches the detector after passing through the holes, it has a certain momentum which we can measure with arbitrary accuracy. Any measurement means new quantum mechanical interactions, i.e. new wavefunctions and boundary conditions on them. In any case momentum is measured also within the Heisenberg uncertainty bounds. Also here: So we can know the electron's position (as it was inside the hole) The width of the hole is the accuracy we can experimentally have, there is no way it can be better.

Functional determinants I wish to know what is the result of this Gaussian Functional Integral $$Z[\chi] = \int[\mathcal{D}\phi]~e^{-i\int d^dx ~\phi^2\chi}$$ where $\phi, \chi$ are position dependent fields. Now, my question is whether $$Z[\chi] = (\det\chi)^{-1/2} ?$$ But, since, $\chi$ is not an operator just a scalar field is $Z[\chi] = \chi^{-1/2}$ then? What is the correct answer?

$\det \chi$ is a slight abuse of notation. You are actually computing the determinant of the multiplication operator $m_\chi:\phi \mapsto\chi\phi$. As @CosmasZachos pointed out in the comment, this is the product integral : $$\det m_\chi = \prod_{x\in\mathbb R^d} \chi(x)^{\text d^dx} = \exp\left(\int_{x\in\mathbb R^d}\ln(\chi(x))\text d^dx\right) \tag 1$$ Edit The multiplication operator is diagonal, as it can be written as an integral kernel proportional to $\delta(x-y)$ : $$m_\chi(x)\phi(x) = \int_{\mathbb R^d} \text d^dy \ m_\chi(x)\delta^{(d)}(x-y)\phi(y)$$ Then, the determinant is just the product of the diagonal entries, which gives equation $(1)$ above. Another way to see this, more rigorously, is to discretize space on a finite lattice $\Lambda$ of size $L^d$ and lattice step $a \to 0$ and take the limit $a\to 0, L\to \infty$. Then, a scalar field is just an element of $\mathbb R^{L^d}$ and a multiplication operator is a diagonal $L^d \times L^d$ matrix. Therefore : $$\det m_\chi |_{\Lambda,a} = \prod_{x\in \Lambda}\chi(x)$$ To get a regular limit, we take it to the power $a$, to get : \begin{align} \det m_\chi &= \lim_{a\to 0,\Lambda \to \infty}\det m_\chi|_{\Lambda,a } \\ &= \lim \exp\left(a\sum_{x\in\Lambda}\ln \chi(x) \right) \\ &= \exp\left(\int_{x\in\mathbb R^d}\ln\chi(x)\text d^dx\right) \end{align}

Why does the electron not spin? The goto answer to that question is that the electron is a pointlike particle and cannot spin. The electron is not pointlike though. It is described by a wavefunction. One can prepare the wavefunction to describe a very small electron, but not a point-like electron. Is there a genuine answer to the problem?

I want to address this: The electron is not pointlike though. It is described by a wavefunction. One can prepare the wavefunction to describe a very small electron. The standard model electron is a point particle. The wavefunctions used in the quantum mechanical models to model an electron , call it $Ψ$, which defines the probability $Ψ^*Ψ$ to find the electron at an (x,y,z). Probabilities are measured by taking many events with the same boundary conditions and have no connection with the size of the electron which is assumed axiomatically in the calculation of $Ψ$ . Edit after comment: to elaborate on "which is assumed axiomatically in the calculation of $Ψ$ ." In the quantum field theory of the standard model all the elementary particles in the table define a field in all points of space time,( fermions represented by the plane wave solution of the Dirac equation, bosons the Klein Gordon, photons the quantized Maxwell) on which fields creation and annihilation differential operators act. The calculations are done with Feynman diagrams where all particles are treated at the vertices as point particles.

Why aren't all quarks clumped together in one giant hadron? As far as I am aware, the strong interaction is attractive only, and its carrier, the gluon, is massless meaning it has unlimited range. If this is the case, how come we only observe quarks in pairs and triplets? What's preventing every quark in the universe from attracting every other quark and forming one ever-growing mass of colored particles?

The strong interaction is not "only attractive". A $qgq$-vertex has a color factor associated with it that depends on quark color and the gluon color/anti-color. The total color factor for a 2 vertex diagram is: $$ C=\frac 1 2 c_1c_2 $$ A positive (negative) color factor is attractive (repulsive). Interactions are as follows: So here is $rr\rightarrow rr$, with two possible gluon states. The red quarks only interact with the red part of the gluon. This interaction is repulsive: Color swapping is attractive: Now you have to combine red, green, and blue quarks, and also anti-red, anti-green, and anti-blue quarks. This process is governed by the representation theory of $SU(3)$, where $(r, g, b)$ are the fundamental representation, ${\bf 3}$, while $(\bar r, \bar g, \bar b)$ form ${\bf \bar 3}$. They combine according to: $${\bf 3} \otimes {\bf \bar 3} ={\bf 8} \oplus {\bf 1}$$ Graphically: Meanwhile, a $qq$ baryon/meson hybrid doesn't work: There is no singlet combination. The Baryon sector is as follows: where the red-dot on the far right represent the baryon color wave function: $$\psi^c_{qqq} = \frac 1{\sqrt 6}(rgb-rbg+gbr-grb+brg-bgr) $$ At this point, you can compute the color factor for singlet and non-singlet states. It comes out attractive for singlets, and repulsive for other states. Nevertheless, it is a postulate that only color-singlet states are observed in nature.. Searches for more complicated singlets such as the pentaquark ($qqqq\bar q$) are on-going. (Illustrations are from https://www.hep.phy.cam.ac.uk/~thomson/lectures/partIIIparticles/Handout8_2009.pdf).

Antiunitary operators and compatibility with group structure (Wigner's theorem) From Wigner's theorem, we get that a physical symmetry can be described either by a unitary or antiunitary operator, eventually with a phase factor, as in here. However we have to respect the group structure, so we need to have: $$O(f)\circ O(g)=e^{i \cdot\phi(f,g)}\cdot O(f*g)\tag{1}$$ where $*$ is the group operation. For unitary operators this is fine, but for antiunitary operators there is a problem: the left side of the equation is linear while the right side is antilinear. Here my doubt comes. The argument of linearity is used to show that only unitary operators can describe continuous symmetries (e.g. spatial translations), but why does it not work for symmetries associated with a finite group (e.g. time reversal)? I'm not asking why time reversal needs to be described by an antiunitary operator, I'm asking how come an antiunitary operator can satisfy $(1)$ if it's the representation of a finite group.

Time reversal is a implemented at the level of the Hilbert space $\mathcal H$ by a (projective) representation $\rho$ of $Z_2$, with $\rho(0) = \mathbb I$ and $\rho(1) = T$. Note that $$\rho(0+0)=\rho(0)=\mathbb I = \mathbb I\circ \mathbb I = \rho(0)\circ \rho(0)$$ $$\rho(0+1)=\rho(1)=T = \mathbb I \circ T = \rho(0)\circ \rho(1)$$ $$\rho(1+1)=\rho(0)=\mathbb I = e^{i\theta} T\circ T = e^{i\theta}\rho(1)\circ\rho(1)$$ where $\theta=0$ for bosonic systems and $\theta=\pi$ for fermionic systems. The second line of the above is an antiunitary operator, while the first and third are unitary operators.

Why is everything not invisible if 99% space is empty? If every object is $99$% empty space, how is reflection possible? Why doesn't light just pass through? Also light passes as a straight line, doesn't it? The wave nature doesn't say anything about its motion. Also, does light reflect after striking an electron or atom or what?

The solar system is also mostly empty space. So why doesn't an asteroid from outside the solar system pass straight through? The solar system has many massive objects with wich the asteroid could interact gravitationally. If an asteroid would pass through the solar system it would at least be deflected by the sun a little. Similarly light waves interact electromagnetically. The particles in atom are very small; how small depends on if you view them as point particles or probability clouds. But since they have a charge their effect can be felt at large distances. When light waves interact with the charged particles it produces disturbances in the wave pattern and when these disturbances from multiple atoms combine they can change the wave in such a way to produce effects like reflection, refraction and diffraction. For example reflection is explained in this answer

If air is a bad conductor, how does fire heat up a room? If air is a bad heat conductor, how does fire heat up a room? Could someone help me, as I really don't get this?

There are three mechanisms at play: conduction, convection and radiation. Radiation is the most immediate. Your environment irradiates you with black body radiation at room temperature (assuming that you are in a room at 20 C/ 293 K). As soon as your fire burns it emits black body radiation at a temperature of a 600 C (900 K), that is mostly in the infrared. The power emitted by a black body is proportional to $T^4$ (Kelvin), according to the Stefan-Boltzmann law, and this radiation is quite intense. Also the air will be heated by your fire and hot air will reach you by convection. The least important effect is air conduction.

A pure rolling sphere problem Question: A sphere $S$ rolls without slipping, moving with a constant speed on a plank $P$. The firction between the upper surface of plank $P$ and the sphere is sufficient to prevent slipping. Friction between plank and ground is negligible. Initially plank $P$, is fixed to ground with help of a pin $N$. How will motion of sphere and plank be affected if, the pin $N$ is suddenly removed? According to me the plank should move in opposite direction to that of sphere, since centre of mass must remain at rest. But according to my teacher's explanation, the motion of plank and sphere will remain unaffected because since in question, it is given that that sphere is moving with constant speed, which means no acceleration. This implies there is no force acting hence, there would be no friction acting between plank and sphere. This explanation seems odd to me. Please give appropriate explanation and point out any faults in any of the explanation mentioned above. Any suggestions are massively appreciated.

It might come as a surprise to you that the frictional force between the sphere and the horizontal plank when the sphere is rolling without slipping along the plank at constant velocity is zero. This must be so as the only possible horizontal force on the sphere is the force of friction on it due to the plank and this horizontal frictional force on the sphere, if it existed, would mean that the sphere would not be moving at constant velocity. So there is no horizontal force of interaction between the sphere and the plank.

States on finite dimensional Hilbert spaces In my quantum theory lecture we talked about states on finite dimensional Hilbert spaces and had the following statement: Let $\mathcal{H}$ be a finite dimensional complex Hilbert space, $\mathcal{L}(\mathcal{H})$ the set of bounded linear operators, $\omega \colon \mathcal{L}(\mathcal{H}) \to \mathbb{C}$ a state and $H\in \mathcal{L}(\mathcal{H})$ be self-adjoint. Then $\omega$ is a ground state of $H$, i.e. $$ \omega(H)\leq \tilde{\omega}(H)\quad \text{for all states } \tilde{\omega} \text{ on } \mathcal{L}(\mathcal{H})$$ if and only if $$ \omega(A^*[H,A])\geq 0\quad \text{for all }A\in\mathcal{L}(\mathcal{H}).$$ I tried to compute both sides and get to the other, but it absolutely didn't work... I have no idea how to see that this statement is true. I would be greatful for tips or help!

