Recent content by Petr Matas

I dropped the relativity altogether after finding that it does not resolve the paradox.

Yes, but the phase space has two dimensions (##z## and ##v##), so we must integrate over both, $$ \begin{align} \langle z \rangle &= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, z \nonumber \\ &= C \int_0^h dz \int_{-\infty}^{+\infty} dv \...

Let me explain. Assuming that particles currently at the floor have thermal spectrum of velocities ##v_0## and that they move without collisions, I derived the spectrum of velocities ##v## at altitude ##z##. I found that the spectrum is the same as at the floor and that the density is lower. But...

I am aware that ##v_0^2## characterizes the entire trajectory. So you have a population of trajectories. Where does its distribution come from?

I see, ## t_{\rm P}(v_0) = 2T ##, where ##T## is the time to reach maximum. It seems to me that you incorporated it correctly into your formula for average height of individual particle ## \langle z(v_0) \rangle = h f(r) ##. The problem is averaging this average not only over the particles...

I didn't evaluate the modified formula, because it is too complicated. The approach ##\rho(0, v_0) \mapsto \rho(z, v)## stands on the same assumptions (i.e. non-interacting bouncing balls) as your approach, it is easier to apply and it leads to a result in accordance with equilibrium. Although...

Maybe I have found a mistake. If I read this expression correctly, you are integrating over the particles, which are currently at the floor. I agree, that every particle bounces off the floor regularly, but they do so with different time periods ##t_{\rm P}(v_0)## and with different vertical...

I think that ##m## is missing at the end of the formula. Hamiltonian of a point mass in homogeneous gravitational field is $$H({\bf x},{\bf p})=\frac{p^2}{2m}+mgz.$$

@anuttarasammyak: Why are you taking such a complicated approach? Doesn't post #104 answer your question?

I see, no problem. Thinking about this property, there is also a related quantity called mean free path.

You're right. The density ##\rho(z)## as a function of altitude ##z## will be the same up to a multiplicative constant. We should just remember that we are talking about an ideal gas. Did you mean many collisions between particle and particle? Exactly.

I meant something else: Interactions push the system towards thermodynamic equilibrium (they only don't for identical particles in 1D, as you have shown) by redistributing energy. If the system has not reached equilibrium yet, then the interactions will be important. In my analysis, I assume...

That means that in 1D, particle interactions with each other cannot redistribute the kinetic energy and thus equalize the energy spectrum and thus make the system evolve towards the thermodynamic equilibrium. Only collisions with the floor or ceiling will be able to do that. The situation is...

Let us continue. The particle mass $$ m = M / N_{\rm A}, $$ where ##M## is the molar mass, ##N_{\rm A} \approx 6.022 \times 10^{-23} \, \text{mol}^{-1}## is the Avogadro constant. $$ \beta m = \frac {M / N_{\rm A}} {k_{\rm B} T} = \frac {M} {R T} $$ where ##R = N_{\rm A} k_{\rm B} \approx 8.314...

Let us have a vertical column of ideal gas. ##\rho(z)## is the density of particles at altitude ##z##, ##\beta = \frac {1} {k_{\rm B} T}## is the thermodynamic beta, ##k_{\rm B}## is the Boltzmann constant, ##T## is the thermodynamic temperature, ##m## is the particle mass, ##h## is the column...