antimatter - How detectors in particle colliders can differentiate neutrons from antineutrons?

Their mass is the same. None of them interacts with EM fields. And their decay (around 1000s) is far too slow to see their decay products yet in the detector.

How is it then possible to differentiate them?

Answer

Detectors at particle colliders are layered like onions around the collision vertex.

The CMS detector at CERN

First there are charged particle sensitive detectors where charged particles leave tracks because of ionisation, but mass density is low so strong interactions do not happen often; their momentum can be measured by the curvature in the imposed magnetic field.

Then there are electromagnetic calorimeters, where photons leave their energy and charged particles continue as tracks.

Then come the hadronic calorimeters with a lot of mass so that strongly interacting particles, hadrons, protons neutrons antiprotons antineutrons deposit their energy. Protons will have a continuous path up to the hadronic calorimeter due to their charge. Antiprotons will have negative charge. Neutrons will deposit energy without a previous track trace. Antineutrons will also deposit energy without a track, except due to the annihilation with matter the shower will be more energetic.

At LHC energies the difference in the multiplicity due to the annihilation for antineutrons will not be distinguishable. At low energies , antineutrons have higher multiplicity showers.

Generally in colliders the existence of antineutrons might be guessed at by conservation of charge and baryon numbers, in low multiplicity events.

pattern - Four-digit quadruples

A friend gave me this riddle, stating that it would be easy for children but hard for grown-ups.

Any combination of four digits has an determined value. Here are some examples (note, the position of the example is of no relevance):

0000 = 4

0001 = 3
1111 = 0
1234 = 0
2266 = 2
4068 = 4
6666 = 4
6088 = 6
7654 = 1
8888 = 8
8901 = 4

9966 = 4
9998 = 5

You may ask for any other four-digit number and there is a determined value for it as well. What is the value-function based upon?

Answer

The value is:

How many circles appear within the numbers

Explanation:

Each number corresponds with a value, and they are additive for the final value of the 4 digit string. 0, 6, and 9 map to 1. 8 maps to 2, everything else maps to 0. This value for each digit is the number of circles within the drawing of the digit

quantum field theory - Are $mathrm U(1)$ charges quantised?

Consider a theory with a $\mathrm U(1)$ symmetry, i.e., such that there exists a unitary operator $U\equiv\mathrm e^{iQ}$ that commutes with the $S$ matrix (or the Hamiltonian). The hermitian operator $Q$ represents a conserved charge, which we may refer to as the electric charge (or baryonic charge, etc.).

The states of the theory are classified according to unitary representations of the symmetry group, which in this case contains a $\mathrm U(1)$ factor. Now, the unitary representations of this group are of the form $z\mapsto z^n$ with $z\in\mathbb C-\{0\}$ and $n\in\mathbb Z$, which means that states are labelled according to \begin{equation} Q|n,\dots\rangle=n|n,\dots\rangle \end{equation} where "$\dots$" refers to other labels. From this I would conclude that electric charge, or any other $\mathrm U(1)$ charge, is always quantised. There exists a minimal charge, say $q$, such that the charge of any other state is $nq$ for some $n\in\mathbb Z$. This seems to be in agreement with what we observe experimentally.

Now comes my question: I would have expected that we should allow for projective representations rather than regular ones. This means that we may now allow $z\mapsto z^n$ with $n\in\mathbb R$, i.e., the quantum number is no longer quantised. Any charge should be observed rather than only those a scalar multiple of some minimal one. This does not seem to agree with what we observe experimentally. Why is this? Why must we only consider regular representations rather than projective ones? Should or should not $\mathrm U(1)$ charges be quantised?

Answer

To be specific, I'll consider here the case of the electric charge.

There are two models which can explain the quantization of the electric charge.

1. 
   

   When the electric charge generator is one of the generators of a broken large simple group, (such as in the Georgi–Glashow model) then the electric charge is represented by a matrix acting on a vector wave function. This transformation is not projective, thus in order to have a proper action, the charge needs to be quantized.

   
   


2. 
   

   In the Kaluza-Klein scenario. here the charge is actually the velocity in the fifth dimension. If the fifth dimension is a circle, then a quantization of the electric charge will take place similar to the quantization of the momentum of a particle moving on a circle.

   
   

entropy - What is information?

We're all familiar with basic tenets such as "information cannot be transmitted faster than light" and ideas such as information conservation in scenarios like Hawking radiation (and in general, obviously). The Holographic Principle says, loosely, that information about a volume of space is encoded on its two-dimensional surface in Planck-sized bits.

In all these contexts, I can take "information" to mean predictive or postdictive capability, i.e. information is what enables us to state what the outcome of a measurement was or will be (locally). But what is information, exactly? Do we have any kind of microscopic description of it? Is it just a concept and, if so, how can we talk about transmitting it?

I suspect this is probably as unanswerable as what constitutes an observer/measurement for wave function collapse, but I'd love to know if we have any formulation of what information is made of, so to speak. If I'm talking nonsense, as I suspect I may be, feel free to point this out.

quantum mechanics - Eigenvalue of $L_z$

In section 4.3 of Griffths' "Introduction to Quantum Mechanics", just below Figure 4.6, the sentence begins

Let $\hbar \ell$ be the eigenvalue of $L_z$ at this top rung...

Why is this valid? In the previous pages, there is no derivation of this fact. It's not surprising that this eigenvalue has $\hbar$ in it, but I don't see why I should expect it to be an integer multiple of $\hbar$.

Answer

When you initially set the eigenvalue at the top rung to $\hbar l$, you don't need to assume that $l$ is an integer, you can think of it as any multiplicative constant. Clearly there is no loss of generality there. The beautiful aspect of the ladder operator approach is that you can use it to prove that $l$ must be a non-negative integer or half-integer.

This argument is presented clearly in Griffiths, at least in the second edition (perhaps you are using the first edition?). Using the ladder operators $L_+$ and $L_-$, and the conditions that there must be a top rung and a bottom rung for the ladder of eigevnalues, you automatically find that

the eigenvalues of $L_z$ are $m \hbar$, where $m$ ... goes from $-l$ to $+l$ in $N$ integer steps. In particular, it follows that $l = -l + N$, and hence $l = N/2$, so $l$ must be an integer or a half-integer.

So, the nature of $l$ is discovered as a conclusion - there is no initial assumption.

cosmology - How did the universe shift from "dark matter dominated" to "dark energy dominated"?

In order to get dark energy to dominate, wouldn't you first need another form of energy to push the expansion until dark energy could dominate? Otherwise I don't understand how the universe could shift from having a decelerating expansion to an accelerating expansion. Is there any analogy that could help understand this?

Answer

The title and the text actually ask two different questions. While Kyle Oman and Thriveth answer the title excellently, I'll address the question in the text which asks "Why did the Universe expand in the first place, before dark energy (DE) started to dominate".

The answer to this is inflation (we think). The first fraction of a second after the creation of space, it was dominated by "something" that mimicked the effect of DE, causing space to expand by a factor of $\sim e^{60}$. The epoch of inflation lasted until the Universe was some $10^{-32}\,\mathrm{s}$ old.

The expansion continued, but were slowed down by the mutual attraction of radiation, and later matter. If the ratio of DE-to-matter had been smaller, this attraction might have slowed it down sufficiently to halt the expansion before DE started dominating, but that was just not the case in our Universe.

Now what caused the inflation is another question, which someone else than me is better at answering. But I think the most accepted theory, or rather hypothesis, is some scalar field consisting of inflatons.

Analogy

You request an analogy. I can give you the following:

Throw a rock into the air. Your push is inflation. The distance from Earth to the rock is the size of the Universe. The gravitational force between Earth and the rock is the mutual attraction between various forms of energy in the Universe. The speed of the rock is the expansion rate of the Universe. Now if your pitch was too weak, the rock will eventually fall back (Big Crunch), while if you throw hard enough (11 km/s), the rock will escape Earth's pull (Big Freeze). But even if the initial speed was less than 11 km/s, if the rock comes sufficiently close to the Moon (dark energy), it will start picking up speed and eventually escape.

general relativity - Has the curvature of spacetime been measured at the human scale?

The curvature of spacetime has been observed many times from the deflection of light around massive astronomical objects. But has it been observed around small objects in a lab?

In the Cavendish experiment, the gravitational attraction between two masses does sufficient effect for it to be measured on Earth. Thus, it raises the question whether light deflection from curved spacetime is also measurable in the lab.

If it has not been achieved, how far are we from it? How much precision would be needed?

Answer

Yes. A relatively direct measurement of curvature of spacetime has recently been performed using a 16 cm atom interferometer setup: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.118.183602 (paywalled).

The basic idea of this type of experiment is that a packet of atoms is prepared in a quantum superposition of locations, then recombined much like light on a beamsplitter, and the phase change is read out. The two spatially separated parts of the wavefunction experience difference phase shifts due to the small changes in the local gravitational field. Rather amazingly, the curvature that they measure is not due to the Earth's field- they need something they can move, so that they can show a phase change that depends on its position. So, they measure the curvature from an 84 kg test mass placed near the interferometer. As a result, this is a measurement of curvature on the human scale both in terms of size and in terms of the amount of mass generating the curvature.

Note that the authors make a distinction between this measurement, which equivalently can be seen as a measurement of the (non-local) gravitational tidal force, and the (local) measurement of acceleration at a point or gravitational redshift. This curvature measurement, unlike those two types of measurements, is coordinate-independent and in particular can not be replicated by going into an accelerated frame of reference. To do this, their device actually involves two separate interferometers, whose outputs are themselves interfered to effectively make a comparison between two spatially separated measurements of the gravitational shift.

See here for more details at a relatively accessible level (open access).

Electromagnetic waves and photons

In water waves, the wave is transmitted through vertically moving water particles which face no displacement nonetheless the wave is moving. So does the same happen with photons in electromagnetic waves or does photons move along the wave?

