swansont

When m goes to infinity, the mass of the gas is irrelevant. The CoM will be the same, no matter what happens to the gas. It’s a non-issue for this explanation

The collisions are elastic and the box mass is infinite. Ideal system, remember? If there is no displacement, there is no work. These collisions would increase, not decrease, the volume of the box, so no.

When the ball is traveling, the CoM is moving, so this isn’t necessarily a constraint but it’s an idealized system (we start with an ideal gas) so the container has a mass >> the mass of the gas. Effectively infinite mass. edit: xpost with Ghideon

V doesn’t change. No, that does not follow Start with one ball, with some energy E, under the conditions of an ideal gas, in some box with rigid walls. Obviously, it can be anywhere in the box. Same thing for two. Wile they will occasionally collide and exchange energy and momentum, there is no mechanism that requires one ball be on each side. Two balls being on one side does not require a change in volume. Three balls, four balls - still no mechanism. Any conclusion about their average position is statistical. It’s like a coin toss of multiple coins - it becomes less likely to get all heads, or all tails, but nothing prevents it. Now increase it to an arbitrary number N. At what point does a mechanism manifest that prevents all the balls being on one side? What is that mechanism?

The piston was your introduction - which violates the conditions of the problem - so this clarifies nothing.

A piston means V isn’t constant. position ≠ momentum Why does it break these laws? It’s not a low entropy condition, as such. Did they say that, or is that your interpretation? Saying that doesn’t make it true. How does this allow a jump to absolute zero?

What does equilibrium mean in terms of a gas? So the parameters are not well-defined in that situation - they can show large deviations. Also note I responded to your question about the 2nd law being statistical, not the OP. The “small number” as applied to the OP is fabricated You did, though. Your gas became regularly-spaced, with no relative motion. It’s true there’s no path to get to that state, but AFAIK, nobody is claiming that state exists They are not “representative” if the videos you are decrying do not use them. You are trying to use a response to one very specific question and apply it to a broader question, which does not necessarily share the same assumptions. Do you want to discuss the issue you brought up in the OP, fine - do that. You want to discuss issues arising in stat mech, fine - open a new thread and do that. Don’t mix them.

Who said they were non-equilibrium? You didn’t indicate N=42 before, and regularly-spaced is a new addition, as well as zero relative motion. Where did these come from? You can’t go changing the parameters like that. Wikipedia articles like this are more like a textbook, but there’s more in the link. Not sure why you are thinking the system is low entropy.

It comes from statistical mechanics “Statistical mechanics postulates that, in equilibrium, each microstate that the system might be in is equally likely to occur, and when this assumption is made, it leads directly to the conclusion that the second law must hold in a statistical sense. That is, the second law will hold on average, with a statistical variation on the order of 1/√N where N is the number of particles in the system. For everyday (macroscopic) situations, the probability that the second law will be violated is practically zero. However, for systems with a small number of particles, thermodynamic parameters, including the entropy, may show significant statistical deviations from that predicted by the second law. Classical thermodynamic theory does not deal with these statistical variations.” https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Statistical_mechanics

Or, maybe we don’t compare police and the military - especially the highly-specialized, elite forces within the military. We do NOT expect a given police officer to manage violent confrontations on a daily basis. Over the scope of all police, a few will be faced with a violent situation on a given day, but there are ~800,000 police officers in almost 18,000 departments. Special forces in the military are less than 1/10 of that number. It’s unreasonable to expect to train that many people, especially absent the selection criteria we have for special forces that the police lack. If you want these to be closer to analogous, you need to select a subset of police for this training.

Is this accurate? Are non-supervisory police on salary, or are they wage employees, who make a lot of their money on overtime (time and a half)? Not sure it’s fair to say we “expect” them to work long hours. Also, on the list of dangerous jobs, there are only a couple that pay much more than policeman (on average). Most pay significantly less https://www.monster.com/career-advice/article/high-paying-dangerous-jobs

I think the US police problem is also linked to our incarceration problem, another difference between the US and Europe. (which may be in that video, which I can't watch ATM)

