User3141

@Cleonis Btw, thank you for your answer to an earlier question of mine related to the anti-gravity wheel by Veritasium, it was indeed really helpful.

@Cleonis Right, I did know that, but I thought that the amount of deflection would be too small as the bob on the pendulum moved just a tiny amount east-west. But as Matteo points out, the Coriolis force is indeed small (that's why we usually don't have to think about it), but during a long time span, that does lead to some deflection.

Yeah, I found a really good pdf article on the Foucault pendulum which went through the math, and then I saw how the Coriolis force was just given by twice the cross product between omega and the velocity of an object. And as you say, this means there is a Coriolis force when something moves east-west. It's just that I thought that there would be no force in this case because all I really knew of Coriolis was the demonstration with a canon ball getting shot due north from the equator, and how that resulted in a deflection. I imagined this wouldn't happen if something got shot east-west.

@user2723984 Alright, thank you for all the help and time you've provided, and I think I understand it better now. I don't think I have the right to waste any more of your time ;)

@user2723984 Yeah, sure. So just to be clear (to maybe confirm what I thought I knew till now): the correction to the partition function (and entropy) which fixed extensivity is a purely quantum mechanical effect (derived from the assumption of indistinguishability), and cannot be explained through classical physics?

@user2723984 Alright, that was a bit better. I thought there were purely theoretical reasons for believing in indistinguishability (aside from the extensivity property of entropy, as you mentioned). Meaning that indistinguishability for an ideal gas could be "derived" from simpler principles, instead of being based on a more experimental basis.

@user2723984 But then the particles won't be fermions or bosons, I guess?

@user2723984 So it's more of an experimental fact that the gases we find are bosonic/fermionic, rather than a purely theoretical fact? I mean, surely it would have been possible to imagine a gas where the wave function is separable into single-particle wave functions, in which case the particles are distinguishable.

@user2723984 So you're saying that the particles in all ideal gases we might ever come over is automatically indistinguishable because we assume they're fermions/bosons?

@user2723984 But syrely two particles (fermions, say) don't HAVE to be indistinguishable in QM? For instance, if the combined wave function is unentangled and separable into the product of various single-particle wavefunctios, then it makes sense to talk about the identity of a single particle (hence, distinguishability).

@user2723984 I actually didn't know there were two kinds of ideal gas. So I actually can't be sure which one it is I am referring to, but yes, I think it's related to the Gibbs paradox (though I haven't really got that far yet, but it's the next topic in my stat mech book).

My point is that the second kind of indistinguishability mensioned in the article requires some special properties of the combined wavefunction (as discussed in the next section on symmetry and antisymmetry). Why do we assume that the combined wave function of the gas has these properties, ie. why do we assume the particles are completely indistinguishable?

If the link doesn't work, then it's the "Identical particles" article on Wikipedia.

Look above

I gave the link in the second comment

@user2723984 Maybe I'm just terrible at explaining what I mean, but if you go to this article and read the section ""Distinguishing between particles", then maybe you get what I mean. In particular, the second and third paragraph gets at the heart of what I mean.

"they are only characterized by their state, unless you explicitly label them". Exactly. If they are not completely indistinguishable, then it's possible to distinguish them in principle using their quantum states (or their position and momentum, in the classical case). But complete indistinguishability is established if there is no way to distinguish in this case, since the probability that particle 1 and 2 is in state 1 and 2 respectively is the same as the probability that particle 1 and 2 is in state 2 and 1 respectively.

Yes, but I am of course talking about the quantum indistinguishability, not the classical one you referred to in your last comment. As you said, the combined wave function will be given as a superposition of all the permutations of the particles in the system, when we talk about quantum indistinguishability. My question is why we assume the particles in the gas are indistinguishable in this manner, when there are other possible states for the combined system.

And surely two fermions don't have to be indistinguishable. If you have an unentangled system of a pair of electrons, then you can distinguish the particles by their single-particle wavefunctions (because of the Pauli principle, these wavefunctions must be different, and hence you can distinguish the particles).

I am not sure how this explains why the particles in the gas are indistinguishable. If indistinguishability defines the notions of bosons and fermions, then why not just assign the particles in the gas some other name/properties? In the classical model, the particles in the gas are distinguishable. You can distinguish them by their respective trajectories. To my understanding, particles are indistinguishable in QM only if their "trajectories" have mixed, such that the probability that particle 1 is at x_1 and 2 is at x_2 is equal to the probability that particle 1 is at x_2 and 2 is at x_1.

@RobJeffries Alright, I see. Btw, I found a really great pdf presentation regarding the sun's spectrum, including how a negative temperature gradient yields an absorption spectrum and a positive gradient yields an emission spectrum, and a lot that was previously unclear to me has now been cleared up. Thank you for mentioning this effect, as I was completely unaware of it before reading your answer. Your answer has also been accepted :)

@RobJeffries Thank you very much the extra information in the answer. I can't say I am familiar with this formula, but assuming it's true, I think I understood the rest. The only thing I don't understand now is why the hot chromosphere and corona isn't considered here. The chromosphere, and especially the corona, tends to be much hotter than the photosphere, in which case there should be emission lines. Shouldn't that cancel out the absorption? Or are these emission lines mostly in other regions of the electromagnetic spectrum, not in the visible range?