Every pure state $\psi\in\cal H$ gives rise to a linear operator $\omega:\cal O\mapsto \omega(\cal O)$ by the familiar $\cal O\mapsto\langle \cal O\psi,\psi\rangle\,.$ For two arbitrary pure states $\psi$ and $\tilde\psi\,,$ there always exists a unitary operator $A$ such that $\tilde\psi=A\psi\,.$ Therefore, restricted to pure states $\omega,\tilde\omega\,,$ the first statement becomes $$ \langle H\psi,\psi\rangle\le\langle HA\psi,A\psi\rangle\mbox{ for all unitary }A\,. $$ Since the LHS equals $\langle AH\psi,A\psi\rangle$ it is easy to see that the statment is equivalent to $$ \langle [H,A]\psi,A\psi\rangle\ge 0\mbox{ for all unitary }A\,, $$ resp. to $$ \langle A^*[H,A]\psi,\psi\rangle\ge 0\mbox{ for all unitary }A\,. $$ Perhaps your theorem is stronger than that but I hope the tips are good enough.

Blackbird debate: Case when speed of car smaller than speed of downwind I am puzzling my head around the Veritasium famous video right now: https://www.youtube.com/watch?v=yCsgoLc_fzI Also the Physics Olympiad problem Part B1 https://www.aapt.org/physicsteam/2019/upload/USAPhO-2013-Solutions.pdf I fully understand (I hope) the case when the car has a speed v higher than the speed of wind w it will accelerate until it reaches an constant speed where the power transferred from the wheels is bigger than the power of the propeller. What is puzzling me is the following experiment: Suppose we have a very long treadmill (so we can simulate the case of constant wind). If we were to just place the car in the treadmill without rolling the wheels. The car would go initially backwards with constant speed the one of the belt backwards. It will create some drag with the air of the room which will cause the car to slowly start rolling forward. When it starts rolloing forward the propeller will rotate as well, but the air is still coming from behind the propeller. Is it possible that the car exceeds the one of the belt (assuming the aerodynamic shape of the car does not change)? Will it start rolling forward with speed v greater than w, for any speed w of the belt?

Is it possible that the car exceeds the one of the belt Yes, that is possible. Check out the treadmill videos that they have made, where the small model they built, was able to do this

What is the size of the magnetosphere of a neutron star? More precisely, what are the factors that influence that size (mass of the star, spin velocity, accretion disk around it, ... do any of those things matter?) I tried reading this. I cannot: [paper about relative size of neutron stars' magnetospheres] Is there an equation for the size of neutron-stars magnetospheres? If so, can someone with just barely passable knowledge of differential calculus solve it?

The size of a neutron star's magnetosphere (in-so-far as it can be approximated as something spherical) is, to within a small numerical factor, given by the Alfven radius. This is where the magnetic energy density equals the kinetic energy density of the surrounding gas/plasma. Using Gaussian units, the energy density is $B^2/8\pi$, where $B$ is the magnetic field. If we assume that the magnetic field is dipolar and the magnetic dipole moment is $m$, then $B \simeq m/r^3$, where $r$ is the radial distance from the neutron star centre. If the neutron star is accreting then the kinetic energy density is $\rho v^2/2$, where $\rho$ is the mass density and $v$ the inflow velocity. If we further assume spherically symmetric inflow and that the velocity would just be given by gravitational freefall, then $v = \sqrt{2GM/r}$ and the density is given by the conservation of mass equation: $$ \dot{M} = 4 \pi r^2 \rho v\ ,$$ where $\dot{M}$ is the total mass accretion rate. If you now equate the magnetic and kinetic energy densities at $r=r_A$ $$ \frac{m^2}{8\pi r_{A}^{6}} = \rho \frac{GM}{r_A} = \frac{\dot{M}}{4\pi r_A^2}\left(\frac{GM}{2r_A}\right)^{1/2}\ ,$$ then the Alfven/magnetosphere radius is given by $$r_A = \left( \frac{m^4}{2G M \dot{M}^2}\right)^{1/7}\ .$$ The paper you refer to contains this formula but also notes that it is only good to a factor of two or so. That is because of various factors - the field may not be dipolar (and a dipolar field is of course not spherically symmetric!), the accretion may not be spherically symmetric and so on. These in turn can depend on how fast the neutron star is spinning. So the magnetospheric radius is quite strongly dependent on the magnetic dipole moment of the neutron star and quite weakly, and inversely, dependent on the mass of, and the mass accretion rate onto, the neutron star. This is all that matters to first-order.

Why does running spend more energy than walking? The study energy expenditure of walking and running concludes that running spends more energy than walking. My understanding is that although running makes one feel more tired, that only indicates that the power was higher (since the time of displacement was shorter), but at the end of the day the total energy dispensed to move oneself forward by friction should be the same. Given the study shows otherwise, what could be the flaw in my reasoning?

When running, muscles require a higher consumption of oxygen, so it's expected that a significant part of the ATP synthesis takes the fermentation route, which is knowingly less efficient than the usual cellular respiration.

Does real life have "update lag" for mirrors? This may sound like a ridiculous question, but it struck me as something that might be the case. Suppose that you have a gigantic mirror mounted at a huge stadium. In front, there's a bunch of people facing the mirror, with a long distance between them and the mirror. Behind them, there is a man making moves for them to follow by looking at him through the mirror. Will they see his movements exactly when he makes them, just as if they had been simply facing him, or will there be some amount of "optical lag"?

Even without a mirror there's a lag, because it takes time for light to go from the man to the audience. Divide the distance from the man to the audience by the speed of light to get the delay. With the mirror, the light has to flow from the man to the mirror and then reflect to the audience members. Therefore, the time between the man making the moves and the audience seeing them is essentially the same as if his distance from the audience were the sum of his distance from the mirror and the audience's distance from the mirrow. If he's right behind the audience, going through the mirror doubles the lag. The process of the mirror reflecting the light will also take some tiny amount of time, but I don't know how to quantify this. I suspect it's negligible compared to the differences in distance for different audience members. And as others have pointed out, the speed of light is so fast compared to the reaction time of the human vision system that all these times are essentially instantaneous as far as our perception is concerned. But we can create devices that can detect these delays (RADAR and LIDAR work by measuring the time it takes for a radio or light transmission to reflect off something).

Linear combination of group generators In Matthew Robinson's book Symmetry and the Standard Model he explains that we have generators for rotations $J$ and for boosts $K$. To analyse the group structure, we will look at $N^\pm = J \pm i K$ though and find, that both $N^+$ and $N^-$ form a Lie algebra, which means they generate a Lie group. Why can I look at a linear combination of generators? Is any linear combination of generators a new generator?

Given a Lie group $\mathcal{G}$ the generators of the transformations in the group $\mathcal{G}$, lives in the Lie algebra $\mathfrak{g}$ associated. A Lie algebra as you can tell, is a particular kind of an algebra. An algebra $A$ over a field $\mathbb{K}$ is a $\mathbb{K}$-vector space with an additional structure, that is a $\mathbb{K}$-bilinear map (in the Lie algebra case is the Lie bracket [,], with its additional properties). So the algebra has the structure of a vector space, which allows to sum its elements. The non trivial part is that usually the Lie algebra is defined over a field $\mathbb{R}$, and here in defining $N^+$ $N^-$ you are multiplying the generators by $i$. Basically you are complexifying the Lie algebra, and finding the following chain of isomorphisms. $\mathfrak{so}(3,1)_{\mathbb{C}}\simeq\mathfrak{su}(2)_{\mathbb{C}}\oplus\mathfrak{su}(2)_{\mathbb{C}}\simeq\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$, where the underlying $\mathbb{C}$ in $\mathfrak{g}_{\mathbb{C}}$ means you have complexified the Lie algebra.

Why is centrifugal force called fictitious? When an object undergoes rotation, from the object's reference frame, which is a non-inertial reference frame, the object feels there is a radially outward force, a centrifugal force, acting on it. However, from an inertial reference frame, this force doesn't exist at all. That's why it is called a fictitious force. My argument is, who are we to say what is fictitious or not. The object at the non-inertial frame really feels the centrifugal force! So, it is a real force for the object. Suppose, there are two inertial reference frames $S$ & $S'$ and $S'$ is moving with a velocity v that is a significant fraction of the speed of light. From $S$ it would seem that time is going slower for $S'$. Surprisingly, it would seem from $S'$ that time is going slower for $S$ as well. Now, who is right? Answer: Both of them are right. So, is it really right to call centrifugal force fictitious just because it doesn't exist in an inertial reference frame?

According to the basic Newtonian formulation of mechanics, "real" forces come in couples: a force (action) and its reaction acting on the source of the action. Furthermore "real" forces are independent of the used reference frame. Fictitious forces, as centrifugal or Coriolis one, violate both conditions. This is the reason why they are called in that way. Maybe that is not a good name because they have concrete effects (though the interpretation of these effects may be given without using fictitious forces by just describing the phenomenon in an inertial reference frame).

Why propagator pole is associated to the mass? We say that the pole of the all-orders photon propagator, $$\frac{1}{q^2[1+\Pi(q^2)]}$$ doesn't shift if $\Pi(q^2=0)$ is regular. This amounts to say that the photon remains massless to all orders in perturbation theory. Conversely, the fermion propagator, $$\frac{i}{\not p -m+\Sigma(p)+i\varepsilon}$$ has $\Sigma(p)$ which shifts the pole hence the mass is corrected and must be renormalized. Why the propagator pole is associated to the mass?

Here is a heuristic argument. * *It's a fact that a full connected propagator/2-point correlation function $\tilde{G}_c$ of a Lorentz-invariant theory tend to have a simple pole $\tilde{G}_c \propto \frac{1}{p^2-m^2}$, where for the sake of the argument $m$ is some constant. *Given that a correlation function/scattering amplitude is a measure of how likely a process occur, it is natural to associate an infinity/a pole with the production of a particle with physical mass $m$. *See also the Källén–Lehmann spectral representation. *For unstable/quasi-particles, see also the relativistic Breit–Wigner distribution.

When does a falling flowerpot reach maximum speed? I’m interested to know when a flowerpot of begonias might reach terminal velocity after falling off a high balcony. I’ve read that that a penny reaches terminal velocity after just 15 metres. I think then, it would take longer (farther) for a filled flowerpot but don’t really have a precise idea. Could anyone please provide an approximate distance?

