Paradox: Thermodynamic equilibrium does not exist in gravitational fields
- Context: Undergrad
- Thread starter Petr Matas
- Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
135 replies · 12K views
- 23,752
- 5,953
In other words, each molecule experiences a multitude of collisions and energy transfers per unit time, which translates into significant heat conduction within the gas.Petr Matas said:
Petr Matas
- 91
- 16
I agree, but I believe that my approach is independent of the assumption about collisions.Chestermiller said:
- 37,019
- 16,030
I certainly don’t think so. How does your objection even exist if the gas molecules collide with each other?Petr Matas said:
Petr Matas
- 91
- 16
I think that in the equilibrium, the velocity distribution must be the same whether collisions among the particles occur or not. Let's assume there is a small solid object somewhere in the gas column. Every particle will hit this object from time to time. Even if the particles don't interact with each other, collisions with this object will mediate energy transfer among the particles. I think this allows me to ignore the collisions among the particles in the analysis of the equilibrium state.Dale said:
- 37,019
- 16,030
With the collisions the distribution of the velocity is uniform in direction and Maxwell-Boltzmann in speed. Can you show that is the case without collisions? I am highly skeptical. I think this is the key. Without this proof you do not have thermal equilibrium and therefore don’t have a well defined temperature in the first place.Petr Matas said:
Petr Matas
- 91
- 16
How about the energy transfer mediated by the small solid object that I mentioned? Isn't that enough to allow reaching the equilibrium (after very long time, of course)?Dale said:
Last edited:
- 37,019
- 16,030
Not that I can see. Can you derive the distribution from that? I don't think that I could.Petr Matas said:
Also, if you can derive the usual distribution, doesn't that invalidate your whole argument? Isn't your argument predicated on the distribution being different from the standard thermal equilibrium? I don't see how you can have it both ways.
- 23,752
- 5,953
Have you had a course yet in fluid mechanics or Transport Processes?Petr Matas said:
unless there is no object present.Petr Matas said:
Petr Matas
- 91
- 16
Unfortunately not. However, I am going to assume the gas is still.Chestermiller said:
I am going to assume a usual thermal distribution at certain altitude and from there I expect to conclude that it is such everywhere.Chestermiller said:
No, the standard thermal distribution is the result that I am expecting to obtain.Dale said:
- 23,752
- 5,953
Like I said, using continuum fluid mechanics and heat transfer would be much simpler.Petr Matas said:
- 35,014
- 21,725
Tell us again how you're not using an AI?Petr Matas said:
- 37,019
- 16,030
I don’t see how. Can you show it?Petr Matas said:
I also don’t see the point of adding the little block. What does it do that the walls don’t already do besides make the geometry more complicated?
Last edited:
Petr Matas
- 91
- 16
This was the second timeVanadium 50 said:
. I prefer leaving tedious but simple work to AI (while checking its answers of course) to save time for tasks which are currently beyond the AI capabilities. I mark the AI answers clearly whenever I use them. Or would you prefer me to conceal the use of AI? Or not use AI at all? What is the point?
Petr Matas
- 91
- 16
I must shut up and calculate.Dale said:
Walls suffice.Dale said:
Petr Matas
- 91
- 16
You mean using the law that heat flows from higher to lower temperature? I want to avoid using this law. Actually, I want to prove (using laws of motion) that it is valid even in gas in gravitation without assuming its validity in the first place.Chestermiller said:
- 23,752
- 5,953
Good luck.Petr Matas said:
anuttarasammyak
Gold Member
- 3,166
- 1,597
BTWDale said:
1 . When the particles are photons, they do not interact each other but only with walls.
2 . I revisited Laudau Statistical Mechanics chapter II section 27 for relativistic region (27.5)
[tex]T=constant \ (1-\phi/c^2)[/tex]
The temperature is higher at points in the body where ##|\phi|## is greater. So the things seem to change in relativity.
Petr Matas
- 91
- 16
Theorem:
A gas column in thermodynamic equilibrium in a classical homogeneous gravitational field has the same temperature everywhere.
Proof using the laws of motion:
Let at time 0 a point particle with mass ##m## is located at height 0 and its velocity is ##\mathbf{v}_0 = (v_{\text x 0}, v_{\text y 0}, v_{\text z 0})##. Its total energy is
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z 0}^2). $$
Let us assume the particle moves without collisions. At time ##t## it is at height ##z##, its velocity is ##\mathbf{v} = (v_{\text x 0}, v_{\text y 0}, v_{\text z})## and it has the same total energy
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z}^2) + mgz,
$$ where ##g## is the gravitational acceleration.
