quantum field theory - Is the world $C^infty$?

While it is quite common to use piecewise constant functions to describe reality, e.g. the optical properties of a layered system, or the Fermi–Dirac statistics at (the impossible to reach exactly) $T=0$, I wonder if in a fundamental theory such as QFT some statement on the analyticity of the fields can be made/assumed/proven/refuted?

Take for example the Klein-Gordon equation. Even if you start with the non-analytical Delta distribution, after infinitesimal time the field will smooth out to an analytical function. (Yeah I know, that is one of the problems of relativistic quantum mechanics and why QFT is "truer", but intuitively I don't assume path integrals to behave otherwise but smoothing, too).

Answer

This is a really interesting, but equally beguiling, question. Shock waves are discontinuities that develop in solutions of the wave equation. Phase transitions (of various kinds) are non-continuities in thermodynamics, but as thermodynamics is a study of aggregate quantitites, one might argue that the microscopic system is still continuous. However, the Higgs mechanism is an analogue in quantum field theory, where continuity is a bit harder to see. It is likely that smoothness is simply a convenience of our mathematical models (as was mentioned above). It is also possible that smooth spacetime is some aggregate/thermodynamic approximation of discrete microstates of spacetime -- but our model of that discrete system will probably be described by the mathematics of continuous functions.

(p.s.: Nonanalyticity is somehow akin to free will: our future is not determined by all time-derivatives of our past!)

How is it possible that consciousness-causes-collapse interpretations of QM are not falsified by the Quantum Zeno effect?

As I understand it, consciousness-causes-collapse (CCC) theories, although not very popular among physicists, have not been falsified (e.g. https://arxiv.org/abs/1609.00614).

This confuses me because my understanding of wavefunction collapse is that, at least some of the time, it must happen without a conscious observer present. The quantum Zeno effect, for instance, involves frequently "measuring" a radioactive element and thus preventing it from decaying. Each of the "measurements" in a quantum Zeno experiment are done by the measurement device (pulses of UV light).

While it is the case that no observer will become aware of these measurements until someone is conscious of them, it is still the case that a whole succession of collapses have occurred between conscious observations. This succession of collapses have had a measureable effect on the time evolution of the radioactive element, and the system would look different depending on whether they occurred or not.

My question is then: how do you maintain CCC theories on light of this? Doesn't this mean a single conscious measurement must be able to collapse a whole chain of multiple dependent collapse events far into the past? Or can it still be maintained as a single collapse at the moment of "measurement"? Or am I completely off base?

Thermodynamics: heat transfer

I hope this question isn't too simplistic but I've been in a discussion with someone who claims that no energy is transferred between a cool object and a warm one (either by radiation or conduction) because the 2nd Law of Thermodynamics states that heat always flows from hot to cold.

My intuitive understanding is that energy actually flows in both directions, but the net flow conforms to the 2nd Law as more goes from hot to cold than vice-versa. In effect, the 2nd Law is an emergent property of many interactions. Is this the case?

EDIT: I should add that I've researched as much as possible, but no sources seem to explicitly state anything but the standard 2nd Law statement. If the answer is can be found in a resource I can be pointed towards, I'd be hugely grateful and apologise for wasting anyone's time.

Answer

To see that you are correct, look to radiative heat transfer amongst black bodies. Consider two black bodies, call them bodies A and B, arranged as flat plates facing one another. Suppose body A has a temperature $T_A$ and body B has a temperature $T_B$. Both plates are black bodies, so each radiates energy at a rate given by the Stefan-Boltzmann law: $dE/dt = A\sigma T^4$, where $A$ and $T$ are the surface area and temperature of the body in question.

Since a black body absorbs all incoming radiation, body A transfers energy to body B, even if $T_A

event horizon - Can you have a giraffe shaped black hole?

My reasoning being - lets say a rock is approaching a black hole. It would essentially stop in time for an outside observer once past the event horizon but since it would also bring along some new mass by itself, some of it should stop before reaching the event horizon, becoming the new edge of the event horizon.