This is really a comment, as your question can only be answered with a formula. the terminal velocity formula, v = the square root of ((2mg)/(ρAC)). m = mass of the falling object. g = the acceleration due to gravity. ... ρ = the density of the fluid the object is falling through. A = the projected area of the object. ... C = the drag coefficient. It is only possible to calculate it for the specific pot. To get the distance of free fall before terminal velocity is reached more calculations paricular to the specific flowerpot are necessary. There exist calculators,

Topologically, is a curvature singularity just a hole? Topologically speaking, a hole can be introduced into a manifold and it will still be a manifold, e.g. remove points within a 2-sphere of some radius from the cartesian plane and you'll still have a manifold. Penrose's singularity theorems prove the existence (mathematically) of incomplete null geodesics, but not of curvature singularities, per say. So Im wondering if, in my naive view, maybe a topological perspective is better suited to describe curvature singularities? IF so, I'm not sure how to view curvature singularities topologically. My question: is a curvature singularity, e.g. a black hole, in general relativity simply a topological hole of the spacetime manifold, or is it more topologically complicated? Or instead, is the singular structure not a part of the manifold, as suggested in @benrg's answer to this question, and thus is not a topological hole? Or is it not this simple, and there's some nuance(s) that I'm missing? EDIT: I suppose I can phrase my confusion like this: how is it logically consistent to say that the physical singularity is not a part of the spacetime manifold (like how $\infty$ is not a point on the real line) AND that we can have a description of the singular structure from the metric itself (e.g., the Kerr metric has a ring singularity as can be shown from the metric)? Or do we avoid this confusion if the singularity is a topological hole?

Or instead, is the singular structure not a part of the manifold, as suggested in @benrg's answer to this question, and thus is not a topological hole? These are not two different interpretations. A topological hole is precisely something that is not part of the manifold. Im wondering if, in my naive view, maybe a topological perspective is better suited to describe curvature singularities? Again, this is not an "or."

Proton Electron Merger Can somebody explain what would happen if an electron & a proton, very close to each other are left to "fall" to each other in a straight line?

There is a probability that they will form a neutron, a hydrogen atom in some s state or an unbound electron proton system. Each of these possibilities can occur with a relative probability depending on the initial state. So it is not correct to say that a hydrogen atom must result, even without specifying in which state. Quantum mechanics tells us that a very localised electron centered at a proton corresponds to a superposition of hydrogen bound and ionised states. A very localised electron has very high kinetic energy which may exceed the potential energy. The quickest way to see this is to use Heisenberg's uncertainty principle for position and momentum. HUP tells you that a strongly localised electron wave function requires a superposition of very high momentum waves. Very high momentum means very high kinetic energy as well. Note that if components with high enough kinetic energy are present to overcome the mass difference of neutron and proton and to create an electron neutrino of sufficient energy and momentum, also a neutron plus neutrino may be formed.

Adiabatic evolution of superposition of states I am supposed to find a specific superposition of eigenstates of a time-dependent hamiltonian $H(t)$. The hamiltonian is of the form : $H(t) =\sum_i \left(J\vec{\sigma}_i\cdot \vec{\sigma}_{i+1} + h(t) {\sigma}_i^z \right)$. Periodic Boundary Condition. I have taken the following state at $t=0$, $$\vert \Psi(0) \rangle = \vert 0(0) \rangle + e^{i\phi} \vert 1(0)\rangle$$ Now, the hamiltonian is changing with time, I know that at $t = T$, this state will evolve to: $$\vert \Psi(T) \rangle = e^{i\Phi_0}\vert 0(T) \rangle + e^{i\phi}e^{i\Phi_1} \vert 1(T)\rangle$$ where $\Phi_i$ are the sum of dynamical and Berry phases. The trouble I'm having is that the states $\vert 1(T) \rangle$ and $\vert 0(T) \rangle$ that I'm obtating through some numerics (Lanczos algorithm) are having an arbitrary phase that is not same for all diagonalisations (I have to do multiple diagonalizations for different values of $t$). Because of this, I'm not exactly getting the evolution of the same state for all the times $t$. I want the superposition state for all times $t$.

Your states have an arbitrary global phase due to $U(1)$ gauge symmetry. To remove it, you need to fix a common gauge across all your states. A convenient way to do this is to iteratively impose parallel transport, i.e. modifying the phase of your $|n_i\rangle$ state such that the condition $\text{Im}\langle n_{i-1}|n_i\rangle = 0$ is fulfilled. Note that in this case the (discretized) Berry phase will not be given as a sum of phase differences between your subsequent states along your closed loop in Hilbert space, but simply as the phase difference between your initial and final states, $\phi=\text{Im}\ln\langle n_{0}|n_N\rangle$.

True or False: energy is conserved in all collisions Using introductory physics, how would you answer this question? (I have a disagreement with my instructor and I’m curious to hear your input) One of us says true because the question doesn’t specify “kinetic energy,” or a “system” and all energy is always conserved. The other says false because “only perfectly elastic collisions conserve energy. Otherwise energy will be lost to sound or light” What’s your opinion?

Sadly, the correct answer is "Test-setter has failed their practical exam". The question is worded ambiguously, and any "correct" answer will be determined by psychology not physics - namely what interpretation of the ambiguous question was in the test-setter's mind. There is no way to answer it purely based on coursework or accurate knowledge of the topic, because there will always be that guess element regarding the ambiguity. This happens disappointingly often in tests, and you kind of have to ride it and accept when it does. I tend to write in margins so its on my answer the basis and critique of the question and answer, so I can show I knew my topic. You don't get scored for it, but you pick up "moral win" points, and maybe a good examiner will reflect and award if you show them afterwards.

What is the difference between circular motion and rotation? I've tried so hard to understand the difference, yet no progress. There is a lot of answers here, on Quora, on Youtube,... but everyone give a different answer. So can you please give a simple yet satisfactory answer? Someone says that rotation is only about an axis that oass through the center of mass, other say that the axis can be anywhere inside the body but outside no because if it's outside it will be circular motion, but then if you search Wikipedia about Parallel Axis Theorem, they'll say : If the body "ROTATES" about an axis outside of it, you can use the Parallel Axis Theorem to... So who's right? And one more question : In circular motion, the kinetic energy formula for a body is $\frac{1}{2} MV^2$ or $\frac{1}{2} Iω^2$ (like in rotation)? I mean can we use the equation $x=\frac{1}{2} at^2 + Vt + X$ or $θ=\frac{1}{2}θ"t^2 + θ't + Θ$ (like in rotation)? So many questions yet no one gives me a good answer, I hope that someone can here. And what about this picture here, is it rotation? "The disc (D) can oscillate freely around a horizontal axis (A), perpendicular to its plane and passing through a point O of its periphery." https://i.stack.imgur.com/iBodB.jpg https://i.stack.imgur.com/AJMhI.jpg

A rotary motion is one in which a body turns around its centre of mass. A purely circular motion is one in which a body follows a circular path but does not turn around its own centre of mass. If you walk around a circle, you rotate. Although the centre of the circle is your centre of motion, your body also turns around its own centre of mass once every time it completes a lap of the circle. If instead of walking normally around a circle, you sidle round so that your body always stays facing in one direction (north, say), then your motion is circular, as you do not turn about your own centre of mass as you complete each lap- you remain facing in one direction the entire time.

Cathode Ray Oscilloscope Is Cathode Ray Oscilloscope (CRO) a device that measures and displays using the electron beam from cathode rays? Are cathode rays its working principle? Does it mean that the line or point showed on CRO screen is the result of electron beam from cathode rays strike on the CRO screen?

Yes, that is true. You can think of a cathode-ray oscilloscope as an AC voltmeter, where you can actually see the waveforms and measure their peak values and study their shapes. The oscilloscope responds to very high frequencies because the electron beam that "writes" the waveform onto the screen can move very much faster than any mechanical needle in a moving-coil AC voltmeter.

What has Euler's number $e$ to do with exponential decay? I know how to derive the formula for "quantity at time $t$" for some decaying materials. You can see the derivation here. But, what I don't get is that what the number $e$ is doing here? We get the value of $e$ from the equation of compound interest. Compound interest and decay (like, radioactive decay) are two different things. Is there any intuitive way to see what the Euler's number doing here?

Compound interest and radioactive decay both vary exponentially with time. That simply means that in any set period of time, the value changes by the same fractional amount. Any exponential function can be written using any base. If the number of remaining atoms in a sample is given by $A * e^{-Bt}$, then it is also given by $A * 2^{-Ct}$, for the appropriate value of $C$. Regardless of the base you choose, the number $e$ pops out when you calculate the rate of change of the exponential function (which is another exponential function). So, assuming you are interested in, say, both the number of atoms and how fast that number is changing, you will be stuck with an $e$ somewhere or other anyway. Choosing it as the base eliminates a constant that would otherwise appear in the expressions.

Working of liquid lens we were recently introduced to the liquid lens in school, but i havent quite understood the working of it. I realize that the curvature of the lens is changed by some process, but i dont exactly know what, and how it would work. Itd also be great if you could mention what factors would affect the focal length/ working of a liquid lens.

Considering the physical principles behind it, a liquid lens is no different from a solid one. The focal length $f$ is affected by the refractive index $n$ of the lens material (assuming a refractive index of 1 in the surrounding medium) and the curvatures $R_1$ and $R_2$ of the two surfaces of the lens, as well as it's thickness $d$. It holds $$ \frac 1f = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)d}{nR_1R_2} \right), $$ which can also be found at wikipedia and is known as the lensmaker's equation. For liquid lenses, it is possible to vary the focal distance by changing at least one of the radii and possibly also the thickness. This can be achieved in multiple ways: * *This site describes a mechanism that pushes a water droplet on a surface together to decrease the corresponding radius of curvature and increase the thickness at constant lens volume. *Here liquid is pushed into a cavity formed by a membrane, again increasing thickness and curvature, but at increasing volume. *On the page referenced by the last link, another technique is described which uses electrical attraction to move the border between two liquids, again changing the curvature.

Why is a liquid-vapour mixture at thermal equilibrium called saturated? If we have a liquid-vapour mixture at thermal equilibrium, apparently this means the mixture is saturated. How? From what I understand, saturated means it is on the verge of converting to either liquid or gas. But it also apparently means that the thermodynamic properties of the steam and water are the same. How does the fact the system is in thermal equilibrium tell us this?

