newtonian mechanics - For a massless pulley moving upwards with acceleration, is the upward force equal to the downward force?

Imagine a massless and frictionless pulley with two weights hanging either side of the pulley by a massless string.

Like this except not attached to a ceiling

Rather than being fixed to a ceiling, the pulley is being pulled upward by an external force F, with the weights and string still attached.

Due to Newton's 2nd Law,

$\Sigma F_y=F-2T=ma$,

where $T$ is the tension in the string on either side of the pulley and $a$ is the vertical acceleration of the pulley.

Clearly, since there is a net upward force, the pulley itself will accelerate upwards.

But because the $m=0$,

$F-2T=0$.

Does this not then suggest that the pulley has a constant velocity?

Answer

In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small.

The above reasoning is the justification for the usual assumption that low-mass objects transmit forces unchanged, e.g., that the tension in a rope is the same value throughout the length of the rope.

newtonian mechanics - Why does a surface always exert force normal to it?

In whichever angle an object is thrown at a surface, the surface always exert force normal to it. But why? According to Newton's third law, if an object hits a surface at an angle, the reaction force provided by the surface must be equal and opposite to the applied force by the object. But why does the surface always exert force normal to it?

special relativity - What's the difference between space and time?

I'm having a hard time understanding how changing space means changing time. In books I've read people are saying "space and time" or "space-time" but never explain what the difference is between the two concepts or how they are related.

How are the concepts of space, time, and space-time related?

Answer

Suppose you move a small distance $\vec{dr}$ = ($dx$, $dy$, $dz$) and you take a time $dt$ to do it. Pre-special relativity you could say three things. Firstly the distance moved is given by:

$$ dr^2 = dx^2 + dy^2 + dz^2 $$

(i.e. just Pythagorus' theorem) and secondly the time $dt$ was not related to the distance i.e. you could move at any velocity. Lastly the quantities $dr$ and $dt$ are invarients, that is all observers will agree they have the same value.

Special relativity differs by saying that $dr$ and $dt$ are no longer invarients if you take them separately. Instead the only invarient is the proper time, $d\tau$, defined by:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

In special relativity all observers will agree that $d\tau$ has the same value, but they will not agree on the values of $dt$, $dx$, $dy$ and $dz$.

This is why we have to talk about spacetime rather than space and time. The only way to construct laws that apply to everyone is to combine space and time into a single equation.

You say:

I'm having a hard time understanding how changing space means changing time

Well suppose we try to do this. Let's change space by moving a distance ($dx$, $dy$, $dz$) but not change time i.e. $dt$ = 0. If we use the equation above to calculate the proper time, $d\tau$, we get:

$$ d\tau^2 = \frac{0 - dx^2 - dy^2 - dz^2}{c^2} $$

Do you see the problem? $d\tau^2$ is going to be negative so $d\tau$ is imaginary and has no physical meaning. That means we can't move in zero time. Well what is the smallest time $dt$ that we need to take to move ($dx$, $dy$, $dz$)? The smallest value of $dt$ that gives a non-negative value of $d\tau^2$ is when $d\tau^2$ = 0 so:

$$ c^2d\tau^2 = 0 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

or:

$$ dt^2 = \frac{dx^2 + dy^2 + dz^2}{c^2} $$

If we've moved a distance $dr = \sqrt{dx^2 + dy^2 + dz^2}$ in a time $dt$, the we can find the velocity we've moved at the dividing $dr$ by $dt$, and if we do this we find:

$$ v^2 = \frac{dr^2}{dt^2} = \frac{dx^2 + dy^2 + dz^2}{\frac{dx^2 + dy^2 + dz^2}{c^2}} = c^2 $$

So we find that the maximum possible speed is $v = c$, or in other words we can't move faster than the speed of light. And all from that one equation combining the space and time co-ordinates into the proper time!

homework and exercises - Effective resistance of a weird looking electric circuit

A electric circuit (in the picture) is given where all the resistance are of 1 ohm. I have to find its equivalent resistance. (9,10, 11th points are conncections)

My attempt: I think electron flow will follow 2 paths: 1 2 9 3 10 4 11 5 6 and 1 8 7 6. So the resistances of these 2 paths are in series combination. So, equiavalent resistance of path 1 and 2 respictively are 2 and 3 ohm. As these 2 paths are in parallel the equivalent resistance would be 6/5 ohm.

Am I right? I am assuming that 3 short-circuits are present (2 9 3, 3 10 4, 4 11 5) in those 3 subcycles of the whole circuit.

I think my attempt is wrong. Any hint?

Answer

Current will take all possible paths. There are more than 2 possible paths, and they are not connected in series or parallel.

First try simplifying the circuit. The resistors on the top row are all shorted out, so they can be removed without affecting the circuit. The diagonal resistors are connected at the top and bottom to the vertical resistors, so these are in parallel; there are 2 resistors in parallel at the LH and RH branches, and 3 in the middle two branches.

You can then apply Kirchhoff's Rules to the simplified circuit, making use of symmetry. Alternatively, a combination of 3 resistors connected to the same node can be replaced by 3 resistors arranged in a triangle, using the $Y-\Delta$ Transformation. All of the resistors are then in series or parallel.

cosmology - Did our Universe experience a curvature dominated phase?

So, my question is simple: did our Universe experience a curvature dominated phase?

Or, rather, could our Universe have experienced a curvature dominated phase?

This seems quite shruggish, at first glance, as the Universe has been measured to be pretty locally (one horizon scale) flat. However, within the experimental error, the Universe could be slightly curved. So, I'm thinking if the Universe was positively curved then the curvature could have dominated between the matter dominated phase and the current $\Lambda$ dominated phase? Is that correct?

Answer

Let's analyse the evolution of the curvature in the $\Lambda\text{CDM}$ model. If $\rho_R$, $\rho_M$, and $\rho_\Lambda$ are the densities of radiation, matter and dark energy, and $$ \rho_c = \frac{3H^2}{8\pi G} $$ is the critical density, then we can define $$ \Omega_{R} = \frac{\rho_{R}}{\rho_{c}},\quad \Omega_{M} = \frac{\rho_{M}}{\rho_{c}},\quad \Omega_{\Lambda} = \frac{\rho_{\Lambda}}{\rho_{c}}, $$ and the quantity $$ \Omega_{K} = 1 - \Omega_{R} - \Omega_{M} - \Omega_{\Lambda}, $$ which can serve as a measure of the curvature: if $\Omega_{K} = 0$ the universe is flat, if $\Omega_{K} < 0$ the curvature is positive, and if $\Omega_{K} > 0$ the curvature is negative. We can write these quantities in terms of their present-day values (indicated by subscripts "$0$") as follows: $$ \rho_R = \rho_{R,0}\, a^{-4},\quad \rho_M = \rho_{M,0}\, a^{-3},\quad \rho_\Lambda = \rho_{\Lambda,0}, $$ where $a$ is the scale factor with present-day value $a=1$, so that $$ \Omega_{R} = \Omega_{R,0}\frac{H_0^2}{H^2}a^{-4},\quad \Omega_{M} = \Omega_{M,0}\frac{H_0^2}{H^2}a^{-3},\quad \Omega_{\Lambda} = \Omega_{\Lambda,0}\frac{H_0^2}{H^2}. $$ From the Friedmann equations, we also find (see this post for details) that $$ H^2 = H_0^2\left(\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}\right), $$ so that $$ \Omega_{K}(a) = \frac{\Omega_{K,0}\,a^{-2}}{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}. $$ From this we learn the following:

• If $\Omega_{K,0}=0$, then $\Omega_{K}\equiv 0$. That is, if the universe is exactly flat today, it has always been flat and always will be.


• As $a\rightarrow\infty$, the term $\Omega_{\Lambda,0}$ dominates, so that $\Omega_{K}\rightarrow 0$. In other words, if $\Omega_{K,0}\ne 0$, the curvature of the universe will go to zero in the future under the influence of dark energy.


• As $a\rightarrow 0$, the term $\Omega_{R,0}$ dominates, and again $\Omega_{K}\rightarrow 0$. So in the distant past, the curvature of the universe was also very close to zero; this is known as the flatness problem, and one of the motivations for the existence of an inflationary epoch.

Since $|\Omega_{K}|$ vanishes in the past and in the future, it must have had a maximum value at some intermediate time if it is nonzero today. This maximum occurs when the derivative of $\Omega_{K}(a)$ is zero. After some algebra, this reduces to solving $$ 2\,\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} - 2\,\Omega_{\Lambda,0} = 0. $$ Incidentally, this is also the moment at which the expansion of the universe transitioned from deceleration to acceleration (i.e. when $\ddot{a}=0$, see the previous link for details). Using the values $\Omega_{R,0}\approx 0$, $\Omega_{M,0}\approx 0.3$ and $\Omega_{\Lambda,0}\approx 0.7$, we find the solution $$ a_m \approx \left(\frac{\Omega_{M,0}}{2\,\Omega_{\Lambda,0}}\right)^{1/3} \approx 0.6, $$ and the corresponding curvature $$ \Omega_{K,m} \approx \frac{\Omega_{K,0}\,a_m^{-2}}{\Omega_{M,0}\,a_m^{-3} + \Omega_{K,0}\,a_m^{-2} + \Omega_{\Lambda,0}} \approx \frac{\Omega_{K,0}}{\Omega_{K,0} + (3/2)\,\Omega_{M,0}\,a_m^{-1}} \approx \frac{\Omega_{K,0}}{\Omega_{K,0} + 0.75}. $$ Observations indicate that the present-day curvature is $$-0.02 < \Omega_{K,0} < 0.02,$$ so that the minimum/maximum curvature would have been $$-0.027 < \Omega_{K,m} < 0.026.$$ In other words, the curvature of the universe has always been small.

quantum mechanics - Question on Total, Orbital and Spin Angular momentum

I am reading about the total, orbital and spin angular momentum, and I am not clear as to what these generators actually do after exponentiating.

