quantum mechanics - Associating a Unitary operator to proper Lorentz transformations?

If one reads eg page 32 of Srednicki where he says:

In quantum theory, symmetries are represented by unitary (or antiunitary) operators. This means that we associate a unitary operator U(Λ) to each proper, orthochronous Lorentz transformation Λ. These operators must obey the composition rule...

Where does all this come from? Is it Wigners theorem? Probability conservation?

For instance equation 2.15 in the same link:

$$U(\Lambda)^{-1}P^{\mu}U(\Lambda) = \Lambda^{\mu}_{\nu}P^{\nu} $$

I understand the right hand side no problem: The Lorentz matrix acts on the four vector and "rotates it" end of story. But why do we need the left hand side? What does the left hand side even mean, what is $U$, except being a unitary operator?

electric circuits - Case where potential difference causes no current

why current will not flow from battery if I connect positive terminal of battery to ground even when there is potential difference

Answer

I connect positive terminal of battery to ground...

[The] negative terminal is not connected to anything.

If the negative terminal of the battery is not connected to anything, then no current can flow through the battery. Current can only flow around a complete circuit (i.e., it can only flow around a closed loop).

Quick non rogorous way to obtain Feynman rules from a Lagrangian in a non abelian theory

I have been told that a quick way to get the Feynman rules from a Lagrangian is to take an interaction term, forget about the fields and multiply an $i$. This works perfectly for example for QED but I wonder if there is an analogous non rigorous quick way to obtain the feynman rules of for example non abelian gauge theories.

homework and exercises - Angular Velocity expressed via Euler Angles

On the top of the fourth page from here, the author trivially derives the components of angular velocity, expressed via Euler angles of the system. I fail to understand how the components of angular velocity were derived. May you please enlighten me!

Answer

I assume you know about rotation matrices, and so for a sequence rotations about Z-X-Z with angles $\phi$, $\theta$ and $\psi$ repsectively you have

$$ \vec{\omega} = \dot{\phi} \hat{z} + T_1 \left( \dot{\theta} \hat{x} + T_2 \left( \dot{\psi} \hat{z} \right) \right) $$

The logic here is apply a local spin of $\dot{\phi}$, $\dot{\theta}$ and $\dot{\psi}$ on the local axes in the sequence.

1. Apply spin $\dot{\phi}$ about local Z and then rotate by $T_1$


2. Apply spin $\dot{\theta}$ about local X (rotated by $T_1$) and then rotate by $T_2$


3. Apply spin $\dot{\psi}$ about local Z (rotated by $T_1 T_2$).

Update

There is a way to formally derive the above using the identity $\dot T = \vec\omega \times T$ but it is rather involved for 3 degrees of freedom.

For two degrees of freedom it goes like this. With a rotation matrix $T= T_1 T_2$ (defined as above) the time derivative is

$$ \begin{aligned} \frac{{\rm d} T}{{\rm d} t} & = \dot{T}_1 T_2 + T_1 \dot{T}_2 \\ & = \left( (\dot{\psi} \hat{z}) \times T_1 \right) T_2 + T_1 \left( (\dot{\theta} \hat{x}) \times T_2 \right) \\ & = (\dot{\psi} \hat{z}) \times \left( T_1 T_2 \right) + (T_1 (\dot{\theta} \hat{x})) \times \left( T_1 T_2 \right) \\ & = \left(\dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) \right) \times (T_1 T_2) \\ & = \left(\dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) \right) \times T = \vec{\omega} \times T \end{aligned} \\ \vec{\omega} = \dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) $$

using the distributed property $T (\vec{a} \times \vec{b} ) = (T \vec{a}) \times (T \vec{b} )$.

general relativity - Lagrangian for relativistic massless point particle

For relativistic massive particle, the action is $$S ~=~ -m_0 \int ds ~=~ -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} ~=~ \int d\lambda \ L,$$ where $ds$ is the proper time of the particle; $\lambda$ is the parameter of the trajectory; and we used Minkowski signature $(+,-,-,-)$. So what is the action for a massless particle?

Answer

I) The equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is

$$ \tag{A} \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, $$

where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is

$$ \tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2, $$

where $\lambda(\tau)$ is a Lagrange multiplier. This answer (B) may seem like just a cheap trick. Note however that it is possible by similar methods to give a general action principle that works for both massless and massive point particles in a unified manner, cf. e.g. Ref. 1 and eq. (3) in my Phys.SE here.

II) The corresponding Euler-Lagrange (EL) equations for the action (B) reads

$$ \tag{C} 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2, $$

$$ \tag{D} 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}. $$

III) The action (B) is invariant under world-line reparametrization $$ \tag{E} \tau^{\prime}~=~f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\qquad \dot{x}^{\mu}~=~\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\qquad$$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the familiar geodesic equation.

References:

1. J. Polchinski, String Theory Vol. 1, 1998; eq. (1.2.5).

statistical mechanics - Can one stimulate emission of a photon with an energy different from the emitted photon?

Suppose I have a three-level system with $E_0$ the ground level, $E_1$ the intermediate and $E_2$ the upper level. In thermal equilibrium they will have a certain probability distribution according to the Boltzmann Statistic, in a laser one needs a population inversion, but that doesn't matter for my question.

My question is this: 3-level laser uses the $E_1 \rightarrow E_0$ transition because the $E_2$ level decays quickly (by design), i.e., emission of an $E_1$ photon is stimulated with an $E_1$ photon. But is there also a stimulated emission of energy $E_1$ for an incident photon of energy $E_2-E_1$? My thinking is that a photon energy of $E_2-E_1$ kicks the electron from $E_1$ in the upper state $E_2$, which will then decay, and every once in a while it should decay into the ground level, not back into the $E_1$ state. Is that correct? And if that is so, is the momentum of the emitted photon with energy $E_2$ aligned with the momentum of the incident photon of energy $E_2-E_1$?

homework and exercises - A block on a wedge

The system is as follows -

Friction exists only between the 2 blocks.

I am trying to find out the accelerations of $m_1$ and $m_2$.

Let $a_2$ be acceleration of $m_2$, and $a_x$ and $a_y$ be the accelerations of $m_1$ in the respective directions. Let $R$ be the normal reaction between the 2 blocks, and $N$ be the normal reaction between $m_2$ and floor. Balancing components across the axes, I get the following equations - $$N = m_2g + R\cos\theta \tag{1}$$ $$m_2a_2 = R\sin\theta \tag{2}$$ $$a_x = R(\sin\theta + \mu_s\cos\theta) \tag{3}$$ $$a_y = R(\cos\theta + \mu_s\sin\theta) – m_1g \tag{4}$$

I don’t think $(1)$ is necessary, since friction is not involved between the blocks and the ground. Leaving that aside, I have 3 equations in 4 variables: $a_x, a_y, a_2, R$.

Is there are any way I could perhaps get a 4th equation so that the system of equations could be solved? I can get $|a_1|$ in terms of $R$ from the expressions for $a_x$ and $a_y$, but I don’t think that would help.

quantum mechanics - Feynman's path integral and energy discretization?

Feynman’s path integral formulation of quantum mechanics is based on the following two postulates:

1. 
   

   If an ideal measurement is performed to determine whether a particle has a path lying in a region of spacetime, the probability that the result will be affirmative is the absolute square of a sum of complex contributions, one from each path in the region.

   
   


2. 
   

   The paths contribute equally in magnitude but the phase of their contribution is the classical action, i.e. the time integral of the Lagrangian taken along the path.

   
   
   

Suppose that those postulates are as natural as it can be, i.e. they aren't a distilled version of something more elementary. If this is the case, then how to explain, in layman's terms, that how the discrete energy levels in, e.g. the harmonic oscillator arise from those two postulates? Are there qualitative answers?

Answer

I'm not sure what you expect for qualitative examples in "layman's terms". The time-evolution amplitude from one point to another, $$\langle x_f|U(T)|x_i\rangle=K(x_f,x_i;T)~,$$ evaluated from the path integral is the propagator, which, for the oscillator, happens to be the celebrated 1866 Mehler kernel: a causal Green's function of the oscillator equation. Not coincidentally, this was available 60 years before QM. The point being that the path integral propagator is mostly classical. The quantization of energy levels is actually a feature of the compact time domain, as @Qmechanic comments.

Specifically, the classical action for the oscillator, from the above WP reference, amounts to $$ \begin{align} S_\text{cl} & = \int_{t_i}^{t_f} \mathcal{L} \,dt = \int_{t_i}^{t_f} \left(\tfrac12 m\dot{x}^2 - \tfrac12 m\omega^2 x^2 \right) \,dt \\[6pt] & = \frac 1 2 m\omega \left( \frac{(x_i^2 + x_f^2) \cos\omega(t_f - t_i) - 2 x_i x_f}{\sin\omega(t_f - t_i)} \right)~. \end{align} $$

The propagator, then, the above amplitude, can be evaluated from the functional integral as $$K(x_f, x_i;T) = \Large e^\frac{i S_{cl}}{\hbar} \small ~ \sqrt{ \frac{m\omega}{2\pi i \hbar \sin\omega(t_f - t_i)}}~~, $$ where $T=t_f-t_i$. Effectively, it is the exponential of the classical action, with a minor normalization correction due to quantum fluctuations, which, nevertheless is not that important.

• The quantized energy levels are already in the discrete harmonics predicated by the periodicity of the classical action--themselves oblivious of $\hbar$.