Answer

Really, photons are the wave. What makes a wave in the classical sense is a large number of photons all averaging together.

Your question is an obvious guess to make -- other wave phenomenon is a result of local interactions within some medium, so electromagnetic waves must be the same. For a while, people guessed that there was a medium that carried electromagnetic waves, and they called it the aether. Turns out, it doesn't exist.

error analysis - Has my textbook given the incorrect equation for calculating uncertainty in multiplication?

I am 99% sure that the calculation to obtain the uncertainty of two multiplied values given by my textbook and this university (http://web.uvic.ca/~jalexndr/192UncertRules.pdf) is incorrect.

They both say this:

$$(A \pm a)(B \pm b) = AB \pm (\varepsilon_A + \varepsilon_B)$$

Which should be equivalent to (correct me if I'm wrong):

$$(A \pm a)(B \pm b) = AB \pm (AB (\frac{a}{A} + \frac{b}{B})) = AB \pm (Ba + Ab))$$

Lets consider: $(4 \pm 1)(2 \pm 1)$ This should equal (8 \pm 6) right?

meaning the smallest value possible is 2 while the largest is 14. However we can see that the largest possible value is actually 15 (5 * 3) and the smallest is 3 (3 * 1). If this is plot on a 3D graph you will see there is no way to obtain lower or larger values than 3 and 15. Therefore there must be a problem with the formula. After some math I came up with the following equation which I believe is the correct equation:

$$(A \pm a)(B \pm b) = (AB + ab) ± (Ab + Ba))$$

As long as $A ≥ a$ and $B ≥ b$ and $a, b ≥ 0$.

Is the equation in the textbook (and given by this university) incorrect or have I just missed something?

action - Why does reparameterisation invariance lead to gauge-fixing?

In Becker, Becker and Schwarz, the point particle action is given in terms of an auxiliary field $e(\tau)$ as: \begin{align} \tilde{S}_0 = \frac{1}{2}\int \,d\tau \left(e^{-1}\dot{X}^2 - m^2e\right) \end{align}

It is then shown that under infinitesimal reparametrizations of $\tau$, the action is unchanged. This allows us to pick a gauge, in particular $e(\tau) = 1$.

I'm not sure I'm understanding this right, but I have a few issues with this.

1. 
   

   Doesn't this assume that $e$ takes the value $1$ somewhere?

   
   


2. 
   

   Although $\tilde{S}_0$ (sorry, not $e(\tau)$) may be reparametrization invariant, I don't see how you could pick a reparametrization that can leave $e$ constant. Such a reparametrization $\tau'(\tau)$ would need to map all $\tau$ to a constant, but then $\frac{d\tau'}{d\tau} = 0$, which can't be the case.

   
   



3. 
   

   Invariance is only under infinitesimal transformations. This is related to 2: how do we know that an infinitesimal reparametrisation could make $e$ constant?

   
   

terminology - What are the differences and similarities between dynamical tunneling and quantum tunneling?

In case of a double well potential, particle can tunnel from one well to another and this process is known as quantum tunneling or tunneling in general. I want to know about dynamical tunneling and how it is different from the simple tunneling.

classical mechanics - Where do atoms go after collision?

[I edited the question according to Mark's and Grisha's answers.]

Consider two point-like particles of equal mass colliding centrally in 2D. The final directions of the momenta of the two particles are not determined by conservation of energy and momentum: they can go anywhere.

Under which circumstances the direction of the two particles is determined?

Answer

(Edited to reflect updated question).

The solution has a single parameter, which we can take to be the angle the impacting particle goes after the collision. A particular solution is determined only after this parameter is given. Usually we would talk about the collision of, for example, two spheres, in which case this parameter could be related to whether the spheres are colliding head on or glancing.

For example, in the frame where one particle is initially moving and the other is not, one solution is for the moving particle to come to a halt and for the stationary particle to take up all its motion. Another solution is for both particles to fly off from the impact site at 45 degree angles to the original motion, each with $\sqrt{2}/2$ the initial velocity. Both of these scenarios conserve momentum and kinetic energy.

There are many other solutions. In this frame, the conservation of momentum says

$\vec{p} = \vec{p_1} + \vec{p_2}$

with $\vec{p}$ the momentum before the collision and $\vec{p_1}$ and $\vec{p_2}$ the momenta after. Squaring,

$p^2 = p_1^2 + 2\vec{p_1}\cdot\vec{p_2} + p_2^2$.

Kinetic energy is proportional to $p^2$, so conservation of kinetic energy gives

$p^2 = p_1^2 + p_2^2$.

These last two combine to give

$\vec{p_1}\cdot\vec{p_2} = 0$,

so the momenta point at a 90-degree angle. However, any one particle can choose any angle within 90 degrees of the initial direction of motion to go after the collision. Once this direction is chosen, the direction of motion of the other particle, as well as their momenta, are fixed.

harmonic oscillator - What's a good reference for this classical picture Feynman's talking about?

I have a mathematics background but am trying to educate myself a little about physics. At the beginning of Feynman's QED book (not the popular one) is the following:

Suppose all of the atoms in the universe are in a box. Classically the box may be treated as having natural modes describable in terms of a distribution of harmonic oscillators with coupling between the oscillators and matter.

I guess this is something that physicists learn, but I have never heard of it. What is Feynman talking about and where can I learn more about it? The Wikipedia article on harmonic oscillators gives no indication that physicists do this.

quantum field theory - Dirac equation as Hamiltonian system

Let us consider Dirac equation $$(i\gamma^\mu\partial_\mu -m)\psi ~=~0$$ as a classical field equation. Is it possible to introduce Poisson bracket on the space of spinors $\psi$ in such a way that Dirac equation becomes Hamiltonian equation

$$\dot{\psi}~=~\{ \psi,H\}_{PB}~?$$

Of course, such Poisson bracket would be graded (super Poisson bracket), but if it exists this would explain on classical level why $\frac{1}{2}$-spinors correspond to fermions.

black holes - Why is information indestructible?

I really can't understand what Leonard Susskind means when he says in the video Leonard Susskind on The World As Hologram that information is indestructible.

Is that information that is lost, through the increase of entropy really recoverable?

He himself said that entropy is hidden information. Then, although the information hidden has measurable effects, I think information lost in an irreversible process cannot be retrieved. However, Susskind's claim is quite the opposite. How does one understand the loss of information by an entropy increasing process, and its connection to the statement “information is indestructible”.

Black hole physics can be used in answers, but, as he proposes a general law of physics, I would prefer an answer not involving black holes.

special relativity - Is Einstein's formula $E=mc^2$ the same as the kinetic energy formula $dfrac{1}{2} mv^2$?

If the kinetic energy formula is the same as Einstein's formula then $E=mc^2=\dfrac{1}{2} mv^2$. Or, $\dfrac{1}{2}v^2=c^2$. What does this prove? Does it prove that $\dfrac{1}{2}$ of the velocity squared of a moving object is always equal to speed of light squared?

waves - Speed of a particle in quantum mechanics: phase velocity vs. group velocity

Given that one usually defines two different velocities for a wave, these being the phase velocity and the group velocity, I was asking their meaning for the associated particle in quantum mechanics.

And is one of them more representative for a particle?

Answer

Much like the position itself, the velocity in quantum mechanics isn't just a single number; it is an operator with different probabilities of different outcomes that may result from the measurement of the velocity.

The operator of velocity in the simplest quantum mechanical model is $$ v = p/m = -\frac{i\hbar}{m} \frac{\partial}{\partial x} $$ You may Fourier-transform your wave function to the momentum representation and then you see different values of the momentum, and therefore velocity, and the probability densities of different values are given by $|\tilde \psi (p)|^2$.

If you consider a simple plane wave, $$ \psi (x,t) = \exp( ipx/\hbar - iEt /\hbar ) $$ then the operator $v$ above has an eigenstate in the vector above and the eigenvalue is $p/m$. On the other hand, the phase velocity is given by $$v_p = \omega / k = \frac{E}{p} = \frac{pv}{2p} = \frac{v}2 $$ so the velocity of the particle is equal to twice the phase velocity, assuming that your energy (determine the change of phase in time) is only given by the non-relativistic piece, without any $mc^2$. One may also calculate the group velocity of the wave $$ v_g = \frac{\partial \omega}{\partial k } = \frac{\partial E}{\partial p} = \frac{p}m = v$$ which is exactly the velocity of the particle. The advantage of this relationship is that it holds even in relativity. If $E=\sqrt{p^2+m^2}$, then the derivative of $E$ with respect to $p$ is $1/2E\cdot 2p =p/E = v$ which is exactly the right velocity, too. It's not too surprising because if a wave packet is localized, the group velocity measures how the "center of mass" of this packet is moving but the packet's position coincides with the particle's position, so the two velocities must be equal.

newtonian mechanics - Is there a way for an astronaut to rotate?

We know that if an imaginary astronaut is in the intergalactic (no external forces) and has an initial velocity zero, then he has is no way to change the position of his center of mass. The law of momentum conservation says: $$ 0=\overrightarrow{F}_{ext}=\frac{d\overrightarrow{p}}{dt}=m\frac{d\overrightarrow{v}_{c.m.}}{dt}$$

But I don't see an immediate proof, that the astronaut can't change his orientation in the space. The proof is immediate for a rigid body (from the law of conservation of angular momentum). But the astronaut is not a rigid body.

The question is: can the astronaut after a certain sequence of motions come back to the initial position but be oriented differently (change "his angle")? If yes, then how?