Comparison of the two equations yields
$$
v_{\text z 0}^2 = v_{\text z}^2 + 2gz, \tag 1
$$ $$
v_{\text z 0} = \pm \sqrt{v_{\text z}^2 + 2gz}. \tag 2
$$
Differentiation of equation ##(1)## yields
$$ v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}. \tag 3 $$
Let ##\rho(z, v_\text z)## is the density of particles at height ##z## with vertical velocity ##v_\text z##.
Consider at time 0 a bunch of particles at height 0 with velocities in range from ##v_{\text z 0}## to ##(v_{\text z 0} + dv_{\text z 0})##; they contribute to ##\rho(0, v_{\text z 0})##. The number of such particles, which cross a unit-area horizontal boundary at this height in unit time, is
$$ dN = \rho(0, v_{\text z 0}) v_{\text z 0} dv_{\text z 0}. \tag 4$$
At time ##t##, the same particles are at height ##z##, their velocities range from ##v_{\text z}## to ##(v_{\text z} + dv_{\text z})## and they contribute to ##\rho(z, v_{\text z})##. The same number of particles will cross a unit-area horizontal boundary at this height.
$$ dN = \rho(z, v_{\text z}) v_{\text z} dv_{\text z}. \tag 5$$
Equations ##(2)##, ##(3)##, ##(4)##, ##(5)## together yield
$$ \rho(z, v_{\text z}) = \rho\left(0, \pm \sqrt{v_{\text z}^2 + 2gz}\right). \tag 6$$
Let the velocity distribution at height 0 is Maxwell–Boltzmann (i.e. thermal) with temperature ##T##. That means the density at that height is
$$ \rho(0, v_{\text z}) = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right), \tag 7
$$ where ##C## is a constant.
From equations ##(6)## and ##(7)## we get
$$ \begin{align}
\rho(z, v_{\text z}) \nonumber
& = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2 + 2gz}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot \rho(0, v_{\text z}). \nonumber
\end{align} $$
We can see that the velocity distribution is the same at all heights ##z## (i.e. the temperature is the same everywhere) and the density decreases exponentially with height ##z##. Interaction between the particles cannot change this distribution, because they are already in equilibrium.
End of proof.
This result may be generalized to non-homogeneous classical gravitational fields by replacing ##V(z) = gz## with a generic gravitational potential ##V(x, y, z)##.
A gas column in thermodynamic equilibrium in a classical homogeneous gravitational field has the same temperature everywhere.
Proof using the laws of motion:
Let at time 0 a point particle with mass ##m## is located at height 0 and its velocity is ##\mathbf{v}_0 = (v_{\text x 0}, v_{\text y 0}, v_{\text z 0})##. Its total energy is
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z 0}^2). $$
Let us assume the particle moves without collisions. At time ##t## it is at height ##z##, its velocity is ##\mathbf{v} = (v_{\text x 0}, v_{\text y 0}, v_{\text z})## and it has the same total energy
$$ E = \tfrac{1}{2} m (v_{\text x 0}^2 + v_{\text y 0}^2 + v_{\text z}^2) + mgz,
$$ where ##g## is the gravitational acceleration.
Comparison of the two equations yields
$$
v_{\text z 0}^2 = v_{\text z}^2 + 2gz, \tag 1
$$ $$
v_{\text z 0} = \pm \sqrt{v_{\text z}^2 + 2gz}. \tag 2
$$
Differentiation of equation ##(1)## yields
$$ v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}. \tag 3 $$
Let ##\rho(z, v_\text z)## is the density of particles at height ##z## with vertical velocity ##v_\text z##.
Consider at time 0 a bunch of particles at height 0 with velocities in range from ##v_{\text z 0}## to ##(v_{\text z 0} + dv_{\text z 0})##; they contribute to ##\rho(0, v_{\text z 0})##. The number of such particles, which cross a unit-area horizontal boundary at this height in unit time, is
$$ dN = \rho(0, v_{\text z 0}) v_{\text z 0} dv_{\text z 0}. \tag 4$$
At time ##t##, the same particles are at height ##z##, their velocities range from ##v_{\text z}## to ##(v_{\text z} + dv_{\text z})## and they contribute to ##\rho(z, v_{\text z})##. The same number of particles will cross a unit-area horizontal boundary at this height.
$$ dN = \rho(z, v_{\text z}) v_{\text z} dv_{\text z}. \tag 5$$
Equations ##(2)##, ##(3)##, ##(4)##, ##(5)## together yield
$$ \rho(z, v_{\text z}) = \rho\left(0, \pm \sqrt{v_{\text z}^2 + 2gz}\right). \tag 6$$
Let the velocity distribution at height 0 is Maxwell–Boltzmann (i.e. thermal) with temperature ##T##. That means the density at that height is
$$ \rho(0, v_{\text z}) = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right), \tag 7
$$ where ##C## is a constant.