This expression also equals to the conventional Hilbert space propagator in terms of Hermite functions,
$$ \begin{align} K(x_f, x_i;T ) & = \left( \frac{m \omega}{2 \pi i \hbar \sin\omega T } \right)^\frac12 \exp{ \left( \frac{i}{2\hbar} m \omega \frac{ (x_i^2 + x_f^2) \cos \omega T - 2 x_i x_f }{ \sin \omega T } \right) } \\[6pt] & = \sum_{n = 0}^\infty \exp{ \left( - \frac{i E_n T}{\hbar} \right) } \psi_n(x_f) ~\psi_n(x_i)^{*}~, \end{align} $$ with which you'd propagate your ket to the bra of the amplitude in conventional Hilbert space.(This is the expression Mehler summed in 1866.) But you might wish to pretend you are a Martian, unaware of that formulation, or Mehler's wonderful, "prescient", formula.

Rewrite this as $$ = \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T} 2 \left( 1 - e^{-2 i \omega T} \right)^{-\frac12} \exp{ \left( - \frac{m \omega}{2 \hbar} \left( \left(x_i^2 + x_f^2\right) \frac{ 1 + e^{-2 i \omega T} }{ 1 - e^{- 2 i \omega T}} - \frac{4 x_i x_f e^{-i \omega T}}{1 - e^{ - 2 i \omega T} }\right) \right) }\\ \equiv \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T } 2 ~ R(T)~. $$

The $e^{-in\omega T }$ Fourier modes of R(T) then multiplying this 0-point energy prefactor may be compared to the standard Hilbert space eigenstate expansion of the resolvent, to reassure you of the standard quantized spectrum of the quantum oscillator, $E_n = \left( n + \tfrac12 \right) \hbar \omega~. $

Here, faced with discreteness, you only need appreciate the essential periodicity of the system, the compactness that forces you to a harmonic structure: the waviness of the system; and that most of it is traceable to the classical action in this (somewhat exceptional) quadratic hamiltonian paradigm.

In their elementary textbook, Feynman and Hibbs work it out nicely in Probs 2-2, 3-8, (eqns 2-9,3-59) and "spike" in eqns (8-12), (8-13). (They even go ludicrously further than that, trying to make you "see" the Mehler kernel deconstruct itself to Hermite polynomials, taking requests of your type too far, in my opinion.) In any case, following the simple math is actually less obscure than summarizing it in "code" verbiage.

newtonian mechanics - Why do rockets jettison fuel tanks?

I'm trying to understand why rockets have multistages releasing their fuel tanks. Say a rocket $R$ has two fuel tanks $A$ and $B$, which respectively have masses $m_a$ and $m_b$, and the mass of the fuel in the respective tanks is $m_{fa}$ and $m_{fb}$. Let $M$ be the total initial mass, and suppose the rocket starts from rest. We use the equation $v-v_0=v_{ex}\ln(m_0/m)$, where $v_{ex}$ is the (constant) speed of the expent fuel.

First suppose the rocket spends its fuel all at once. Then $$v_f=v_{ex}\ln\left(\frac{M}{M-(m_{fa}+m_{fb})}\right).$$

Now suppose the rocket first spends its fuel in tank $A$: $v_1=v_{ex}\ln(\frac{M}{M-m_{fa}}).$ Now the rocket detaches tank $A$, so the mass is now $M-(m_{fa}+m_a).$ Consequently, after spending the fuel in tank $B$, we have \begin{align} v_f&=v_1+v_{ex}\ln\left(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\right)\\&=v_{ex}\left[\ln\left(\frac{M}{M-m_{fa}}\right)+\ln\left(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\right)\right].\end{align}

Now I'm not seeing how to show $v_f$ is greater in the second case, so I'm thinking I made some mistake in the argument. Can anyone clear things up?

Answer

The answer is right there in your own math.

You derived that the delta V that results from using the rocket as a single stage rocket is $$\Delta V_{\text{single stage}} = v_{ex}\ln\Bigl(\frac{M}{M-(m_{fa}+m_{fb})}\Bigr)$$ while the delta V from using the rocket as a two stage rocket is $$\Delta V_{\text{two stage}} = v_{ex} \Biggl( \ln\Bigl(\frac{M}{M-m_{fa}}\Bigr) +\ln\Bigl(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\Bigr)\Biggr)$$

So the problem at hand is to determine the conditions needed to make $\Delta V_{\text{two stage}} > \Delta V_{\text{single stage}}$.

First, let's rewrite the two stage result using the fact that $\ln a + \ln b = \ln(ab)$: $$\Delta V_{\text{two stage}} = v_{ex} \ln\Bigl(\frac{M}{M-m_{fa}}\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\Bigr)$$

Since the natural logarithm is monotonically increasing function, the condition needed is that $$\frac{M}{M-m_{fa}}\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})} > \frac{M}{M-(m_{fa}+m_{fb})}$$ Clearing the denominators yields $$M(M-(m_{fa}+m_a))\;(M-(m_{fa}+m_{fb})) > M(M-m_{fa})\;(M-(m_{fa}+m_a+m_{fb}))$$ This nicely reduces to $$m_a m_{fb} > 0$$ In other words, given a rocket designed as a two stage rocket, it's always better to use that rocket as a two stage rocket than as a single stage rocket.

This does not quite mean that it's better to build a two stage rocket than a single stage rocket. If it did, the same logic would dictate that a hundred stage rocket would be better than a ten stage rocket, and so on. There are penalties in building that hundred stage rocket (and in a two stage rocket). A multi-stage rocket has extra engines, extra fuel tanks, and extra structure, one set per extra stage. That's extra mass, and that extra mass means the lower stages also need extra structure to support that extra mass in the upper stages. Each additional stage also introduces extra places where things can go wrong. At some point, dividing a vehicle into ever more stages becomes counterproductive.

In most cases, that break even point is a two or three stage launch vehicle, possibly carrying a payload that is itself a rocket. (A single stage rocket that can carry a reasonable sized payload to orbit is beyond the capability of currently available engineering, physics, and chemistry.) The most complex vehicle ever built was the rocket that brought humans to and back from the Moon, which comprised eight stages in total:

1. S-IC first stage, 2,300,000 kg gross mass, 131,000 kg empty mass.


2. S-II second stage, 480,000 kg gross mass, 36,000 kg empty mass.


3. S-IVB third stage, 120,800 kg gross mass, 10,400 kg empty mass.


4. Launch escape system (thankfully never used for its intended purpose), 4,200 kg gross mass.


5. Service module, 24,500 kg gross mass, 6,100 kg empty mass.


6. Lunar descent stage, 10,300 kg gross mass, 2,100 kg empty mass.



7. Lunar ascent stage, 4,547 kg gross mass, 2,213 empty mass.


8. Command module, 5,560 kg gross mass, including 120 kg fuel.

classical mechanics - What does it mean, when one says that system has N constants of motion?

For example for an isolated system the energy $E$ is conserved. But then any function of energy, (like $E^2,\sin E,\frac{ln|E|}{E^{42}}$ e.t.c.) is conserved too. Therefore one can make up infinitely many conserved quantities just by using the conservation of energy.

Why then one can usually hear of "system having N constants of motion"? "System having only one constant of motion"?

mathematical physics - What are Wightman fields/functions

Simple question: What are Wightman fields? What are Wightman functions? What are their uses? For example can I use them in operator product expansions? How about in scattering theory?

Answer

In axiomatic approaches to quantum field theory, the basic field operators are usually realized as operator-valued distributions. That's what Wightman fields are: operator-valued distributions satisfying the Wightman Axioms.

Wightman functions are the correlation functions of Wightman fields, nothing more. There's a nice theorem that says if you have a bunch of functions that look like they're the Wightman functions of QFT, then you can actually reconstruct the Hilbert space and the algebra of Wightman fields from it.

Neither of these concepts is anything new physically. They're just more precise ways of speaking about things physicists already know. (Thinking of fields as operator-valued distributions instead of operator-valued functions lets you make precise what goes wrong when you multiply two fields at the same point.) You can talk about OPEs or scattering theory in this language, but it won't gain you anything, unless you're trying to publish in math journals.

If you're interested in the Wightman Axioms, they're explained nicely in the first of Kazhdan's lectures in the IAS QFT Year.

thermodynamics - Derivation for the temperature of Reissner-Nordström (charged) black hole

A lot of the text for this is from "How does one correctly interpret the behavior of the heat capacity of a charged black hole?" but this concerns a different question. The Reissner-Nordström black hole solution is: $$ds^2=-(1-\frac{2M}{r}+\frac{Q^2}{r^2})dt^2+(1-\frac{2M}{r}+\frac{Q^2}{r^2})^{-1}dr^2 +r^2d\Omega_{2}^2$$

Let us define $f(r)\equiv (1-\frac{2M}{r}+\frac{Q^2}{r^2})$. Clearly, the solutions to $f(r)=0$ are $r_{\pm}=M\pm \sqrt{M^2-Q^2}$, and these represent the two horizons of the charged black hole. If we are considering a point near $r_+$, we can rewrite $f(r)$ as follows: $$f(r_+)\sim \frac{(r_+ -r_-)(r-r_+)}{r_{+}^2} $$

What I don't understand is how can derive the temperature of the black hole from this relation. The temperature is given by $$T=\frac{r_+-r_-}{4\pi r_+^2}=\frac{1}{2\pi }\frac{\sqrt{M^2-Q^2}}{(M+\sqrt{M^2-Q^2})^2}$$

I couldn't find a reasonable answer as to how we can obtain the temperature from $f(r_+)$. What are the steps and reasoning that are missing when making this jump?

Answer

The temperature of a black hole is related to its surface gravity. For stationary black holes, the surface gravity is given by $$ \kappa^2 = - \frac{1}{2} D^a \xi^b D_a \xi_b \big|_{r=r_+} $$ where $\xi = \partial_t$ is the time-like Killing vector of the space-time and $r=r_+$ is the horizon. For space-times that have metric of the form $$ ds^2 = - f(r) dt^2 + \frac{dr^2}{f(r)} + r^2 d\Omega^2 $$ This quantity works out to be $$ \kappa = \frac{1}{2} f'(r_+) $$ Then the temperature of the black hole is given by $$ T = \frac{\kappa}{2\pi} = \frac{1}{4\pi} f'(r_+) $$ For the RN black hole $$ f(r) = \frac{1}{r^2} ( r - r_+ )( r - r_- ) \implies f'(r_+) = \frac{r_+-r_-}{r_+^2} $$ which reproduces your formula.

astronomy - What are the prerequisites for considering any other planet to be habitable?