Answer

The astronaut can change his or her orientation in the same way that a cat does so whilst falling through the air. After the transformation, the astronaut is still and angular momentum is conserved. There is a rather beautiful way of understanding this rotation as an anholonomy i.e. a nontrivial transformation wrought by the parallel transport of the cat's (or astronaut's) state around a closed loop in cat configuration space. I'll write a bit more about this when I have some more time, but for now, one can give a simple explanation with an idealized "robot cat" (or astronaut) which I made up for the thought experiment:

Above I have drawn a simplified cat. I am a very aural person, so this is good enough for me so long as I can imagine it mewing!

Now our "cat" comprises two cylindrical sections: the "forecat" (F), "hinder-cat" (H) and two legs (L) which can be drawn in so that they are flush with the hinder-cat's surface. With the legs drawn in, the forecat on one hand and hinder-cat + legs assembly on the other have the same mass moment of inertia about the axis of the body. Here is how the cat rotates:

1. Deploy legs symmetrically, i.e. spread them out as shown in the drawing. Now the hinder cat + legs has a bigger mass moment of inertia than the fore cat. Note that, if the legs are diametrically opposite and identical and are opened out symmetrically, the cat undergoes no motion;


2. With an internal motor, the forecat and hinder cat exert equal and opposite torques on one another to accelerate, then stop. Owing to the differences between the moments of inertia, the forecat undergoes a bigger angular displacement than the hind cat;


3. Pull the legs. Again this begets no motion if done symmetrically;


4. Use the internal motor again with an acceleration / deceleration sequence to bring the forecat and hinder cat back to their beginning alignment (i.e. with the line along the cylinders aligned). Now the two halves have the same mass moment of inertia, so when the cat is aligned again, the rotation angles are equal and opposite.

Since the rotation angles are different in step 2, but the same in step 5, our robot cat's angular orientation has shifted.

If you want to know more about the "Berry phase" explanation and the anholonomy of the cat configuration space before I get around to expanding on this, see Mathematics of the Berry Phase by Peadar Coyle. This is not peer reviewed, but looks sound and is in keeping with similar treatments along these lines that I have seen.

electromagnetism - Do electromagnetic fields gravitate?

It's well known that electromagnetic fields contains energy but do they gravitate ?

When we talk about the composition of the universe it's now accepted that the 74 % is dark energy , the 22 % is dark matter and then the 4 % is the rest of ordinary matter/energy that we can see or measure. Are the electromagnetic fields considered in this 4 % ?

general relativity - Relative velocity with respect to event horizon

Consider a cloud of particles falling into a black hole. How does the relative velocity between two such particles, one which is already at the event horizon and one that is still some distance away from the event horizon, depend on the distance between them, as considered in the reference frame of the particle that is still some distance away from the event horizon?

Answer

Based on the fact the nobody even tried to answer this question, I'll venture to answer it myself. I think I understand it well enough at this point to do so, but I'm not perfectly sure.

The relative velocity between a particle approaching the event horizon and one that is at the event horizon in the reference frame of the former is the speed of light. This relative velocity is valid regardless of the separation in time or space between the two particles, provided that we consider this separation in the reference frame of the former particle. In other words, even if the first particle is only a fraction of an atto-second away from crossing the event horizon, the difference in velocity between itself and the particle at the event horizon is the speed of light.

Quantum eraser and coincidence counter

My understanding that In quantum eraser experiment we can interpret the past only when we got the data from the coincidence counter. My question is why we need coincidence counter? Is this becuase the BBO produce entanglement photon and normal photon together ? i.e. if it is efficent 100% ( entanglement photons only ) then we can observe directly that our choice in the future really will change the inteference pattern ? Is there anything in quantum mechanic prevents that in the BBO? So we need always the coincidence counter?

Answer

First, there is no "entanglement photon" and "normal photon". There are simply two photons that will give correlated measurement results. Your question really doesn't need any quantum eraser setup (which doen't bring anything new to the quantum mechanics) but requires better understanding of what quantum entanglement actually is.

Consider photons in a Bell state, \begin{equation} |\psi\rangle=\frac{1}{\sqrt{2}}|\uparrow_z\rangle_A\otimes|\downarrow_z\rangle_B+\frac{1}{\sqrt{2}}|\downarrow_z\rangle_A\otimes|\uparrow_z\rangle_B \end{equation} where $|\uparrow_z\rangle,|\downarrow_z\rangle$ are spin up, spin down states in $z$ axis. Consider for a moment the measurements on just one photon, be it $A$ or $B$ doesn't matter. Then no matter what happens with the other photon you will always get half a time spin up, half a time spin down. One photon of this (maximally) entangled pair is indistinguishable from the photon being with the $50%$ classical (without quantum superposition) probability in the state $|\uparrow_z\rangle$ and with the $50%$ probability in the state $|\downarrow_z\rangle$ (and funny thing is that for other axis it's the same) Where the quantum entanglement appears is in the correlations of the measurement results for $A$ and $B$.

E.g. you may ask with what is probability $P\Big(\uparrow_{\theta,B}|\downarrow_{\phi,A}\Big)$ that if you measure spin of $B$ in axis $\theta$ you get "up" provided that the measurement of spin of $A$ in axis $\phi$ resulted in "down". This correlations are very easily described in quantum theory without introduction of any magic. However if you try to assume the hidden variables - that the nature is actually classical and probabilistic quantum description is simply a rough aproximate picture, then you get into all these troubles with superluminal signals and signals back into the past. Contrary to what for some reason most of the popular sources will tell you nothing of that exists in the quantum mechanics and the quantum eraser experiment and its variations proves nothing new except closing another hidden variable loophole. All the quantum mechanics tells you is that you get probabilistic measurements of $A$ and $B$ that don't influence each other (i.e. measurements on $A$ are the same if you don't compare them with $B$) and there are certain correlations between them.

As I said not about quantum eraser but simpler experiment where you just measure spins of $A$ and $B$ without any extra tricks. Let's first consider the idealized situation - there is no noise and photons $A$ and $B$ always hit their respective detectors. In reality the coincidence counter also helps you significantly reduce the noise so that you more surely detect the entangled pair you are interested in.

Consider now one of the photons. The measurement of its spin is usually done in the following way - you have a detector that catches the photon irrespective of its spin but in the photon's path you put a polarization filter. So when the filter is for example is oriented in the axis $z$ and the photon is in the state $|\uparrow_z\rangle$ it is detected but if it's in the state $|\downarrow_z\rangle$ it is absorbed by the filter and there's no detection signal. If it's in the superposition e.g. in the state $|\uparrow_x\rangle$ it will come through half a time.

Remembering what I said about the Bell state if you are interested only in one photon, be it $A$ or $B$ and don't put any filter no matter what happens with the second photon you will get the same detection rate $n$. If you put a filter in any direction you will get the detection rate $\simeq n/2$, again no matter what happens with the second photon. So $P\Big(\uparrow_{\theta,A}\Big)=\frac{1}{2}$.

But now let's put the coincidence counter. Then you will get signal only when BOTH photons are measured with spin up for their corresponding detector axis. So if you get in this case the detection rate $m$, \begin{equation} \frac{n}{m}\simeq P\Big(\uparrow_{\theta,A},\uparrow_{\phi,B}\Big) \end{equation} And from here you can derive the correlations you are interested in. I think from that presentation it's obvious that it's not that when you measured spin for $A$ you somehow influenced the spin of $B$ and because of that detector at $B$ gives you different detection rate, it's just that you ignore those of the measurements when $A$ had spin down.

Now how it works in case of the quantum eraser (all images are taken from the Wikipedia). You have photon going to the double slit experiment at $D_s$ and another going directly to the detector at $D_p$. At first there's no polarization filter at $D_p$,

and at $D_s$ you get no interference (the measured picture is green).

All the coincidence counter does in this case is ensuring that the photon you measure at $D_s$ indeed belongs to the entangled pair and not just a random noise photon. There will be of course lucky noise photons that will hit both detectors simultaneously but this is rare.

In the second case you put some polarization filter before $D_p$,

and at $D_s$ you get some interference picture. But what will happen if you turn the filter at $D_p$ in the opposite way? You will get another interference picture. And when you sum up those picture you will get (surprise!) the original picture without interference!

That's because in the second case you ignore half of the measurements done on $D_p$ leaving only those that were made provided that the detector at $D_s$ produced a signal.

So here is the final answer: with the coincidence counter you measure a certain correlation between measurements for the different entangled photons AND reduce the noise coming from the non-entangled photons.

electromagnetism - Why is the Maxwell Stress Tensor symmetric?

What is the physical meaning of the Maxwell Stress tensor symmetry?

Answer

No, it's not symmetric. Let me explain:

Say for instance that you only take the magnetic part of the Maxwell stress tensor (let's ignore the electric part). Then you would have the outer product $BB$ + (diagonal tensor). A lot of textbooks usually write it as 1/$\mu BB+$ (diagonal tensor), which is wrong and misleading, since it assumes that the material has a linear behavior $B = \mu * H $.

The right expression is $BH+$ (diagonal tensor), where $B = \mu_0 (M + H)$ Therefore if M is not colinear with $H$ you will get a non-symmetric tensor. However if M is colinear with H then you will get a symmetric one. This colinearity between M and $H$ holds true when the magnetization can be described by a linear and isotropic relationship ... that is $M = $some_physical_constant * $H$.

quantum mechanics - linear response for a simple harmonic oscillator

Really sorry for this simple question, but I think it will be useful/interesting in general.

Consider a quantum simple harmonic oscillator. Add a perturbation $H_I = -\lambda \hat{x}$

Calculate $\langle \psi| \hat{x} |\psi \rangle$ in the new ground state, to first order in perturbation theory in $\lambda$, and relate it to $\langle 0| x^2 |0 \rangle $ where $| 0 \rangle$ is the unperturbed ground state.

Can someone help me on this

optics - Why aren't rainbows blurred-out into nothing after they are produced?

I understand how a prism works and how a single raindrop can scatter white light into a rainbow, but it seems to me that in normal atmospheric conditions, we should not be able to see rainbows.