There is some variation in the use of the terms, but in general: The term "saturated" refers to vapor in equilibrium with liquid at or above the normal boiling point (boiling point at one atmosphere, in the case of water). Adding or removing heat heat increases the vapor component or increases the liquid component, respectively. The term saturated vapor refers to 100% vapor on the verge of either being converting to liquid (if heat removed) or superheated vapor (if heat added). The term saturated liquid refers to 100% liquid being on the verge of either converting to vapor (if heat added), or lowering in temperature (if heat removed). Hope this helps.

What does the absence of these Goldstone boson interactions mean physically? I have read that in several statistical models exhibiting spontaneous symmetry breaking, the resulting Goldstone bosons do not interact with each other via $\theta^{2n}$ terms — only via derivative terms like $(\nabla\theta)^2$. For instance, in the XY model, the free energy has terms like $$F\sim\frac\gamma2 \int (M_0^2+2M_0\delta M)(\nabla\theta)^2.$$ Or in the Heisenberg model, there are terms like $$F\sim\frac\gamma2 \int M_0^2[(\nabla\theta)^2+\sin^2\theta(\nabla\phi)^2],$$ for the two Goldstone modes $\phi$ and $\theta$. My question is, what is the physical interpretation of the absence of $\theta^{2n}$ interactions, and only derivative ones?

When there's a single $\nabla \theta$ term, this tells you that the theory has a shift symmetry of $\theta \mapsto \theta + a$ which moves you between different vaccua (each with a different $U(1)$ charge). Expressions can become logner with additional Goldstones but the principle is the same. Your expression with $(\theta, \phi)$ is invariant under $SO(3)$ rotations which treat these fields as angles on $S^2$. This structure appears because the potential in the original theory is a function on $\mathbb{R}^n$ and Goldstones parameterize the submanifold in $\mathbb{R}^n$ which minimizes it. The theory which describes their fluctuations must therefore preserve the interpretation that each Goldstone field is a co-ordinate on a target space. The way we do this is through the sigma model \begin{equation} S = \int dx \, G_{ij}(\phi) \nabla \phi^i \cdot \nabla \phi^j \end{equation} where $G_{ij}$ is the metric of the target space. You can check that the examples above have this form. If there were additional terms like $G_{ij} \phi^i \phi^j$ without derivatives, $S$ would no longer be similar to the action used to derive the geodesic equation and therefore not describe the "length of a path".

Is there cosmological redshift within the Milky Way? Cosmological redshift is based on the idea that the universe is expanding. When the universe doubles in size, or scale factor, the wavelength of light doubles. But the Milky Way is not expanding so my guess is that there is no cosmological redshift within the Milky Way? There is of course Doppler redshift. Having seen the first 5 answers, which seem to confirm my guess, I am now going to add a corollary. The Doppler effect has almost nothing to do with cosmological red shift, outside our local cluster of galaxies. Is that also correct? Which makes Hubble very lucky.

Even if the Milky Way is expanding with the Hubble flow (and most cosmologists believe that it isn't), the expansion would be difficult to measure. The size of the Milky Way $d$ is approximately $10^{21} \;\text{m}$. Using $$v=Hd$$ with Hubble's constant in SI units of about $2\times 10^{-18} \;\text{s}^{-1}$ means that even a star far away, in the Milky Way, would have a redshift corresponding to a motion of about $2000 \;\text{m}\,\text{s}^{-1}$. However peculiar velocities of stars are typically a hundred times higher than this, for example the sun is thought to be moving relative to the Milky Way at $250 \;\text{km}\,\text{s}^{-1}$. Cosmological redshifts are used to determine distances of stars etc... that are far enough away that the peculiar velocities can be ignored. It would be interesting to see the results of a future experiment that attempted to 'average out' the peculiar motions of millions of stars within the Milky Way, to see if there is a cosmological redshift. It seems that at the moment the consensus amongst cosmologists is that it doesn't exist.

What units is strong nuclear charge measured in? Do particles have a strong nuclear charge in the same way as they have electric charge? If so what unit would be used to measure this? Would it be measured in Coulomb for instance?

The electromagnetic force couples to electric charge, the gravitational force couples to mass charge and the strong nuclear force couples to color charge. Photons are the exchange bosons in the electromagnetic force, gravitons - if they exist- are the exchange bosons in the gravitational force. The strong nuclear force is mediated by the exchange bosons called gluons. Do particles have a strong nuclear charge in the same way as they have electric charge? No, not really, since (as below), all particles (that exist as natural particle states) that take part in the strong nuclear force, will always have a zero net color charge, but can have net electric charge. Would it be measured in Coulomb for instance? Color charge is not really analogous to electric charge in this sense. Electric charge has one value (which is some number multiplied by the electron charge). But color charge can be one of six possible values (red, blue, green, and antired, antiblue, antigreen). These are quantum numbers we assign to the six different flavor of quarks which combine to form hadrons. While particles like $\pi^-$, K$^-$, protons, $\Delta^-$, $\Delta^{++}$ etc., can have a net electric charge, all particles containing quarks must always have zero, or neutral, net color charge due to quark confinement For more about this, see quantum chromodynamics which describes the behavior of the strong nuclear force and strong interactions.

Why does water contract on melting whereas gold, lead, etc. expand on melting? My book mentions that water contracts on melting, but the book doesn't give any reason why it does so. It is mentioned that: $1\,\mathrm g$ of ice of volume $1.091\,\mathrm{cm}^3$ at $0^\circ\mathrm C$ contracts on melting to become $1\,\mathrm g$ of water of volume $1\,\mathrm{cm}^3$ at $0^\circ\mathrm C$. I searched on the internet but I failed to find any useful insight. Could someone please explain why water contracts on melting?

The melting phase transition transforms the long-range ordered-crystalline solid structure into the short-range-ordered average liquid structure. Looking at the process from the solid side, melting can be seen as the dramatic effect of a collective building-up of defects in the solid over a limited interval of temperature, eventually destroying the long-range order at the melting point. The change of density accompanying the melting can be explained in terms of the kind of dominant defects, and these, in turn, depend on the solid structure. In particular, materials characterized by almost isotropic interactions between the molecules crystallize into compact three-dimensional structures, like the face-centered cubic (fcc) structure. In such a case, the dominant effect of the exponential growth of defects at the melting transition is a sudden decrease of density going from the solid to the liquid. The situation is reversed in materials like water and elements like Bismuth or Silicon, where the molecular or atomic interactions are highly anisotropic. In the case of water, the shape of the molecule, and the important role of the hydrogen bond favor an open structure of the crystal in the same way as the anisotropic interaction between Silicon atoms favors the open diamond structure. At the melting point, the most frequent defects in such open structures induce a local and global collapse of the crystal structure, creating the conditions for a liquid phase at a higher density than the coexisting solid. In summary, the reason for the increase of density at the melting point of some materials like water can be traced back to the presence of dominant anisotropic interactions favoring open crystalline structure in the solid phase.

Mass of All the Neutrinos I have read that the Sun produces $2 \times 10^{38}$ Neutrinos per second weighing in at approximately 8 MeV. I have 2 questions. * *Is there any way to calculate how many neutrinos have been produced in the past 13.5B years? *Is this mass part of the calculation of the "visible" mass of the universe? Or is it part of the dark matter?

To add to ProfRob's excellent answer, an excerpt from Simon D.M. White's 2018 essay, Reconstructing the Universe in a computer: physical understanding in the digital age: A possible solution, that dark matter might be made of neutrinos, was greatly encouraged by a 1980 tritium decay experiment which claimed an electron neutrino mass of 30 eV. A critical question was then whether the growth of structure in a neutrino-dominated universe could be consistent with the large-scale structure seen in the present-day galaxy distribution. [Neutrino dominated universes are predicted to have more extensive voids than observed.] This discrepancy led to the abandoning of the known neutrinos as potential dark matter candidates, even though it would be another two decades before they were finally excluded by experimental upper limits on their masses. A recent experimental upper limit is 1.1 eV from An improved upper limit on the neutrino mass from a direct kinematic method by KATRIN, 2019.

Why does twisting a cork make it easier to remove from a bottle? When we want to remove a cork from a bottle first we turn the cork. Turning in one direction makes it easier to remove in the axial direction. Does anyone know something more about this?

Why does twisting a cork make it easier to remove from a bottle? When you apply an axial force and, possibly, a torque to the cork (and, opposite, to the bottle), you generate a state of tangential stress between the cork and the bottle neck. With $A$ being the area of contact, $N$ the axial force and $W$ the torque, the longitudinal component is $\tau_{rz}=N/A$ and the tangential component is $\tau_{r\phi}=W/(rA)$, $r$ being the inner radius of the neck. Because the cork begins to move when the resultant of the tangential stress between the cork and the bottle's neck $$ \tau=\sqrt{\tau_{rz}^2+\tau_{r\phi}^2} \tag{1} $$ exceeds the friction bond, you can recognize that applying a torque (twisting the cork) reduces the axial force $N$ that you need to exert. After establishing that applying a torque reduces the $N$ needed to start moving the cork, when the cork starts moving it's all down hill because the friction coefficient is different (static vs dynamic friction) and also the surface of contact is continuously reduced.

Phase transition vs. critical phenomena Just trying to get some clarity in terminology: is phase transitions synonymous with critical phenomena? At the first glance they mean the same thing, but I am not sure whether phase transitions really include such phenomena as Anderson localization and percolation, which are not thermally driven. However, what then about quantum phase transitions, which are not thermally driven either? Clarification To clarify the above, here are the options: * *phase transitions and critical phenomena are the same thing *phase transitions are a subset of critical phenomena - e.g., we may consider Anderson localization, percolation, topological transitions, Mott transition to be critical phenomena, but not phase transitions - in this case one needs to define the difference between the two. *Finally, there may be simply no clearly established terminology, as is suggested by expressions such as quantum phase transition, which likely should not be called phase transition under the classification proposed in the previous bullet.

In thermodynamics, phase transition means the transition from one phase (solid, liquid, gas, or other phase) to another phase. Also in thermodynamics, critical point is the transition from where two separate phases exist to where only one phase exists. Beyond the critical point, only one phase exists. Approaching the critical point, the two phases become ever more similar, at pressures or temperatures beyond the critical point, there is no longer a dividing line, no longer two phases but one. https://en.m.wikipedia.org/wiki/Critical_point_(thermodynamics) Quantum phase transitions are completely different and unrelated. For example, solid and liquid are not quantum phases.

Why can't the speed of gravitational waves be greater than the speed of light if the universe can expand faster than the speed of light? Since the expansion speed of the universe can be greater than the speed of light, why can't gravitational waves, which also uses space as the medium, travel faster than the speed of light?