Well, there is a measure of how a planet could be considered like Earth, called Planetary habitability. Based on this measure, what are the prerequisites needed to consider a planet to be a habitable one?

Answer

Habitable by whom? There are conditions that are uninhabitable by humans, however, many "extremophiles" survive perfectly happy.

Although, if you are talking about humans, here is a small list (all the rest are probably more "nice to have" requisites):

• Approximately 20% oxygen (more or less depending on the pressure)


• Temperatures that allow for liquid water


• Adequate access to water (and food)


• Adequate protection from radiation.


• Then a whole host of conditions that wouldn't end human life. Such as deadly pathogens on chemicals in the atmosphere.

All this said, since we only have a sample size of one currently for planetary life, we really don't know what is possible, or how to bound the problem. ANYTHING is just speculation drawn from this one sample. That said, we have found life on our own planet where we never suspected it to be. Life has proven to be nearly unstoppable in propagating throughout every niche on this planet. So, the better question I think would be what are the requisites for abiogenesis? I think once life manages to start on any planet, it will adapt to whatever conditions the planet presents (to within a reasonable degree).

electromagnetic radiation - How does energy transfer between B and E in an EM standing wave?

I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.

Using a standing wave as an example, I came up with the equations

$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$

$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$

I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:

$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$

Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?

Answer

The energy conservation is written $\dfrac{\partial u}{\partial t} + div \vec S=0$, where $u$ is the energy density $\vec E^2+\vec B^2$, and $\vec S$ is the Poynting vector $\vec E \wedge \vec B$ (skipping irrelevant constant factors).

If you choose $x,t$ such as $\sin \omega t=0$ and $\cos k x=0$, both $E$, $B$, the energy density $u$ and the Poynting vector $S$ will be zero.

We have $\vec S \sim \sin 2k x \sin 2\omega t ~\hat{x}$. The divergence of the Poynting vector will be $div \vec S\sim \cos 2k x \sin 2\omega t $, so it it zero too (because of the $t$ dependence ), and so $\dfrac{\partial u}{\partial t} = 0$. However, the first time derivative of $div(\vec S)$is not zero : $\dfrac{\partial (div \vec S)}{\partial t}\sim \cos 2k x \cos 2\omega t $, so, from the energy conservation equation, the second derivative of the density energy $\dfrac{\partial^2 u}{\partial t^2} = - \dfrac{\partial (div \vec S)}{\partial t}$ is not zero.

So, you may write :

$u(x,t+\epsilon) = \frac {\epsilon^2}{2} \dfrac{\partial^2 u}{\partial t^2} + o(\epsilon^2)$

So, at infinitesimal times after $t$, the energy density is not zero.

electromagnetism - Metal Sphere in a Uniform Electric field

An uncharged metal sphere of radius $R$ is placed in a uniform electric field $\vec{E} = E_0 \, \hat{z}$. The field will push the positive charges to the "northern" surface of the sphere, and the negative charges to the "southern" surface.

Griffiths in Example $3.8$ says that the sphere is an equipotential. But how do I know that?

Why does time-reversal mean negation of time i.e. $tmapsto -t;?$

Disclaimer : This is a follow-up to this question.

---

For long time, I've been pondering of this but couldn't come to a stern conclusion to the question:

Why is time-reversal the negation of time $t\;?$ Meant to say, how does negation of time mean the backward flow of time?

Dumb though the query maybe; but still I want to ask it.

For instance, this is forward flow of time: $0,10,20,30,40,50\;;$ when it is reversed, the time sequence would look like $50,40,30,20,10,0\;;$ is it the negative of the forward time flow $0,10,20,30,40,50\;?$ No. But still it is the backward flow of time, isn't it?

Why has time-reversal to do with the negation of time, for $50,40,30,20,10,0$ represents the backward flow of time $0,10,20,30,40,50$, although the former is not the negation of the later?

I am sure I'm missing something but couldn't point it out.

So, can anyone explain it to me why time-reversal means $t\mapsto -t\;?$

Answer

The flow of time is not simply modelled by the real line. The real line with sum $(\mathbb{R},+)$ is an abelian group that is commonly used to model the time coordinate, and the fact that it is possible to "shift the origin" of time (choose which time should represent the starting point).

The flow of time could be modelled by a function $f:\mathbb{R}\to \mathbb{R}$ that works as follows. Its domain is a time variable, as it is its co-domain (both measured in seconds); and to a time $t\in\mathbb{R}$ we associate the flow $f(t)$ of $t$ seconds by $$f(t)=t_0 +t\; ,$$ where $t_0$ is a fixed reference time of origin, that is customary chosen to be zero. If $t_0=0$, the function $f$ reduces to the identity function.

Now what does it mean to take the reversed flow $f_\textrm{inv}$, i.e. "reverse time"? It is quite intuitive within the model above: $f_\textrm{inv}:\mathbb{R}\to\mathbb{R}$ such that $$f_\textrm{inv}(t)=t_0 - t\; .$$ In this way a time lapse of $t$ seconds has flown "backwards" (in the inverse direction wrt the group operation of the coordinates).

Now I would say that the time reversal is the mapping $TR:f\mapsto f_\textrm{inv}$ between the flow and the inverse flow, so it is a map between functions, and not between numbers. If $t_0=0$, it is easy to justify the abuse of notation $TR: t\mapsto -t$. However it remains a map of functions (time flows), and not simply of numbers.

[ following the comments ].

To get the two sequences of time the OP mentioned using the flows, we should think as follows:

We start from $t_0$, and look at a sequence of 6 "forward" discrete time steps, each of 10 seconds. Using the flow $f$, we may describe them by $t_0+t$, for any $t\in \{0,10,20,30,40,50\}$. If $t_0=0$, we get the sequence of forward times $0,10,20,30,40,50$. Now we suppose that at the final counted time $t_0'=t_0+50$ we reverse time, and look at a sequence of 6 "reversed" discrete time steps of 10 seconds each. We have now to use the inverse flow $f_{\text{inv}}$, and we get $t'_0-t=t_0+50-t$, for any $t\in\{0,10,20,30,40,50\}$. With $t_0=0$, this gives $50,40,30,20,10,0$.

particle physics - How are neutrino beams emitted at CERN?

As far I know they come from accelerator collisions, but I have read confusing things like magnetically focused. How could neutrinos be guided magnetically if they aren't affected by the electromagnetic field?

I would like to have a better idea of how neutrinos are emitted.

Answer

The basis of all neutrino beams is a less exotic (protons most of the time) beam smashing some mundane target and making scads of assorted particles---many of them charged. Those charged particles are focused (and possible subjected to a second filtering for energy by using collimators and more magnets, though this step is not being done at CERN), then they are allow to fly for a while during which time many of them decay, and the decay products include neutrinos which are well collimated in the beam direction by the Lorentz focussing. The un-decayed particles are stopped with a thick pile of something that the neutrinos go right through.

The particles that are most interesting for this purpose are those that decay only by weak processes. Both because it takes more time to focus them that strong-decays allow, and because weak interactions are necessary to make neutrinos.

So mostly we have $$ k^- \to \mu^- + \bar{\nu}_\mu $$ $$ \pi^- \to \mu^- + \bar{\nu}_\mu $$ and several other channels (or the charge conjugates, of course (or not and because the horn selects for one sign)); end-state muons subsequently decaying as $$ \mu^- \to e^- + \nu_\mu + \bar{\nu}_e ,$$ but we arrange the decay beam line so that few of them do this before reaching the beam stop (which means that few of the products end up in the final beam as decays from rest are isotropic).

homework and exercises - Explanation: Simple Harmonic Motion

I am a Math Grad student with a little bit of interest in physics. Recently I looked into the Wikipedia page for Simple Harmonic Motion.

Guess, I am too bad at physics to understand it. Considering me as a layman how would one explain:

• What a Simple Harmonic motion is? And why does this motion have such an equation $$x(t)= A \cos(\omega{t} + \varphi)$$

associated with it? Can anyone give examples of where S.H.M. is tangible in Nature?

Answer

This is all about potential; it is common that a particle movement is described by a following ODE:
$m\ddot{\vec{x}}=-\nabla V(\vec{x})$,

where $V$ is some function; usually one is interested in minima of $V$ (they correspond to some stable equilibrium states). Now, however complex $V$ generally is, its minima locally looks pretty much like some quadratic forms, and so the common assumption that $V(x)=Ax^2$... this makes the last equation simplify to:

$\ddot{x}=-\omega^2x$,

with solution in harmonic oscillations.

The common analogy of this is a ball in a paraboloid dish resembling potential shape; it oscillates near the bottom.

standard model - N=2 SSM without a Higgs

In arXiv:1012.5099, section III, the authors describe a supersymmetric extension to the standard model in which there is no Higgs sector at all, in the conventional sense. The up-type Higgs is a slepton in a mirror generation, and other masses come from "wrong Higgs" couplings to this particle. I'm wondering if this approach can revive the fortunes of N=2 supersymmetric extensions of the standard model, where there are three mirror generations, but which run into trouble with oblique electroweak corrections.

I realize that this may be, not just a research-level question, but a question which will require original research to answer! However, if anyone has an immediately decisive argument about the viability of such a theory, now's your chance. :-)

thermodynamics - Difference between irreversibility and entropy?