When multiple raindrops are side-by-side, their emitted spectra will overlap. An observer at X will see light re-mixed from various originating raindrops. The volume of rain producing a rainbow typically has an angular diameter at least as wide as the rainbow itself, does it not?

So why can we still see separate colours?

EDIT: To emphasise the thing I am confused about, here is a rainbow produced from a single raindrop...

...here are the rainbows produced by two raindrops, some significant distance apart...

...so shouldn't many raindrops produce something like this?

I will accept an answer which focuses on this many-raindrops problem, I will not accept an answer which goes into unnecessary detail as to how a single raindrop produces a rainbow.

classical mechanics - Hamilton-Jacobi Equation

In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and hence I do not understand how we can partial differentiate $S$ with respect to time? A simple example would also be helpful.

units - Why can fuel economy be measured in square meters?

With help from XKCD, which says

Miles are units of length, and gallons are volume — which is $\text{length}^3$. So $\text{gallons}/\text{mile}$ is $\frac{\text{length}^3}{\text{length}}$. That's just $\text{length}^2$.

I recently realised that the units of fuel efficiency are $\text{length}^{-2}$ (the reciprocal of which would be $\text{length}^{2}$) and I can't work out why this would be, because $\mathrm{m}^2$ is the unit of area, but fuel efficiency is completely different to this. The only reason I could think of for these units is just that they were meant to be used as a ratio; but then again, ratios are meant to be unitless (as far as I know, e.g: strain).

Please could someone explain why these units are used.

Answer

Imagine that you have a tube laid along some path and that the tube is completely filled with the fuel that you would spend to cover that path.

Area of the cross-section of that tube is the area you're asking about.

Now, if this area is bigger, the tube is thicker, which means more fuel. That is, more fuel to cover the same distance, which means lower efficiency.

Therefore, efficiency is proportional to the inverse of the area of that tube and that's why it can be measured in inverse square meters.

wavefunction - Interpretation of Dirac equation states

In Pauli theory the components of two-component wavefunction were interpreted as probability amplitudes of finding the particle in particular spin state. This seems easy to understand.

But when talking about Dirac equation, we have four-component wavefunction, two of which correspond to usual spin components of Pauli electron, and another two... How do I interpret positron-related components of Dirac electron? Are they probability amplitudes for the particle to appear to be positron? Or maybe to appear to not be positron (taking Dirac sea picture into account)?

Answer

The interpretation of the Dirac equation states depend on what representation you choose for your $\gamma^\mu$-matrices or your $\alpha_i$ and $\beta$-matrices depending on what you prefer. Both are linked via $\gamma^\mu=(\beta,\beta\vec{\alpha})$. Choosing your representation will (more or less) fix your basis in which you consider the solutions to your equation (choosing another representation will rotate your entire solution).

The representation that I will choose is the Dirac-Pauli representation, given by: $$\beta=\left(\begin{array}{c c}\mathbb{I}_{2\times2}&0\\0&-\mathbb{I}_{2\times2}\end{array}\right) \quad\text{and}\quad \alpha^i=\left(\begin{array}{c c}0&\sigma^i\\\sigma^i&0\end{array}\right),$$ where $\sigma^i$ are the Pauli-matrices.

If you would solve the Dirac-equation in this representation, you will find 4 independent solutions: $$ \psi_1(x)=N_1\left(\begin{array}{c}1\\0\\\frac{p_z}{E+m}\\\frac{p_x+ip_y}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu) $$ $$ \psi_2(x)=N_2\left(\begin{array}{c}0\\1\\\frac{p_x-ip_y}{E+m}\\\frac{-p_z}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu) $$ $$ \psi_3(x)=N_3\left(\begin{array}{c}\frac{p_z}{E-m}\\\frac{p_x+ip_y}{E-m}\\1\\0\end{array}\right)\exp(ip_\mu x^\mu) $$ $$ \psi_4(x)=N_4\left(\begin{array}{c}\frac{p_x-ip_y}{E-m}\\\frac{-p_z}{E-m}\\0\\1\end{array}\right)\exp(ip_\mu x^\mu) $$

The way to interpret these states is to look at them in the rest-frame, so the frame in which they stand still $p^\mu=(E,0,0,0)$, the states will become simply the following: $$\psi_1=N_1\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-iEt}, \psi_2=N_2\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)e^{-iEt}, \psi_3=N_3\left(\begin{array}{c}0\\0\\1\\0\end{array}\right)e^{iEt}\text{ and } \psi_4=N_4\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)e^{iEt},$$ by inspection of the time-evolution of the phase factor we can already see that $\psi_1$ and $\psi_2$ represent positive energy states (particles) and the $\psi_3$ and $\psi_4$ represent negative energy states (so anti-particles).

In order to know the spin you should use the helicity-operator, given by: $$\sigma_p=\frac{\hat{\vec{p}}\cdot \hat{S}}{|\vec{p}|},$$ In the case of the Dirac-equation the spin operator is given by the double Pauli-matrix: $$\hat{S}=\frac{1}{2}\left(\begin{array}{cc}\vec{\sigma}&0\\0&\vec{\sigma}\end{array}\right),$$ if we let this one work on the spinors $\psi_1$, $\psi_2$, $\psi_3$ and $\psi_4$, we find that their spin is respectively up, down, up, down. So looking at electrons the Dirac-spinor can be interpreted in the Pauli-Dirac representation as (for example for the electron): $$ \psi=\left(\begin{array}{c}e^-\uparrow\\e^-\downarrow\\e^+\uparrow\\e^+\downarrow\end{array}\right). $$ When the momentum is NOT equal to zero these different states mix up and you can't make such a simple identification. Usually one says that the electron becomes a mixture of an electron with positrons when it starts moving.

What properties are entangled in quantum entanglement?

When two particles are entangled, one can measure the properties of one of the particles and instantaneously know the properties of the other. This is because the two particles possess the same properties when entangled, no matter how far apart.

Which properties are entangled?

Is it just fundamental properties, such as spin, or can it be instantaneous/changing properties, such as the speed of the particles?

It may seem very odd to suggest that speed could be entangled, but I think it's a reasonable question to ask as quantum entangles is plenty weird anyway.

homework and exercises - How do I find the energy stored in a capacitor that has unequal charges

So, while discussing problems with friends, I came across a capacitors problem which looked something like this:

So, my questions are:

1. 
   

   Can this be called a capacitor even though it same polarity of charge on the plates?

   
   


2. 
   

   Charge – After closing the switch, will there be equal chares on both plates (each ($Q_1 + Q_2$)/2) in order to minimise repulsion? — Update : I’ve figured this out using Guass’s law. You can ignore this part.

   
   


3. 
   

   How do I find the energy in the first configurations? Can I find the electric field due to one plate, integrate it to get potential and then multiply the charge on the other plate to get the energy? Or is there any other method? — Update : I have found a method a for this. Find the voltage (charge on inner surface)/C and then use $(1/2)(C)(V^2)$. But, I am not sure whether this is correct because it only takes into account the charges on the inner surfaces.

   
   
   


4. 
   

   In the second case, will the energy stored be zero because the potential difference is zero? Then again, there are some positive near each other, repelling each other and held together by by the metal plates only. Shouldn’t these have a potential energy just like two point positive charges kept some distance apart? This is why I think that the method in question 3 (i.e, the formula $(1/2)(C)(V^2)$) could be wrong when the plates have unequal charges.

   
   


5. 
   

   If energies found in part 3 and 4 are not equal, where would the difference go? I think it can't be heat since there is no resistance (assume ideal wires) — Update : I have learned through a comment on my answer to a similar question that this energy is lost in the form of EM radiations.

   
   

Your help would be appreciated.

PS: Sorry if there are too many questions. I felt that they are all related to each other, and so I put them in a single post.

Answer

1) I would not call this a capacitor. Your typical parallel plate capacitor has two charged plates kept at some potential difference (by being hooked up to opposite terminals of a battery, for example). These are just two charged plates that end up being connected and the charges balance out on each side. In otherwords, I would not say you are "storing" charge here like what you would expect a capacitor to do. You could still define a capacitance for the system, but it would not take the general form $C=\frac QV$, since we do not have a single $Q$ to reference.

In general, you can define a capacitance matrix $C_{ij}$ such that $$Q_1=C_{11}V_1+C_{12}V_2$$ $$Q_2=C_{21}V_1+C_{22}V_2$$

Of course, this is more useful when the potentials of the plates are given. However, there is such thing as an "elastance matrix" $P_{ij}$, which is the inverse of the capacitance matrix:

$$V_1=P_{11}Q_1+P_{12}Q_2$$ $$V_2=P_{21}Q_1+P_{22}Q_2$$

These matrices are symmetric so that $C_{12}=C_{21}$ and $P_{12}=P_{21}$. These terms are related to the mutual capacitance between the plates. The diagonal terms deal with the self capacitance.

---

2) Due to symmetry and the fact that we are dealing with perfect conductors, the charge on each plate must be equal

$$Q_1'=Q_2'=\frac{Q_1+Q_2}{2}$$

---

3) You can still figure out the potential energy difference between the two plates. If the plate separation is small, then between the plates we are looking at distances very close to plates, so we can treat them as infinite planes of charge. Using Gauss's law, we get that $E_1=\frac{\sigma _1}{2\epsilon _0}$ and $E_2=\frac{\sigma _2}{2\epsilon _0}$. Therefore, in between the plates, the field is

$$E=E_1-E_2=\frac{\sigma _1-\sigma _2}{2\epsilon _0}$$

Therefore, the potential difference between the plates is just

$$V=Ed=\frac{\sigma _1-\sigma _2}{2\epsilon _0}d$$.

However, you cannot express this in terms of $U=\frac 12 \epsilon _0 E^2$ for $E$ just inside the plates because the field is not $0$ outside of the plates.