While gravitational waves and cosmic expansion are both gravitational phenomena, they are not the same. The universe "expanding faster than the speed of light" is not a local issue. The Hubble parameter describes the expansion rate now $H_0 \sim 70$ km/s/Mpc (kilometers per second per megaparsec). So for every megaparsec of empty space between two points the expansion makes them appear to recede from each other at a speed of $70$ km/s. For reference the Milky Way has a radius of about $30$ kiloparsecs ($0.03$ Mpc, but it's not really empty). This is apparent motion. The two points don't actually move. A person at each point would perceive themselves to be at rest. Each point is stationary with respect to its local spacetime. The addition of new space between the points makes the distance between them get bigger, but nothing is actually moving. We can calculate the current separation required for the apparent recessional velocity to be the speed of light $c$ $$ d_\mathrm{horizon} = \frac{c}{H_0} = \frac{300\,000 \,\,\mathrm{km/s}}{70\,\,\mathrm{km/s/Mpc}} = 4300\,\,\mathrm{Mpc} $$ Something $4400$ Mpc away appears to move faster than the speed of light away from us. The light it emits will never reach us. But its speed relative to the local spacetime will always be less than the speed of light. (We can see things farther away than this distance, because they were closer when they emitted the light that we see, but that's a different question) Gravitational waves (GWs) are another aspect of spacetime. The waves move across spacetime, but they don't carry spacetime with them. The wave speed of GWs is a consequence of general relativity and is a fundamental property of spacetime itself (like the wave speed on a string depends on the properties of the string like its density and tension).

Why don't we use absolute error while calculating the product of two uncertain quantities? I've found a rule that says, "When two quantities are multiplied, the error in the result is the sum of the relative error in the multipliers." Here, why can't we use absolute error? And why do we've to add the relative errors? Why not multiply them? Please give me an intuition to understand the multiplication of two uncertain quantities.

It basically comes from calculus (or more generally just the mathematics of change). If you have a quantity that is a product $z=x\cdot y$, then the change in this value based on the change of $x$ and $y$ is$^*$ $\Delta z=x\Delta y+y\Delta x$. So then it is straightforward that $$\frac{\Delta z}{z}=\frac{x\Delta y+y\Delta x}{xy}=\frac{\Delta x}{x}+\frac{\Delta y}{y}$$ The reason you don't use absolute uncertainty or multiply the relative uncertainties is the same reason why $(a+b)^2\neq a^2+b^2$. It's just not the result you get when you do the math. $^*$We are neglecting the term $\Delta x\cdot\Delta y$ in $\Delta z$, since ideally $\Delta x<x$ and $\Delta y<y$ to the extent that $\Delta x\cdot\Delta y\ll xy$ such that $\Delta x\Delta y/xy$ is much less than both $\Delta x/x$ and $\Delta y/y$.

Why do little chips break off so easily from strong neodymium magnets? I have some strong toy neodymium magnets. Typically after a while little chips start breaking off, unlike from most other small metal objects, like in this image. It could of course be that neodymium is more brittle than metals used for other objects, or that they often hit each other much harder than in a fall due to their magnetism, or that they are just low-quality, but I was wondering if it could have to do with internal tensions that are not present in non-magnetic objects, maybe due to adjacent domains of different magnetization? Does anyone know what could cause this?

I was wondering if it could have to do with internal tensions that are not present in non-magnetic objects, maybe due to adjacent domains of different magnetization? That you are right. The point is that the material is not only sintered but this time it is under the influence of a strong external magnetic field which align domains and the involved subatomic particles.

The faster you move, does it take more and more energy to increase your speed at the same rate? I'd like to confirm this somewhat counterintuitive result. Starting with the definition of kinetic energy: $$E = \frac{1}{2} mv^2$$ Assume a vacuum, no external forces, and starting from rest. Adding some energy $E$ to the system by burning some fuel (by firing a rocket, etc.), the following should be true. $$\frac{1}{2} m{v_1}^2 = \frac{1}2 m{v_0}^2 + E$$ Solving for $v_1$, $$v_1 = \sqrt{{v_0}^2+\frac{2E}m}$$ If $E$ remains constant, by burning fuel at a constant rate, $\Delta v$ decreases as $v$ increases.

The algebra is correct, but the interpretation may not be what you intend. In your description, there is a contradiction between "no external forces" (conserved momentum) and changing velocity from $v_0$ to $v_1$ (non-conserved momentum). You refer to a "rocket" that is "adding some energy $E$ to the system by burning some fuel". A rocket is more complicated to analyze than what you are showing, because the "system" (on which there are no external forces) is the rocket plus its propellant. The system is not moving as a rigid body and cannot be described by a single mass and velocity, so your equation does not apply. The energy $E$ released by burning is the net increase in kinetic energy of rocket plus propellant, but this may be distributed in different ways between the two. A simpler application of your approach would actually be when a vehicle is subject to external forces, but the thing that momentum is exchanged with is so massive that its energy change is negligible -- say, the Earth. Take an idealized wheeled land vehicle with only static friction (tires gripping the road), with no rolling resistance or drag, and with a perfectly efficient engine and transmission. Then your formula does apply directly: The kinetic energy of the vehicle increases by the energy $E$ of fuel burned.

Proving that flux contribution due to rotation is zero for a triangular loop welded to infinite wire set up Consider a triangular loop attached at the vertex to an infinitely long wire which has time varying current flowing in the +x direction Adapted from JEE advanced paper-1 2016 One may find that the contribution due to the current varying is given as: $$ V= \frac{\mu_o d}{\pi} \frac{di}{dt}$$ But now, suppose the set up is rotated about the axis of wire, what extra EMF would be generated? Apparently the answer is zero but I find it a bit tricky to understand. So far, I understand that this is due to $\vec{B}$ by the wire being a function of $r$ and hence all point at same distance of axis is equivalent (at least to the magnetic field). I need to show some how prove that, if we were to rotate the set up about the vertex, then the $\vec{B}$ field at all points in the interior of the triangle would be the same as the original unrotated configuration for all possible rotation angles. Preferably, I wish for a mathematical explanation but a physical one is fine as well if it detailed.

I have drawn the set up in four different configurations upon rotation. The black arrow is the area vector, and x is a point marked inside the triangle. We see that under rotations about the axis, the perpendicular distance from the axis to points inside the triangle remains fixed eg: the distance to x is fixed. This needs a proof but it's geometrically intuitive. For more intuition, see that each point in the triangle moves in a circle cantered at the axis, and the B vector field is tangent to this circle at every point.

How to understand the ambiguity of vector resolvation? When we solve problems where there is a pendulum suspended using a tight, inextensible string and the question asks about the tension developed in the string at the highest point of the bob's swing. The following is the conventional approach to solve the problem. As you can see, even I resolved the tension and $mg$ into their respective components. The confusion I had was - here $T=mg\cos\theta$ and $mg=T\cos\theta$. How do I know which one to consider? Because they both make equal sense (to me at least) - their directions match perfectly.

The mg sin(θ) produces a torque causing an angular acceleration. The T - mg cos(θ) is not zero. It must provide a centripetal acceleration.The mg – T cos(θ) gives a downward component of acceleration.

Computing the longitudinal and traceless part of the left hand side of Einstein's equation I am reading a textbook on cosmology. Consider $G^i_j$, the left hand side of Einstein's equation. If $\Psi$ and $\Phi$ are first order perturbations to the time and spatial components respectively of the metric, $G^i_j$ can be written as $$G^i_j = F(\Phi,\Psi)\delta^i_j + k^ik_j\frac{\Phi+\Psi}{a^2}$$ where $F$ is a function of $\Phi$ and $\Psi$ The textbook then tries to consider only the longitudinal and traceless part of $G^i_j$ by contracting $G^i_j$ with the operator $\hat{k}_i\hat{k}^j -\frac{1}{3}\delta^j_i$. I have a few questions here. What does "longitudinal" mean? Also, how does one concoct the operator $\hat{k}_i\hat{k}^j -\frac{1}{3}\delta^j_i$ and why does it pick out the longitudinal and traceless part of $G^i_j$?

We are working in Newton gauge which is also called longitudinal gauge. The perturbed metric can be written as $g_{\mu\nu} = g_{\mu\nu}^{(0)} + g_{\mu\nu}^{(1)}$ with, \begin{equation} g_{\mu\nu}^{(1)} = \left(\begin{matrix} 2\psi & v_i \\ v_i & 2\phi \delta_{ij}+h_{ij} \end{matrix}\right) \end{equation} Unlike in synchronous gauge, here, due to presence of the potential, the observers experience the velocity. The velocity components can be separated into longitudinal ($v_L$) and transverse components ($v_T$) such that, the former is curl free ($\nabla \times v_L =0 $) and the later is divergence free. Since $v_L$ is curl free it can be written as gradient of the scalar; and the scalar here would be the potential. We knew this beforehand as we are in Newtonian gauge there has to be velocity associated with the potential. We can carry the same analogy to the spatial perturbations $h_{ij}$ which can be decomposed into transverse, longitudinal and trace part. These modes are independent of each other (-you can take the diveregnce of the metric to get only the transverse part, similarly for the longitudinal part you can take the curl) - this is the decomposition theorem. We take vectors $\partial^ih_{ij}^L, \partial^ih_{ij}^T$, which are longitudinal and transverse respectively (analogous to our arguments for a vector). \begin{equation} \epsilon_{ijk}\partial_i\partial_kh^L_{ij} = 0 \implies h^L_{ij} = \left( \partial_i\partial_j - \frac{1}{3}\delta_{ij}\nabla^2\right)S \end{equation} where $S$ is a scalar. The last term in momentum space turns out to be a projection operator you are looking for (of course in your case you are acting it on the Ricci tensor). Sorry for the late reply, I completely forgot. Let me know if you need more.

Interesting answer as a range of tension in pulley-block-plane system I want some intuitive understanding on why there will be a range in tension in the below question. (On solving we will get that the system is at rest ($a=0$) and since its starts from rest the blocks will be stationary). Now, this seems experimentally feasible to have a unique absolute tension and I wonder why this is not the case (I initially thought friction might change but as the blocks are stationary it's not the case). A system of two blocks and a light string are kept on two inclined faces (rough) as shown in the figure below. All the required data are mentioned in the diagram. Pulley is light and frictionless. (Take $g=10\,\text m/\text s^2$, $\sin37^\circ=3/5$) If the system is released from rest then what is the range of the tension in the string? Note: Please assume the wedge to be at rest even though not mentioned.