Assuming that is true, if I were to feed a black hole from a single direction, it should start growing a spike, right? Going further with the same technique, you should be able to shape the black hole into a giraffe if you wanted to. Are there any flaws in this future business plan of mine?

Answer

You are quite correct that if we drop an object into a black hole and watch it fall then we'll see it freeze at the event horizon. But this freeze occurs very close to the event horizon. In fact so close that it's barely distinguishable from the horizon. So dropping things into the black hole creates only a tiny perturbation and we couldn't use this trick to build any shape significantly different from a sphere.

If we consider the simplest case of a non-rotating black hole and drop an object from a long way away then the velocity of the infalling object is given by:

$$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$

I've discussed this before, in Will an object always fall at an infinite speed in a black hole?, and borrowing the graph from that post the velocity as a function of distance looks like:

Note that:

1. 
   

   the infall velocity peaks at about three times the event horizon radius

   
   


2. 
   

   the peak velocity is about $0.385c$ or about $115,000$ km/sec

   
   

Integrating equation (1) to get the distance as a function of time is rather messy, but we can do a quick back of the envelope calculation. If we take a Solar mass black hole then the event horizon is at about $3$ km so the peak velocity is at $9$ km. That means the infalling object is only $9$ km away and moving inwards at $115,000$ km/sec, so you'll appreciate that it's going to cross most of the $6$ km towards the event horizon pretty quickly. In fact if I do a quick and dirty numerical integration I get the following graph for time taken as a function of distance:

The infalling object gets to within 1% of the event horizon radius in less than a millisecond.

This is the problem with your idea. Even though strictly speaking we never see the objects pass through the event horizon they very quickly get so close to it that to a distant observer they appear to have merged with it. The end result is that the horizon remains effectively spherical and we can't use your idea to build interesting shapes.

This isn't just theoretical, because we have actually observed the merger of two black holes at the LIGO gravitational wave observatory. The black holes were rotating around each other not falling directly towards each other, but even so the merger was effectively complete after about $150$ ms - that is, after $150$ ms the merged object was indistinguishable from a single spherical black hole even though the two black holes technically take an infinite time to fully merge.

Photon emission and absorption by atomic electrons

Assume a photon is produced by an atomic electron making a transition down from a certain energy level to another.

Can that photon only be absorbed by another atomic electron making exactly the opposite transition?

Is there any chance that the photon could be absorbed by an atomic electron undergoing a transition with a slightly different energy difference?

homework and exercises - Change in entropy of two isolated systems merged into one system

From Statistical Physics, 2nd Edition by F. Mandl:

Two vessels contain the same number $N$ molecules of the same perfect gas. Initially the two vessels are isolated from each other, the gases being at the same temperature $T$ but at different pressures $P_1$ and $P_2$. The partition separating the two gases is removed. Find the change in entropy of the system when equilibrium has been re-established, in terms of the initial pressures $P_1$ and $P_2$. Show that this entropy change is non-negative.

I'm a little confused about a few things.

1. Is there a temperature change in this process? Intuitively, I would say no because $(T+T)/2=T$. My other guess would be that the temperature must change because we now have a third pressure, $P_3$, that is different from the pressure of the other two and also because we have increased the volume.


2. I believe this is an irreversible process, correct? Because you can't realistically separate the gasses into that which came from vessel A and that which came from vessel B.



3. 
   

   Can the change in volume simply be called $V_A+V_B$? I thought it would be that simple but thinking more about it I feel as though the change in pressure and possible change in temperature might change things.

   
   


4. 
   

   When the partition is removed, is there a heat exchange between the 2 gases? My intuition says no because heat can only flow when there is a temperature difference and in this case both vessels are at temperature $T$.