---

4) There is no energy stored in the system, at least in the sense of energy typically stored in a typical capacitor. There is potential energy since the excess charges on each plate are interacting, but it would take no work to move one charge from one plate to the other since a perfect conductor is an equipotential surface. (Once you move that charge though, then moving another charge would require work, but this would involve some external force keeping the first charge in place, like from a battery, which then makes the system not an ideal conductor). Typically when you talk about energy being stored on a capacitor, you are talking about the energy needed to separate the charges and maintain that separation.

---

5) The electric field does work to move charges from one plate to the other. This is where the energy goes.

---

I am not an expert on capacitance, so anyone can correct my reasoning here if something is off. I think something we take for granted in relating the energy stored in the capacitor to the energy in the fields is that in the typical parallel plate capacitor the field is $0$ outside of the system so that the potential difference and the energy and the field are easily to relate. I think in your initial set up you have to be careful in thinking about if you just want to consider the field between the plates or the overall field.

general relativity - Calculating the determinant of a metric tensor

Suppose the line element is $$ds^2 = -A(t,r)^2dt^2+B^2(t,r)dr^2+C^2(t,r)d\theta^2+C^2\sin^2\theta d\phi^2.$$
Since the metric is diagonal, to find the determinant I can multiply the diagonal entries, $$\det g_{ab} = g = -A^2B^2C^4 \sin^2\theta.$$ I have a few questions about this.

1. First off, why do we call the metric determinant $g$?



2. Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?


3. When will it be true that $g = 1$?

kinematics - Galileo's law of odd numbers

The Galileo’s law of odd numbers states that the distances traveled are proportional to the squares of the elapsed times. In other words, in equal successive periods of time, the distances traveled by a free-falling body are proportional to the succession of odd numbers($1, 3, 5, 7,$ etc.).

I clearly understand from kinematics equation that the distances traversed in a time interval are $-\frac{1}{2} gt^2$ so it will be proportional to squares of elapsed times. But what I don't get is how is it proportional to the succession of odd numbers?

Answer

This is because $(n+1)^2-n^2=2n+1$, which is odd. Hence, as in the 1st second, $d_1=-\frac{g} {2} $, in the 2nd second, $d_2=-\frac{4g} {2} $, and so on, we get $\frac{d_2 - d_1}{d_1}=\frac{4-1}{1}=3=2.1+1$

quantum mechanics - Does the Breit–Wigner formula indicate "violation" of energy conservation?

In the quantum mechanical derivation of Breit–Wigner formula, for example, in the particle physics book of Martin & Shaw, we assume if the resonance particle $X$ is in an initial energy state with energy $E_0$ in its rest frame, it can decay into a state with energy $E_f$ not being equal to $E_0$. This seems strange to me.

For examplem In the $Z$ boson decay we have$$ Z \rightarrow l^+ l^-$$Shouldn't we always have $E(l^{+})+E(l^{-})=M_{Z}$ in the rest frame of $Z$? If so why can we have a distribution as a function of the invariant mass of the decay products?

ps: I fully understand the derivation of Breit–Wigner formula, it's this "violation" of energy conservation that I don't understand.

This is the process in the book. How does one know if $X^{-}$ is off-shell? The word off-shell is again confusing, because as far as I know it's a mathematical way of calculation and not real particle. What's the relation between off-shell particles and unstable particles?

Answer

For examplem In the $Z$ boson decay we have$$ Z \rightarrow l^+ l^-$$Shouldn't we always have $E(l^{+})+E(l^{-})=M_{Z}$ in the rest frame of $Z$? If so why can we have a distribution as a function of the invariant mass of the decay products?

There is an intrinsic width in all resonances, i.e. in particles that decay, they do not have a fixed mass and the Breit Wigner gives a good estimate of this width. As in all quantum mechanical calculations the computation gives the probability of finding the invariant mass of the decay particles at that value of the energy. Over all energy and momentum conservation are guaranteed by the Lorenz invariance built in the calculations, whether simple ones or full QFT ones.

In your example above the Z is not long lived enough to be a "real" particle, it is always virtual and the balance of energy and momentum is taken up by the interaction that produced it. If it is electron+positron as in LEP, one can run the experiment changing the energy of the beams and thus see the width of the Z. See the report here . It is a mathematical continuation of the width of spectral lines, it can be related to the Heisenberg Uncertainty Principle . Basically it is due to the fact that at the quantum level one only measures probabilities, i.e. accumulated distributions. Each event is conserving energy and momentum, in the theory by construction of the theory to be Lorenz invariant. The theory fits the data very well.

In the recent measurement of the decays of the Higgs, the theoretically calculated width is much less than the measurement error width, sophisticated methods need to be used . The HEP community expects to go to a lepton collider in order to get enough accuracy for testing the partial decay widths.

What's the relation between off-shell particles and unstable particles?

It depends on whether we can assume that the unstable particle can exist in space time long enough for our measurement apparatus to be able to measure it. The muon for example lives long enough to act like a classical particle, and any quantum mechanical calculation,taking as input particles the ones in the reaction that produced it, will always give an extremely small width for the mass Breit Wigner, unmeasurable.

One can safely write down $μ+p$ and treat the muon as a real particle in the diagram calculation. This is not true for the Z or other particles in the standard model . They appear as internal lines in the Feynman diagram calculations, just carrying the quantum numbers, their mass in the propagator.

What one should keep in mind is that energy and momentum conservation are built in the mathematics of the theory by making sure it is Lorenz invariant.

quantum gravity - Can two observers on different components of the wave function, interact gravitationally?

In a semiclassical gravity theory, the Einstein tensor is proportional to the expected value of the energy-momentum operator on a state.

Now if the state is a combination of two eigenstates, each one contains an "observer", can they detect the existence of the other by measuring the Einstein tensor? Is there an experiment or an argument to forbid this?

Say, if I setup a Cavendish-type experiment where the configuration depends on the answer of a quantum random number generator. Then the whole experiment will be in a superposition as being observed by people from the outside. The question is, should we expect a deviation from the classical gravity theory be observed?

Why is charge = nALe

Sorry if this question is a bit broad but I can't find any info on this by just searching.

The equation q = neAL where L is the length of a conductor, A is the area of cross section, n is the number of charges and e is the fundamental charge. This equation clearly does not make sense from a dimensional point of view at first glance. So why is it true and why is it assumed true in most derivations for charge density in terms of drift velocity?

Answer

$n$ is the particle density: Number of particles per cubic meter in SI units. $e$ would be the charge on a particle (Coulombs), here equal to the elementary charge (they could be protons). With that I think that you will see the the dimensions are correct.

soft question - Are there any good audio recordings of educational physics material?

I am studying to return to school in physics and would like to start spending as much time as possible on that task. Most of my small amount of free time, however, I am either doing house work or commuting to work. While it is difficult or impossible to read while doing these chores, I think I could get great benefit out of listening to educational physics material. I know that there are textbook reading services for the blind, but I have not been able to locate audio versions of physics texts available to the public. I know that several schools put their lecture material on YouTube, but I would love to get something that is specifically oriented to an audio only audience.

Are there any good resources to find educational material on physics in the form of audio?

general relativity - maximum rotational speed

I am wondering if there is a limit to rotational speed of an object just like there is one for translation speed ? what are the implications of general relativity for rotating objects ?

homework and exercises - When is tension constant in a rope?

Suppose we have a massless rope with pulling forces applied at each end. In which scenarios is the tension in the rope constant throughout? For example if there is a knot in the rope the tension is not constant throughout (why?. Similarly if the rope is hung over a cylindrical pulley of non-neglible radius, the tension is not constant (why?). But if there isn't anything touching the rope, for example 2 people tugging on the rope at each end, the tension is constant (why?).

Lagrangian to Hamiltonian in Quantum Field Theory

1. 
   
   

   While deriving Hamiltonian from Lagrangian density, we use the formula $$\mathcal{H} ~=~ \pi \dot{\phi} - \mathcal{L}.$$ But since we are considering space and time as parameters, why the formula $$\mathcal{H} ~=~ \pi_{\mu}\partial^{\mu} \phi - \mathcal{L}$$ is not used?

   
   


2. 
   

   Is there any particular book/lecture notes dealing with these kind of issues in theoretical physics, I would love to know them?

   
   

Answer

Vladimir's answer has the right essence but it is also misleading, so let me clarify.

The formula $$ H = \sum_i p_i\dot q_i - L $$ relating the Hamiltonian and the Lagrangian is completely general. It holds in all theories that admit both Lagrangians and Hamiltonians, whether they're relativistic or not, whether or not they have any other symmetry aside from the Lorentz symmetry.

When you have field theory, both the Hamiltonian and the Lagrangian may be written as spatial integrals of their densities. $$ H = \int d^3x \, {\mathcal H}, \quad L = \int d^3x\, {\mathcal L} $$ Combining that with the first formula, we get the relationship $$ \mathcal{H} = \sum_i\pi_i \dot{\phi_i} - \mathcal{L} $$ Now, you proposed a different formula and I guess that the reason why you proposed it is that it looks more Lorentz-invariant to you, as appropriate for Lorentz-invariant field theories. That's a nice motivation.

However, what's wrong about your reasoning is the assumption that both the Hamiltonian density and the Lagrangian density are Lorentz-invariant. While the Lagrangian density is a nice scalar, so it is Lorentz-invariant (the density at the origin, at least), and it's because the integral of it is the Lorentz-invariant action which should be stationary, the same is not true for the Hamiltonian and its density.

The Hamiltonian is intrinsically linked to the time direction: it is the generator of the translations in time (the spatial counterparts of the Hamiltonian are the spatial components of the momentum); it is the energy, the 0th component of a 4-vector, $H\equiv p^0$. So the argument that this formula should be Lorentz-covariant is invalid, your proposed formula is wrong, and the right formula was justified at the beginning of my comment.

quantum electrodynamics - What would the collision of two photons look like?