Cedric Villani recently wrote an article on Landau damping in collisionless plasmas, where at least one topic discussed confused me. Besides discussing the issue of how a process can be microscopically reversible and macroscopically irreversible, he mentions something regarding a "entropy-preserving relaxation mechanism" that is irreversible.

Am I missing something? I think the article is suggesting that something can preserve entropy but still be irreversible.

This begs the following questions:

• Are entropy and irreversibility synonymous?


• Or are they distinct entities each with their own fundamental properties?

electromagnetic radiation - Why doesn't there exist a wave function for a photon whereas it exists for an electron?

A photon is an excitation or a particle created in the electromagnetic field whereas an electron is an excitation or a particle created in the "electron" field, according to second-quantization.

However, it is often said in the literature that the wave function of a photon doesn't exist whereas it exists for an electron.

Why is it so?

Answer

Saying that a photon doesn't have a wavefunction can be misleading. A more accurate way to say it is that a photon doesn't have a strict position observable. A photon can't be strictly localized in any finite region of space. It can be approximately localized, so that it might as well be restricted to a finite region for all practical purposes; but the language "doesn't have a wavefunction" is referring to the non-existence of a strict position observable.

An electron doesn't have a strict position observable, either, except in strictly non-relativistic models.

In relativistic quantum field theory, nothing has a strict position observable. This is a consequence of a general theorem called the Reeh-Schlieder theorem. The proof of this theorem is not trivial, but it's nicely explained in [1].

Relativistic quantum field theory doesn't have strict single-particle position observables, but it does have other kinds of strictly localized observables, such as observables corresponding to the magnitude and direction of the electric and magnetic fields inside an arbitrarily small region of space. However, those observables don't preserve the number of particles. Strictly localized observables necessarily turn single-particle states into states with an indefinite number of particles. (Actually, even ignoring the question of localization, "particle" is not easy to define in relativistic quantum field theory, but I won't go into that here.)

For example, relativistic quantum electrodynamics (QED) has observables corresponding to the amplitudes of the electric and magnetic fields. These field operators can be localized. The particle creation/annihilation operators can be expressed in terms of the field operators, and vice versa, but the relationship is non-local.

Technically, the Reeh-Schlieder theorem says that a relativistic quantum field theory can't have any strictly-localized operator that annihilates the vacuum state. Therefore, it can't have any strictly-localized operator that counts the number of particles. (The vacuum state has zero particles, so a strictly-localized particle-counting operator would annihilate the vacuum state, which is impossible according to the Reeh-Schlieder theorem.)

Strictly non-relativistic models are exempt from this theorem. To explain what "strictly non-relativistic" means, consider the relativistic relationship between energy $E$ and momentum $p$, namely $E=\sqrt{(mc^2)^2+(pc)^2}$, where $m$ is the single-particle mass. If $p\ll mc$, then we can use the approximation $E\approx mc^2+p^2/2m$. A non-relativistic model is one that treats this approximate relationship as though it were exact. The most familiar single-particle Schrödinger equation is a model of this type. Such a model does have a strict position operator, and individual particles can be strictly localized in a finite region of space in such a model.

Since photons are massles ($m=0$), we can't use a non-relativistic model for photons. We can use a hybrid model, such as non-relativistic QED (called NRQED), which includes photons but treats electrons non-relativistically. But even in that hybrid model, photons still can't be strictly localized in any finite region of space. Loosely speaking, the photons are still relativistic even though the electrons aren't. So in NRQED, we can (and do) have a single-electron position observable, but we still don't have a single-photon position observable.

"Wavefunction" is a more general concept that still applies even when strict position observables don't exist. The kind of "wavefunction" used in relativistic quantum field theory is very different than the single-particle wavefunction $\psi(x,y,z)$ familiar from strictly non-relativistic quantum mechanics. In the relativistic case, the wavefunction isn't a function of $x,y,z$. Instead, it's a function of more abstract variables, and lots of them (nominally infinitely many), and it describes the state of the whole system, which generally doesn't even have a well-defined number of particles at all. People don't use this kind of wavefunction very often, because it's very difficult, but once in a while it's used. For example, Feynman used this kind of "wavefunction" in [2] to study a relativistic quantum field theory called Yang-Mills theory, which is a simplified version of quantum chromodynamics that has gluons but not quarks.

In this generalized sense, a single photon can have a wavefunction.

In the non-relativistic case, the $x,y,z$ in $\psi(x,y,z)$ correspond to the components of the particle's position observables. When physicists say that a photon doesn't have a wavefunction, they mean that it doesn't have a wavefunction that is a function of the eigenvalues of position observables, and that's because it doesn't have any strict position observables.

Also see these very similar questions:

Can we define a wave function of photon like a wave function of an electron?

Wave function of a photon?

EM wave function & photon wavefunction

---

References:

[1] Witten, "Notes on Some Entanglement Properties of Quantum Field Theory", http://arxiv.org/abs/1803.04993

[2] Feynman (1981), "The qualitative behavior of Yang-Mills theory in 2 + 1 dimensions", Nuclear Physics B 188: 479-512, https://www.sciencedirect.com/science/article/pii/0550321381900055

Causality and Simultaneity in special relativity

I am a little confused about the implications of special relativity on causality and simultaneity.

Are the following two statements true?

1. For two events A and B that are close enough in space and time such that A could possibly cause B, no matter how what inertial reference frame, A will always occur before B.

and

2. For two events A and B that are far enough in space and time such that neither could not possibly cause the other, depending on your inertial reference frame, A could occur before B, or B before A.

Answer

Both statements are true.

It would be more precise to say that for timelike separated points the temporal order is always preserved while for spacelike separated points it is not.

Interpretation of $phi(x)|0 rangle$ in interacting theory

I'm trying to relearn quantum field theory more carefully, and the first step is seeing what facts from free field theory also hold in the interacting theory.

In free field theory, the state $\phi(x) | 0 \rangle$ contains exactly one particle, which is localized near $\mathbf{x}$ at time $t$ where $x = (\mathbf{x}, t)$. However, typical quantum field theory books are silent on the question of what $\phi(x)$ does in the interacting theory. It seems like it's implicitly assumed to still create a particle localized near $\mathbf{x}$, because correlation functions are said to be the amplitude for particle propagation, but I haven't seen any explicit justification.

• Does $\phi(x)$ still create a field excitation localized near $x$? If so, how can we see that? (I'm already aware that the localization is not perfect in the free case, but that's a separate issue.)


• Is $\phi(x) | 0 \rangle$ still a one-particle state? It's definitely not a one-particle state using the free theory's number operator, but is it in some sense a state with one 'dressed' particle? If so, how would we formalize and prove that?

More generally, how should I think of the action of $\phi(x)$ in the interacting theory? How about for a weakly interacting theory?

Answer

All this can be answered by the same object: the Green`s function $\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle$ and the polology of it. First, we need to do a fourrier transform:

$$ G(q_1,...,q_n)=\int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}\,e^{-iq_1.x_1}...e^{-iq_n.x_n}\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle $$

Poles at on-shell $q=q_1+...+q_r$, i.e. $q²=-m²$, indicates the existence of a particle of mass $m$, the existence of a one-particle state $|\vec{p},\sigma\rangle$ in the spectra. The residue of that pole measure the projection of this one-particle state $|\vec{p},\sigma\rangle\langle\vec{p},\sigma|$ with the state: $$ \int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_r}{(2\pi)^d}e^{-iq_1.x_1}...e^{-iq_r.x_r}\langle 0 |\mathcal{T}\phi(x_1)...\phi(x_r) $$ and the state $$ \int \frac{d^dq_{r+1}}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}e^{-iq_{r+1}.x_{r+1}}...e^{-iq_n.x_n}\mathcal{T}\phi(x_{r+1})...\phi(x_n)|0\rangle $$ In particular, one can show the LSZ reduction formula by putting each $q_i$ on-shell. You can see all this in the chapter $10$ of the first volume of Weinberg QFT textbook.

Now, if you want to know how to make sense of the state $\phi(x)|0\rangle$ you just need to plug this state into an arbitrary Green's function and work out the residue of various poles. Note that just Lorentz symmetry fixes the projection of this state with one-particle state, up to a overall normalization:

$$ \langle \vec{p},\sigma|\phi_{\alpha}(x)|0\rangle = \sqrt{Z}\,u_{\alpha}(\vec{p},\sigma)e^{-ipx} $$ where this $u$-function is the polarization function, responsible to match the Lorentz indice $\alpha$ with the little group indice $\sigma$. In interacting theories, $\phi_{\alpha}(0)|0\rangle$ is not parallel to $|\vec{p},\sigma\rangle$, since by the polology and LSZ reduction formula above, this would mean that the Green's function of more than two points are zero, implying that the S-matrix are trivial.

Answering both your questions:

1. Yes, the state $\phi_{\alpha}(x)|0\rangle$ correspond to a local preparation, a local field disturbance if you like, because it has the right properties under Lorentz-Poincaré transformations.


2. No, the state $\phi_{\alpha}(x)|0\rangle$ is not a one-particle state because this would imply a trivial S-matrix by the LSZ reduction formula and polology of the Green's function, contradicting the assumption of an interacting theory.

quantum mechanics - Norm preserving Unitary operators in Rigged Hilbert space

If we take the free particle Hamiltonian, the eigenvectors (or eigenfunctions), say in position representation, are like $e^{ikx}$. Now these eigenfunctions are non-normalisable,so they don't belong the normal $L^2(\mathbb R^d)$ but the rigged Hilbert space.

My question hereto is that any unitary operator defined as a map in Hilbert space preserves the norm. But in the case of free particle, although the operator $e^{iHt}$ is unitary (since $H$ is hermitian), there is (atleast) no (direct) condition of norm preserval, as the norm cannot be defined for these eigenfunctions.