That's the whole point of "defund the police" - they are not trained for many of the jobs they are being asked to do, and money would be better spent on people trained to do those jobs. And we run into problems when people trained in the use of guns are asked to do jobs where guns are way down on the list of solutions to the problem. (When all you have is a hammer, and all that). You use the money to hire people trained for these other tasks. That way, fewer police are needed. Your bit about specialization points to this. I was reading recently how this has already happened. Once upon a time police were summoned for medical emergencies, and transported people to the hospital. But police are not trained to deal with medical emergencies. When we replaced police and police cars with EMTs and ambulances, we got a better result. https://twitter.com/JamieFord/status/1272273637173637120 "Defund the police" is applying that same rationale to other aspects of what is now police work. Don't send untrained people with guns into situations where guns won't help, and might make it worse. Send people trained in that particular situation. (BTW, I do more than 5 hours of training per year as a federal employee. If police aren't doing more than 5 hours then that's a broken system)

You don’t have to stop counting until you bring them back together. No timing required, though that’s not a big issue (if you do a measurement that lasts a day, who cares if you have a few nanoseconds of dilation. It has a negligible effect on the answer, if you do the experiment properly). You start and stop with the two systems next to each other. Yes, this has been done with clocks. Tom van Baak’s version was to bring some clocks up on Mt Rainier for few days while the family went camping, and compare to the clocks left at home. http://www.diyphysics.com/2012/03/15/tom-van-baaks-family-friendly-relativistic-time-dilation-experiment/ (I think someone linked to this recently, but I don’t recall which thread)

How does the rate of radioactive decay depend on Planck’s constant? Why not address the experiment I described, instead of trying to come up with a more complicated experiment?

We’ve done the equivalent experiment with clocks, several times.

That’s not how clocks work. That’s not even how these measurements work - we don’t care about this “rectangular pulse” We just count the decays. Discrete values. We have a radioactive sample. We measure 1000 dps at the reference system. Now we move the clock up such that the frequency changes, according to your theory, to 998 dps. GR predicts 1001 dps. Let the system sit there for an hour. Then we compare to the reference. From what you’ve said, your prediction is 120 decays fewer, what GR predicts is 60 more.

I’m talking about an atomic clock. What are you talking about? What relevance do those drawings have? If I have a clock running at a frequency F, and I count the number of “ticks”, I measure time. Now we take a clock, running at F, to some new height, H, above the reference clock. According to your earlier post, the frequency will be lower by a relative value 2gH/c^2. I can count the number of ticks at this new height, return it to the original height and compare it to a reference clock. GR predicts a higher frequency at H, and you predict a lower one. This is easily checked and trivially falsified.

How about addressing the clock experiment I have brought up twice.

So this implies that you think these other terms are constant, but hbar and c are variable. Why is this? Why aren’t you saying charge is variable, too? Why not? The effect is on rates, so one clock evolving at a faster rate will accumulate more phase. (i.e. time) This can be directly compared to a reference clock.

A shift in the frequency emitted by a photon means the transition in question has shifted by that amount, and you're saying the shift is in the opposite direction predicted by GR (you say the lower system has a higher frequency). Is that correct? 2 problems with this: 1. If the frequency shift is predicted to be two times as large, then what happens if you just don't send a signal, so there is no photon. You move a clock to a certain height and let it accumulate a difference in time, and then move it back to the reference point, and check. No photon to worry about, and there's a factor of 2 difference in the predicted effect. Easy to observe. (Yes, the experiment has been done.) Therefore, easy to disprove the conjecture. 2. I don't think this having asymmetric shifts is consistent with the Pound-Rebka effect. You have the photon shift of one value, but the resonance of the atom has shifted by a different value.

That's what you're trying to show, so it does not "follow" in that what joigus stated is part of standard electrodynamics. The speed of light can vary for other reasons, such as from the change in index when it's in a medium. It also deviates from the invariant value if you are in an accelerated frame of reference. OK, so what's the prediction about the net effect on time? How is gravitational redshift an "illusion"? Why do we get good agreement with the GR formula if there is this other effect of opposite sign?

As joigus said, you have given the definition of the fine structure constant. You haven’t presented any connection between GR and QM. Did they measure this? Did they do it in free space? The key here is “irrefutable proof” Science doesn’t deal in proof, and is never irrefutable. They will never pay out, since they have a way out of doing so.

You can’t summarize?

The devil’s in the details. How do you do this? Why are existing experiments insufficient to confirm relativity? Where does Yanchilin's formula come from? Why is it necessarily correct if GR is correct? How do you arrive at that interpretation? They measured the magnetic field near a black hole binary, not the permeability of free space.