Could you give me a physical picture of what happens to the quantum set, after being acted upon by the operator obtained after exponentiating these?

For example does the total angular momentum rotate the ket in a circle, about the normal? What about the other two?

Answer

Angular momenta are sets of three operators $\{\hat{L}_{i}\}$ such that

$$[\hat{L}_i,\hat{L}_j]=i\hbar\epsilon_{ijk}\hat{L}_{k} $$

where $\epsilon_{ijk}$ is the Levi-Civita symbol. Orbital angular momentum and spin are angular momenta because they satisfy that commutation relation. (That commutator can be derived if you define $\vec{L}=\hat{\vec{r}}\times\hat{\vec{p}}$, a formal vector product between vector operators)

As in classical physics, angular momentum is useful when you have spherical symmetry. Given the case, instead of using Cartesian coordinates $x,y,z$ you may want to use $r,\theta,\phi$ to represent the wave function $\Psi(\vec{r})$.

The eigenstates of these operators $|l,m\rangle$ can be used to write your wave function as

$$\Psi(\vec{r})=R(r)\varphi(\theta,\phi) $$

with

$$ \varphi(\theta,\phi)=\displaystyle\sum_{l,m}c_{lm}|l,m\rangle$$

You can write $|l,m\rangle$ as the spherical harmonics.

Angular momentum is related to rotations, they are the infinitesimal generators of rotations. What does that mean? It means that you can write a rotation of angle $\alpha$ over an axis $\vec{n}$ exponential of the angular momentum about that axis $\hat{L}_{n}$

$$\hat{R}(\vec{n},\alpha)=\exp(-i\frac{\alpha}{\hbar}\vec{n}\cdot\vec{L}) $$

where the dot product is understood as $n_{1}\hat{L}_{1}+n_{2}\hat{L}_{2}+n_{3}\hat{L}_{3}$. As you said the rotation operator $\hat{R}(\vec{n},\alpha)$ does rotate the quantum state represented by $\Psi(\vec{r})$.

Spin case is a little more tricky because the introduction of spin (e.g. $1/2$) in the formalism requires the Hilbert space of states of the system, say $\mathcal{H}$, as well as a two dimensional complex vector space $\mathbb{C}^2$ to the variables of spin. And this lead us to the introduction of quantities called spinors, usually represented by column vectors. The idea of the spin is the same as discussed above, the spin-space rotation operator can be written as exponentials of the spin operators and they do rotate the spinor of the state.

An arbitrary rotation operator acting upon an arbitrary state vector may be a little hard to write, but you can expand the state as a series of the angular momentum operator eigenstates and the spin operator eigenstates to simplify the calculations. The answer of course depends on the particular rotation and state, but as I said, it is always a rotation (in $\mathcal{H}$, in $\mathbb{C}^2$ or both).

Reference: http://students.washington.edu/tkarin/rotations.pdf it a crash course but it may be helpful

thermodynamics - If black and white object are in isolated space will the black absorb heat from the white?

This question has come to me from my friend in fact: he noted that the heating in the pub is painted black. I replied that it's better for heat emission.

I don't know where did I know that from. And he disagreed, asking me: "Why would black emit more heat than white?" I didn't know. We could, however, agree on fact, that black absorbs more heat than white.

I quickly created a thought experiment to proof by contradiction that the black must emit more light than white:

Assume that white and black both emit the same amount of light. Put a black and white object in an area. Assume that any light (or heat) emitted by the objects can only be absorbed by them. Imagine that the white object emits light and heat. The black one will absorb considerable amount of it. When the black object emits the light the light one will reflect a great portion of it - which can be then absorbed back by the black one.

Written like this, it seems that second law of thermodynamics is being broken by this concept. However, my friend has also thought something to oppose me:

If black objects emits more heat than the white one, why the black objects are hotter when put in the sun? Shouldn't they emit the extra light they absorbed?

I understand that it's not all that simple. There's an article that says I'm right but it doesn't really satisfy me.

For start, infrared is no color at all - how could it have to do anything with black, white or any other color? Shouldn't an "infrared" painted stuff reflect most heat?

I ask for an answer that sufficiently explains why the black and white things absorb/emit heat as they do. And, if questions post by both mine and my friend's arguments are asked, I will be very happy.

Answer

You can't argue with a black object and a white object alone, as I think you partially understand in trying to build your thought experiment. You need a little bit more to define things properly. See whether the following helps.

Imagine a black object at a temperature $T_0$ and a white object also at $T_0$ inside a perfectly isolating box full of blackbody radiation at some higher temperature $T_1>T_0$ (i.e. without the black and white objects, this radiation is in thermodynamic equilibrium).

To understand exactly what would happen, you would have to describe the "colour" of your objects with emissivity curves that show emissivity as a detailed function of frequency. So your "black" and "white" would need to be defined in much more detail. You would also have to define the surface areas of the two objects and what they are made of (i.e. define their heat capacities). But all of this only effects the dynamics of how the system reaches its final state, i.e. these details only influence how the system evolves. What it evolves to is the same no matter what the details: the box would end up with everything at the same temperature such that the total system energy is, naturally, what it was at the beginning of the thought experiment. "Blacker" as opposed to "Whiter in this context roughly means "able to interact, per unit surface area, with radiation more swiftly": the blacker object's temperature will converge to that of the radiation more swiftly than does that of the whiter object, but asymptotically the white object "catches up". Blacker objects absorb more of their incident radiation its true, but they also emit more powerfully than a whiter object at the same temperature. The one concept emissivity describes the transfer in both directions. Think of emissivity as being a fractional factor applied to the Stefan-Boltzmann constant for the surface as well as being the fraction of incident light absorbed by the surface relative to a perfect blackbody radiator.

This description is altogether analogous to that of the situation where $T_0

Maybe the following will help thinking about what is a really quite a complex question: it would be a fantastic last question for an undergrad thermodynamics exam BTW: You can abstract detail away by saying lets define object $A$ to be blacker than object $B$ if, when both objects are made of the same material, are the same size and shape, the temperature of $A$ converges to the final thermodynamic equilibrium temperature more swiftly than that of $B$ when they are both compared in the box-radiation-object thought experiment above.

Thinking about this now, I am not sure whether the above definition would hold for every beginning temperature of the radiation. Maybe there are pairs of surfaces whose relative blackness is different at different beginning temperatures such that $A$ is blacker than $B$ with some beginning temperature whilst the order swaps at a different beginning temperature. I think it is unlikely, but that is probably a different question altogether.

By the way, which pub do you drink in? I might come along.

You ask by implication what is the best colour to paint a heater. This is not a simple question and involves the dynamics of the heater system. It's really an engineering question. I suspect in general it is better for them to be blacker rather than whiter. Here's a glimpse of the kind of factors bearing on the situation.

If you can say a heater has a constant nett input of $P$ watts, then at steady state that's going to be its output to the room, altogether regardless of its colour. There may be a materials engineering implication here: if you paint the heater whiter, and if its dominant heat transfer to the room is by radiation (rather than by convection or conduction), then it has to raise itself to a higher temperature than it would were it blacker so as to radiate $P$ watts into the room. So its materials might not be as longlasting, and it might be more of a fire hazard than it would be were it blacker.

If the heater is the hot water kind, and again if radiative transfer is significant, then the heating system has to run hotter to output power at a given level if the heater is whiter. At a given flow rate and given temperature of heating water, the heat output of heater is lower if it is whiter. You're trying to design the heater to be an "anti-insulator": you want the heat to leak out of the flow circuit in at the heater, not through the lagging on the hot water pipes outside the building channelling the water from the boiler to the heaters. If the hot water pipes leak heat in the same room, then that's no problem.

Recall the quartic dependence of the Stefan Boltzmann law. At room temperatures with a low temperature heater (the hot water kind) $\sigma\,T^4$ is likely to be pretty small compared with other heat transfer mechanisms, in contrast to my idealised scenarios above. So the heater's colour is likely to be pretty irrelevant.

homework and exercises - Free complex scalar field - showing operators are creation and annihlation?

Using the method of canonical quantization we can show that for a free scalar field we have: $$\phi(x)=\int d\tilde p (a(\vec p) e^{-ipx}+b^\dagger(\vec p) e^{ipx})$$ where $a(\vec p)$ and $b^\dagger(\vec p)$ are two operators. It turns out that $a(\vec p)$ and $b^\dagger(\vec p)$ are creation and annihilation operators, but every argument I have seen for this has been fairly hand-wavy.

So my question is: What is the easiest method to show that $a(\vec p)$ and $b^\dagger(\vec p)$ have to be creation and annihilation operators and can be nothing else?

The definition of creation and annihilation operators is the Fock definition, as given e.g. on page 12 here.

optics - Wavelength-dependent refractive index

I read in a book about optical fibers that the different spectral components of a light pulse transmitted in the fiber propagate with different velocities due to a wavelength dependent refractive index. Can someone explain that? Why is that silica refractive index depends on the wavelength/frequency of the wave?

Answer

The fundamental reason for the wavelength dependance of refractive index ($n$), in fact the fundamental description of refraction itself, is the domain of quantum field theory and is beyond my understanding. Hopefully somebody else can provide an answer on that subject.