Now, how can one connect the unitarity of $e^{iHt}$ and norm preserval in this context ?

PS : I know one can use box-normalised wavefunctions and do away with the calculations and then take $L \rightarrow \infty$ limit. But I am rather interested in the actual question of unitarity and norm in rigged Hilbert spaces.

Answer

The so-called rigged spaces are made with a triple $(S,\mathscr{H},S')$; where $\mathscr{H}$ is the usual Hilbert space, $S$ is a dense vector subspace of $\mathscr{H}$, and $S'$ the dual of $S$.

Usually when $\mathscr{H}=L^2(\mathbb{R}^d$, then $S$ is taken to be the rapidly decrasing functions, and $S'$ the tempered distributions. If this is the case, then there is no notion of norm for the "extended" eigenvectors in $S'$, since the latter is not a Banach (or metrizable) space.

So even if $e^{-itH}e^{ikx}=e^{-itk^2/2m}e^{ikx}$, and therefore the evolution indeed acts as a phase, there is no norm to be preserved.

The point is that rigged Hilbert spaces are, as far as I know, simply a mathematical convenience to justify the emergence of "generalized eigenvectors" for some (very special) self-adjoint operators that have purely continuous spectrum. If you want to do (meaningful) quantum mechanics, you have to consider states of the Hilbert space, where the evolution is indeed unitary and everything works.

statistical mechanics - How to compute entropy of networks? (Boltzmann microstates and Shannon entropy)

I also asked in SO here a few days ago, thought it may be also interesting for physics-related answers.

I would like to model a network as a system. A particular topology (configuration of edges between vertices) is a state-of-order of the system (a micro-state). I am trying to compute the entropy of a specific topology as a measure of the complexity of information embedded in that topological structure.

I don’t have a degree in physics, I would like to have answers that can help in creating a concept of entropy applied to networks (particularly small-world networks), as systems embedding information in their topology.

Below, I share my reasoning and doubts.

1. 
   

   I first thought to make an analogy with Shannon entropy applied to strings: Here entropy is a measure of the randomness of a string as a sum of probability to have certain digits happenings. Similarly, I then thought that entropy may hold for an Erdős–Rényi random network, and the measure could reflect the randomness of an edge between a pair of vertexes.

   
   
   
   
   • Does Shannon entropy hold for non-random types of networks?
   
   
   
   


2. 
   

   As a second approach, I thought that according to Boltzmann’s definition, entropy is the multiplicity of equivalent states.

   
   
   
   
   
   • 
     

     How could equivalent topologies be modelled (or how to can we compute similarity between two networks)?

     
     
   
   
   • 
     

     How to measure how much a state of order of a particular topology is “uncommon”, with respect to all other possible configurations?

     
     
   
   
   • 
     

     Should I attempt to model a topology as a probability over all possible distributions of edges (complete network)?

     
     
   
   
   
   
   

Answer

For all definitions of entropy, you have to define an ensemble of states to define the respective probabilities (this is related, if not equivalent to the macrostate). For example when you calculate the Shannon entropy of a string, you assume an ensemble of possible strings (and their likelihood) given by the probability of certain letters in your language of choice. For a sufficiently long string, you can estimate those probabilities from the string itself and thus “bootstrap” your ensemble.

So, to do something similar for networks, you first have to define an appropriate ensemble of networks that you want to consider. These would be your “equivalent topologies”. What makes sense here depends on how you want to interpret your entropy or, from another point of view, what properties of the network you consider variable for the purpose of encoding information. One way you may want to consider are network null models a.k.a. network surrogates. There are several methods available for obtaining such¹, but note that the properties of the underlying ensemble differ and are not always obvious.

Some further remarks:

• 
  

  Assuming that each network is its own microstate, the Shannon and the Gibbs entropy should be equivalent, except for a constant factor.

  
  



• 
  

  You may want to take a look at Phys. Rev. Lett. 102, 038701, which applies thermodynamic concepts to networks, though I never found out what ensemble they are considering.

  
  


• 
  

  
  

  how to can we compute similarity between two networks

  
  

  
  

  There are several proposals of a distance metrics for network, starting with the Euclidean distance of the adjancency matrix. Unfortunately, I do not have a good citation at hand.

  
  


• 
  
  

  For any reasonable ensemble, you end up with a gargantuan amount of networks/microstates. Therefore it is usually not feasible to empirically estimate the probability of all microstates or even a given microstate. Instead, you have to estimate the probability density for a neighbourhood of microstates. For this you need the above distance.

  
  

---

^{¹ I tried to cite all of them in the introduction of this paper of mine, but that’s not entirely up to date.}

Why isn't spent nuclear fuel used as a heat source?

We all know spent fuel rods taken out from a reactor core keeps generating tremendous amount of heat and needs to be kept cool by running cool water. It is also known that if cooling system fails water will evaporate quickly leading to a blast and meltdown

If the fuel can still produce so much heat, why isn't it used as heat source for some low temperature systems, e.g. power generation via thermocouples, or home heatings for entire blocks or cities?

Why is all that spent fuel kept as a waste?

particle physics - What is the difference between a neutron and hydrogen?

Differences? They are both an electron and a proton, since the neutron decays to a proton and an electron, what's the difference between a neutron and proton + electron? so is it just a higher binding energy between the two?

general relativity - Entering a black hole, jumping into another universe---with questions

I'm quite familiar with SR, but I have very limited understanding in GR, singularities, and black holes. My friend, which is well-read and is interested in general physics, said that we can "jump" into another universe by entering a black hole.

Suppose that we and our equipments can withstand the tidal forces near black holes. We jumped from our spaceship into a black hole. As we had passed the event horizon, we couldn't send any information to the outside anymore.

1. 
   

   Can this situation be interpreted as that we were in another universe separate from our previous universe?

   
   


2. 
   

   Is there corrections or anything else to be added to above statement?

   
   

Answer

Let me attempt a more "popular science" answer (Ron please be gentle with me!).

In GR a geodesic is the path followed by a freely moving object. There's nothing especially complex about this; if you throw a stone (in a vacuum to avoid air resistance) it follows a geodesic. If the universe is simply connected you'd expect to be able to get anywhere and back by following geodesics.

However in a static black hole, described by the Schwartzchild geometry, something a bit odd happens to the geodesics. Firstly anything following a geodesic through the event horizon can't go back the way it came, and secondly all geodesics passing through the event horizon end at a single point i.e. the singularity at the centre of the black hole.

If you now rotate the black hole, or you add electric charge to it, or both, you can find geodesics that pass through the event horizons (there are now two of them!), miss the singularity and travel back out of the black hole again.

But, and this is where the separate universes idea comes in, you now can't find any geodesics that will take you back to your starting point. So you seem to be back in the normal universe but you're in a region that is disconnected from where you started. Does this mean it's a "separate universe". That's really a matter of terminology, but I would say not. After all you just got there by coasting - you didn't pass through any portals of the type so beloved by SciFi films. And there's no reason to think that physics is any different where you ended than where you started.

If you're interested in pursuing this further I strongly recommend The Cosmic Frontiers of General Relativity by William J. Kaufmann. It claims to be a layman's guide, but few laymen I know could understand it. However if you know SR you shouldn't have any problems with it.

schroedinger equation - $nabla$ and non-locality in simple relativistic model of quantum mechanics

In Wavefunction in quantum mechanics and locality, wavefunction is constrained by $H = \sqrt{m^2 - \hbar^2 \nabla^2} $, and taylor-expanding $H$ results in:

$$ H = \dots = m\sqrt{1 - \hbar^2/m^2 \cdot \nabla^2} = m(1-\dots) $$

While the person who asked this question accepted the answer, I was not able to understand fully.

Why would $\nabla^{200}$ in the taylor-expansion of $H$ be so problematic (200 can be replaced by any arbitrary number) - resulting in non-locality? Isn't this just some exponentiation of dot product of gradient?

Answer

Non-locality comes from presence of infinite many terms in that expansion. To see that, lets assume we are applying the non-polynomial function $f(\vec{z})$ of $i\nabla$ on any function $\psi(\vec{x})$: $f(i\nabla) \psi(\vec{x})$. Assuming that $f(z)$ is "nice enough" to have the Fourier representation $f(\vec{z}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot \vec{z}}$ for suitable transform function $F(\vec{k})$. Then:

$f(i \nabla) \psi(x) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{i \vec{k}\cdot (i \nabla)} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) e^{- k\cdot \nabla} \psi(\vec{x}) = \int_{-\infty}^{+\infty} \frac{d^3\vec{k}}{(2\pi)^3} F(\vec{k}) \psi(\vec{x}-\vec{k})$

This is the superposition of values of $g$ calculated at points different than $\vec{x}$. that is the non-locality.

resource recommendations - Books for general relativity

What are some good books for learning general relativity?

Answer

I can only recommend textbooks because that's what I've used, but here are some suggestions:

• Gravity: An Introduction To General Relativity by James Hartle is reasonably good as an introduction, although in order to make the content accessible, he does skip over a lot of mathematical detail. For your purposes, you might consider reading the first few chapters just to get the "big picture" if you find other books to be a bit too much at first.



• A First Course in General Relativity by Bernard Schutz is one that I've heard similar things about, but I haven't read it myself.


• Spacetime and Geometry: An Introduction to General Relativity by Sean Carroll is one that I've used a bit, and which goes into a slightly higher level of mathematical detail than Hartle. It introduces the basics of differential geometry and uses them to discuss the formulation of tensors, connections, and the metric (and then of course it goes on into the theory itself and applications). It's based on these notes which are available for free.