   
   

My attempt:

So all in all I need to solve $\Delta S= \int \frac{dQ}{T}$

$$\Delta S= \int \frac{dQ}{T}$$ $$=\int \frac{dE+dW_{by}}{T}$$ We know that $dE=0$ and that $dW=PdV$

$$=\frac{1}{T}\int PdV$$ $$=\frac{P\Delta V}{T}$$

This is where I'm stuck - I don't think there is a valid thing to put in for $\Delta V$ because there were 2 systems that formed into 1 bigger system. If the final system is $V_1+V_2$, then what was its previous size? $V_1$ or $V_2$? Or can I say that it was $\frac{V_1+V_2}{2}$?

Answer

The total volume of the two rigid containers does not change, so the combined system does no work W on the surroundings. The two containers are presumably insulated, so no heat Q is exchanged with the surroundings. So, from the first law of thermodynamics, the change in internal energy of the combined system is zero. Since, for an ideal gas, internal energy is a function only of temperature, the final temperature of the combined system is equal to the initial temperature of the separate systems.

The process is irreversible, but not for the reason you gave. Since the same gas is present in both containers, the system can be returned to its original state, but not without incurring a change in the surroundings, involving heat transfer.

Quarky Quanta's intuition was correct with regard to the final equilibrium pressure of the combined system, provided n is the total number of moles of gas in the two original containers.

COMPLETION OF PROBLEM SOLUTION: $$V_1=\frac{NkT}{P_1}$$ $$V_2=\frac{NkT}{P_2}$$ $$V_1+V_2=\frac{NkT(P_1+P_2)}{P_1P_2}$$ $$P_F=\frac{2NkT}{(V_1+V_2)}=\frac{2P_1P_2}{(P_1+P_2)}$$ $$\Delta S=Nk\ln{\frac{P_1}{P_F}}+Nk\ln{\frac{P_2}{P_F}}=2Nk\ln{\left[\frac{(P_1+P_2)/2}{\sqrt{P_1P_2}}\right]}$$ So the change in entropy is determined by the ratio of the arithmetic mean of the initial pressures to their geometric mean (a ratio which is always greater than 1).

What's the meaning of the Feynman propagator for the driven quantum harmonic oscillator?

Consider a quantum harmonic oscillator that is driven for a finite time by a force $J(t)$, and work entirely in Heisenberg picture. Then we may define the 'in' and 'out' vacua $$|0_{\text{in}} \rangle, \quad |0_{\text{out}} \rangle$$ to be the ground states of the Hamiltonian at early and late times. In Schrodinger picture, the 'in' vacuum corresponds to a state in the usual QHO ground state before the driving starts, while the 'out' vacuum corresponds to a state that ends up in that state when the driving ends.

In Mukhanov and Winitzki's book the retarded Green's function is defined as a matrix element between 'in' states, $$\langle 0_{\text{in}} | \hat{q}(t) | 0_{\text{in}}\rangle = \int J(t') G_{\text{ret}}(t, t') \, dt', \quad G_{\text{ret}}(t, t') = \frac{\sin \omega(t - t')}{\omega} \theta(t - t').$$ This makes perfect sense thinking semiclassically, as $\langle \hat{q}(t) \rangle$ is just the average position of the particle, given that it was at rest in the far past; that's basically the definition of what a retarded propagator is. Similarly, one can define the advanced propagator using $|0_{\text{out}}\rangle$.

Finally, Mukhanov and Winitzki define the Feynman propagator by $$\langle 0_{\text{out}} | \hat{q}(t) | 0_{\text{in}}\rangle \propto \int J(t') G_{F}(t, t') \, dt'.$$ Now, I've been searching for an intuitive understanding of the Feynman propagator for years. Typical explanations in quantum field theory speak of "negative energy solutions" and antiparticles (e.g. here) which I've always been confused about, since they don't exist in ordinary quantum mechanics (as I asked here). But above we have a Feynman propagator for an exceptionally simple non-QFT system! So if there's an intuitive explanation at all it'll be right here, but I can't quite see what the matrix element means physically.

I have two questions: first, how is this equivalent to the usual definition of the Feynman propagator, involving a particular contour choice? Second, are there intuitive words one can drape around this definition? Does it provide any additional physical insight?