However, I can state that it isn't just silica that has a wavelength dependent $n$. In fact, every material has some wavelength dependence, and this property is called dispersion. In optical materials, the dispersion curve is very well approximated by the Sellmeier Equation: $$ n^2(\lambda) = 1 + \sum_k \frac{B_k \lambda^2}{\lambda^2 - C_k} $$

usually taken to $k=3$, where $B_k$ and $C_k$ are measured experimentally. As far as I know this equation is not derived from theory; it is completely empirical.

quantum mechanics - What is an antiunitary operator?

What is an antiunitary operator? In field theory one can define a time reversal operator $T$ such that $T^{-1} \phi (x) T = \phi (\mathcal T x)$. It is then proved that $T$ must be antiunitary: $T^{-1} i T = -i$.

How is this equation to be understood? If $i$ is just the unit complex number, why don't we have $T^{-1} i T = i T^{-1} T$ which is just the identity times $i$?

Answer

If I correctly understood your misunderstanding, the answer is: operator is not always a matrix. Technically, action of time inversion operator contains complex conjugation. E.g., in spin up/spin down basis it is written as $-i\sigma_y\mathcal{K}$, where $\mathcal{K}$ is complex conjugation.

What's the definition of distance in curved space-time in general relativity?

While I'm learning general relativity, the definition of the distance really confuses me. For example, we observe the distance between the Earth and the Sun (usually by a transit of Venus), what does the distance mean? When we say the PSR1913-16's semi-major axis is 1,950,100 km, what does it mean?

In Schwarzschild space-time, we can be a stationary observer because the space-time is stationary. Then we have a clear definition of the simultaneity surface, so we can define the distance as the proper length in the simultaneity surface between two objects. (However,I'm not sure whether we can do so if we are not a stationary observer). It is the same as we do in flat space-time.

However, in general, the space-time isn't always stationary. We may even NOT have a time-orthogonal coordinate system and NOT have any Killing vector fields. So we may have NEITHER special coordinate systems NOR special vector field. We cannot give a suitable and unique definition of the simultaneity surface and the distance.

We have 3+1 formalism, but when the certain observer is given, it also gives arbitrary simultaneity surfaces so we have arbitrary definitions of the distance. I know little about this, so I'm not sure.

In short, I just want to know:
1. Given a certain observer(such as human in the earth),is there a suitable and unique definition of the distance in curved space-time in general relativity? If the answer is yes,what is it?
2. If the answer of the first question is no, what does the PSR1913-16's semi-major axis mean? What does the distance between the Earth and the Sun mean?

Thanks for your help.

Answer

As far as I know, the answer to the first question is: No, there is no unique definition of a spacial distance in a curved space-time. Already in simple cases like the Friedmann metric there are several definitions of distance: https://en.wikipedia.org/wiki/Distance_measures_(cosmology) However, for small distances and small curvatures (which applies in the case of the Earth and Sun example) all the definitions give approximately the same result (the Minkowski one). As soon as the curvature or the distance (e.g. points next to a black hole or points far apart in a Friedmann universe respectively) the deviations of the different definitions become large.

renormalization - Spinor field normalisation from poles in the propagator

In the theory of free scalar bosons (KG field) it is a basic result that the propagator $\Delta(p)$ has poles at $p^2=m^2$, with residue $1$ (or any other constant, depending on conventions). Thinking of $p^2$ as a variable is acceptable because it can be shown that the propagator is a function of $p^2$, even in the interacting case.

On the other hand, in the theory of free spinor fields (Dirac field), we can show that the propagator is not a function of $p^2$, but of $\not p$, so we can't use the $p^2$ trick. Conventions aside, the propagator is $$ S(p)=\frac{\not p+m}{p^2-m^2+i \epsilon} $$

It is clear that we cannot think of poles as $p^2\to m^2$, because $S(p)$ depends on $p$ in a non-so-trivial way. The usual solution is to think of $S$ as a function of $\not p$, as if $\not p$ was a number instead of a matrix, thus writing $$ S(\not p)=\frac{1}{\not p-m+i\epsilon} $$ and, again, we find a pole at $\not p=m$, with residue $1$. I think this is not mathematics. This is just nonsensical to me (may be I'm too skeptic, and we can give a meaning to this last formula).

In an interacting theory, we define the field strength normalisation $Z$ as the residue of the propagator at the poles: $$ \Delta(p^2)=\frac{Z}{p^2-m^2+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\rho(\mu^2)\frac{1}{p^2-\mu^2+i\epsilon} $$ $$ S(\not p)=\frac{Z}{\not p-m+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\frac{\not p\rho_1(\mu^2)+\mu \rho_2(\mu^2)}{\not p^2-\mu^2+i\epsilon} $$

Now, I don't understand what's the proper definition of $Z$ as a residue. In what sense is it a residue? What's the variable? I just can't accept it is a residue as $\not p\to m$. Do we really take this "think of $\not p$ as a variable'' seriously? Is it possible to formalise these matters, by thinking of a residue for an actual variable (as $p_0$)?

Perhaps the $\not p$ trick is just that: a trick, which may simplify calculations, but such that it can be shown that the actual result, found by a more standard procedure, is the same. Is this the case? If so, what is the correct procedure?

(I would really appreciate if the answers don't assume that we can boost to the rest frame of the particle, as I'd like it to be as general as posible. I want to take into account the possibility of $m=0$, so please don't boost into $k=(m,\boldsymbol 0)$ if it's not really necessary)

Answer

Very good question, OP! Your scepticism is most certainly justified. The good news is, someone already has addressed your concerns. You can find the answer in Ticciati's Quantum Field Theory for Mathematicians, section 10.13.

Long story short: given \begin{equation} S(\not p)=\frac{1}{\not p-m-\Sigma(\not p)+i\epsilon} \end{equation} you can always parametrise the matrix $\Sigma$ as \begin{equation} \Sigma(\not p)=a(p^2)1+b(p^2)\not p \end{equation} for a pair of scalar functions $a,b$. This general expression is the result of Lorentz- and parity-invariance. But you already know that.

With this, you can rationalise the denominator into \begin{equation} S(\not p)=\frac{i(\not p+\alpha)}{(1-b)(p^2-\alpha^2)+i\epsilon} \end{equation} where \begin{equation} \alpha(p^2)\overset{\mathrm{def}}=\frac{m+a(p^2)}{1-b(p^2)} \end{equation}

The required pole at $p^2=m^2$ implies \begin{equation} {\color{red}{\alpha(m^2)=m}},\tag1 \end{equation} that is, \begin{equation} S(\not p)=\frac{1}{(1-b)(1-2m\alpha')}\frac{i(\not p+m)}{p^2-m^2+i\epsilon}+\mathcal O(1) \end{equation}

If the residue is to be equal to that of a normalised field, that is, \begin{equation} S(\not p)=\frac{i(\not p+m)}{p^2-m^2+i\epsilon}+\mathcal O(1) \end{equation} then we must have \begin{equation} {\color{red}{(1-b(m^2))(1-2m\alpha'(m^2))=1}}\tag2 \end{equation}

So far so good: we have rigorously characterised the normalisation conditions of a bispinor field.

The key point is the following: if we introduce the formal complex variable $\not p\in\mathbb C$, we can formally write $(1),(2)$ as \begin{equation} \begin{aligned} \Sigma(m)&=0\\ \Sigma'(m)&=0 \end{aligned} \end{equation} as can be checked by a straightforward computation. This justifies the introduction of the formal complex variable $\not p$: the formal manipulations of this variable are equivalent to the more correct procedure of introducing the pair of scalar functions $a,b$, and the end result is the same. All's well that ends well I guess.

special relativity - Twin paradox on hypertorus

I will not describe the twin paradox again. But let's suppose we have two twins, one stationary and the other moving with uniform velocity $c/2$ for instance. And let's suppose that they live in a hypertorus space, which is topological equivalent to an flat plane, so no curvature. But the hypertorus is a finite space, it means the moving twin will meet again at an instant $t$ in time in the future. What will be their age when they cross each other again?

Answer

The answer is in Time, Topology and the Twin Paradox by J.-P. Luminet

The twin paradox is the best known thought experiment associated with Einstein's theory of relativity. An astronaut who makes a journey into space in a high-speed rocket will return home to find he has aged less than a twin who stayed on Earth. This result appears puzzling, since the situation seems symmetrical, as the homebody twin can be considered to have done the travelling with respect to the traveller. Hence it is called a "paradox". In fact, there is no contradiction and the apparent paradox has a simple resolution in Special Relativity with infinite flat space. In General Relativity (dealing with gravitational fields and curved space-time), or in a compact space such as the hypersphere or a multiply connected finite space, the paradox is more complicated, but its resolution provides new insights about the structure of spacetime and the limitations of the equivalence between inertial reference frames.

The inertial frames for the twins are not symmetric.

In Special Relativity theory, two reference frames are equivalent if there is a Lorentz transformation from one to the other. The set of all Lorentz transformations is called the Poincaré group – a ten dimensional group which combines translations and homogeneous Lorentz transformations called “boosts”. The loss of equivalence between inertial frames is due to the fact that a multiply connected spatial topology globally breaks the Poincaré group.

In a multiply connected spatial topology, there are more than one straight path to join 2 points.

quantum mechanics - Matrix representation angular momentum

We are supposed to give a matrix representation of $L\cdot S$ for an electron with $l=1$ and $s=\frac{1}{2}$.

I read $L\cdot S$ as $L \otimes S$. Is this correct? Then we would have e.g. for

$L\otimes S (|1,1\rangle \otimes |1/2,1/2\rangle) = L |1,1\rangle \otimes S|1/2,1/2\rangle $ $= \sqrt{2} \hbar |1,1\rangle \otimes \sqrt{\frac{3}{4}} \hbar |1/2,1/2 \rangle = \sqrt{\frac{3}{2}}\hbar^2 |1,1\rangle \otimes |1/2,1/2\rangle $.

