- #1
darkrystal
- 4
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Something strange showed up while I was thinking about the following problem: I'd like to prove, by the only mean of the principle of maximum entropy, that the thermodynamic destiny of a block sliding on an infinite horizontal plane is to stop and come to thermal equilibrium with the plane itself.
Let [tex]T_a, T_b, v, F[/tex] be respectively the temperature of the block, of the plane, the speed of the block and the friction force.
Now the first principle of thermodynamics tells us that
[tex]\displaystyle dQ = T_a dS_a = dU_a + dL = C_v dT_a + m v dv + F dx[/tex]. Now, solving for [tex]dS_a[/tex], we find that [tex]\frac{\partial S_a}{\partial v}=\frac{mv}{T_a}[/tex] and [tex]\frac{\partial S_a}{\partial T_a}=C_v[/tex]
Oddly enough, I then noticed that [tex]\frac{\partial^2 S_a}{\partial T_a \partial v}=-\frac{mv}{T_a^2}[/tex] and [tex]\frac{\partial^2 S_a}{\partial v \partial T_a}=0[/tex], which is something we wouldn't expect because of Schwarz's theorem.
If one goes on mindlessly, integration (which is something we couldn't actually do, because of the problem with mixed partial derivatives) gives
[tex]\Delta S=C_v \log \left( \frac{T_a}{T_0} \right)+C_b \log \left( \frac{T_B}{T_{0B}} \right)+\frac{m}{2} \left(\frac{v^2}{T_a}-\frac{v_0^2}{T_0}\right)[/tex]
which is maximum for [tex]v=0, T_a=T_b[/tex], as I'd like to prove...
So, what am I missing here? I'm sure it's something stupid, but at the moment I can't really see it.
Thank you all!
Let [tex]T_a, T_b, v, F[/tex] be respectively the temperature of the block, of the plane, the speed of the block and the friction force.
Now the first principle of thermodynamics tells us that
[tex]\displaystyle dQ = T_a dS_a = dU_a + dL = C_v dT_a + m v dv + F dx[/tex]. Now, solving for [tex]dS_a[/tex], we find that [tex]\frac{\partial S_a}{\partial v}=\frac{mv}{T_a}[/tex] and [tex]\frac{\partial S_a}{\partial T_a}=C_v[/tex]
Oddly enough, I then noticed that [tex]\frac{\partial^2 S_a}{\partial T_a \partial v}=-\frac{mv}{T_a^2}[/tex] and [tex]\frac{\partial^2 S_a}{\partial v \partial T_a}=0[/tex], which is something we wouldn't expect because of Schwarz's theorem.
If one goes on mindlessly, integration (which is something we couldn't actually do, because of the problem with mixed partial derivatives) gives
[tex]\Delta S=C_v \log \left( \frac{T_a}{T_0} \right)+C_b \log \left( \frac{T_B}{T_{0B}} \right)+\frac{m}{2} \left(\frac{v^2}{T_a}-\frac{v_0^2}{T_0}\right)[/tex]
which is maximum for [tex]v=0, T_a=T_b[/tex], as I'd like to prove...
So, what am I missing here? I'm sure it's something stupid, but at the moment I can't really see it.
Thank you all!