Could someone explain to me what the collision of two photons would look like? Will they behave like,

1. Electromagnetic waves: they will interfere with each other and keep their wave nature


2. Particles: they will bounce like classical balls

I assume that energy of that system is too small to make creation of pairs possible.

Answer

Your assumption that pair production is ruled out, rules out* that two photons interact through higher-order processes. Quantum electrodynamics tells us that two photons cannot couple directly. That leaves us with classical electromagnetism, which tells us that electromagnetic waves pass through each other without any interference.

*Edit. The photons can interact through higher-order processes. As pointed out in the comments (and I hope I'm getting this right), there is a (quite small) probability amplitude for two photons to get absorbed in, and two photons be emitted by, e.g. an fermion-antifermion virtual pair (which is the leading contributor to the combined amplitude of all such processes). Whether (and this is my cop-out) the emitted photons can be considered the same photons as the absorbed photons, I leave to the, certainly more knowledgeable, commentators.

electrostatics - Why is electric flux defined as $Phi = E cdot S$?

Flux, as I understand it, is the amount of substance passing through a particular surface over some time. So, from a simple perspective, considering photons that go through some virtual surface $A$ (or $S$, doesn't matter). They have a fixed speed in vacuum, $v=299,792,458$ $\text m/\text s$. To simplify even further, they're all hitting the surface head-on. So, if we wanted to figure out how many photons go through the surface, we conclude that at a constant velocity they will only pass through the surface if they are in the volume bounded by sweeping the surface area along the velocity vector (perpendicular to the surface, the opposite of its normal) a distance $d$ in the alloted time $t$: $d = vt$

So the flux volume is well-defined as $V(t) = Avt$. We could just look for a period of unit time and "drop" the dependency on time. But even then, it's useless if we cannot sample photon volume density $\rho_p$ to determine how many photons occupy a unit volume in order to determine the actual flux. That makes sense:

$$\Phi = \rho_pV(t)$$

And now comes along the electric flux and thwarts my understanding of the whole notion completely. An electric field is generated when a charge is dropped somewhere in space. Any other charge, especially idealized point charges, placed in its vicinity would experience a force exerted on them by the source charge, its magnitude modulated by the amount of charge. So, the electric field maps points in space with force vectors (ie. a vector field) whose direction and magnitude is parametrized by the interaction between the source charge and the point charge.

And this electric flux is defined as $\Phi = E \cdot S$ (I'll use $\Phi = E \cdot A$) and I just cannot interpret the semantics of this dot product, the product's dimensionality is not what I've come to expect from the notion of flux ($\text{Vm}^{-1}$ or any other). How does this in any way show how much electric field flow goes through a surface? Furthermore, what is this vague thing called electric field flow? It seems like it is completely disconnected.

I've tried expressing it in different ways from the derivation of the expression of an electric field, which makes sense (non-vector form, dropped unit vector):

$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$

I've intentionally separated the inverse square of the distance and the $4\pi$ which is, I presume, a part of the normalization factor (steradians of sphere) -- but I noticed that together they forge the area of a sphere $A_r = 4\pi r^2$. This way, I could see the expression as the uniform charge density distribution on the surface of a sphere, scaled by the vacuum permittivity.

$$E = \frac{Q}{A_s\varepsilon_0}$$

And then, due to presumed uniformity, by multiplying by an arbitrary area, I could get the flux, the amount of charge(?) flowing through a particular surface in unit time(?):

$$\Phi = E \cdot A = \frac{Q}{A_s\varepsilon_0} A $$

I could see that as flux, but I'm really not sure can I really reinterpret parts of the normalization factor and the inverse-square of the distance into the area of a sphere. From the perspective of voltage over distance it makes absolutely no sense to me.

Any help would be appreciated.

Answer

When the notions of electric and magnetic fields were conceptualized, they imagined that there was an invisible fluid being pushed around by charges, and they leveraged some of the equations and terminology of fluid mechanics.

The modern understanding of fields has largely gotten rid of this picture, but some colorful langauge like "electric flux" remains. If you want to picture positive charge as "amount of fluid added to region per unit time" and negative charge as "amount of fluid removed from region per unit time", you can, but this thinking only gets you so far. Safer to just think of it as an abstract mathematical definition.

homework and exercises - Gravity on flat object

I was wondering how gravity would behave on object of different shapes.

1. 
   

   If the Earth was squeezed into a thin disk what would the gravitional acceleration be at the center of the flat surface? Would it be really low because the amount of matter beneath me would be small? If I stood on the edge would the gravitational acceleration be enormous because the amount of matter beneath me was huge?

   
   


2. 
   

   If an object is lowered into the Mariana trench will the effect of gravity increase because it gets closer to the center of the Earth?

   
   

Answer

In the Newtonian framework, you just need to solve the integral $$\int{ \frac{ G \rho_{(r)}}{|\vec{r} - \vec{r}_o|^2} \frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dV$$ for the volume in question, where $\vec{r}$ is the distance from an arbitrary reference point to the element of matter where density is $\rho_{(r)}$, and $\vec{r}_{o}$ the distance from the same reference point to the position where you are finding the gravitational force value. And the answers would be:

1) Assuming you mean squeezing the sphere-like Earth into a disc-like shape, you would get: $$\int{ \frac{ G\sigma_{(r)}}{|\vec{r} - \vec{r}_o|^2}\frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dS$$ where if you are near the center of this disc, you could assume is an infinite surface and disregard border effects. Also assuming constant $\sigma_{(r)}=\sigma_o$, choosing our reference point in the surface, and the point of interest at a height $h$ over the plane, the symmetry conditions with respect to $\theta$ yields: $$\int_{h}^{\infty}{ \frac{ \pi\sigma_oGh }{r^2} }dr = \pi\sigma_oG $$ A constant value! This is typical from fields depending on the inverse squared distance. Furthermore since $\sigma_o = \frac{M_T}{\pi R_T^2}$, where $M_T$ and $R_T^2$ are respectively the Earth's mass and radius (of the disc which I intentionally chose the same as sphere radius), we get

$$\pi\sigma_oG = G \frac{M_T}{R_T^2}$$

The force in the surface of Earth-disc (and unlike in the Earth-sphere, in every point over it) is the same as the value in the surface of the of the Earth-sphere!

2) I will leave to you the calculations, using the same first formula, but you will see that indeed in the bottom of the Marianna Trench you should feel a smaller attraction force. In fact if you open a hole through the Earth the intensity of the force would decrease linearly with the the distance to to the center. The reason for this is that inside mass shells the attraction from its different parts compensate among them, as you can see for yourself if you solve the integral to find the force in a point inside a mass shell. But from symmetry only, you see that in the center of a spherical mass shell the force should be zero, right?

electromagnetic radiation - Radio antenna producing waves in the visible spectrum

If a radio could produce waves in the visible light spectrum, what would the result be?

This is a thought experiment that I've pondered for a few years now. I realize there are a few/many real-world constraints, but if we lifted these constraints for the sake of thought, what could we expect?

Personally, I don't see why we wouldn't observe visible light emitting from the antenna, disregarding any light from Blackbody Radiation.

Answer

You might want to have a look at Does light induce an electric current in a conductor?. It's probably impossible for a radio aerial to emit visible light as the frequency of light is around the plasma frequency of the metal that the aerial is made of.

We're not really supposed to address hypothetical questions, but if you could find some material with a high enough plasma frequency (remember this applies to everything in the circuit) then the emitted light would just be light. You'd see the aerial glowing.

gauge theory - How can a massless boson (Gluon) mediate the short range Strong Force?

I thought massless particles were mediators for long range forces such as electromagnetism and gravitation. How can the massless gluon mediate the short range strong force?

Answer

Unlike photons, gluons carry the "charge" of the strong force (confusingly, it's called "color"). This means that, unlike photons, gluons interact with each other. The effect is that, rather than spreading out in all directions (as photons do), gluons tend to stick together and form strings. For example, two quarks (which have color) are not connected by a spread-out field, but by a string. The effect is that the force between the quarks doesn't weaken with distance, but is approximately constant. For an imperfect analogy, think of two balls connected by a rubber band.

Now consider trying to pull two quarks apart. Because the force (the tension in the "string") is roughly constant, the energy required is proportional to the separation between the quarks. (Work equals force times distance.) The tension is huge, about 50 tons (!! yes !!), so it requires an enormous amount of energy to separate the two quarks even by the size of an atomic nucleus. Thus, every object we see that is significantly larger than a nucleus is color neutral (no strings attached ;). Therefore, no long range forces - all because gluons aren't color neutral.

voltage - Are square wave harmonics real-life phenomena or just mathematical abstractions?

Based on my limited knowledge, it is my understanding that square waves can be mathematically broken down into an infinite sum of sinusoidal waves (of different amplitudes, frequency, etc) . This is very interesting, and mathematically it makes sense, however, this concept starts breaking down for me when I start hearing about how all square waves (no matter how they are generated) abide by that rule.

For example, sometimes, I stumble upon articles about electronics where the author mentions how a square voltage wave being impressed into a circuit will be felt by the circuit as a bunch of sinusoidal waves of multiple amplitudes, frequencies, etc. Really? How can this be? I understand that if a square wave is built using sinusoidal waves then it makes sense that the circuit will feel all the sinusoidal waves that the square wave is made of, however, some square waves are made from just “on” / “off” transitions (nowhere do we inject sinusoidal waves to that square wave) so I don’t see how those particular square waves can be felt by the circuit as any type of sinusoidal waves.

So what is going on here? Could someone please help me understand this a little better?

cosmology - How fast am I moving?

Given the speed of light is 299,792,458 meters per second and that it is constant through out the universe, (i.e a person who measures the speed of light while standing still will get the same result as a person moving through space in a rocket)

My question is: if I'm standing still, what is the speed of me moving through the universe? Compared to the speed of light?