Is this correction correct? In that case should I proceed in this way with all the other basis vectors and write the eigenvalues down the diagonal in a matrix?

Answer

There are two problems to deal with which must be disentangled to solve problems like these.

• 
  

  Both angular momentum operators are vector operators, so in some sense they "take values" in $\mathbb R^3$; you are being asked for their dot product, which should be taken within that copy of $\mathbb R^3$. You would have the same problem if you were asked to calculate the dot product $\mathbf r\cdot\mathbf p$ for a single particle without spin.

  
  


• 
  

  The orbital and spin angular momentum operators act on the two different factors of a tensor product of Hilbet spaces. Thus any (operator) product of a scalar orbital operator with a scalar spin operator should be interpreted as a tensor product. You would have the same problem if you were asked to calculate the product $L^2S^2$, which would need to be interpreted as $L^2\otimes S^2$.

  
  
  

Thus, in your case, you must read $L\cdot S$ as $$ \mathbf{L}\cdot \mathbf{S}=\sum_{i=1}^3L_iS_i=\sum_{i=1}^3L_i\otimes S_i. $$ To compute the matrix representation of this, you should begin with the matrix representation of each $L_i$ and $S_i$. You then compute the tensor product matrices $L_i\otimes S_i$. Finally, you add all of those matrices together to get the final result.

This is all much clearer with an example. The $z$ component, for example, is easy, since each matrix is given by $$ L_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \quad\text{and}\quad S_z=\frac\hbar 2 \begin{pmatrix}1&0\\0&-1\end{pmatrix}, $$ in the bases $\{|1\rangle,|0\rangle,|-1\rangle\}$ and $\{|\tfrac12\rangle,|-\tfrac12\rangle\}$ respectively. The tensor product matrix, then, in the basis $\{|1\rangle\otimes|\tfrac12\rangle ,|0\rangle\otimes|\tfrac12\rangle ,|-1\rangle\otimes|\tfrac12\rangle , |1\rangle\otimes|-\tfrac12\rangle ,|0\rangle\otimes|-\tfrac12\rangle ,|-1\rangle\otimes|-\tfrac12\rangle \}$, is given by $$ L_z\otimes S_z=\frac{\hbar^2} 2 \begin{pmatrix} 1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} & 0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \\ 0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} & -1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \end{pmatrix} =\frac{\hbar^2} 2 \begin{pmatrix} 1&0&0& 0&0&0\\0&0&0&0&0&0\\0&0&-1&0&0&0\\ 0&0&0&-1&0&0\\0&0&0&0&0&0\\0&0& 0&0&0&1 \end{pmatrix}. $$ This procedure should be repeated with both the $x$ and the $y$ components. Each of those will yield a six-by-six matrix (in this case). To get your final answer you should add all three matrices.

Can we observe an edge of the universe?

Firstly, I apologise if my thinking is completely incorrect - I am by no means a physicist (yet!), I have included layman illustrations to hopefully assist in explaining my thinking.

I am imagining the big bang to be, for lack of a better word, an explosion, thus the universe to be progressively expanding:

Therefore, to me it seems logical to suggest that there is an edge.

If this is the case, then is it reasonable to suggest that explosion + edge = 'bright edge'? Where light projects in all directions at the edge, not just outwards into the 'nothingness'.

So, this brings me to my question: why, when someone looks deeply enough 'at the edge' do they see an absence of light?

Again, apologies if I'm jumping to irrational conclusions & am way off.

Answer

You are thinking that the big bang happened in a particular point in space and then expanded outwards from that point. This is not true. The big bang happened at all points in space. This is because space itself expanded in the actual bang. Therefore each point in space has its own "horizon" of 13.7 billion light years across.

This edge is due to light simply not having enough time to reach that point, therefore nothing can be observed beyond that point. It is not a literal edge.

spacetime - How do we know we live in 3 dimensions?

How do we know that we are really living in 3 dimensional world?

We can draw 3D in a paper that is one dimension. So, maybe the world around us just looks 3D! how can we prove that we're living in 3 dimensional world not one dimensional?

Answer

The simple proof is that you can tie your shoelaces.

This is a proof because you can only tie a knot in an (approximately) 1D object in three dimensions. In lower than three dimensions it's impossible because that would require the shoelace to self intersect. In higher than three dimensions it's impossible because there's always a way for the knot to untie itself.

electricity - Electrical flow in a simple parallel circuit

I'm having trouble understanding something in one of my text books:

Let’s have a look at the implications of each circuit configuration. Figure 3.13 shows the Conventional representation of a parallel circuit. If you assume that the resistance of the wires can be neglected, then the voltage drop across each bulb is equal to the e.m.f. of the source, the dynamo. The Current flowing from the source is divided between each bulb depending on its resistance (remember I = E/R), Removing one of the bulbs would not affect the voltage drop across the other bulb and would therefore not affect the Current in it, although the overall Current from the dynamo would drop as the demand has been reduced.

Specifically, the line about removing one of the bulbs not affecting the voltage drop. My understanding is that the sum of voltage drops in a circuit must be equal to the output of the emf, but the way I'm reading the text suggests that if you have three bulbs in a parallel circuit and you remove one of them, the voltage drop remains the same - this is what I don't understand, though. If you a lamp is removed, then that's one less lamp consuming emf. Does it not get redistributed to the other two lamps?

Likewise, the text mentions the overall current from the dynamo dropping because demand has been reduced.

My initial understanding was that current from an emf is only affected by the emf - not the demands of components further along the circuit, as in this case. What have I misunderstood?

quantum mechanics - Do electrons follow wave like path or become waves?

when an electron starts moving with high velocity does it move in a wave like path or become a wave itself?

Answer

when an electron starts moving with high velocity does it move in a wave like path or become a wave itself?

In the present day model for elementary particles, the standard model, the electron is a point particle with the given mass and other attributes shown in the table, which have been experimentally determined.

It is not a wave. What has a wave behavior is the wavefunction $Ψ$ determining the probability of finding the electron at (x,y,z,t) by the value of $Ψ^*Ψ$ , dependent on the boundary conditions of the specific problem.

The image you show is wrong. The electron, since there are no boundary conditions, will move in a straight line within the Heisenberg uncertainty principle constraints of its given momentum.

particle physics - Does an electron accelerated in a linear accelerator lose any energy?

In a circular one, it would lose energy due to bremsstrahlung (synchrotron emission), but if it is accelerated from rest to ~GeV energies, does it lose any energy in this acceleration process?

Answer

According to Larmors' formula in Lorentz invariant form the power loss $P$ of a accelerated particle is given by (cgs-units are used, $e$ is the electron charge, $c$ speed of light, $m_0$ the rest mass of the electron):

$$ P = - \frac{2}{3}\frac{e^2 c}{(m_0 c^2)^2}\left[ \frac{dp^\mu}{d\tau} \frac{dp_\mu}{d\tau}\right] $$

where $\tau$ is the proper time, and $p^{\mu}= (E/c, \vec{p})$ are the 4 components of the 4-momentum (E is the energy of the electron and $\vec{p}$ its 3-momentum). If the 4-momentum change rate is written out in its components we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{d\tau}\right)^2 - \frac{1}{c^2} \left( \frac{d E}{d\tau}\right)^2\right] $$

Using $\frac{dE}{d\tau} = v \frac{dp}{dt}$ and replacing the proper time of the particle by the observer time $t$ by $d\tau = \frac{1}{\gamma}dt $ gives ( $\frac{1}{\gamma} = \sqrt{1-\beta^2}$ with $\beta =\frac{v}{c}$):

$$ P = \frac{2}{3}\frac{e^2c \gamma^2}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{dt}\right)^2 - \frac{v^2}{c^2} \left( \frac{d p}{dt}\right)^2\right]\,\,\, (*)$$

In case of a linear acceleration we can set $\vec{p} = p\cdot \vec{e}_p$ where $p$ is the length of the momentum vector and $\vec{e}_p$ its time-independent (!) unit direction vector ($\frac{d \vec{e}_p}{dt}=0$). Then we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \gamma^2(1-\beta^2) \left( \frac{d p}{dt}\right)^2\right]$$

The term $ \gamma^2(1-\beta^2) =1$.

The parameter which in linear accelerators measures the acceleration typically is the energy gain per length, i.e. $\left(\frac{dE}{dx}\right)$ which is equal to: $\frac{dE}{dx} = \frac{dp}{dt}$. With this in mind we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2} \left(\frac{dE}{dx}\right)^2 $$

The energy gain per meter in a typical linear accelerator is $15$MeV/m, with this value one gets a radiation power in order of $10^{-17}$W (changing the formula to SI-units). So the answer is: Yes, principally there is energy loss in a linear accelerator, but it is so small, that it can be neglected right away.

In the opposite case one is only interested in the radiation on a circular path, term $\frac{dp}{dt}$ can be set to zero in formula (*), as the length of the momentum vector does not change. However we can use:

$$ \left(\frac{d\vec{p}}{dt}\right)^2 = \vec{F}^2 = (m\cdot a)^2 = \left(\frac{mv^2}{R}\right)^2 $$

where for $a$ the centripetal acceleration $a= \frac{v^2}{R}$ was used.
If however the centripetal acceleration is plugged in we will finally get the well-known formula for the radiation power in a circular accelerator which is tremendously much larger than in case of the linear accelerator.

thermodynamics - Cup of water with oil on top

I was given a homeowork where i need to explain why did water cool down slower when it had a coat of oil on top of it. We had a cup with just water and a cup with same amount of water but some oil on top. I know the starting temperatures and the temperatures after 10 min in the cups, i know the volume of water and oil, i know the size of the cups, i know the mass of the mixture. But i don't know how to explain this problem with formulas. I would really like to know where to start, what to calculate. Thank you.

astrophysics - What is the temperature of the surface and core of a neutron star formed 12 billion years ago now equal to?

In what part of the spectrum is it radiating? In the infrared, in the microwave? Or is not radiating anymore at all?

In russian:

Чему сейчас равна температура поверхности и ядра нейтронной звезды, которая образовалась 12 миллиардов лет назад? В каком диапазоне она сейчас излучает? В инфракрасном, микроволновом? Или не излучает вообще?

Answer

This depends a lot on the heating and cooling mechanisms you have in your model. Certainly initially neutron stars cool when neutrinos created in nuclear reactions escape. However, this is highly sensitive to such effects as superconductivity of the nucleon species (which both suppresses nuclear reactions and reduces heat capacity) and the density profile (which is in turn determined by the endlessly debated equation of state for the matter). At some point, the star likely reaches a thermal equilibrium of sorts, cooling by blackbody radiation from the surface.

Of course cooling is not the only process. Just a few of the heating ideas that have been floated over the years include:

• Friction between superfluid and non-superfluid layers sapping energy out of the star's rotation;


• The latent heat of crystallization of the crust being released whenever the crust "cracks" due to its equilibrium obliquity being reduced as the star spins down;


• Conversion of magnetic field energy into heat via, e.g., pair production;


• Redistribution of species as the centrifugally-induced chemical potentials change with spindown;


• Interaction with the interstellar medium; and even


• Colliding and interacting with the occasional free-floating magnetic monopole, if you believe in such things.

A rather extensive review of all this and more can be found in this review by Tsuruta, if you have access. The paper is replete with cooling curves for all sorts of models - too many to summarize here. Extrapolating to $12~\mathrm{Gyr}$ leads to temperatures¹ of anywhere from a few Kelvin (obviously the lower limit based on the temperature of the CMB reservoir in which the system sits) to $10^4~\mathrm{K}$ (blackbody peaks in the UV) and higher.

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¹ Often the axes are labeled by luminosity rather than temperature, the conversion being simple enough given the model's radius. Also note that all observables in the paper are properly gravitationally redshifted; if you want to know about the radiation field just above the surface, you have to blueshift the photons back into potential well.

quantum electrodynamics - Why is the spinor field anti-commutator not made gauge invariant?

When we introduce minimal coupling for the Dirac spinor field, we introduce terms into the Lagrangian, by the substitution $i\frac{\partial}{\partial x^\mu}\mapsto i\frac{\partial}{\partial x^\mu}+eA_\mu$, so that the Lagrangian is invariant under arbitrary changes of phase at every point of space-time, $$\psi(x)\mapsto e^{-ieG(x)}\psi(x),\qquad A_\mu\mapsto A_\mu+\partial_\mu G(x),$$ where $G(x)$ is a scalar function. We don't make this substitution, however, in the anti-commutator $$\left\{\psi_\xi(x),\overline{\psi_{\xi'}(x')}\right\}=\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}i\Delta(x-x'),$$ which, it seems, is therefore not invariant under the transformation $\psi(x)\mapsto e^{-ieG(x)}\psi(x)$, $\psi(x')\mapsto e^{-ieG(x')}\psi(x')$. Presumably the path-integral formalism doesn't care about this, but does this have consequences in subsequent derivations in the canonical approach?

I expect that in fussing about this I'm missing something very obvious. Am I? What is it?

EDIT: So it appears from the Answers, for which Thanks, that the anti-commutator given is valid for both space-like and time-like separation for the free Dirac field, and it's valid for space-like separation in QED, but it's nae valid for time-like separation in QED. Put somewhat loosely, this is OK in the canonical formalism because this anti-commutator is valid for the phase space. I'd rather like to know what a valid expression for the anti-commutator is at time-like separation in QED, but since that would, in a sense, be a solution of the theory I guess I'll have to whistle for it. I'll be grateful for Comments if this is an obviously obtuse reading of the Answers, though I ask that you consider that I would like a manifestly covariant version (so to speak, perhaps obscurely) of the canonical formalism before leaping in.

Answer

Because $\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}i\Delta(x-x')$ is a symbolic expression for a given analytical right-hand side. It is written so for convenience (not yet calculated) but it is a specific expression like $\delta(x-x')$ or $\delta(x-x')'$. It should not acquire any "gauge extension" by definition. This expression does not contain a "particle momentum".

gravitational redshift - Can a photon exiting from a gravity well ever reach a frequency of zero / wavelength of $infty$?

In reading another question about gravity's effects on a photon, I wondered if it were possible for a photon to ever be redshifted to zero wavelength.

I know that black holes have a gravity field strong enough to keep light from escaping but what about the affect on the photon that is escaping. Say we have a single photon escaping on a path directly away from a non-spinning singularity. Also conjecture that we can release a single photon at various distances from the singularity (still on a trajectory directly away) so as to select the strength of the gravity field it must traverse. I know the distance has already been calculated (the event horizon I believe) but what of the photon itself?

Is there a point where the amount of redshifting (e.g. gravitational) causes the wavelegth of the photon to become infinitely large? This answer mentions the change in potential and kinetic energy. Is it possible for all of a photon's kinetic energy be transformed to potential energy?

What are the implications, both if it were possible and/or impossible? More importantly if impossible, what makes it impossible? This would beg the question, what is the smallest amount of energy possible (or quanta) for a photon? If that minima-photon attempted to escape from within a gravity field, it can't slow down so what does it do? As the wavelength increases approaching infinity, so must it's frequency decrease to approaching zero.

[EDIT] Thanks to @david-z for an excellent answer but I was hoping for more depth for the case where the photon is emitted at the precise event horizon. He shows that the photon is indeed 'stuck' there but I'm interested in more of the math describing such a photon. If such a photon has a zero wavelength then what else does it imply about the physical existence of the photon? If a photon has zero wavelength then how can you calculate it's energy? Using an equation from @david-z 's answer to the above question,

An electromagnetic wave has a total energy given by $E_\text{total} = > \langle N\rangle hf$, where $\langle N\rangle$ is the number of photons in the wave

If the number of photons $\langle N\rangle$ = 1 and the wavelength is zero, it doesn't matter what the frequency is, the total energy goes to zero. If the total energy of the photon is zero then can we claim that the photon must not exist? I am making the statement in order to be corrected but also so readers understand the core of the question that holds my interest.

I'm thinking about it like in calculus where a curve has no value at a specific point either due to an asymptote, a hole (like caused by division by zero), or other type of discontinuity. Considering that a photon can only have the velocity of c, what can we say about the velocity of a photon caught at the event horizon? Does it become undefined or is there more exotic physics that need to be brought in to describe the situation? I realize that this may end up as a meaningless question but am open to why it is a meaningless question.

Basically, what weirdness happens when we force a photon to achieve zero wavelength?

[EDIT] Ok, I figured out a way to describe the scenario in which I'm interested. Let's do it like Einstein would using a thought experiment.

Say you were in orbit around a black hole with a size of 10 solar masses. The exact size is not that important but for this scenario we need the point where spaghettification happens is inside the event horizon so objects that fall through the event horizon will stay intact in that immediate region of space. Let's also say you brought a flashlight with you and thinking about Einstein and the speed of light you realize you can do an experiment simply by dropping the flashlight into the black hole. Now, there is something special about the flashlight in that it has the quality that allows it to always point away from the black hole and it emits a stream of individual photons. As the flashlight falls through space it continuously emits photons at the speed of light but since the curvature of space is redshifting the photons, we could graph the changing frequency of the light coming from the flashlight as a function of distance from the event horizon using the X axis to represent distance from the event horizon and the Y-axis being the photon's frequency. There is likely to be a limit at X=0 but what does that tell us about the curve of space there. Do photons get stuck at the event horizon and just build up over the lifetime of the hole? When the hole gets larger as it gobbles up more nearby mass, the event horizon also gets larger but what happens to those photons that were caught at the previous horizon? This may or may not be related to recent talk about the event horizon being a hologram that stores the inbound information but it seems to me that there is still something intriguing about the region near the horizon that can be used to investigate some profound physical concepts.

In summary, when David Z says,

"In other words, by emitting a photon close enough to the event horizon, you can arrange for it to be redshifted to as large a positive wavelength (and thus as small a frequency) as you want. But there's no value of ϵ that will actually give you an infinite wavelength (zero frequency).

If you were to go all the way down to ϵ=0 , i.e. emit the photon from right on the event horizon, then it would just be stuck there, since there are no outgoing null geodesics."

For the photon emitted right on the event horizon, yes it may be stuck from a null geodesic point of view but what can we say about its wavelength/frequency? If it is stuck, does that mean the frequency is zero? If the frequency can be zero wouldn't that make the energy of the photon zero? If that is somehow impossible, then are we saying that there may be some natural limit involved that hasn't been contemplated?

How do we know what type of gauge field to add to a theory?

I've been watching Leonard Susskind's particle physics lectures and in one lecture, he discusses a very simple gauge theory. We have a complex scalar field $\phi(x)$ with Lagrangian

$$\mathscr{L} = \partial_\mu \phi^* \partial^\mu \phi - V(\phi^* \phi)$$

and we want to do the gauge transformation $\phi(x) \to e^{i\theta(x)} \phi(x)$. But the derivative term in the Lagrangian is not invariant if $\theta$ varies from place to place, so we add a new vector field $A_\mu$, and define it to transform like $$A_\mu \to A_\mu + \partial_\mu \theta$$ under the gauge transformation. Then we change the Lagrangian to use covariant derivatives $D_\mu = \partial_\mu + iA_\mu$ instead of ordinary ones; now when the gauge transform is done all the derivatives of $\theta$ cancel out, and the result is invariant.