From equations ##(6)## and ##(7)## we get
$$ \begin{align}
\rho(z, v_{\text z}) \nonumber
& = C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2 + 2gz}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot C \exp\left(-\frac{m}{k_\text B T} \cdot \frac{v_{\text z}^2}{2} \right) \nonumber \\
& = \exp\left(-\frac{m}{k_\text B T} \cdot gz \right) \cdot \rho(0, v_{\text z}). \nonumber
\end{align} $$
We can see that the velocity distribution is the same at all heights ##z## (i.e. the temperature is the same everywhere) and the density decreases exponentially with height ##z##. Interaction between the particles cannot change this distribution, because they are already in equilibrium.
End of proof.
This result may be generalized to non-homogeneous classical gravitational fields by replacing ##V(z) = gz## with a generic gravitational potential ##V(x, y, z)##.
Last edited:
anuttarasammyak
Gold Member
- 3,166
- 1,597
By differentiation of (1) I got (3)Petr Matas said:
[tex]v_{\text z 0} dv_{\text z 0} = v_{\text z} dv_{\text z}+gdz. \tag 3[/tex]
Am I wrong ?
Petr Matas
- 91
- 16
At height 0, let us have a narrow velocity interval from ##v_{\text z 0}## to ##(v_{\text z 0} + dv_{\text z 0})##. The goal is to find what velocity interval at height ##z## it is mapped to. Therefore ##z## is a constant. Any ideas how to express it more clearly?anuttarasammyak said:
Last edited:
anuttarasammyak
Gold Member
- 3,166
- 1,597
Thanks. So have you conquered your losing-speed intuition by your calculation which tells same velocity distribution but different number density by height ?
Petr Matas
- 91
- 16
Exactly.anuttarasammyak said:
anuttarasammyak
Gold Member
- 3,166
- 1,597
Congratulations. Let me understand more. Say a bouncing ball goes up and down between floor and ceiling. By gravity
ball speed near floor > ball speed near ceiling
time duration ball staying in 10cm layer from floor < time duration ball staying in 10cm layer from ceiling
ball number desity in 10cm layer from floor < ball number density in 10cm layer from ceiling
This contradicts the result. Where am I wrong ?
ball speed near floor > ball speed near ceiling
time duration ball staying in 10cm layer from floor < time duration ball staying in 10cm layer from ceiling
ball number desity in 10cm layer from floor < ball number density in 10cm layer from ceiling
This contradicts the result. Where am I wrong ?
Petr Matas
- 91
- 16
There are many balls.anuttarasammyak said:
anuttarasammyak
Gold Member
- 3,166
- 1,597
Each ball has high probability to be near ceiling and the sum of them, though there is no interaction between, show high number density near floor. Very interesting to me.
Petr Matas
- 91
- 16
Another observation: Trajectories of individual particles are parabolas. A particle never appears above the top of its parabola.
Petr Matas
- 91
- 16
Another observation 2: As a bunch of particles rises and loses speed, they come closer together in space, but farther apart in velocity. These opposite effects on density cancel out, which can be seen from equation (6): It is an equality between densities of the same bunch at different heights; all multiplicative factors disappeared.
Last edited:
anuttarasammyak
Gold Member
- 3,166
- 1,597
Statistical/thermal mechanics explains that maximum entropy state of vertical column gas has throughout same temperature as Maxwell said. Newton Mechanics of bouncing balls seem to have different nature to me where energy conservation is applied individually to the balls not to the whole system and with no concept of entropy.anuttarasammyak said:
Anyway it is your thread not mine. If you have no ambiguity I should shut up.
Petr Matas
- 91
- 16
Analysis using Newton mechanics may be too difficult to apply on large ensembles, but it must give the same results as thermal laws. In our case its application was rather easy and it showed me where exactly my intuition went wrong. This allows me to attain deeper understanding.anuttarasammyak said: