Paradox: Thermodynamic equilibrium does not exist in gravitational fields

[anuttarasammyak]

I didn't evaluate the modified formula, because it is too complicated.

Obviously, your normalization constant mβ2π has to change accordingly – it is replaced with the denominator above.

I am afraid that these properties suggest your formula as well as mine does not coincide with the thermal equilibrium.

My argument to your point:

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 tP(v0) and with different vertical velocities v0 (i.e. the time spent at 0–1 μm above the floor depends on v0). Both of these affect the particle's contribution to the velocity distribution at the floor and I think that they have to be compensated for in your integral by giving weight

We do not have to think of bouncing time period here because time average in period has been already considered in getting f(r).

time to reach maximum
T=v0g
<z>=1T∫0Tz(t)dt
=v023g

 

[Petr Matas]

We do not have to think of bouncing time period here because time average in period has been already considered in getting f(r).

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 currently at the floor, but over those in the entire volume:
$$
\langle z \rangle = \frac 1 N \int_N \langle z(v_0) \rangle \, dN,
$$ where $N$ is the number of particles in the volume.

The Boltzmann distribution for $z = 0$ is the distribution of $v_0$ for particles currently at the floor, not in the entire volume.

 

[anuttarasammyak]

The problem is averaging this average not only over the particles currently at the floor, but over those in the entire volume:

Why do you think that I have focused on the particles currently at the floor ? I used v_0^2 is to express energy of a trajectory wherever on the trajectory the ball is.

 

[Petr Matas]

Why do you think that I have focused on the particles currently at the floor ? I used v_0^2 is to express energy of a trajectory wherever on the trajectory the ball is.

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

 

[anuttarasammyak]

Where does its distribution come from?

It is coming from you. You stated that energy of bouncing-balls have same distribution spectrum as Maxwell's.

 

[Petr Matas]

It is coming from you. You stated that energy of bouncing-balls have same distribution spectrum as Maxwell's.

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 the derived spectrum isn't the spectrum of velocities $v_0$ at the bottom of the trajectory, but of velocities $v$ at altitude $z$. Velocity $v_0$ is greater than $v$. At altitude $z$ (unlike at the floor), there are no particles with $|v_0| < \sqrt{2gz}$. In other words, different altitudes have the same spectra of $v$, but different spectra of $v_0$.

 

[anuttarasammyak]

. Assuming that particles currently at the floor have thermal spectrum of velocities v0 and that they move without collisions, I derived the spectrum of velocities v at altitude z.

Recalling that for each trajectory
v0^2=v^2+2gz
this integral is exressed as

Is the spectrum
Ce^{-(\frac{1}{2}mv^2+mgz)\beta}=Ce^{-\frac{1}{2}mv_0^2\beta}
specified by total energy of the trajectory $v_0^2$ ?

 

[Petr Matas]

Is the spectrum
Ce^{-(\frac{1}{2}mv^2+mgz)\beta}=Ce^{-\frac{1}{2}mv_0^2\beta}
specified by total energy v_0^2 ?

Yes, but the phase space has two dimensions ($z$ and $v$), so we must integrate over both,
$$
\begin{align}
\langle z \rangle &= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, z \nonumber \\
&= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, \langle z(v_0) \rangle, \nonumber
\end{align}
$$ or compensate for it with the weight.

 

[Petr Matas]

Yup, that's peace for ya. Relativity

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

 

[anuttarasammyak]

Yes, but the phase space has two dimensions (z and v), so we must integrate over both,
<br /> \begin{align}<br /> <br /> \langle z \rangle &amp;= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, z \nonumber \\<br /> <br /> &amp;= C \int_0^h dz \int_{-\infty}^{+\infty} dv \, e^{-\frac{1}{2}mv_0^2\beta} \, \langle z(v_0) \rangle, \nonumber<br /> <br /> \end{align}

=C \int_0^h dz \ e^{-mgz\beta} \int_{-\infty}^{+\infty} dv \ e^{-\frac{1}{2}mv^2\beta} &lt;\mathbb{z}(\sqrt{v^2+2gz})&gt;
which is equivalent to my

this integral is exressed as
\sqrt{\frac{m\beta}{2\pi}}\int_{-\infty}^{+\infty}dv\int_0^h dz \ f(\frac{2gh}{v^2+2gz})e^{-(\frac{1}{2}mv^2+mgz)\beta}
where integration wrt z and integration wrt v are not separatively done.

Now we get the same result which is different from thermal equilibrium.

 

[pines-demon]

=C \int_0^h dz \ e^{-mgz\beta} \int_{-\infty}^{+\infty} dv \ e^{-\frac{1}{2}mv^2\beta} &lt;z(\sqrt{v^2+2gz})&gt;
which is equivalent to my

Now we get the same result which is different from thermal equilibrium.

Now that the problem is solved with statistical mechanics, one could check if we can exhaust the methods and see if one gets the same solutions with Langevin equations of with Fokker-Planck equation.

 

[DrDu]

The problem is also how you define temperature. In thermodynamics, systems in thermal equilibrium are defined to have the same temperature. However, this is not the temperature you measure, say, with a mercury thermometer which is the "local temperature". Hence you cannot run a carnot engine between two states at different heights, if they are in thermal equilibrium, although they have different local temperatures.

 

[anuttarasammyak]

The problem is also how you define temperature. In thermodynamics, systems in thermal equilibrium are defined to have the same temperature. However, this is not the temperature you measure, say, with a mercury thermometer which is the "local temperature". Hence you cannot run a carnot engine between two states at different heights, if they are in thermal equilibrium, although they have different local temperatures.

Maxwell refferred in post #26 says that temperature is same at high and low positions. What you mention is the progress after him ?
I find the relation in (27.2) of Landau-Rifshitz Statistial Mechanics
T\sqrt{g_{00}}=const.
So in thermal equillibrium
T_{@high}&lt;T_{@low}

 

[DrDu]

I would call $T\sqrt{g_{00}}$ the thermodynamical temperature and $T$ the local temperature. thermodynamical temperature is a consequence of the zeroth law of thermodynamics (transitivity of thermal equilibrium). There is no reason to think that it does not hold in gravitational systems. The situation is comparable to the concept of mass, given your interest, in special relativity there is rest mass, longitudinal mass and transversal mass. All three concepts coincide in the non-relativistic Galilean limit. So in gravitational fields, it might be useful to introduce several temperature concepts. This is ok, so long as you make clear which one you are using.

 

[anuttarasammyak]

I would call $T\sqrt{g_{00}}$ the thermodynamical temperature and $T$ the local temperature. thermodynamical temperature is a consequence of the zeroth law of thermodynamics (transitivity of thermal equilibrium). There is no reason to think that it does not hold in gravitational systems.

It resembles with
\omega\sqrt{g_{00}}=const.
along world line of a photon which means red shift of going up light. They have the same reason.
Photon gas in vertical cylinder at equilibrium show blueish at bottom end and reddish at top end.

 

[Demystifier]

The temperature involving the factor $\sqrt{g_{00}}$ is known as the Tolman law. I have written a short insight article about it: https://www.physicsforums.com/insights/tolman-law-in-a-nutshell/