I initially thought friction might change but as the blocks are stationary it's not the case It's not that friction is changing over time, it's that the specific value for friction (in the static case) is unknown, so the specific value for tension is also unknown. You know the maximum possible value for static friction on the blocks, but not the specific value. Let's go to a more extreme case. Imagine that the coefficient of friction is so high that the blocks can remain in place on the ramp without a rope. When you set them down, what is the tension on the rope? The answer is that it depends. You could put them down with 0 tension and they would stay in place. You could put them down with very high tension and they would stay in place. For the problem given, the low coefficient of friction limits the possible values for tension, but doesn't indicate a unique value.

Can an observer in a double tidal locked system figure out it is orbiting? Tidal locking, when the spin rate of a body matches the orbital rate so that it always faces the other body with the same side, usually occurs for just one of the bodies in orbit. However, there are also systems in which both bodies are tidally locked to one another (e.g. Charon around Pluto), see animation. Consider an universe where these two bodies are the only bodies, so that there are no external reference points. If an observer is on one of the bodies, the system will appear frozen in its reference frame (i.e. the other body is always in the same spot in the sky showing the same face). Is there a way the observer can deduce that the bodies are in orbit without simply arguing "gravity exists, we are not falling towards each other, hence we must be in orbit" * I can imagine a civilization in such a scenario would have a harder time understanding gravity in the first place ;)

There would be Coriolis and centrifugal forces due to the rotation of the reference frame. These could be detected. For example, a Foucault pendulum with its vertical axis perpendicular to the plane of the orbit would precess once per “year”.

Can nothing other than light have speed independent of source? Is it impossible for there to be some phenomenon that travels at a different speed than light to have speed independent of the source? Because if there were such a phenomenon there would be competing formulas for time dilation, correct? To be clear, this is not a faster-than-light question; the phenomenon could have constant speed lower than light and still be problematic if I'm not mistaken.

All waves in a medium (e.g. sound waves, surface waves in water) travel with a speed that is independent of that of the source (that is ignoring dispersion, which would create a velocity dependence via the Doppler effect) In a vacuum, it is only light (i.e. electromagnetic waves), gravitational waves, or other massless objects that have a speed independent of that of the source (namely the speed of light c)

How can the change of the Earth's temperature be determined with more than 1/10 K accuracy as IPCC suggests? How can one determine the extent of global warming (expressed as a temperature difference) with such precision? The latest IPCC report states temperature changes to fractions of one degree - without actually knowing the Earth’s absolute temperature! What is the statistical concept that allows for such precise determination of a temperature difference without having an absolute reference (i.e. the Earth’s absolute temperature)? If we would know the earth`s effective heat capacity one could try and set up an energy balance (using data from space observation), but I doubt that this could be done with such precision? Some guidance would be most appreciated (note: I just entered high school, so please not too complicated). Thanks. Marie

Temperatures are recorded at many weather stations in lots of countries. This data has been recorded accurately for many years, then an average is done over 30 years, for example, to get the yearly change to $0.1K$ accuracy.

Torques acting on a unicyclist When a unicyclist rounds a bend, he or she has to tilt in order to generate a frictional force, the friction acting as the centripetal force constraining the unicyclist to a circular path. However, this tilting causes the combined weight of the unicyclist and the unicycle to have a torque about the point of contact of the tyre of the unicycle and the ground. What balances this torque? It cannot be the frictional force nor can it be the normal reaction on the tyre since both forces act on the point of contact and thus have no torque about it. On the other hand, if the torque remains unbalanced, the unicyclist will eventually crash to the floor yet this doesn't happen in many cases. I am assuming the motion is observed by a non-accelerating observer, so there is no centrifugal force or any other fictitious force.

Why do you need to lean forward when standing on a train when it is accelerating? You displace your CG to be out of line with the normal force on your feet. In that case, the torque about your CG due to the normal force balances that due to the static friction on your feet. The case of the unicyclist may be similar, though things are a little more complicated with the angular momentum of the wheel.

Why is there an $i$ in the definition of hadronic decay constants? The decay of (for example) a pion can be parameterized by a decay constant $f_\pi$ defined via $$ \langle 0 | \bar d \gamma_\mu \gamma^5 u |\pi^+(p) \rangle = i f_\pi p_\mu $$ $$ f_\pi \approx 131 \text{ MeV.}$$ My question is why do we include the $i$ on the RHS? In other words how do we know that the matrix element on the LHS is purely imaginary?

Well, you may define your conventions, and physicists are perverse enough to actually do that (watch them...), any way you want. This is the dominant convention, and the logical "chain of custody", so to speak, is, schematically, $$ \pi^+\mapsto \pi^+ + f_\pi \theta^+, \qquad \leadsto \\ J_{5}^{\mu~~+} = \frac{δ{\mathcal L}}{δ\partial_\mu \pi^+} \frac{δ\pi^+}{δ\theta^+}=f_\pi \partial^\mu \pi^+ ~~~(\sim -\bar d \gamma^\mu \gamma^5 u ),~~~\leadsto \\ \langle 0 | \bar d \gamma_\mu \gamma^5 u |\pi^+(p) \rangle = i f_\pi p_\mu ~. $$ I've been cavalier/dyslexic with signs and factors, but this is the choice for a real weak iso-raising matrix $\tau^+$. So the i pops up in the conversion from p to -i∂ . For the neutral pion (hermitean), you'd get a hermitean current, which is the standard convention.

What are quantum fields made up of? If quantum fields are mathematical entities made up to explain nature, what they explain is definitely something physical and is made up of something. So why can’t there be an answer to what these mathematical quantum fields are made up of? I mean, if physicists are making them up, then they definitely would have certain criteria as to what these mathematical fields are made up of, won't they?

I'm going to give a bit more literal of an answer with a bit less story since that's my taste. What we call a quantum field is an operator as a function of position and/or time, i.e. at every $x, t$ there is an operator $\hat{\phi}$. Just like other operators, this operator acts on elements of the hilbert space. In mathematical physics, they are even more precise and formulate the fields as operator-valued distributions. That is the simplest quantum field; in general it can have multiple operator components. For example, in QED the vector potential $A^\mu$ has components $A^0, A^1, A^2, A^3$ which are each operators. It is a quantum field. We call them fields because in physics, the word "field" is used for (possibly time-dependent) objects which depend on position. These mathematical objects are part of our theories which we use to make predictions for experiments like the ones at the Large Hadron Collider. However, we have in no way proven that they are fundamental to reality - they could just be part of our human formulation of (our best progress so far at) the fundamental theory of nature. So there is a sense in which we don't even know that they really exist, much less that they are made of one thing or another.

Why does a fluid follow the wing? When air is moving above and under a wing that is curved, why does the air at the top of the wing follow the wings shape and go downwards when it could just go in a straight line? It doesn't make sense to me.

When air flows over a wing (or more generally, when any fluid flows over a solid surface), viscosity tends to make the air stick to the surface and creates a region of sheared flow (the boundary layer) where the speed varies between zero at the surface and a larger value farther out in the flow. Assuming sufficiently large Reynolds numbers, the boundary layer is thin and its effects on the surrounding flow can be modelled by distributed vorticity on the wing's surface. Because of the vorticity, the previously uniform flow (before encountering the wing) is modified, typically producing low-pressure regions forward on the upper surface and adverse pressure gradients farther aft as the flow approaches the trailing edge, where it has to match the pressure from the lower surface. For sufficiently streamlined cases and mild enough flow conditions (i.e. large enough Reynolds numbers and small enough lift coefficients), the inertia of the flow is enough to allow the flow to overpower the adverse pressure gradients and continue to the trailing edge without separating from the surface. A similar process is responsible for the so-called Coanda effect, in which a jet passing over a flap is deflected downward so it contributes to the lift as well as the thrust. In summary: the viscosity of the flow is what allows it to follow the wing's surface. Without viscosity, airfoils would have no lift or drag (in 2D, at least).

If the experimentalist does not look at the recorded data in the 'which-way' experiment, will it affect the inference? Related question quantum eraser question In https://en.wikipedia.org/wiki/Double-slit_experiment#%22Which-way%22_experiments_and_the_principle_of_complementarity "Which-way" experiments and the principle of complementarity A well-known thought experiment predicts that if particle detectors are positioned at the slits, showing through which slit a photon goes, the interference pattern will disappear.[6] My questions are: (1) if the experimentalist does not look at the data of the detector, despite the detector looks like working, will the interference still disappear at the screen? (2) if the answer to the question (1) is, still disappear, then what would be the definition(s) of measurement/observer? Because in step (1), the experimentalist will not know the result of measuring the electron goes to which slot, the measurement will proceed up to the device, not the experimentalist (if observer is the human, then it seems lead to subtly).

It is enough to put detectors on the slits to measure which way was taken. The interference pattern will dissapear. You don't need a human being to notice the results. The point is interaction (an event happening somewhere on slit A or B, not that someone notices it). Things happen without someone spending attention to it all the time.

Index Notation Question We're busy doing a GR course, and index notation has always been something that confuses me. In particular, is there a difference between the following, and if so, what is it? $A^\mu_\nu$; $A^\mu{}_\nu$ and $A_\nu{}^\mu$

Let's say we start with a general (not symmetric or anti-symmetric) tensor $A_{\mu\nu}$. Note $A_{01} \neq A_{10}$ (and similarly for other combinations). Then \begin{equation} A^\mu{}_\nu = g^{\mu\rho} A_{\rho\nu} \neq g^{\mu\rho}A_{\nu\rho} = A_\nu{}^\mu \end{equation} which you could see by explicitly writing out both sides of the not equals sign for a specific value of $\mu$ and $\nu$ (say $\mu=\nu=0$). If $A_{\mu\nu}$ is symmetric however, in the sense that $A_{\mu\nu}=A_{\nu\mu}$, then $A^\mu{}_\nu = A_\nu{}^\mu$. In that case, out of sheer laziness, some people write $A^\mu_\nu$ instead of one of $A^\mu{}_\nu$ or $A_\nu{}^\mu$. However the notation $A^\mu_\nu$ is just a shorthand and is meaningless if $A$ is not a symmetric tensor.

Poynting Vector Perpendicular to Surface In Spherical coordinates if we have a source at the origin generating $E$ in the $\hat{r}$ direction and $H$ in the $\hat{\phi}$ direction then our Poynting Vector will be in the $\hat{\theta}$ direction. When considering the power of such a source we know that the total power is the flux of the Poynting vector through a surface that encloses the sphere. If we take this surface to be a sphere of radius $R$, then when taking the flux at every point it is zero since $\hat{\theta} \times \hat{n}=0$ where $\hat{n}$ is the unit vector normal to the surface given by $\hat{r}$, so is the power of the source $0$? This doesn't seem entirely correct to me since there is obviously fields being generated, and the magnetic and electric field phases can be switched to alter the power factor, but here it seems as though regardless of what the fields are their directionality ensures $0$ power.

