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Paradoxes and contradictions

Anonymous Wed 13 May 05:25:49 2020 No.11666434 View Reply Original Report

Quoted By: >>11666441 >>11666465 >>11666478

tl;dr:
Ask about and discuss paradoxes and contradiction in any STEM field.

I'm an Ms.c. student in physics and sometimes find myself asking questions that have a contradictive or paradoxically nature to them.

Some have an answer such as:
>In fluid mechanics, you pull a piston away from a one-dimensional gas, a rarefaction wave starts and spread throughout the gas. This is an isentropic process, the entropy is constant.
>In statistical mechanics when we pull a piston and the gas start s to expand, we know that the final state has a larger entropy.
Both statements have an expanding gas, but with different outcomes. How is this possible? I will give the answer if people would like.

Anonymous

Anonymous Wed 13 May 2020 05:28:58 No.11666441 Report

Quoted By: >>11666451

>>11666434
>one-dimensional gas
No such thing.

Anonymous

Anonymous Wed 13 May 2020 05:30:27 No.11666448 Report

Quoted By:

Some don't:
>To ensure an extensive entropy in a classical ideal gas and solve Gibbs paradox we present particle permutation symmetry i.e. identical particles. The Hamiltonian is invariant under this symmetry so in the original calculation (with $Z_N=Z_1^N$ ) we over-count the number of states. When adding a factorial denominator ( $Z_N=Z_1^N/N!$ ) to the overall partition function we get an extensive entropy (for large number of particles).
>We look at a set of harmonical oscillators with the Hamiltonian
$H=\sum_i\frac{p^2_i}{2m}+\frac{m\omega^2(x_i-a_i)}{2}$
>Here, the Hamiltonian is not invariant to particle permutation since the potential is localized. Each small oscillator is more likely to be found at a small region near its center $a_i$ . When calculating the partition function, indeed we get an extensive entropy without adding the Gibbs factor ( $Z_N=Z_1^N=(k_bT/\hbar\omega)^N$ ).
>But, when we calculate the partition function there is integration over the spatial space ( $Z=\int\limits_\infty^\infty e^{-\beta H}dxdp$ ). The result of this integration is independent of the oscillators center, so we can arbitrarily take them to be equal, all the oscillators are centered around the same point (which we can taken to be $a_i=0<script type="math/tex">. >Now the Hamiltonian is invariant under particle permutation, and the particles are identical. This means we need to add the Gibbs factor to the partition function. This Gibbs factor now make the entropy non extensive. So the question is what counts for identical particles and when should we add the Gibbs factor? If you have encountered problem like this in all ares of stem, feel free to post and discuss.$

Anonymous

Anonymous Wed 13 May 2020 05:31:30 No.11666451 Report

Quoted By:

>>11666441
okay, so take a three-dimensional gas in a slab symmetric chamber.

Anonymous

Anonymous Wed 13 May 2020 05:40:46 No.11666465 Report

Quoted By: >>11666555

>>11666434
I suppose that the "fluid mechanics" answer does not impose a slowly expanding cylinder. Processes in thermodynamics described by these laws tend to require that the system stays in thermodynamic equilibrium. I don't know shit about fluids, but I assume that the wave can only form if the piston is pulled away at a speed that breaks thermodynamic equilibrium?

Anonymous

Anonymous Wed 13 May 2020 05:45:24 No.11666478 Report

Quoted By: >>11666489 >>11666555

>>11666434
>In statistical mechanics when we pull a piston and the gas start s to expand, we know that the final state has a larger entropy
What? Adiabatic expansion doesn't increase entropy.

Anonymous

Anonymous Wed 13 May 2020 05:50:56 No.11666489 Report

Quoted By: >>11666555 >>11666585

>>11666478
I guess it's implied that drawing the piston is done slowly to allow for a quasi-equilibrium, thus being isothermic?

Anonymous

Anonymous Wed 13 May 2020 06:21:27 No.11666555 Report

Quoted By: >>11666585

>>11666465
No. We have a simple slab symmetry chamber with one wall pulled away.
In fluid mechanics you assume a local thermodynamic equilibrium everywhere. The rarefaction wave can be formed by pulling at any speed. There is a critical value for the pulling speed that is equal to the speed of expansion into a vacuum, if you pull at larger speed the wall detaches from the fluid and the fluid expands (locally) into a vacuum.

>>11666478
>>11666489
Let me re-state the problem. Instead of being pulled, we take out the piston. Now the process is irreversible and will have an increase in entropy. For the fluid problem it just means the fluid expands into a vacuum and this is still an isentropic process.