@RobJeffries Could you provide some reading material so that I can read more about this? I can't say I really understand how a temperature gradient solves any of this, but whenever there's something you don't understand, then it's always better to read about it from several different sources.

Doesn't the second and third paragraph contradict each other? In the second, you say that a constant temeprature means no absorption lines, and in the third you say that a constant temperature would induce an absorption spectrum.

Your answer is probably correct, as you seem to have a lot of upvotes, but I can't seem to fully understand it. Some analogies would be really great, if there is any to explain this further.

proton, as far as I have understood it so far, Bob's wave function can't collapse at the moment of Alice's measurement, because Bob doesn't know about the measurement. Therefore, as far as he is concerned, there is still a 50-50 chance that he will measure spin up/spin down. But of course, ALICE knows that Bob will measure spin down, but Bob doesn't know that.

So as far as Bob is concerned, neither Alice's nor his wave function collapses at the moment of Alice's measurement.

Charles, one thing I do find odd, though, is that your answer seems to be referring to relational quantum mechanics, not the conventional Copenhagen interpretation. But surely the Copenhagen interpretation must have something to say here too, as I can't imagine that I am the first who thought about this thought experiment, so somebody must have surely come up with a solution, right?

Yeah, I see. The only place where I have ever seen this mentioned before is in one of Carlo Rovelli's popular science books. But I haven't even seen it being mentioned in my quantum physics text book.

Charles, right, so the state of a particle only makes sense compared to some observer or object, and when I say "the wave function has collapsed", then I should really specify who it collapsed with repsect to?

Wait, so what Brian really meant is that the wave function of B collapses some time after the measurement of A compared to an observer by B?

Okay, this was an interesting answer, and I will have to think about this for a moment. But first let me just clarify: so there is actually no way of talking about the state of a particle compared to some reference frame, only the state of the particle compared to some other particular object?

I should also mention that it doesn't make sense to identify the location of particle A and the time of the collapse as the event where the collapse happened, because in the general case, the particle has no well-defined position, and its position wave function might be spread out across space.

Therefore, it can simultaneously collapse for both particles for all reference frames, without violating special relativity. This would also solve the entire problem. But if this is all true, then why did Brian Greene, a leading theoretical physicist, say that there must be a time-delay?

The important point here is that all the rules of special relativity only apply to EVENTS in spacetime, which is a collection of a location in space and a moment in time. But if the collapse has no location, then it doesn't happen in a particular event either.

What Brian seems to have done is identify the location of particle A, say, and the time of the wave function collapse as the EVENT where the collapse happened, and likewise for B, and in that case, then sure, there might be a time-delay according to the rules of special relativity.

I mean, the wave function collapse is just a name of a reduction of the spin wave function from being in the superposition UD+DU to being of the form UD or DU. But this wave function is a ket vector in Hilbert space, so it simply doesn't exist in ordinary 3D space.

Alright, but what I really wanna know at this point is what Brian Greene is referring to when he says there can be some time-delay between the collapse of the wave function of A and the one for B. I thought a lot about this yesterday, and the conclusion I got (which also solves my problem) is that the wave function collapse doesn't actually happen at a particular location in space.

So now the theory has to explain this result.

And from what I know, this isn't an "imaginary" experiment. This seems like something you could do in practise, unlike making an object float in air. So now quantum theory has to defend itself and show why this isn't a flaw in the theory, because, like you said, we can't observe in reality two entangled particles having the same spin at the same time.

Okay, Will, let me put this another way too. If somebody asks you about the twin paradox in special relativity, then I bet you wouldn't say, " this isn't a paradox, because we can observe that a twin can't be older and younger than the other twin at the same time." Instead, you would defend the theory and explain why there is no problem in it. Just like the twin paradox, my question is a doubt about the theory itself. So it is the theory that has to defend itself, not the universe.

But that isn't my question though. All I am asking is what I am missing in my reasoning in the original question. I feel like you didn't really answer that, you just stated that this never actually works (namely that you can't have that both particles end up having spin up) without telling me where I have made a mistake in my reasoning. I mean, there has got to be an explanation here.

But explain how. I mean, there's no invisible force or something stopping us from doing the experiment in practise and violate this. I would like you to take a look at Charles' answer, for I think he is into something there. What do you think about it?

Will, right, and that is exactly what I don't understand. I don't understand why this never happens. I mean, what stops us from doing the expreiment I describe and violate the correlation?

I don't know about your last question, but it still seems to me that there is a problem here. For if I measure spin up for particle B, then I would say that the spin of A is down. But if someone else has already measured A and found that it is spin up, then there is a conflict here about what I think of A and what the other one thinks of A.

But if the wave function of B hasn't collapsed, then as I say, there is still some chance that a measurement of B will yield spin up. In other words, even though the wave function of A has collapsed, the wave function of B hasn't, so B is still in a superposition. Thus it seems like this is a counterexample of your claim that "all observers will predict the same correlation".

Also, what I was trying to do, was to give an example where it seems that both particles might have spin up, so that they won't be "perfectly anti-correlated".

No, I didn't say that. It might be that the probability of getting spin up is much greater than the probability of getting spin down. My logic still applies.

I don't really see how this answers my question. You just state that different observers will make the same predictions, but you don't explain how that can possibly be in the particular example I gave.

Shay, okay that's a good point lol. So maybe what I said in the first of the two comments above is true.