The following variables should be taken into consideration

1. 
   

   the speed at which the earth is rotating,

   
   


2. 
   

   The speed at which the earth is moving around the sun,

   
   


3. 
   

   The speed at which out solar system is moving around our galaxy,

   
   



4. 
   

   The speed at which our galaxy is moving in our universe.

   
   

particle physics - Observed composition of UHE cosmic rays

How much is known about the composition of ultra high energy cosmic rays (say $E>10^{20}\text{ eV}$)? I get the impression that the particles are often assumed to be protons or other heavier nuclei, but what empirical basis do we have for this? Is there experimental evidence that excludes photons or leptons as candidates? Do we know if the UHE's are matter or antimatter?

I understand that there is quite a bit of literature examining the possible mechanisms of production of UHE's as nuclei of various masses. For example, it could be easier for Iron nuclei to get up to high energies in a magnetic field since they would have to be accelerated for less time than protons.

However, I don't want to put the cart before the horse; the fact is that we don't know where they come from, and so we should not base our knowledge of what they are made of on these theories.

Answer

To address the actual question of how we know the composition of UHECR without relying on source information (of which we have none), we have to look at their extensive air showers (EAS). After an UHECR hits the top of the atmosphere an EAS is created in the air, but p and Fe will create EAS with different shapes. Properties of hadronic interactions are measured at the LHC and these are then extrapolated to the higher energies (50 TeV and above) of UHECR. Then, shower simulations are done in the atmosphere (with varying atmospheric density, the earth's magnetic field, and everything) to predict the shape of the shower. As these showers propagate through the atmosphere, they fluoresce. Telescopes at the Pierre Auger Observatory (largest CR observatory in the world, is in Argentina) and High-Res (in Utah, provides northern hemisphere information, although is much smaller) measure this fluorescence and plot something they call $X_{max}$ - the depth of the shower from the initial interaction point (in units of $X$: g/cm$^2$ - such that when it is divided by a density it gives a length) at which the maximum radiation is observed. The average and RMS values are then calculated over a series of events and then compared to the simulations.

Auger and High-Res continue to disagree on this point. You can see the Auger data here (arxiv abs) on page 11 by the author or page 16 of the pdf. The red lines are for the predictions for protons across various hadronic interaction models and the blue lines for iron. Auger seems to clearly favor a heavy (or at least heavier) composition at the highest energies.

High-Res data favors proton composition up to the highest energies. Of course, they have significantly less data, but they still claim a significant result. Their data is presented here (arxiv abs) along with pretty plots of representative shower events.

One known problem in the hadronic simulation is that low energy muons aren't being taken into account correctly. It is a problem that CR and LHC physicists are working together to correct, but it will probably be at least a year or two before it is incorporated into the necessary models. Also, there is a working group comprised of members of both Auger and High-Res that is working on sorting out of number of ongoing discrepancies between the two experiments. While the largest has to do with the energy spectrum, I am sure that this is on their list as well.

A rather comprehensive overview from Auger can be found here (arxiv abs).

Conclusion: this is very much an open problem. Future telescopes such as JEM-EUSO on the ISS may be able to improve the data here and solve this problem. In addition, improved data from the LHC may bring the experiments into agreement, or improving the experiments' systematics may resolve the problem. Finally, JEM-EUSO may provide enough data that could be confined with galactic and extra-galactic magnetic field information (that would have to be dramatically improved) to put some limits on charge once one or more source(s) is(are) identified.

homework and exercises - Solving a problem using Newtonian mechanics and D'Alembert principle

yI have to solve that problem with two methods (applying Newtonian mechanics and the D'Alembert principle.

The problem consists in two balls inside a spherical cylinder, it consists in determine the minimum value of $M$ making the tube not to knock down (where $M$ is the mass of the cylinder and $m$ the masses of the two spheres).

I have issues with both methods. With Newtonian method, I don't know what influence has $M$ on the problem, because I can choose a reference point in the center of the cylinder and there will be no torque. With D'Alembert principle, the problem is I have no idea what virtual displacement I have to choose.

The Newtonian process brings me to this meaningless expression if the normal force acts on the lower right corner.

Why are the laws of thermodynamics "supreme among the laws of Nature"?

Eddington wrote

The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.

and Einstein wrote

[classical thermodynamics] is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts.

Why did they say that? Is it a very deep insight they had, or is it something one can be convinced of quite easily? Or even, is it trivial?

particle physics - hadron jets from quark-antiquark in colliders

The observation of hadron jets from electron-positron collisions (LEP) is explained (e.g. Wilczek, The Lightness of Being, p 55) as follows- e,p collide and produce a virtual photon. the photon goes into a quark, antiquark pair, and one jet of hadrons comes out of the quark, and an opposing jet comes out of the anti-quark.

Question: How can a quark with fractional electric charge produce a jet of particles that all have integral electric charge? What happens to electric charge conservation in the process?

conformal field theory - Question about correlation functions of 2d CFTs

I have a question regarding equation (2.22) in Ginsparg's lecture notes on CFTs. Equation (2.22) is $$ \langle T(z) \phi_1(w_1, {\bar w}_1) \cdots \rangle = \sum_{i=1}^n \left( \frac{h_i}{(z-w_i)^2} + \frac{1}{z-w_i} \frac{\partial}{ \partial w_i} \right) \langle \phi_1(w_1, {\bar w}_1) \cdots \rangle $$ Here, $T(z)$ is the stress tensor of the CFT and $\phi_i$ is a primary operator of weight $(h_i,0)$ which transforms under conformal transformations as $$ \delta_\epsilon \phi_i = \left( h_i \partial \epsilon + \epsilon \partial \right) \phi_i $$ He derives (2.22) from (2.21) which reads $$ \langle \oint \frac{dz}{2\pi i} \epsilon(z) T(z)\phi_1(w_1, {\bar w}_1) \cdots \rangle = \sum_{i=1}^n \langle \phi_1(w_1, {\bar w}_1) \cdots \delta_\epsilon\phi_i(w_i, {\bar w}_i) \cdots \rangle $$ by setting $\epsilon(x) = \frac{1}{x-z}$.

My question is - Is (2.22) correct?

Here are my reasons to believe that it is not -

1. 
   

   I believe he derives (2.22) from (2.21) by setting $\epsilon(x) = \frac{1}{x-z}$ in (2.21). (2.22) is then derived if the following holds $$ \langle \oint \frac{dx}{2\pi i} \frac{T(x)}{x-z} \phi_1(w_1, {\bar w}_1) \cdots \rangle = \langle T(z)\phi_1(w_1, {\bar w}_1) \cdots \rangle $$ This would be true if the integrand on the LHS had only a pole at $x-z$. However, it has also has poles at each $x = w_i$, but those contributions aren't considered.

   
   


2. 
   

   I can try and derive (2.22) in a different way - namely via contractions. I start with the LHS of (2.22) and contract $T(z)$ with each $\phi_i$. Each contraction is replaced with the operator product $$ T(z) \phi_i(w_i {\bar w}_i) = \frac{h_i \phi_i(w_i {\bar w}_i) }{ ( z - w_i )^2 } + \frac{ \partial \phi_i(w_i {\bar w}_i) }{ z - w_i } + : T(z) \phi_i(w_i {\bar w}_i) : $$ Again, if I only consider the singular terms, I reproduce the RHS of (2.22). But what about $: T(z) \phi_i(w_i {\bar w}_i) :$?? In a general CFT, conformal normal ordering $:~:$ is not equivalent to creation-annihilation normal ordering ${}^\circ_\circ~{}^\circ_\circ$. The latter would vanish in a correlation function, but not the former. So, I believe in general there would be extra terms on the right of (2.22).

   
   

What am I misunderstanding?

electromagnetism - How to prove the following relation $H=frac{Etimes v}{c}$

Using CGS units, how can we prove the following relation $$H=\frac{E\times v}{c}$$ where H is magnetic field, E is electric field and v is velocity

Answer

What you have here is basically the B-field seen in the frame of a charge with velocity $\mathbf{v}$, moving in an electric field that is $\mathbf{E} \perp \mathbf{v}.$ You can derive this expression by considering the relativistic field transformations of $\mathbf{E}$ and $\mathbf{B}$ in a moving frame, I'll only show you the most important steps, for a complete walkthrough see chapter 26 of Feynman Lectures Vol. 2. Starting with the magnetic field $\mathbf{B}$ expressed in terms of the vector potential $\mathbf{A}$,: $\mathbf{B} = \mathbf{\nabla}\times \mathbf{A},$ with the four-vectors in component being(using the short notation for partials): \begin{align*} \nabla_{\mu} &= (\partial_t,-\partial_x,-\partial_y,-\partial_z)\\ \mathbf{A}_{\mu} &= (\phi,A_x,A_y,A_z) \text{ where } \phi = A_t \end{align*}

Now we can write the outer product results for $\mathbf{B},$ to simplify we use a short-hand term like $B_y=F_{xz}= \partial_z A_x - \partial_x A_z,$ which in a generalized form is: $$ F_{\mu \nu} = \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu} $$

Next we Lorentz transform the $F_{\mu \nu}$ tensors, e.g. for $F'_{xy}$ we have: $$ F'_{xy} = \frac{F_{xy}-vF_{ty}}{\sqrt{1-v^2/c^2}} $$ for shortness I will only write the magnetic field terms: \begin{align*} B'_x &= B_x \\ B'_y &= \frac{B_y+vE_z}{\sqrt{1-v^2/c^2}} \\ B'_z &= \frac{B_z-vE_y}{\sqrt{1-v^2/c^2}} \end{align*} With these transformations you can now calculate the B-field in a moving frame (with speed $v$ here). In order to get closer to your expression, we still have to express these in terms of vector products, for this we assume the velocity vector is directed in the positive $x$-direction, so you can e.g. rewrite $B_y +vE_z$ as the y-component of $(\mathbf{B}-\mathbf{v}\times \mathbf{E})_y,$ and so on so forth. Finally we just rewrite everything in terms of $\parallel$ (to $\mathbf{v}$) and $\perp$ components, with the field components along x-axis being the parallel ones and y,z the perpendicular ones. With the parallel terms being same in either frames, we have for the $\perp$ term (with $\gamma$ the Lorentz factor): $$ B'_{\perp} = \gamma\left(\mathbf{B}-\frac{\mathbf{v}\times \mathbf{E}}{c^2}\right)_{\perp} $$

