Recent content by Mute

Perhaps you missed Synaptic say (Emphasis mine). So, what exactly wasn't there that I read and responded to? Synaptic has been arguing that China's government is evil and the US, by contrast, is a "a monument to quintessential ethics and virtue". While I still agree that the US has a much...

Of course. But the tone of the discussion thus far is that China is the epitome of EVIL and the US is a shining example of virtue and ethics, which it's not. The US has a much, much better domestic human rights record than China; I certainly don't argue against that. However, the topic of...

American media does talk about how great the US is. American textbooks are accused of being or found to be inaccurate all the time, often portraying a very US-centric view of history and justification of the US's actions. The government does control which information comes out to the public...

Many Americans disagree with you. Many Chinese might feel the same way about the Communist Party's control over China. One might claim those Chinese citizens have been brainwashed by propaganda into feeling that way, but how can you prove that you have not similarly been brainwashed?

Well, given recent events, we may need to rethink some of those assumptions. The situation isn't really new. China has already been spreading its influence. So far, it doesn't seem to be trying to undermine democracy around the world. The Communist party seems to be primarily interested in...

Would the proofs be at all accessible to another reader if you didn't fill in the gaps in explanations of the logic and explaining this or that step? A paper that no one understands is not much better (in terms of communicating research) than no paper at all. If your role is making that...

Note also that you write the indefinite integral as $$\int dx~x\cos(bx) = \frac{bx\sin(bx) + \cos(bx) - 1}{b^2} + C'.$$ You can do because the factor I introduced, ##-1/b^2##, is just a constant which could be re-absorbed into C'. If you now take the limit as ##b \rightarrow 0##, you will find...

g(z) is the initial condition; integrating G(z;z') against f(z') should give the solution in this case, although... Is z the only variable in this DE? How can the initial condition u(0) = g(z) depend on z?

That really depends on you. I took a 3 hour differential equations class in the evening once, and it was fine, but that's me. Any three hour lecture can be tough just because they're so long, but that doesn't mean you won't learn anything. You also usually get a short break halfway through. Try...

Thermodynamic laws such as ##PV = nRT## are only valid in a statistical sense, when there are a large number of particles in the system (because fluctuations in the pressure, volume or temperature are small when there are a large number of particles). In the situations you describe, there are...

If ##x## is the distance from the leftmost edge of the box to the movable wall, then the volume of the gas should just be ##V = Ax##, where A is the surface area of the movable wall. Then, the formula you are given would be ##S \sim N \ln(x T^{3/2})##. If that's the equation the problem gives...

This sounds like the most reasonable interpretation to me. The others that I talked to had the same intuition that ##\dot{p}## was in some sense the more fundamental quantity, but I agree that it probably encapsulates more than just the "force" on the particle, hence why it doesn't boil down to...

The information about the wall's position is contained in the volume in the equation for the entropy. That is, the volume of the gas will be the surface area of the wall times a linear length which contains the variable x. Your attachment doesn't specify what x is measured relative to, so I...

I edited the post to also present the discussion in terms of Fourier transform solution methods, which may make the issue a bit clearer. I hope that also helps!

What's going on, basically, is an abuse of notation in which ##\frac{1}{1-\frac{\partial}{\partial t}}## is a shorthand for the inverse operator ##L^{-1}= \left(1-\frac{\partial}{\partial t}\right)^{-1}##, where ##L = 1 - \frac{\partial}{\partial t}##. Under certain conditions, you can write...