It's possible to pose questions about field configurations that can't be made! An everywhere radial $\mathbf E$ field suggests a net charge with a spherically-symmetrical distribution, which is fine. But an everywhere azimuthal $\mathbf H$ field suggests a current in the $z$-direction, which is inconsistent with a spherical charge distribution (it leads to a time-dependent dipole moment). So I think you should first specify a physically possible field configuration.

Why is Hamilton's principle (or principle of least action) still valid in a relativistic field theory? I am struggling to understand why the principle of least action which is derived in classical mechanics from d'Alembert's principle continues to be valid in a regime that treats a relativistic field. What is it that tells us that the equations of motion can still be obtained from the principle of least action (applied on a lagrangian that is relativistically invariant)?

Here is one possible line of reasoning: * *In the case of relativistic point mechanics, one can still use d'Alembert's principle and the relativistic version of Newton's laws to derive Lagrange equations. (The main difference compared to the non-relativistic case is the form of the kinetic term in the Lagrangian.) *In the case of e.g. a relativistic scalar field with EOM $$ \mp\Box\phi~=~{\cal V}^{\prime}(\phi)$$ with Minkowski signature $(\pm,\mp,\mp,\mp)$, it is not hard to see that a Lagrangian density is given as $$ {\cal L}~=~\pm\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - {\cal V}(\phi). $$ In fact, in relativistic field theory, the Lagrangian formulation is often taken as a starting point/first principle, cf. above comment by Jerry Schirmer.

Is space really expanding? In a book called "Einstein, Relativity and Absolute Simultaneity" there was this sentence by Smith: There is no observational evidence for a space expansion hypothesis. What is observed are superclusters of clusters of galaxies receding from each other with a velocity that is proportional to its distance. He goes on to say space is Euclidean and infinite. Wouldn't this mean Big Bang was a explosion in spacetime rather than a expansion of spacetime as it is often told? Is Smith just wrong or don't we know yet?

Smith is just wrong. He is right to say that there is no direct observational evidence (in a naive sense) that space is expanding rather than space being fixed while galaxies spread apart through it. They both just look like galaxies getting farther apart. But there is a lot of observational evidence that special and general relativity are correct. And, if those theories are correct, the only way to make sense of the observations is to say that space itself is expanding. It's kind of like standing in a highway and seeing a car getting bigger and bigger and saying that there is no evidence it's getting closer because maybe it's just actually growing in size. We know cars just don't do that. The problem is that, at small scales and velocities, the model Smith wants to support really works well, unlike the "growing car" model. So it's not obviously dumb to think that the galaxies are just flying apart through unchanging space. But, when you understand how things work at near-light speeds, it becomes clear that it just can't be right.

Does electron physically move in an interband transition? How does electron move say from ground state energy level to first excited state? Is there any actual displacement in terms of motion?Is there a way to logically think about this by the help of creation and annihilation operators? I know this is a quantum mechanical point of view but is there some way we can think in a classical sense?

The bands are the many-to-one relation between energy and momentum for electrons in a crystal. This relation is not position dependent. (I am not talking about band warping near crystal borders.) To go from one band to another just means that an electron's energy, and possibly momentum, changes. For a so-called vertical transition only the energy changes. There is no movement in real space, only a move in ($E$, $\vec k$) space. For an atomic transition in general the spatial distribution changes upon excitation. In this sense the electron moves.

Is it true that Maxwell equations are interpreted by taking right side of formula as the "origin" and the left part as "consequence"? When books or various references interpret the meaning of Maxwell equations, they typically state that the source (origin of the phenomena) is the right part of the formula, and the resulting effect is on the left part of the formula. For example, for Maxwell-Faraday law, $\vec{\nabla} \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$ one states "a time varying magnetic field creates ("induces") an electric field." (see for example : https://en.wikipedia.org/wiki/Maxwell%27s_equations#Faraday's_law ) It seems to me that this is not true. One could interpret in both direction. For the example above, we could also state that a change of direction of the electric field will create a temporal change of the magnetic field. Is it true that Maxwell equations should be interpreted by taking right side of formula as the "origin" and the left part as "consequence"? Or could we take also the left side as the origin?

You are correct. Maxwell-Faraday states that $\vec{\nabla} \times \vec{E}$ is the same thing as $-\frac{\partial \vec{B}}{\partial t}$. Both quantities express the same phenomenon. There is no cause and effect in either direction. Once you express the fields in terms of derivatives of the vector potential the two expressions become identical. As the Coulomb potential is rotation free we can set it to zero for this excercise. Then $\vec{E} = -\partial_t \vec{A}$ and since $\vec{B} = \vec{\nabla} \times \vec{A}$ both sides of the Maxwell-Faraday are seen to be identical. In this sense Maxwell-Faraday expresses the fact that the fundamental quantity of electromagnetism is the vector potential.

Rabi Hamiltonian with three instead of two Pauli matrices This question was motivated by a question on Mathoverflow, in which a Hamiltonian is considered that looks like the Rabi Hamiltonian, but with three instead of two Pauli matrices: $$H=\omega a^\dagger a+\Delta\sigma_z+g_x\sigma_x(a+a^\dagger)+ig_y\sigma_y(a-a^\dagger).$$ In the literature I have only found the Rabi Hamiltonian with $g_y=0$, so with two Pauli matrices. The case $g_y\neq 0$ with three Pauli matrices seems essentially different, at least if $g_x\neq g_y$ and one does not make the rotating wave approximation. Has it been considered before? Since this model is so fundamental, I am a bit surprised at not having found this generalization, I may be missing something obvious.

I probably should have searched more extensively before posting here, I did eventually find this Hamiltonian in Exceptional and regular spectra of a generalized Rabi model by Michael Tomka, Omar El Araby, Mikhail Pletyukhov, and Vladimir Gritsev. The context is different from quantum optics (which is probably why I could not locate it earlier), it is a model of a two-dimensional electron gas with parabolic confinement in one direction, with a Zeeman field and Rashba and Dresselhaus spin-orbit couplings. The three-Pauli-matrix Rabi Hamiltonian is also referred to as the anisotropic Rabi Hamiltonian (arXiv:1401.5865). In my answer to the Mathoverflow question I made a conjecture for the ground state energy in the special case that $\Delta=2g_xg_y$: The generalised Rabi Hamiltonian $$H_0=\begin{pmatrix} a^\dagger a+\alpha^2&\alpha a+\beta a^\dagger\\ \alpha a^\dagger+ \beta a&a^\dagger a+ \beta^2 \end{pmatrix}$$ has a doubly degenerate ground state at zero energy for any $\alpha,\beta\in\mathbb{R}$. I now see that this conjecture follows from the Bethe Ansatz solution of the generalized Rabi Hamiltonian given in the Tomka et al. paper, section II.B.1. The case $\Delta=2g_xg_y$ corresponds to $\lambda_-=\delta$ in their notation, which is the "exceptional case"$^\ast$ in which the Rabi Hamiltonian has an eigenvalue $E_0=-\lambda_+=-(g_x^2+g_y^2)$. This corresponds to a zero eigenvalue for $H_0$, because the spectrum of $H_0$ is shifted relative to the Rabi spectrum by $(\alpha^2+\beta^2)/2=g_x^2+g_y^2$. The zero eigenvalue is the ground state, because $H_0$ is positive definite, $$H_0=Q^\dagger Q,\;\;\text{for}\;\;Q=\begin{pmatrix} \alpha&a\\ a&\beta \end{pmatrix}.$$ The two ground state wave functions, such that $Q\Psi_\pm=0$, are $\Psi_\pm={{\sqrt\beta}\choose{\mp\sqrt\alpha}}|\!\pm\!\!\sqrt{\alpha\beta}\rangle$, in terms of the coherent state $|\gamma\rangle$ with $a|\gamma\rangle=\gamma|\gamma\rangle$. $^\ast$ In arXiv:1407.5213 the symmetry is identified that enforces the doubly degenerate ground state.

Why the mechanism of everything in the universe has a pattern? why everything in the universe has a pattern which can be identified and understood to determine outcomes, properties, effects of almost everything. I am saying that couldn't the universe be like patternless, non-deterministic and chaotic. For example why the gravitational force between any two objects has a pattern which always obeys universal law of gravitation and can be predetermined. Couldn't be the gravitational force between any two given objects would have no pattern and would be completely random and non-deterministic. Is this property of universe in which everything has a pattern is a complete matter of chance or it is a property of even something fundamental.

Humans like looking for patterns in nature and when we find them we show each other in books, lectures etc... It is remarkable that so many patterns have been discovered. However there are lots of things where patterns haven't been found, we probably don't talk about those as often, especially on 'Physics StackExchange'. Even within physics there are things where there doesn't seem to be a pattern, or at least not a known one. For example the value of the fine structure constant or the masses of the particles, the radii of the planets etc...

Will two bodies initially connected to and revolving around each other, start spinning when disconnected? Two extended bodies are connected with a string and revolve around each other (that is, around the center of mass of this system). No gravity, no external forces. The string is cut, and they start to depart from each other. Will they spin around their own centers of mass? For a (representative) particular case when one of the bodies is a dumbbell (two point masses connected by a light rigid rod), the answer is trivial - each point mass will start its tangential movement with a different velocity, so the pair will be revolving. On the other hand, for the following configuration, there seems to be no rotation. Is there a general approach?

Take a ball of radius $r$ connected to a thin center-post by string of length $R$. The post accelerates up to and stays at a high final rotation speed $\Omega$. Obviously the ball revolves at $\Omega$. Now imagine case 2 where a thin, light ball-bearing is connected to the string and the ball rotates within it. Lookin down along the axis of rotation, draw an arrow ⬆️ on the top of the ball. In case 2 it always points the same way, and in case 1 it rotates once per revolution. At the same angular velocity $\Omega$ of the system: $$ v_1=v_2 =v= \Omega (R+r) $$ $$ \omega_1 = \Omega ~,~ \omega2 = 0$$ By conservation of angular and linear momentum, after the cut: $$v_{1,f}=v= \Omega (R+r) ~,~ \omega_{1,f}= \Omega$$ $$v_{2,f}=v= \Omega (R+r) ~,~ \omega_{2,f}= 0$$ Yes it spins as it travels away no matter what shape it is. In your lower pic the tangent velocities are not parallel for the two balls. By the way, the kinetic energy in case 1 is higher: $$E_1= \frac{1}{2} m v^2 + \frac{1}{2} I_{ball} \omega^2$$ where $I_{ball}$ is the moment of inertia of a sphere around around its axis. The second term is gone for case 2: $$ E_2= \frac{1}{2} m v^2= \frac{1}{2} I_{mass} \Omega^2$$ where $I_{mass}$ is the moment of inertia of a point-mass in revolution, $m(R+r)^2$ in this case. We also just derived the parallel axis theorem where $E_1= \frac{1}{2} I_{tot} \Omega^2$ and $I_{tot}=I_{mass}+I_{ball}$.