From here, if you consider the special case of having no B-field in the rest frame, i.e. $\mathbf{B}=\mathbf{0},$ then the perpendicular components of $B'$ as seen in the moving charge's frame is: $$ B'_{\perp} = -\gamma\frac{\mathbf{v}\times \mathbf{E}}{c^2} $$ Now absorb the Lorentz factor in E and rewrite it as a prime term, then remove the minus sign by using the anticommutative property of vector products and you're done.

visible light - Looking for the actual reason of refraction explained precisely without analogies

I'm a high school teacher trying to teach my students (15year olds) about refraction. I've seen a lot of good analogies to explain why the light changes direction, like the marching band analogy, that the light "choose" the fastest way etcetc, and for most of my students these are satisfying ways to explain the phenomenon. Some students, however, are able to understand a more precise and physically correct answer, but I can't seem to find a good explanation of why the lightwaves actually changes direction.

So what I'm looking for is an actual explanation, without analogies, of how an increase/decrease in the speed of a lightwave cause it to change direction.

Thanks a lot

general relativity - Apparent Horizon vs. Event Horizon

I understand that an apparent horizon is the boundary of trapped codimension-2 surfaces in the spacetime (surfaces from which both ingoing and outgoing light rays must be converging). Meanwhile, an event horizon is the boundary of the past of future null infinity (boundary of points from which a light ray can escape to infinity).

I have a few questions about this:

1. 
   

   I've read that apparent horizons are always inside or coincident with event horizons. If it's inside then doesn't that allow for points from which null rays can travel outwards i.e. points inside the event horizon that are causally connected to infinity?

   
   


2. 
   

   Apparent horizons are observer dependent. Why? Shouldn't all observers agree on light converging or diverging?

   
   


3. 
   

   If apparent horizons are observer dependent, why are they at all useful in GR which is famously diffeomorphism invariant (observer independent)?

   
   
   


4. 
   

   Both definitions seem to me to be a boundary between points that can and points that can't send light to infinity. Can someone explain the difference between them please? I know there are several posts on this but I'm afraid it has not helped clear it up in my head.

   
   


5. 
   

   Does solving $g^{rr}(r)=0$ locate the apparent horizon? Wouldn't we need to find this by solving e.g. the Raychauduri equation or something?

   
   

Answer

It might be easiest to open your idea of coordinates to include lots and lots of different coordinates.

For instance if you took a regular Schwarzschild solution then you can imagine putting almost any kind of coordinate system on it. To make paint a concrete picture lets first draw the solution in a Kruskal-Szekeres coordinate system.

In a Kruskal-Szekeres coordinate system you can draw the $y$ axis as vertical and the $x$ axis as horizontal. And for fixed $b\geq 0$ there is are hyperbola like $y=+\sqrt{x^2+b^2}$ that correspond to surfaces of constant areal coordinate inside the black hole event horizon (it is the event horizon when $b=0$). And there is one value of $b$ that corresponds to the singularity.

Similarly, for fixed $b\geq 0$ there is are hyperbola like $y=-\sqrt{x^2-b^2}$ that correspond to surfaces of constant areal coordinate inside the white hole event horizon (it is the event horizon when $b=0$). And there is one value of $b$ that corresponds to the singularity.

And for fixed $b\geq 0$ there is are hyperbola like $x=+\sqrt{y^2+b^2}$ that correspond to surfaces of constant areal coordinate outside the black hole event horizon (it is the event horizon when $b=0$). And there is no value of $b$ that corresponds to the singularity (assuming the mass of the black hole is positive) because you are outside.

Similarly, for fixed $b\geq 0$ there is are hyperbola like $x=-\sqrt{y^2+b^2}$ that correspond to surfaces of constant areal coordinate outside the black hole event horizon in another universe (it is the event horizon when $b=0$). And there is no value of $b$ that corresponds to the singularity (assuming the mass of the black hole is positive) because you are outside.

That's how the the hyperbolas in the Kruskal-Szekeres coordinates relate to the areal coordinate. And the areal coordinate is the one people sometimes call radial, but only in the outside part far from the horizon does it start to become close to a radial distance. Really it's a surface such that if you keep it fixed at $r$ and keep Schwarzschild time fixed at $t$ you get a surface of area $4\pi r^2.$ So it is an area-al (areal) coordinate, not a radius-al (radial) coordinate.

Next we look at Schwarzschild time $t$. For a fixed $a\in[-1,+1]$ then outside there are curves like $y=ax$ that correspond to curves of constant Schwarzschild time with $a=0$ corresponding to $t=0$ and $y=0.$ And positive $a$ corresponding to positive Schwarzschild time, and negative $a$ corresponding to negative Schwarzschild time.

As you go from $a=0$ to $a=+1$ you go from $t=0$ to $t=+\infty$ and through every positive $t$ in between. And $a=1$ is $t=+\infty$ where $t$ is Schwarzschild time and $y=1x$ is part of the event horizon.

Similarly, as you go from $a=0$ to $a=-1$ you go from $t=0$ to $t=-\infty$ and through every negative $t$ in between. And $a=-1$ is $t=-\infty$ where $t$ is Schwarzschild time and $y=-1x$ is the other part of the event horizon.

Note that all of those lines only cover the outsides of the event horizon ($a\in[-1,+1]$)./To cover the inside you can again draw a series of lines but in this region you still pick an $a\in[-1,+1]$ but the line is the line $x=ay$ and again $a=+1$ is $t=+\infty$, $a=0$ is $t=0$ and $a=-1$ is $t=-\infty$ but inside the event horizon, the Schwarzschild time $t$ is not a timelike coordinate. In particular, it is increasing $y$ that is future directed. So for any point in the event horizon of the balck hole there is a future pointing null ray with increasing $y$ and decreasing $x$ (ingoing future pointing) and another one with increasing $y$ and increasing $x$ (outgoing future pointing). The former has $t$ getting smaller, the latter has $t$ getting larger. Both have the areal coordinate getting smaller.

You can see this because null rays go at 45 degrees in a Kruskal-Szekeres coordinate system. And so the left-and-up going one passes through lines like $x=ay$ with different (and decreasing) $a$ that's the ingoing null ray. And the right-and-up going one passes through lines like $x=ay$ with different (and increasing) $a$ that's the outgoing null ray.

So far I haven't breathed a word about apparent horizons. I'm just drawing a picture of the Schwarzschild solution in the Kruskal-Szekeres coordinate system and making sure you see where the curves of constant Schwarzschild time $t$ are (they are the lines through the origin). And where the curves of constant areal coordinate are (they are hyperbolas like $y=+\sqrt{x^2+b^2}$ or $x=+\sqrt{y^2+b}.$

And we briefly identified different curves as future/past pointing and as ingoing/outgoing. increasing $y$ and decreasing $x.$ Don't gloss over a single bit of this. Nothing is deep. Nothing is complicated. But getting it wrong means we'd just talk past each other, so it is essential that you learn it so that we can communicate.

So now we can be inside or outside and use the Kruskal-Szekeres coordinates or the Schwarzschild time $t$ areal coordinate $r$ and the angles $\theta$ and $\phi.$ But we can make lots of other coordinates systems besides those two. It doesn't change which curves are ingoing, outgoing, future pointing, or past pointing. All of that is given by the Kruskal-Szekeres coordinates becasue it was designed to respect causality.

Any event located inside the black hole event horizon has ingoing and outgoing future pointing null directions (going left-up in Kruskal-Szekeres and going right-up in Kruskal-Szekeres). Inside the event horizon you can make alternative coordinates where one of these null rays can have increasing coordinate value for a bit, even though eventually it comes back.

But those range of coordinates never cover the outside. For instance if you used Schwarzschild time and areal coordinate then the areal coordinate never gets larger than the Schwarzschild radius and the Schwarzschild time never reaches the event horizon either, no matter what finite value it reaches in either direction. So it's a fine coordinate chart for the inside. But do keep in mind that inside the Schwarzschild time is a spacelike coordinate, and the areal coordinate is a timelike coordinate.

But it's just one coordinate system. And there are lots. And you can pick ones where the null rays gets a larger value for a while and then reaches a maximum. Lots of choices. There are even choices of coordinates where there are no apparent horizons.

So you might be used to Schwarzschild coordinates where the ingoing null ray has Schwarzschild time strictly decrease and the outgoing null ray has Schwarzschild time strictly increase and both have areal coordinate strictly decrease (because they are future pointing rays). But in other coordinates the coordinates of those same rays could increase for a while then reach a maximum, then decrease.

Why is it useful? Since it is local (a null ray can get a larger coordinate, a maximum coordinate, then a smaller one) it's easier to find. And if you know it is inside an event horizon, then now you know that somewhere outside that surface is an event horizon and those were hard to find.

So you might want to find an event horizon and it's hard to find. So you try some coordinates and find an apparent horizon. Now you know there is an event horizon and you narrowed down where it is.

That's nice if for instance you are counting on cosmic censorship to hide a singularity and you are doing numerical relativity and want to avoid a singularity, because the outside of the event horizon (what you want to solve for) is outside the apparent horizon, so you know now you've got more than enough to have what you want.

Sure it's a bit of overkill to solve for everything outside the apparent horizon, but if you do that, then you know you got everything you needed.