My question is: was there any freedom in choosing to add a vector field, with that specific gauge transformation law? Are there any other ways we could get gauge invariance here—by using a different transformation law, or with a different type of field (say, a scalar, tensor, or spinor) with some other transformation law?

More generally, how can we tell what type of field is needed and what its gauge transformation law should be, to get some particular Lagrangian to be invariant under some particular gauge transformation?

Answer

We have no choice.

Let $G$ be our gauge group and $\Sigma$ our spacetime. Then, for the theory to actually be gauge invariant, every field must have a defined action of the gauge group upon it, i.e. every field must transform in a representation of this group:

$$\phi : \Sigma \to V_\rho \text{ where there is a group morphism } \rho : G \to \mathrm{GL}(V_\rho)$$

We want a derivative $\mathrm{d}_A : \Omega^k(\Sigma) \to \Omega^{k+1}(\Sigma)$ acting upon the fields (more generally, upon $k$-forms producing $k+1$-forms) such that for every gauge transformation $g : \Sigma \to G$ we have $\mathrm{d}_A(\rho(g)\phi) = \rho(g)\mathrm{d}_A\phi$, i.e. the derivative must also transform in the representation.

Now, the forms only come with two natural operations producing a $k+1$-form out of a $k$-form: The exterior derivative $\mathrm{d}$, which fails miserably on its own, and the wedge product of the $k$-form with some $1$-form. Therefore, the only natural way of searching for the exterior derivative is

$$\mathrm{d}_A \omega := \mathrm{d}\omega + A \wedge \omega$$

for some $1$-form (i.e. dual vector field) $A$. It must be stressed that, though $A$ is, as a $1$-form, indeed a (dual) vector field with regards to the Lorentz group, it is not transforming in a proper linear representation of $G$, since its transformation law is (in order to make the group action and the derivative commute)

$$ A \mapsto gAg^{-1} + g\mathrm{d}g^{-1}$$

It is called a connection form and corresponds to a particular choice of an orthogonal subspace of a tangent space of the underlying principal bundle. From this, it is also seen that $A$ must take values in the Lie algebra $\mathfrak{g}$ (also since else the transformation law above would make litte sense). For all gauge theories, this choice of subspace (called an Ehresmann connection) is in bijection with choosing a $1$-form $A$ on the bundle, which projects down to a (local!) $1$-form on $\Sigma$, giving again the vector field we heuristically found by searching for the covariant derivative.

---

However, it must be remarked that if we relax our notions of what a gauge theory is, then there are "connection forms" that are no vector fields. The best example (and the only one I know) is that of the Christoffel symbols in GR, being sections of the tangent bundle of the frame bundle, the jet bundle (in contrast to the gauge fields being sections of the tangent bundle of the principal bundle), which can be seen as being the connection forms determining the Levi-Civita connection, analogous to the gauge fields determining the Ehresmann connection.

quantum mechanics - Why does $ell=0$ correspond to spherically symmetric solutions for the spherical harmonics?

In quantum mechanics why do states with $\ell=0$ in the Hydrogen atom correspond to spherically symmetric spherical harmonics?

Answer

One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two.

Then, the rotation operator $\hat R(\theta)$ around a direction $\vec n$ with angle $\theta$ is given by $$\hat R(\theta)=\exp(i\theta \,\vec n . \vec{\hat L} )$$ from which we clearly see that the state $|0,0\rangle$ is invariant for all rotations : $\hat R(\theta)|0,0\rangle=|0,0\rangle$ and is thus spherically symmetric.

In this formulation, you see that it is the only state like that. You can also show that the state $|l,0\rangle$ is axially symmetric (along $z$), etc. See for instance this nice picture :

fluid dynamics - Bernoulli's equation for flow between cylinders

Let's consider the situation bellow noting that the water cylinders aren't equal and there will be water flowing to the right (As indicated by the arrow)

Taking Bernoulli's equation at the tip of the arrow and it's root the $\rho g z$ cancel and noting that $v_1=v_2$ as water is incompressible and the two areas can be said to be equall. We are left with:

$$P_1+\frac{\rho v^2}{2}=P_2+\frac{\rho v^2}{2}$$

But we already know that the two pressures aren't equal because the water levels aren't equal and by pascal's principal they are different!

The only resolution I see is to say that Bernoulli's equation doesn't apply in cases when the water is accelerated (such as here) but I found no mention of this supposed limitation on the internet.

Can someone point me in the correct direction for solving such problems.

Answer

I think that it helps to define appropriate control volumes. See the image below where I define surfaces A and B.

Here, we can say that the pressure at A is given by $\rho g h_A$ and the pressure at B is given by $\rho g h_B$, recognizing that $h_a$ and $h_b$ are functions of time. If the tank is open to atmosphere the $P_A$ and $P_B$ terms will be equal to atmosphere and cancel. If one side is open then that side will take on atmospheric pressure and the other will be equal to zero. This pressure difference drives the flow and, you can calculate the flow velocity between $A$ and $B$, noting that $v_A$ does equal $v_B$ and that doesn't violate incompressibility.

Of course, in a real pipe you could calculate the pressure difference and then use Poiseuille's Law to get the flow in that section of the pipe.

fluid dynamics - Hydrostatic pressure on a teapot spout

The phenomenon where water flows on the outside side of a teapot spout is named "The teapot effect", and occurs due to a difference in pressure between water and the atmosphere. Consider the image of a teapot spout below, and consider the pressures at points A, B, C and D.

(Quick translation: "Bico" means "Spout", and "Água" means "Water")

I've been kind of lost in this question due to different theories, so I'm going to arrange them into statements to make it more organized:

1. 
   

   (This doesn't alter the answer, but I wanted to know if it's correct.) The whole stream of water has a lower pressure (if compared to atmospheric) because it has a certain velocity, and fluids with velocity have lower pressure;

   
   


2. 
   

   A and D must be atmospheric pressure, since they're both in direct contact with air;

   
   


3. 
   

   B must be higher than both because of Pascal's Law;

   
   


4. 
   
   

   C must be lower than D, but I don't have a clue why this is true. I'm assuming that because it must be the pressure difference that supports water against gravity (considering that I'm right about D being atm pressure), but that is kind of "doing the problem in reverse", what would the real reason be?

   
   

Answer

This answer is a bit of a long story, but I have split it up for the different statements for your convenience. Having thought about it a bit more after the discussion with @Mephisto I actually believe that Bernoulli's equation is not applicable in points B and C, because it is based on conservation of energy and therefore only applies if wall friction is negligible.

1: False

The pressure at point D will be higher than atmospheric (see answer for statement 2). Given that the liquid is connected to the same reservoir (the tea inside the teapot), Bernoulli's equation would predict that in the low velocity regions, i.e. points B and C, the pressure will be higher according to: $$\frac{1}{2}\rho v^2 +\rho g h + p = \text{constant}$$

However, it should be noted that Bernoulli's equation is not valid in points B and C, because we are in the laminar boundary in which frictional losses are non-negligible. Given that losses to the wall will lower the pressure we don't know whether the pressure is above or below atmospheric in these points so the only statement we can make is:

The pressure in the liquid is at best only at SOME places lower than atmospheric, but not at ALL places

2: False

For the interface of two immiscible fluids (gas-liquid or liquid-liquid) there will be a pressure jump across the interface depending on the curvature of the interface. This is predicted by the Young-Laplace equation: $$\Delta P_c =\gamma \left(\frac{1}{R_1}+\frac{1}{R_2}\right) $$

where $\Delta P_c$ is the capillary pressure jump across the interface, $\gamma$ is the surface tension between the two fluids and $R_1$ and $R_2$ are the two principle radii of curvature. In case of the water running along the bottom of the spout, say at point D, we have by approximation the following situation:

In this case the first radius of curvature $R_1$ has some value whereas the second radius of curvature $R_2$ goes to infinity (straight lines have no curvature). The pressure on the convex side (so in the liquid) will be higher than the pressure at the concave side (gas side) thus the pressure in point D is above atmospheric. This is in fact necessary to keep the liquid from falling down.

For point A I am actually not completely sure. In the drawing of the OP the interface also looks convex there, but I think this is more of an artist impression than the truth. If the interface is flat it would mean that the pressure at point A is atmospheric.

3: True

The pressure in B should indeed be higher than in point A and D because of Pascal's law and the simple fact that the flow is running from A/B towards D. This last addition is necessary, because with sufficient pressure difference it is obviously possible for a liquid to flow up against gravity (e.g. by pumping).

4: True

Indeed the pressure in C should be lower than in D (or at least equal to) otherwise the liquid would simple flow from C to D and thus detach from the solid. Here, the explanation is the adhesion of the solid and the liquid as explained in great detail in this paper (later published in Phys. Rev. Lett.).

In short, if the solid is hydrophilic ('likes' water) there will be an attractive force pulling the water against the solid thus introducing a lower pressure near the wall than further away from it. When the surface becomes more and more hydrophobic, the effect slowly disappears until, at the extreme of full hydrophobicity, it completely dissappears according to: $$We_{crit} \propto \left(\frac{r_i^2}{e_0^2}+\frac{r_i}{2e_0} \right)\left(1+\cos \theta_0\right) $$ where $We_{crit}$ is the critical Weber number (which thus relates the velocity as well), $r_i$ is the radius of curvature at the spout, $e_0$ is the thickness of the liquid stream and $\theta_0$ is the contact angle which indicates the wettability ($\theta_0=0$ for complete hydrophilicity and $\theta_0=180$ for complete hydrophobicity).

classical mechanics - What makes a Lagrangian a Lagrangian?