Pressure in open tube In the L shape tube shown in the image shouldn't the pressure at all points A, B,C,D be same as atmospheric pressure? I have to find relation between A, B, C and D

The "$\omega$" in the drawing suggests that the tube is rotating. In that case there will be a pressure gradient in the horizontal part to provide the centripetal acceleration: $$ \rho_{\rm air}\omega^2 r = \frac{dP}{dr} $$

Issue in deriving Ehrenfest's theorem Working in Schrodinger picture, while deriving Ehrenfest's theorem, we go - $$ \frac{d}{d t}\langle A\rangle=\frac{d}{d t}\langle\psi|\hat{A}| \psi\rangle $$ $A$ is an operator. Expanding RHS- $$ \frac{d}{d t}\langle A\rangle=\left\langle\frac{d}{d t} \psi|\hat{A}| \psi\right\rangle+\left\langle\psi\left|\frac{\partial}{\partial t} \hat{A}\right| \psi\right\rangle+\left\langle\psi|\hat{A}| \frac{d}{d t} \psi\right\rangle $$ My doubt is regarding the second term. Why do we write $\frac{\partial}{\partial t}\hat{A}$ and not $\frac{d}{d t}A$? Of course, this notation wouldn't matter incase there is only an explicit dependence of $t$, if there's any $t$ dependence at all. What if $A$ were composed of other time dependent operator $\hat{O}(t)$, i.e. $\hat{A}(t)=A(\hat{O}(t),t)$. Can we have such operators? In that case $\frac{\partial}{\partial t}\hat{A} \neq\frac{d}{d t}A$.

We us the partial derivative because there are other variables in play --- such as $x$ and $p$, both of which may be time dependent. The partial derivative symbol is used because it implies that we are keeping all the other variables fixed when we vary $t$. Using the "$d$" derivative would imply that $$ \frac{d}{dt}F(x(t),p(t),t)= \frac{\partial F}{\partial t}+ \dot x \frac{\partial F}{\partial x}+\dot p\frac{\partial F}{\partial p}. $$

The relationship between velocity of centre of mass and angular velocity of a rigid body Consider a rotating object with mass $m$, moment of inertia $I$, along an inclined plane of vertical height $h$. Then simply speaking the following conservation law holds. $$\frac{1}{2}(mv_{CM}^2+I\omega^2) = mgh$$ Recognise that all other letters than $v_{CM}$ and $\omega$ are given constants. Thus, solving for $\omega$, $$\omega = \sqrt{\frac{2mgh-mv_{CM}^2}{I}}$$ The result above means that we get the angular velocity as a function of the velocity of the centre of mass. But I personally feel that this contradicts with my intuition, because experimentally if there are given inclination $\theta$, height $h$, moment of inertia $I$, and initial velocity $v_0=0$, then there should be a unique set of $(v_{CM}, \omega)$. What's the problem?

There are a unique set of $\omega$ and $v_{cm}$. They are in fact both related to each other by the equation $$\omega=\frac{v_{cm}}{r}$$

Optimizing for "sudden" change in angular momentum I'm trying to create an elaborate impossible-to-juggle juggling club, by triggering a sudden change in angular momentum via a mechanism inside of the juggling club while it is in mid-air. The mechanism is going to be a torsion spring attached to a flywheel. For the prototype, I will manually wind up the torsion spring ahead of time, and I have a release mechanism that will be triggered via a small hobby servo controlled by an arduino. I would like to do some calculations to make sure that my setup will produce the sufficient unexpected angular momentum change, however, I'm not sure what exactly I should be optimizing for. I could try to just maximize the total change in angular momentum (angular impulse), but that doesn't necessarily account for the "suddenness". I could try to make sure that the total amount of time for the angular momentum change is minimized, however, I was thinking that maybe it would be better to optimize for maximum jerk. Or maybe jerk over time? I've never been able to wrap my head around jerk. What should I be optimizing for?

For maximum effect you will want to maximize the mass and radius of the flywheel and the torque exerted by the spring. You will need to decide whether you want the flywheel to be set spinning or allowed to oscillate. Its axis should be perpendicular to the long axis of the club. Your biggest problem will building a mechanism which can withstand being dropped.

Is the Lagrangian formulation a mathematical inevitability? An analogy with functions: Say, we have a function $f(x)$ and we have an equation to solve, $f(x)=0$. We can always re-formulate the problem of solving $f(x)=0$ with the problem of extremising $F(x)$, where $F(x)$ is the anti-derivative of $f(x)$. This is purely mathematical. In physics, we postulate that the path taken by a particle $x(t)$ is completely determined by the initial conditions $x(t_0)$ and $v(t_0)$. Mathematically, this can be expressed as: $$\frac{d^2 x(t)}{dt^2}=F(x(t),v(t))$$ or equivalently, $$\frac{d^2 x(t)}{dt^2}-F(x(t), v(t))=0$$ This is an equation which takes a path $x(t)$ as input and evaluates to 0. This is similar to $f(x)=0$ in the analogy at the top of the post. So, is it a mathematical certainty that we are able to re-formulate this problem in terms of extremising a functional which takes the path $x(t)$ as input? If yes, it's not obvious to me how.

The cheapest way to make $F(x,\,\dot{x})-\ddot{x}=0$ an ELE is with an auxiliary variable $y$ viz. $L=y(F(x,\,\dot{x})-\ddot{x})$, so $y$ is dynamical. You may prefer to add a total derivative $\frac{d}{dt}(y\dot{x})$ to get an alternative Lagrangian, $L=yF(x,\,\dot{x})+\dot{y}\dot{x}$. If $x$ were complex-valued, the choice $y=x^\ast$ would be a common quantum-mechanical technique, although you'd want to make your Lagrangian Hermitian in the last step by adding a total derivative.

Diffraction conundrum Take any macroscopic crystal of VIS transparent monocrystalline type of material like table salt ($\text{NaCl}$), sugar (sucrose), quartz (silica), calcite ($\text{CaCO}_3$), glass (which is non-crystallin, of course) etc etc. A VIS laser beam travels through these crystals effortlessly. But grind the same materials to a fine powder and they become thoroughly opaque to VIS. Now replace the laser with a collimated X-ray beam and the powders are transparent to it, producing a diffraction pattern (if the set-up is correct) So why are the powders transparent to X-rays but not to VIS? This question is inspired by this Q and A.

The reflectivity, in the visible range, of your samples is nonzero. For large crystals, the light (generally) only passes thru two interfaces, i.e. air-crystal and crystal -air , and thus rather little signal is lost. WHen you grind it up, not only do the crystal faces of each particle point in random directions, but the light passes thru many many particles, getting both attenuated and redirected (prism) along the way. So far as visible vs. X-ray: the wavelengths are rather different, so at the atomic level very different things happen. Roughly speaking, the X-rays see the crystal lattice as a diffraction grid. That same grid spacing is far too small to diffract visible wavelengths.

How is Lorentz force frame-independent? I have studied that the net force on a charged object moving with velocity $v$ under both electric and magnetic fields in given as $\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$. I have also been told by my teachers that the net force given by this equation is frame independent for inertial frames. Regarding this I have two doubts: * *Is the force frame independent in only magnitude or both magnitude and direction? *Why is either magnitude or direction frame independent at all, since the force is dependent on velocity which is not frame independent. For example, if we assume a constant electric field $E$ and a constant magnetic field $B$ and two observers $O_1$ and $O_2$ where $O_1$ is at rest and $O_2$ is moving with constant velocity $v$. Now if we project a charged particle also with velocity $v$, $O_1$ will observe the particle's velocity as $v$ and hence the force on the particle observed by $O_1$ is $q(E+v\times B)$. $O_2$, however, observes the particle to be at rest as there is no relative velocity between O2 and particle. So shouldn't $O_2$ observe the force as only $qE$? The net direction of the forces observed by both observers should also be different. How do we resolve this?

Good question. You have stumbled upon the fact that Maxwell's equations are not invariant under a Galilean Transformation. Galilean Relativity says that two objects separating with a relative velocity $\vec v$ will calculate the same force on a particle: $$x'=x-vt$$ therefore $\ddot x'=\ddot x$ So first of all let us make this clear: The force is in fact unequal since Galilean Relativity is wrong and Maxwell's equations are correct. To prove this: Start with the conjecture that the force is equal in the two frames and you will see that this condition does not reproduce the Maxwell's equations. This was in fact one of the problems that got Einstein interested in Special Relativity back in 1905.

Spontaneous emission laser Why exactly can we not create a 'laser beam' from a medium undergoing spontaneous emission inside a cavity? It seems like the dielectric mirrors are going to cause a standing wave to appear irrespective of whether the medium is doing stimulated or spontaneous emission, and that will produce something looking like a laser beam. Why does it specifically have to be stimulated emission?

For driven harmonic motion, there must be a resonant frequency and an oscillating driver, and they must be somewhat close in frequency. You may think of a pendulum for example and not immediately realize we do have that situation in a way. One way to think of it is that driven oscillation situation is mimicked. We have no such mimicking here. With undriven SHM, the restoring force is proportional to displacement which is related, and the same in some ways, but not identical to driven SHM. You do not have that with standing waves in dielectric mirrors. Nothing equivalent to $F=-kx$. Here’s just one example of actually driven harmonic motion: a hose hanging from the ceiling oscillates sometimes. There are two characteristic frequencies that can reinforce each other, the pendulum frequency $f_p$, and the flow length frequency $f_{uL}$, where $u$ is fluid velocity: $$f_p= \frac{\sqrt{g}}{2\pi \sqrt{L}} \text{ , } f_{uL}=\frac{u}{L}$$ To drive oscillation: $$ \sqrt{gL}=2\pi u \text{ , eg: } L=1m, u=\tfrac{1}{2}\tfrac{m}{s}$$ But what’s happening is that the driver is oscillating.