• General Relativity by Robert M. Wald is a classic, though I'm a little embarrassed to admit that I haven't read much of it. From what I know, though, there's certainly no shortage of mathematical detail, and it derives/explains certain principles in different ways from other books, so it can either be a good reference on its own (if you're up for the detail) or a good companion to whatever else you're reading. However it was published back in 1984 and thus doesn't cover a lot of recent developments, e.g. the accelerating expansion of the universe, cosmic censorship, various results in semiclassical gravity and numerical relativity, and so on.


• Gravitation by Charles Misner, Kip Thorne, and John Wheeler, is pretty much the authoritative reference on general relativity (to the extent that one exists). It discusses many aspects and applications of the theory in far more mathematical and logical detail than any other book I've seen. (Consequently, it's very thick.) I would recommend having a copy of this around as a reference to go to about specific topics, when you have questions about the explanations in other books, but it's not the kind of thing you'd sit down and read large chunks of at once. It's also worth noting that this dates back to 1973, so it's out of date in the same ways as Wald's book (and more).


• Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity by Steven Weinberg is another one that I've read a bit of. Honestly I find it a bit hard to follow - just like some of Weinberg's other books, actually - since he gets into such detailed explanations, and it's easy to get bogged down in trying to understand the details and forget about the main point of the argument. Still, this might be another one to go to if you're wondering about the details omitted by other books. This is not as comprehensive as the Misner/Thorne/Wheeler book, though.


• A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics by Eric Poisson is a bit beyond the purely introductory level, but it does provide practical guidance on doing certain calculations which is missing from a lot of other books.

quantum field theory - What's the difference between helicity and chirality?

When a particle spins in the same direction as its momentum, it has right helicity, and left helicity otherwise. Neutrinos, however, have some kind of inherent helicity called chirality. But they can have either helicity. How is chirality different from helicity?

Answer

At first glance, chirality and helicity seem to have no relationship to each other. Helicity, as you said, is whether the spin is aligned or anti aligned with the momentum. Chirality is like your left hand versus your right hand. Its just a property that makes them different than each other, but in a way that is reversed through a mirror imaging - your left hand looks just like your right hand if you look at it in a mirror and vice-versa. If you do out the math though, you find out that they are linked. Helicity is not an inherent property of a particle because of relativity. Suppose you have some massive particle with spin. In one frame the momentum could be aligned with the spin, but you could just boost to a frame where the momentum was pointing the other direction (boost meaning looking from a frame moving with respect to the original frame). But if the particle is massless, it will travel at the speed of light, and so you can't boost past it. So you can't flip its helicity by changing frames. In this case, if it is "chiral right-handed", it will have right-handed helicity. If it is "chiral left-handed", it will have left-handed helicity. So chirality in the end has something to do with the natural helicity in the massless limit.

Note that chirality is not just a property of neutrinos. It is important for neutrinos because it is not known whether both chiralities exist. It is possible that only left-handed neutrinos (and only right-handed antineutrinos) exist.

electromagnetic radiation - If you view the Earth from far enough away can you observe its past?

From my understanding of light, you are always looking into the past based on how much time it takes the light to reach you from what you are observing.

For example when you see a star burn out, if the star was 5 light years away then the star actually burnt out 5 years ago.

So I am 27 years old, if I was 27 light years away from Earth and had a telescope strong enough to view Earth, could I theoretically view myself being born?

Answer

Yes, you can. And you do not even need to leave the Earth to do it.

You are always viewing things in the past, just as you are always hearing things in the past. If you see someone do something, who is 30 meters away, you are seeing what happened $(30\;\mathrm{m})/(3\times10^8\;\mathrm{m}/\mathrm{s}) = 0.1\;\mu\mathrm{s}$ in the past.

If you had a mirror on the moon (about 238K miles away), you could see about 2.5 seconds into earth's past. If that mirror was on Pluto, you could see about 13.4 hours into Earth's past.

If you are relying on hearing, you hear an event at 30 m away about 0.1 s after it occurs. That is why runners often watch the starting pistol at an event, because they can see a more recent picture of the past than they can hear.

To more directly answer the intent of your question: Yes, if you could magically be transported 27 lightyears away, or had a mirror strategically placed 13.5 lightyears away, you could see yourself being born.

waves - after the reentry of the side booster rockets of the Falcon heavy, why was the sonic boom heard?

what is the underlying principle of this boom here? did the boosters landed with speed greater than the sound barrier ? And they were heard after some time they both landed , how is that possible? is it because sound travels slowly as compared to light? please clarify .

quantum mechanics - What is the difference between the phase in molecular orbitals and the actual complex phase component of the wave function?

You often see in atomic and molecular physics texts that bonding occurs between two atomic orbitals when their wave functions are in phase. These pictures often depict the 'phase' as whether or not the wave function is positive of negative. This is obviously not the same as the complex phase component of the wave function. The $ p_z $ orbital is completely real: $$ \psi (r, \theta)\propto \frac{r}{a_o}e^{-r/2a_0}\cos\theta $$ While the $p_x$ orbital is actually complex: $$ \psi (r, \theta, \phi)\propto \frac{r}{a_o}e^{-r/2a_0}\sin \theta e^{i\phi} $$

But both these p orbitals have the same behaviour when bonding, as if their complex phases are the same. What is the difference between this atomic orbital 'phase' and the actual complex phase of the wave function?

Do the standard cosmology models spontaneously break Lorentz symmetry?

In standard cosmology models (Friedmann equations which your favorite choice of DM and DE), there exists a frame in which the total momenta of any sufficiently large sphere, centered at any point in space, will sum to 0 [1] (this is the reference frame in which the CMB anisotropies are minimal). Is this not a form of spontaneous Lorentz symmetry breaking ? While the underlying laws of nature remain Lorentz invariant, the actual physical system in study (in this case the whole universe) seems to have given special status to a certain frame.

I can understand this sort of symmetry breaking for something like say the Higgs field. In that situation, the field rolls down to one specific position and "settles" in a minima of the Mexican hat potential. While the overall potential $V(\phi)$ remains invariant under a $\phi \rightarrow \phi e^{i \theta}$ rotation, none of its solutions exhibit this invariance. Depending on the Higgs model of choice, one can write down this process of symmetry breaking quite rigorously. Does there exist such a formalism that would help elucidate how the universe can "settle" into one frame ? I have trouble imagining this, because in the case of the Higgs the minima exist along a finite path in $\phi$ space, so the spontaneous symmetry breaking can be intuitively understood as $\phi$ settling randomly into any value of $\phi$ where $V(\phi)$ is minimal. On the other hand, there seems to me to be no clear way of defining a formalism where the underlying physical system will randomly settle into some frame, as opposed to just some value of $\phi$ in a rotationally symmetric potential.

[1] The rigorous way of saying this is : There exists a reference frame S, such that for all points P that are immobile in S (i.e. $\vec{r_P}(t_1) = \vec{r_P}(t_2) \forall (t_1, t_2)$ where $\vec{r_P}(t)$ is the spatial position of P in S at a given time $t$), and any arbitrarily small $\epsilon$, there will exist a sufficiently large radius R such that the sphere of radius R centered on P will have total momenta less than $\epsilon c / E_k$ (where E is the total kinetic energy contained in the sphere).

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Ben Crowell gave an interesting response that goes somewhat like this :

Simply put then : Causally disconnected regions of space did not have this same "momentumless frame" (let's call it that unless you have a better idea), inflation brings them into contact, the boost differences result in violent collisions, the whole system eventually thermalises, and so today we have vast swaths of causally connected regions that share this momentumless frame.

Now for my interpretation of what this means. In this view, this seems to indeed be a case of spontaneous symmetry breaking, but only locally speaking, because there should be no reason to expect that a distant causally disconnected volume have this same momentumless frame. In other words the symmetry is spontaneously broken by the random outcome of asking "in what frame is the total momentum of these soon to be causally connected volumes 0?". If I'm understanding you correctly, this answer will be unique to each causally connected volume, which certainly helps explain how volumes can arbitrarily "settle" into one such frame. I'm not sure what the global distribution of boosts would be in this scenario though, and if it would require some sort of fractal distributions to avoid running into the problem again at larger scales (otherwise there would still be some big enough V to satisfy some arbitrarily small total momentum).

What is the "interaction picture" or "rotating frame" in quantum mechanics?

$\renewcommand{\ket}[1]{\left\lvert #1 \right\rangle}$ In typical quantum mechanics courses, we learn about the so-called "Schrodinger picture" and "Heisenberg picture". In the Schrodinger picture, the equation of motion brings time dependence to the states, $$i \hbar \partial_t \ket{\Psi(t)} = H(t) \ket{\Psi(t)}$$ while in the Heisenberg picture the equation brings time dependence to the operators, $$i \hbar \partial_t A(t) = [A(t), H(t)] \, .$$

When we have a Hamiltonian which can be split into an "easy" part $H_0(t)$$^{[a]}$ and a time dependent "difficult" part $V(t)$, $$H(t) = H_0(t) + V(t)$$ people talk about the "interaction picture" or "rotating frame". What's the difference between the interaction picture and rotating frame, and how do they work?

$[a]$: We assume that $H_0(t)$ commutes with itself at different times.

Answer

$$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$$

Basic idea: the rotating frame "unwinds" part of the evolution of the quantum state so that the remaining part has a simpler time dependence. The interaction picture is a special case of the rotating frame.

Consider a Hamiltonian with a "simple" time independent part $H_0$, and a time dependent part $V(t)$: $$H(t) = H_0 + V(t) \, .$$

Denote the time evolution operator (propagator) of the full Hamiltonian $H(t)$ as $U(t,t_0)$. In other words, the Schrodinger picture state obeys $\ket{\Psi(t)} = U(t, t_0) \ket{\Psi(t_0)}$.