I just wanted to know what the characteristic property of a Lagrangian is?

How do you see without referring to Newtonian Mechanics that it has to be $L=T-V$?

People constructed a Lagrangian in Special Relativity and General Relativity. But is there a general recipe to find a Lagrangian for a theory or is a Lagrangian just chosen so that it works out?

Of course there are some symmetry and invariance properties that play an important role there, but is it possible to put the theory of Lagrangian to a more abstract and general level so that we can say a priori how a Lagrangian for a given theory must look like?

Which information are necessary to construct a Lagrangian for a physical theory?

Answer

As far as I know there is no way of rigorously constructing a Lagrangian for a new physical theory. The point is you just have to guess a Lagrangian (i.e. construct your own theory), check all the invariance/symmetry properties you want to have and hope that the predictions your theory makes will agree with the measurements.

The hard part is guessing 'correctly' - as my professor in QFT put it: So far, that 'correct guess' happenend twice, in both cases earning the authors a Nobel Prize (QED and electroweak Lagrangian).

In electromagnetic radiation, how do electrons actually "move"?

I've always pictured EM radiation as a wave, in common drawings of radiation you would see it as a wave beam and that had clouded my understanding recently.

Illustration on the simplest level:

Which obviously would not make sense (to me), as electrons would collide more likely moving as such.

For example, in a 10 meter (kHz) radio wavelength, do particles electrons move forward and back ten meters? If so, in which direction, and if in one why not any others?

What does wavelength actually have to do with its movement? Does it change the polarity, make it go in reverse or does it continue the same as others, higher frequency just means "more energy"?

Answer

In EM radiation, there are no electrons involved (well, there usually are electrons moving around in the antenna that produces the radiation, but not in the radiation itself).

So... what do these "10 meters" refer to? That's the so-called wavelength. EM radiation travels in waves, but now what does that mean? Let's first go to another type of waves: Water waves.

If you look at a bunch of waves and measure the distance of their crests to each other, you get the wavelength: The picture below shows a snapshot of a wave, and $\lambda$ denotes the wavelength.

If, on the other hand, you would stay in one place and count how often at that specific point the water rises up and down in one complete cycle and if you count the cycles per seconcd, that would give you the wave frequency.

Now, in electromagnetic radiation, what is moving up and down is not actual matter. It is just the strength of the electric and magnetic field at a particular point. Imagine you had some fancy measurement device that would tell you the strength of the electric field. Then if you'd keep it at one point in space, it would oscillate between a maximum and a minimum with a certain frequency. For radio waves, that's usually around $100 MHz$, i.e. 100 Million cycles per second.

If, on the other hand, you could record a snapshot in time of your electric field and compare how far apart two maxima are, you would obtain the wavelength.

So, what's "moving" around are the electric and magnetic fields, not actual charges. Thus, drawings of radio waves as beams of waves are accurate pictures of what's going on, unless you go to very very low intensity radio waves where you have to start thinking about the quantum nature of EM radiation...

particle physics - The decay of the strange quark via weak interactions?

The strange quark can decay via weak charged currents, following the rule that if $|\Delta S|=1$ then $\Delta S=\Delta Q_H$ where $Q_H$ is the hadron charge. The source below states that an $s$-quark can only decay into 'up-type quarks' why is this so? and can a similar thing be said about $\bar s$?

References

(1) Pal, P.B., 2014. An introductory course of particle physics. Taylor & Francis. (p502, link to Google books)

thermodynamics - Poincare recurrence time of the Universe

I've read around a bit, and it seems to be universal that the notion of a Poincare recurrence time for the universe exists. And it seems to be debated that the universe can be given an entropy, as it is contested whether or not it can be viewed as a closed system.

Something about this argument makes me think that the notion of a Poincare recurrence time for the universe also can't exist. Would this depend on the universe being a closed or static system?

quantum mechanics - How does entanglement connect particles with other particles?

How would this work with the ideas of physic we already understand? I would like to see the mathematics written out and please don’t send me to an unfamiliar website to read.

Answer

Rchter65 has stated the fact with few words.

For quantum mechanical states, and particles in this sense are quantum mechanical entities,their behavior in space time is described by solutions of quantum mechanical equations (and by special relativity of course).

These solutions, called wave functions are functions of the fourvectors of the particles , i.e. energy and momentum and of the quantum numbers describing the particles, as is spin. Conservation of quantum numbers is mainly what the term "entanglement" carries in quantum states.

For example, take the $π^0-> γ +γ$ . The $π^0$ has spin 0 in its wavefunction. The wavefunction after the decay for the two $γ$ from conservation of angular momentum has to have the gammas emitted with opposite spins ( a $γ$ has spin + or -1 to its direction of motion). When you measure the spin of one $γ$ you immediately know the spin of the other,because they are entangled with their wavefunction describing the decay of the $π^0$ and because of conservation of angular momentum.

So it is all in the mathematics of wavefunctions and the conservation laws and quantum numbers.

Please note that this is not special to the mathematics of wavefunctions. The same logic can apply in every day situations. Take a pair of twins, John and Harry and you know that one of them works in London and the other in New York. If you meet for work with John in an office in London, you know immediately that the one in New York is Harry.

fluid dynamics - What causes acceleration of particles in the expansion section of a De Laval nozzle?

A De Laval nozzle has a compression section, where the propellant is compressed (and thereby accelerated) as it moves towards a narrow section (the throat). After the throat, the nozzle widens out again. Here's a reference image.

According to what I've been reading, the propellant is moving at subsonic speeds before the throat, and is accelerated to supersonic speeds as it passes the throat. Furthermore, the propellant continues to accelerate as the nozzle expands again. What I don't understand is: what force continues to accelerate the propellant after it passes through the throat? I've been trying to imagine a single molecule of propellant and thinking about the forces from the particles around it, but can't come up with a force "pushing" from upstream (because it's moving at supersonic speeds), or "pulling" from downstream (because at the minimum there would be 0 opposing force in a vacuum, but certainly not a negative attractive force from downstream). Any ideas?

optics - Does switching a laser on create multiple frequencies?

I have been fascinated by a Sixty Symbols YouTube video on the topic of the maximum bandwidth that can be sent down an optical cable. In summary, we are informed of how a laser, which emits red light of one frequency, only does so when it is permanently on. Turning it on or off causes nature to create new frequencies which somehow facilitate the wave being there when it's on, and not when it isn't.

Since I'm a musician, I can place a loudspeaker in a medium such as air, and connect it to a sine wave generator in series with a switch. When the switch is open, no sound is generated, but close the switch and the loudspeaker outputs a sine wave. If I were to close the switch precisely when the phase was zero, surely there are no added frequencies other than the one in question which "turns the sound on", and likewise for "turning it off".

I think that this would not be the case for a sine wave at any other phase, as a 'pulse' or 'jolt' is needed to force the loudspeaker into the correct position when opening or closing the switch. I believe this causes the loudspeaker itself to generate odd harmonics for a brief period of time, as though it were being driven by the rising edge of a square wave. (That click/snap sound when attaching DC to a loudspeaker, also known as a Dirac delta function?)

I do not understand how this fits into using a laser of one frequency and sending 'pulses'. One comment on the video seemed to contradict it by asking what would happen if a filter was applied to the laser for its exact frequency, and whether that would cause the filter to always be letting light through like the laser was on, even if it was actually off.

It sounds like he's trying to describe sending the kind of pulse analogous to a square wave of sound which requires you to drive the transducer with multiple frequencies. It must be easier to detect a pulse if the amplitude ramps up quickly, but surely you're definitely not using a one frequency laser anymore to achieve that.

Q: Please can you shed some light onto how a laser being 'pulsed' actually works with respect to bandwidth used? And is the analogy with sound valid?

Answer

If I were to close the switch precisely when the phase was zero, surely there are no added frequencies other than the one in question which "turns the sound on", and likewise for "turning it off".

I made more or less this same statement to the professor in one of my undergraduate EM classes.

He gently reminded me that this zero crossing switched on/off sinusoid is equivalent to the product of a genuine sinusoid (over all time, never stopping etc.) and a boxcar function.

As soon as he made that statement, I knew how to go the rest of the way since I had already studied Fourier transforms and, in particular, that (1) the Fourier transform of the product of two time domain signals is the convolution of their respective frequency domain representations and (2) the Fourier transform of the boxcar function extends over all frequencies,

Further, and simply put, a time domain signal with compact support (non-zero over a finite time) is necessarily non-zero over all frequencies. See, for example, this answer at the Signal Processing Stack Exchange site:

In other words, a (nonzero) time-limited signal cannot be also band-limited. In other words, a function and its (continuous) Fourier transform cannot both have finite support.

operators - Wigner-Weyl ordering in exponential

If the particle number is $\hat{a}^\dagger\hat{a}\leftrightarrow|\alpha_w|^2-1/2 $, it can be mapped on the Wigner fields by assuming symmetric ordering:$|\alpha_w|^2\leftrightarrow\hat{a}^\dagger\hat{a}+\hat{a}\hat{a}^\dagger$.

My question is: is there a clear way to work with functions of this? I would be interested in how the parity operator

$$\hat{\Pi}=\exp(i\pi\hat{a}^\dagger\hat{a})$$

can be mapped on the $\alpha_w$ fields.

Of course, it would be possible to expand the exponential and perturbatively reorder the leading terms, but perhaps there is a better approach.

Answer

I will perorate on the R Kubo 1964 trick that generically Weyl-orders absolutely any operator systematically, albeit formally. I will rely on Ch 18 of our booklet, including the Exercise at its end, using gothic characters for operators, and mindful of the fundamental algebraic isomorphism with your oscillators, $[\hat a, \hat a ^\dagger ]=1 \leftrightarrow [i\mathfrak{p}/\hbar, ~\mathfrak{x}]=1 $. The correspondence is fleshed out in this WP page and this one.