Anonymous

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Anonymous Wed 13 May 2020 06:35:51 No.11666585 Report

Quoted By: >>11666589 >>11666598

>>11666489
>isothermic
An isothermal expansion would mean there's heat transfer into the cylinder. Obviously, intuitively this increases entropy. Classically:
>change in entropy of the closed system is positive because the increment of heat into the system is positive. or, $\text{d}S>m\delta Q/T_\text{bath}$
statistically:
>probability of state n of canonical ensemble is $\p(n)=e^{-\beta E_n}/Z$
>but entropy is proportional to p*log(p) summed over all states
>temperature is constant but energy of each state increases with heat
>so partition function shrinks and p(n) goes up and so too does S.
>>11666555
>fluid
The problem is that there IS no isotropic rarification wave UNLESS the volume expands adiabatically, in which case of course we expect no increase in the entropy of a normal gas. I see no contradiction or paradox.

Anonymous

Anonymous Wed 13 May 2020 06:38:27 No.11666589 Report

Quoted By:

>>11666585
isentropic***

Anonymous

Anonymous Wed 13 May 2020 06:43:33 No.11666598 Report

Quoted By: >>11666600

>>11666585
The contradiction supposedly is you have a gas that expands into a vacuum with a larger volume. But,
1) rarefaction waves are isentropic, so the entropy is constant.
2) it is an irreversible process, so the entropy increases.

Anonymous

Anonymous Wed 13 May 2020 06:45:51 No.11666600 Report

Quoted By: >>11666603

>>11666598
>rarefaction waves are isentropic
but these waves dont exist?

Anonymous

Anonymous Wed 13 May 2020 06:48:41 No.11666603 Report

Quoted By: >>11666644

>>11666600
They exist. How else will the gas flow into the vacuum?

Anonymous

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Anonymous Wed 13 May 2020 07:03:38 No.11666644 Report

Quoted By: >>11666677 >>11667392

>>11666603
These waves exist under the assumption that the gas expands adiabatically. Classical thermo predicts no increase in entropy for an adiabatic process and statistical mechanics too predicts no increase in entropy since the number of states at an energy in an isolated system doesn't change with volume. If the gas isn't isolated, then there's no isentropic expansion.

Anyway, here's a zinger: throwing a couple identities at euler's equation for a potential flow shows the drag force on a body in a fluid is zero so long as velocity is constant. This is experimentally very false, even at high Reynolds numbers. What gives?

Anonymous

Anonymous Wed 13 May 2020 07:17:05 No.11666677 Report

Quoted By:

>>11666644
>These waves exist under the assumption that the gas expands adiabatically.
You are correct.
>Classical thermo predicts no increase in entropy for an adiabatic process.
Also correct.
>and statistical mechanics too predicts no increase in entropy since the number of states at an energy in an isolated system doesn't change with volume.
Correct that statistical mechanics will give the same results. But you are wrong about the number of states. When the gas expands into a vacuum and comes into an equilibrium the configuration space is getting larger and there are more possible states.
>If the gas isn't isolated, then there's no isentropic expansion.
Correct again but the gas is isolated.

The process is irreversible, its entropy must increase. See http://lectureonline.cl.msu.edu/~mmp/kap12/cd340.htm

The answer to the paradox is:
In the statistical problem the gas comes into equilibrium. It stops expanding when it hits the walls of the chamber. When a fluid hits a wall it creates a shock waves which increase the entropy.
The fluid problem ignores the impact between the fluid and the wall.

Anonymous

Anonymous Wed 13 May 2020 13:59:59 No.11667392 Report

Quoted By:

>>11666644
>Anyway, here's a zinger: throwing a couple identities at euler's equation for a potential flow shows the drag force on a body in a fluid is zero so long as velocity is constant. This is experimentally very false, even at high Reynolds numbers. What gives?

Funny, my research may come in contact with that. Birkhoff suggested the possibility that Prandtl's solution of boundary layers in viscous fluid is not the underlying reason for de'Lambert's paradox.
Hoffman and Jhonson wrote few papers about that, but I've found a mathematical error in it, but still I think it may be interesting to check for the stability of invicid flow.

Anonymous

Anonymous Wed 13 May 2020 21:01:45 No.11668586 Report

Quoted By:

Bumping

Anonymous

Anonymous Thu 14 May 2020 03:10:01 No.11669710 Report

Quoted By:

i still don't know how to make sense of liar sentences.
i'm not buying the claim that "this sentence is false" is nonsense. it's obviously readable and seems to have a truth value, there's something going on that's missing from "asdflasdasdf"

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