The time evolution operator from just $H_0$ is (assuming $H_0$ is time independent, or at least commutes with itself at different times) $$U_0(t, t_0) = \exp\left[ -\frac{i}{\hbar} \int_{t_0}^t dt' \, H_0(t') \right] \, .$$ Note that $$i\hbar \partial_t U_0(t, t_0) = H_0(t) U_0(t, t_0) \, .$$

Define a new state vector $\ket{\Phi(t)}$ as $$\ket{\Phi(t)} \equiv R(t) \ket{\Psi(t)}$$ where $R(t)$ is some "rotation operator". Now find the time dependence of $\ket{\Phi(t)}$: \begin{align} i \hbar \partial_t \ket{\Phi(t)} =& i \hbar \partial_t \left( R(t) \ket{\Psi(t)} \right) \\ =& i \hbar \partial_t R(t) \ket{\Psi(t)} + R(t) i \hbar \partial_t \ket{\Psi(t)} \\ =& i \hbar \dot{R}(t) \ket{\Psi(t)} + R(t) H(t) \ket{\Psi(t)} \\ =& i \hbar \dot{R}(t) R(t)^\dagger \ket{\Phi(t)} + R(t) H(t) R(t)^\dagger \ket{\Phi(t)} \\ =& \left( i \hbar \dot{R}(t) R(t)^\dagger + R(t) H(t) R(t)^\dagger \right) \ket{\Phi(t)} \, . \end{align} Therefore, $\ket{\Phi(t)}$ obeys Schrodinger's equation with a modified Hamiltonian $H'(t)$ defined as $$H'(t) \equiv i \hbar \dot{R}(t) R(t)^{\dagger} + R(t) H(t) R(t)^\dagger \, . \tag{$\star$}$$ This is the equation of motion in the rotating frame.

Useful choices of $R$ depend on the problem at hand. Choosing $R(t) \equiv U_0(t, t_0)^\dagger$ has the particularly useful property that the first term in $(\star)$ cancels the $H_0(t)$ part of the second term, leaving \begin{align} i \hbar \partial_t \ket{\Phi(t,t_0)} = \left( U_0(t, t_0)^\dagger V(t) U_0(t, t_0) \right)\ket{\Phi(t, t_0)} \, . \end{align} which is Schrodinger's equation with effective Hamiltonian $$H'(t) \equiv U_0(t)^\dagger V(t) U_0(t) \, .$$ This is called the interaction picture. It is also known by the name Dirac picture.

electromagnetism - Ampère's law applied on a "short" current-carrying wire

Why doesn't Ampère's law hold for short current carrying wires? Of course, such wires should be part of a closed circuit, but that's a physical fact, and there is a numerous amount of ways to close it. Ampere's law, on the other hand, is merely a mathematical law.

The problem is Ampère's law can't really "tell" the difference between infinite wires and finite wires. Such law states: $$\oint \mathbf{B} \cdot d\mathbf{l}=\mu_0I$$ As far as I know there is no restriction on the integration curve as long as it is closed. Then I could pick a circle away from the wire, centered at its axis, oriented in its direction. The integral then only depends on $\theta$, and $\mathbf{B}$ comes out because it does not (there is azimuthal symmetry). With this logic (which is evidently flawed), $\mathbf{B}$ is the same regardless of its wire axis coordinate (which is true for an infinite wire, but not in this case). My assumptions here were:

1. $\mathbf{B}$ doesn't depend on $\theta$


2. $\mathbf{B}$ goes in the azimuthal direction. Biot-Savart law forces it to be perpendicular to the wire ($I$ goes in that direction) and to the radial component ($\mathbf{r}$ has a component parallel to $\mathbf{B}$ and another one to the radial component for every segment of wire), because the direction of the $\mathbf{B}$ field generated by each small segment of wire is in the direction of $\mathbf{B}\times\mathbf{r}$.



3. The closed amperian loop can be arbitrarily chosen.


4. Ampère's law works for finite, open, line currents.

To finish, in Griffith's Introduction to Electrodynamics (4th ed., p. 225) the author derives a formula for a finite straight current carrying wire using the Biot-Savart law:

$$B=\frac{\mu_0 I}{4\pi s}(\sin{\theta_2-\sin{\theta_1})}$$

This is only the magnitude. But above, he writes:

In the diagram, $(d\mathbf{l'}\times \mathbf{\hat{r}})$ points out of the page(...)

Which, by the azimuthal symmetry of the problem, implies that the magnetic field goes in the azimuthal direction. That would leave assumptions 3 and 4 as the sources of the absurdity.

Answer

There is no such thing as a "finite, open line current". If you did have such a thing, the ends of the line segment would accumulate charge at a linear rate, and you are no longer on a static situation, and you can no longer apply the magnetostatic Ampère's law.

This isn't being nit-picky, either, because you need the current to obey the continuity equation in order for Ampère's law to even make sense. This is most cleanly stated in the integral version with a volumetric current density $\mathbf J(\mathbf r)$, for which Ampère's law reads $$ \oint_\mathcal C \mathbf B\cdot\mathrm d\mathbf l = \mu_0\int_\mathcal{S} \mathbf J\cdot \mathrm d\mathbf S, $$ for any closed curve $\mathcal C$ and any surface $\mathcal S$ whose boundary is $\mathcal C$. For this to be true, you require the integral on the right to be independent of the choice of surface, and by the standard argument this is equivalent to requiring that the flux of the current density out of any closed suface be zero: $$ \oint_\mathcal{S} \mathbf J\cdot \mathrm d\mathbf S=0. $$

This requirement fails for a current-carrying finite line segment, because the current flowing into a sphere about either end of the segment is nonzero. For such a current, then, Ampère's law makes no sense.

(Since you explicitly mention the differential form of Ampère's law, here goes that version of the argument: you have $\nabla \times \mathbf B = \mu_0 \mathbf J$, so therefore $\nabla \cdot \mathbf J = \nabla \cdot(\nabla \times \mathbf B) \equiv 0$, and $\mathbf J$ obeys the differential stationary continuity equation. There is no way to reconcile this with a finite current-carrying segment.)

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So where does this leave the current-carrying line segment? It's a really neat concept, and it mostly doesn't lead us astray, so what gives?

Well, mostly, it doesn't lead us astray because we use finite segments correctly: as part of a multi-part curve $\mathcal C = \mathcal C_1 \oplus \mathcal C_2 \oplus\cdots\oplus \mathcal C_N$, where the endpoints overlap correctly, pair by pair and back again from $\mathcal C_N$ to $\mathcal C_1$, and no charge is either appearing or disappearing - there is no surface with a nonzero current flux.

That's a legit circuit, and Ampère's law holds for it. The way to bring the analysis down to each individual segment is (exclusively!) through the Biot-Savart law, which is provably the solution to the magnetostatic Ampère's law, and which reads for this circuit $$ \mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int_{\mathcal C} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} = \frac{\mu_0}{4\pi} \sum_{n=1}^N \int_{\mathcal C_n} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} . $$

For such a circuit, it is convenient to split the field up into components, defined as $$ \mathbf B_n(\mathbf r) = \frac{\mu_0}{4\pi} \int_{\mathcal C_n} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} , $$ and which we interpret as the field produced by the individual segment $\mathcal C_n$. However, these fields are completely in our heads: there are no physical current configurations that produce the $\mathbf B_n(\mathbf r)$, and there is no way to make sense of them beyond the Biot-Savart law (and certainly no way to make sense of them using Ampère's law).

electromagnetism - What symmetry gives you charge conservation?

This is a popular question on this site but I haven't found the answer I'm looking for in other questions. It is often stated that charge conservation in electromagnetism is a consequence of local gauge invariance, or perhaps it is due to some global phase symmetry. Without talking about scalar or spinor fields, the EM Lagrangian that we're familiar with is: $$ \mathcal{L} = -\frac{1}{4}F^2 - A \cdot J + \mathcal{L}_\mathrm{matter}(J) $$ The equation of motion for $A$ is simply $$ \partial_\mu F^{\mu \nu} = J^\nu $$ From which it follows that $J$ is a conserved current (by the antisymmetry of the field strength). But what symmetry gave rise to this? I'm not supposing that my matter has any global symmetry here, that I might be able to gauge. Then so far as I can tell, the Lagrangian given isn't gauge invariant. The first term is, indeed, but the second term only becomes gauge invariant on-shell (since I can do some integrating by parts to move a derivative onto $J$). If we demand that our Lagrangian is gauge-invariant even off-shell, then we can deduce that $\partial \cdot J = 0$ off-shell and hence generally. But we can't demand that this hold off-shell, since $J$ is not in general divergenceless!

For concreteness, suppose that $$ \mathcal{L}_\mathrm{matter}(J) = \frac{1}{2} (\partial_\mu \phi) (\partial^\mu \phi) \qquad J^\mu \equiv \partial^\mu \phi $$ Then we find that $\phi$ (a real scalar) satisfies some wave equation, sourced by $A$. The equations of motion here constrain the form of $J$, but off-shell $J$ is just some arbitrary function, since $\phi$ is just some arbitrary function. Then it is clear that the Lagrangian is not gauge-invariant off-shell.

And this is a problem, because when we derive conserved quantities through Noether's theorem, it's important that our symmetry is a symmetry of the Lagrangian for any field configuration. If it's only a symmetry for on-shell configurations, then the variation of the action vanishes trivially and we can't make any claims about conserved quantities.

So here's my question: what symmetry does the above Lagrangian have that implies the conservation of the quantity $J$, provided $A$ satisfies its equation of motion? Thank you.

particle physics - Besides the up and down quark, what other quarks are present in daily matter around us?

Protons and neutrons, which are found in everyday matter around us, compose of up and down quarks. Are the other two generations of quarks, i.e. $c,s,t,b$ quarks found in everyday matter around us?

I am learning about these fundamental particles and would like to know how they relate to our daily life. Are they mostly irrelevant to our daily life except in extreme physical conditions, like in the particle colliders?