The key point is that the c-number "Weyl symbol" kernel g(x,p) of any operator $\mathfrak {G}$, in indifferent/arbitrary ordering, is provided by the Wigner map, $$ g(x,p) =\frac{\hbar}{2\pi} \int d\tau d\sigma ~ e^{i(\tau p + \sigma x)} \operatorname{Tr~}\left ( e^{-i(\tau {\mathfrak p} + \sigma {\mathfrak x})} {\mathfrak G} \right ) \\ = \hbar \int dy~ e^{-iyp} \left \langle x +\frac{\hbar}{2}y \right | {\mathfrak G}({\mathfrak x},{\mathfrak p}) \left | x-\frac{\hbar}{2}y \right \rangle . $$ The Weyl symbol is then pluggable into the Weyl map formula (the inverse of the above!) which defines symmetrized Weyl order, $$ {\mathfrak G}({\mathfrak x},{\mathfrak p}) =\frac{1}{(2\pi)^2}\int d\tau d\sigma dx dp ~g(x,p) \exp \Bigl ( i\tau ({\mathfrak p}-p)+i\sigma ({\mathfrak x}-x) \Bigr ) , $$ so you are done—provided you can take all traces and perform all integrals involved.

In practice, I doubt anyone uses it extensively, but it is an "in-principle Weyl-symmetrizer" undergirded by the force of theorem.

• As a lark, and cavalierly with over-all normalizations, let us evaluate the Weyl-ordering of $\bbox[yellow]{\exp (-\pi \mathfrak {xp}/\hbar)}= -i \exp \left ( \frac{-\pi}{2\hbar}(\mathfrak {xp} +\mathfrak {px}) \right )$, by utilizing its Weyl symbol, $$ g(x,p)\propto \int dy~ e^{-iyp} \left \langle x +\frac{\hbar}{2}y \right |\exp (-\pi \mathfrak {xp}/\hbar) \left | x-\frac{\hbar}{2}y \right \rangle \propto \bbox[yellow]{\delta(x) \delta(p)} . $$ (Recall that $\mathfrak{p}|z\rangle= i\hbar \partial_z |z\rangle$, trivial to prove; so the pseudo-dilatation operator merely flips the sign of the space argument of the ket, $\exp (-i\pi z\partial_z)|z\rangle=|-z\rangle$; and thus nets $\delta(x)$ in the dot product.)

Plug this into the Weyl map formula, to net the manifestly Weyl-ordered expression, $$ \bbox[yellow]{\int d\tau d\sigma \exp \Bigl ( i\tau {\mathfrak p}+i\sigma {\mathfrak x}\Bigr )} , $$ in fact, the integral of the generating function of all Weyl-ordered polynomials.

quantum gravity - What specifically does the phrase "continuum limit" mean?

I'm interested in the meaning of the phrase "continuum limit" specifically as it is often used in expressions relating to the ability of a quantum gravity theory to recover GR in the continuum limit.

I believe I understand the meaning but want to make sure I am not missing some important part of the precise definition as my intuition may be off and I have not seen it defined anywhere.

Pointers to a place where it is defined in an online resource would be appreciated. A google search just turned up many references to it being used in papers and articles and such.

Thank you.

Answer

Usually it relates to discrete models where it means "becoming less discrete".

To make this more formal and give a simple example, consider a set $$\Lambda_{\alpha}(M) := \{ x \in {\mathbb R}^d \,|\, x_i = k \alpha, \, 0 \leq k \leq \lfloor {M \over \alpha} \rfloor \}$$

and imagine there is an edge between nearest neighbors so that part of it looks like this

this can be thought of as discretization of a $d$-dimensional cube of side $M$ modelling some crystal. Now by continuum limit one understand the limit $\alpha \to 0$. In this case one would recover the original cube $[0, M]^d$. The reason this is important is that lattice is a much simpler object than a continuous space and we can compute lots of things there directly. In some cases we are then able to carry out limits and recover the original continuous theory.

---

As for your talk about GR, I think there is only one possible interpretation in the context of Loop Quantum Gravity which deals with discrete space-time. I don't know much about this theory but I'd assume that in order to recover GR one would need to perform the continuum limit of the space-time spacing.

But note that LQG doesn't mean quantum gravity in general. It is just one of proposed theories for quantum gravity and other theories (e.g. string theory) don't assume discrete space-time, so there is no continuum limit to carry out.

---

Update: David has made a very good point about the nature of continuum limit, so I decided to elaborate on the matter.

There are two semantically different concepts of continuum limit (although mathematically they are the same):

1. 
   

   One can start with a discrete microscopic theory that is already complete (e.g. lattice of crystals; or with space-time in LQG). In this case the semantics of continuum limit is getting rid of microscopic physics. Once you perform the limit, you obtain a simpler theory.

   
   


2. 
   

   But sometimes the complete theory can already be continuous (e.g. Quantum Field Theory) and one first wants to discretize it in order to be able to perform the calculations more easily. In this case when we are shrinking the scale we are actually making the discrete model a better approximation to the original theory until in continuum limit we recover the original theory completely. So here we also lose the microscopic degrees of freedom but we don't care because the discrete model wasn't physical anyway -- it was just a mathematical tool.

   
   

general relativity - Two "Robertson-Walker observers," velocity of baseball as seen by second observer right before it's caught?

The spacetime metric of a spatially flat ($k = 0$) radiation dominated FLRW universe is given by$$ds^2 = -dT^2 + T[dx^2 + dy^2 + dz^2].$$Consider two "Robertson-Walker observers," i.e., observers with $4$-velocity $(\partial/\partial T)^a$. At time $T = T_1$, the first observer throws a baseball toward the second with velocity $v_1$. The baseball is caught by the second observer at time $T = T_2$.

Now, I am wondering, what is the velocity, $v_2$, of the baseball as seen by the second observer just before it is caught?

Note that $v_1$ and $v_2$ are the physical velocities of the baseball (as would be measured, e.g., by a "radar gun"), not a "coordinate speed" (such as "$dx/dT$"). We are not assuming here that $v_1$, $v_2 \ll c$.

Answer

Throughout the question I will use $p(T_1)$ and $p(T_2)$ to denote the 4-momentum of the baseball at times $T_1$ and $T_2$, $\mathbf{v}_1$ and $\mathbf{v}_2$ to represent the spatial component of its physical velocity, and $a(T_1)$ and $a(T_2)$ to represent the scale factor of the Universe at these times.

The homogeneity and isotropy of the Universe mean that no matter what direction the baseball is thrown in by a comoving observer, it will follow a geodesic in FRW spacetime, which is a 'radial' trajectory in the sense that

\begin{equation} ds^2 = -dT^2 + a^2(T) \: d\chi^2, \end{equation}

and

\begin{equation} \dot{p}_{\chi} = 0, \end{equation}

where $\chi$ is the FRW radial co-ordinate such that $d\chi = dr/\sqrt{1-Kr^2}$ for comoving curvature $K$, and $p_{\chi}$ is the component of the baseball's 4-momentum in this direction. The dot denotes the derivative w.r.t. proper time.

Mathematically, this condition on $p_{\chi}$ can be seen by lowering indices on the geodesic equation $\dot{p}^a+\Gamma^a_{bc}p^bp^c=0$ and relabelling dummy indices to obtain

\begin{equation} \dot{p}_a = \frac{1}{2}(\partial_a g_{bc})p^bp^c. \end{equation}

Since the metric here is independent of $\chi$, we see that $p_{\chi}$ is constant along the geodesic.

Intuitively, since the Universe is expanding away from every point, it is expanding away from observer 1 in all direction, so all directions correspond to throws along a radial trajectory.

With this knowledge, we want to formulate the problem in terms of covariant components of the momentum, so we will use the appropriate line element for a massive baseball,

\begin{equation} g^{\mu \nu}p_{\mu} p_{\nu} = -m^2 = -p_T^2(T_1) + \frac{1}{a^2(T_1)} p_{\chi}^2 \end{equation} \begin{equation} -m^2 = -p_T^2(T_2) + \frac{1}{a^2(T_2)} p_{\chi}^2. \end{equation}

The mass is not low-velocity, so using the special-relativistic mass-shell condition $E^2 = m^2+|\mathbf{p}|^2$, we get

\begin{equation} m^2 = p_T^2(T_1) - |\mathbf{p_1}^2| \end{equation}

\begin{equation} m^2 = p_T^2(T_2) - |\mathbf{p_2}^2|. \end{equation}

Substituting these $m^2$ into the line element, cancelling the $p_T^2$ and taking the ratio of the two equations then gives \begin{equation} \frac{|\mathbf{p_2}^2|}{|\mathbf{p_1}^2|} = \frac{a^2(T_1) p_{\chi}(T_2)}{a^2(T_2) p_{\chi}(T_1)}. \end{equation}

But as previously discussed, the $p_{\chi}$ are conserved along the geodesic, and so they cancel! Finally, since the mass is conserved, we can write the spatial momenta in terms of the spatial velocities as

\begin{equation} \frac{\gamma_1 |\mathbf{v}_1|}{\gamma_2 |\mathbf{v}_2|} = \frac{a(T_2)}{a(T_1)}. \end{equation}

This gives $|\mathbf{v}_2|$ in terms of $|\mathbf{v}_1|$ as required.

This picture of the time-sliced Universe should help to visualise the situation. The red lines are the comoving observers, the blue line is the trajectory of the baseball, and the black arrows are the spatial components of the velocity of the baseball at times $T_1$ and $T_2$.