Answer

Every nucleon has what are called sea quarks in it, in addition to the valence quarks that define the nucleon as a proton or neutron. Some of those sea quarks, especially the strange quarks, have some secondary relevance in practical terms regarding how the residual strong nuclear force between protons and neutrons in an atomic nucleus is calculated from first principles and how stable a free neutron is if you calculate that from first principles. Strange quarks are also found in the $\Lambda^0$ baryon (which has quark structure $uds$), which is present at a low frequency in cosmic rays, but has a mean lifetime of only about two tenths of a nanosecond and is only indirectly detected in the form of its decay products.

Strange quarks are also relevant at a philosophical level that could impact your daily life, because mesons including strange quarks called kaons, are the lightest and most long lived particles in which CP violation is observed; thus, strange quarks are what made it possible for us to learn that the laws of physics at a quantum level are not independent of an arrow of time.

You could do a lot of sophisticated engineering for a lifetime without ever knowing that second or third generation quarks existed, even nuclear engineering. Indeed, the basic designs of most nuclear power plants and nuclear weapons in use in the United States today were designed before scientists knew that they existed. The fact that protons and neutrons are made out of quarks was a conclusion reached in the late 1960s and not widely accepted until the early 1970s, although strange quark phenomena were observed in high energy physics experiments as early as the 1950s. Third generation fermions were discovered even later. The tau lepton was discovered in 1974, the tau neutrino in 1975, the b quark in 1977, and the top quark in 1995 (although its existence was predicted and almost certain in the 1970s).

Otherwise, these quarks are so ephemeral and require such concentrated energy to produce, that they have no real impact on daily life and are basically never encountered outside of high energy physics experiments, although some of them may be present in and influence to properties of distant neutron stars. Second and third generation quarks also definitely played an important part in the process of the formation of our universe shortly after the Big Bang.

The only second or third generation fermion in the Standard Model with significant practical engineering applications and an impact on daily life and on technologies that are used in the real world are muons (the second generation electron). Muons are observed in nature in cosmic rays (a somewhat misleading term since it doesn't include only photons) and in imaging technologies similar to X-rays but with muons instead of high energy photons. Muons are also used in devices designed to detect concealed nuclear isotypes. Muons were discovered in 1937, although muon neutrinos were first distinguished from electron neutrinos only in 1962, and the fact that neutrinos have mass and that different kinds of neutrinos have different masses was only established experimentally in 1998.

quantum mechanics - Virial theorem and variational method: a question

I have an hydrogenic atom, knowing that its ground-state wavefunction has the standard form $$ \psi = A e^{-\beta r} $$ with $A = \frac{\beta^3}{\pi}$, I have to find the best value for $\beta$ (using the variational method). After having included the Darwin correction $$ H_D = D\delta(r) $$ with $D = \frac{\alpha^2\pi Z}{2}$, into the Hamiltonian $$ H_0 = -\frac12\nabla^2-\frac{Z}{r} $$

I calculated $$\langle\psi(\beta)|-\frac{\nabla^2}{2}|\psi(\beta)\rangle = \langle \psi(\beta)|T|\psi(\beta)\rangle = \frac{\beta^2}{2}$$ then I thought to use the virial theorem to calculate the part $\langle V\rangle $: $$ \langle \psi(\beta)|-\frac{Z}{r}|\psi(\beta)\rangle = \langle \psi(\beta)|V|\psi(\beta)\rangle = -2\langle \psi(\beta)|T|\psi(\beta)\rangle = -\beta^2 $$ with the Darwin part left to be easily calculated.

Looking at the solutions that my professor wrote, I've just found a different result:

$$\langle V\rangle = \langle \psi(\beta)|-\frac{Z}{r}|\psi(\beta)\rangle = \frac{Z}{\beta}\langle \psi(\beta)|-\frac{\beta}{r}|\psi(\beta)\rangle = -2\frac{Z}{\beta}\frac{\beta^2}{2} = -\beta Z$$

Why's that? Is this somehow related to the fact that I'm going to use the variational method later? Please, help me understand.

Answer

In summary, you can either find a $|\psi\rangle$ that solves the schrodinger equation (and hence the virial theorem) or does not solve the schrodinger equation and then minimize $E(\beta)=\langle \psi(\beta)|H|\psi(\beta)\rangle$.

If you've solved the schrodinger equation then there is nothing to minimize. If you have guessed the right functional form of $|\psi(\beta)\rangle$ then that function will solve the schrodinger equation only at the minimum value $E(\beta)$.

What you have is (ignoring the constant $A$ for now) the right functional form of $|\psi(\beta)\rangle$ which does not satisfy the conditions the virial theorem for all values of $\beta$ because it does not solve the schrodinger equation for all values of $\beta$. We can illustrate this last point.

Calculating $H|\psi\rangle$ for arbitrary $\beta$ (and ignoring the darwin term),

$$H|\psi\rangle= \left(\frac{\beta}{r}-\frac{\beta^2}{2}-\frac{Z}{r}\right)|\psi\rangle$$

(up to a multiplicative constant.) Since $E$ must be independent of $r$ we find that $\beta$ must equal $Z$. This is what you've found by demanding that the virial theorem holds. This leaves you with the familiar result that

$$E=-\frac{Z^2}{2}.$$

Notice that if you try to apply the virial theorem and then the variational method, the total energy $E(\beta) \sim -\beta^2$, which as you can see is unbounded from below. You can't minimize this! Your professor's approach is correct -- show yourself that you can minimize $\langle T \rangle + \langle V \rangle$ to conclude that $\beta = Z$.

electric circuits - Why is the voltage drop across an ideal wire zero?

I'm having trouble conceptualizing why the voltage drop between two points of an ideal wire (i.e. no resistance) is $0~V$. Using Ohm's Law, the equation is such:

$$ V = IR \\ V = I(0~\Omega) \\ V = 0$$

However, conceptually I can't see how there is no change in energy between these two points.

It is my understanding that the electrical field of this circuit produces a force running counterclockwise and parallel to the wire which acts continuously on the electrons as they move through the wire. As such, I expect there to be a change in energy equal to the work.

Voltage drop is the difference in electric potential energy per coulomb, so it should be greater than $0~V$:

$$ \Delta V = \frac{\Delta J}C \\ \Delta J > 0 \\ \therefore \Delta V > 0 $$

For example, suppose I have a simple circuit consisting of a $9~V$ battery in series with a $3~k\Omega$ resistor:

If the length from point 4 to point 3 is $5~m$, I would expect the following:

$$ W = F \cdot d \\ W = \Delta E \\ F > 0 \\ d = 5 > 0 \\ \therefore W > 0 \\ \therefore \Delta E > 0$$

Since work is positive for any given charge, the change in energy for any given charge is positive -- therefore the voltage drop must be positive. Yet, according to Ohm's Law it is $0~V$ since the wire has negligible resistance.

Where is the fault in my logic?

Answer

The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire).

This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have infinite mobility) rearrange themselves on the surfaces of the conductor in such a way as to perfectly cancel out any internal field.

So, the only fields in your circuit would be 1) in the battery, and 2) in the resistor.

I should also add that this is due to the approximation of the wire as 'perfect'. A real wire has some resistance, or equivalently, its charges don't perfectly reorder so as to perfectly cancel an internal field.

quantum mechanics - How many of which particles are in Hawking radiation?

My understanding is that a black hole radiates ~like an ideal black body, and that both photons and massive particles are emitted by Hawking radiation. So for a low temperature black hole, photons are emitted according to Planck's law, but the peak of this spectrum shifts to higher frequencies as temperature increases.

For sufficiently large temperature, does it follow that massive particles of equivalent energy are also emitted, without preference to other particle properties? If not, I'd be interested to know what physics are involved in determining the form in which the black hole's energy is radiated away.

mathematical physics - Principal value of 1/x and few questions about complex analysis in Peskin's QFT textbook

When I learn QFT, I am bothered by many problems in complex analysis.

1) $$\frac{1}{x-x_0+i\epsilon}=P\frac{1}{x-x_0}-i\pi\delta(x-x_0)$$

I can't understand why $1/x$ can have a principal value because it's not a multivalued function. I'm very confused. And when I learned the complex analysis, I've not watched this formula, can anybody tell me where I can find this formula's proof.

2) $$\frac{d}{dx}\ln(x+i\epsilon)=P\frac{1}{x}-i\pi\delta(x)$$

3) And I also find this formula. Seemingly $f(x)$ has a branch cut, then $$f(z)=\frac{1}{\pi}\int_Z^{\infty}dz^{\prime}\frac{{\rm Im} f(z^{\prime})}{z^{\prime}-z}$$ Can anyone can tell the whole theorem and its proof, and what it wants to express.

Now I am very confused by these formula, because I haven't read it in any complex analysis book and never been taught how to handle an integral with branch cut. Can anyone give me the whole proof and where I can consult.

Answer

The first equation, $$\frac{1}{x-x_0+i\epsilon}=P\frac{1}{x-x_0}-i\pi\delta(x-x_0)$$ is actually a shorthand notation for its correct full form, which is $$\underset{\epsilon\rightarrow0^+}{lim}\int_{-\infty}^\infty\frac{f(x)}{x-x_0+i\epsilon}\,dx=P\int_{-\infty}^\infty\frac{f(x)}{x-x_0}\,dx-i\pi f(x_0)$$ and is valid for functions which are analytic in the upper half-plane and vanish fast enough that the integral can be constructed by an infinite semicircular contour.

This can be proved by constructing a semicircular contour in the upper half-plane of radius $\rho\rightarrow\infty$, with an indent placed at $x_0$, making use of the residue theorem adapted to semi-circular arcs. See Saff, Snider Fundamentals of Complex Analysis, Section 8.5 Question 8.

The third one is the Kramers-Kronig relation, as Funzies mentioned.