Classical mechanics ? Liouville?s Theorem

Factual Questions
1

I don’t understand how my text proves Liouville’s Theorem. I found a website that uses the mass continuity rule to prove it, and I understand how that was done, but here’s how my text does it.

p’ = time rate of change of p.
d = both the partial and full derivative symbol.
rho = density of phase points.
Using a phase space diagram it comes up with the formula for phase points moving into an area (dqdp) per unit of time as rho*(q’dp + p’dq)

Then it says by a Taylor series expansion the number of points moving out of the area per unit time is [rhoq’ + d/dq(rhoq’)dq]dp + [rhop’ + d/dp(rhop’)dp]dq

My question is a Taylor series expansion of what? If it’s the Taylor expansion of the first equation then it would seem the proof is ridiculous. I.e x - x = 0.

2

Sorry - I dug out Goldstein again, but I don’t see how your particular proof works.

Is there perhaps a typo? Something like:

  • Here is the formula for phase points moving into an infinitesimal area in phase space
  • Here is a Taylor expansion for such a formula.
  • That expansion equals 0 (does it? I’m too lazy to go there…)
  • Therefore flux is 0.

I think I’m still missing something before that conclusion, but I’m afraid I can’t help you here…

3

Yes that’s what the first equation is,

They never say what equation they’re doing a Taylor expansion on. They just say that The total increase in phase density equals the first equation minus the second equation. They fool around with the terms for awhile and discover that it’s just the total time derivative of rho, which means that the phase space density remains constant during the motion.

I can’t even figure out how you’d do a Taylor expansion on the first equation. But even if you did wouldn’t you just be subtracting an approximation of something from itself? What would that prove?
I think I’ll forget this and just move on, but thanks for trying to help.

4

Um, that doesn’t really look right; there are no time elements. Here is a proof that doesn’t look too bad:

http://astron.berkeley.edu/~jrg/ay202/node27.html

5

Ok, this is my understanding of the proof, after considering Analytical Dynamics byHand and Finch:

Consider a phase space distribution of identical systems of particles, governed by identical Hamiltonians. The only differences between systems is the initial co-ordianates § and momenta (q).

Now, as the systems evolve, the phase space becomes more “twisted”, or dynamically evolved, the total space filled with particles remains constant. This is essentially what Liouville’s theorem says.

Now, consider that there are sufficiently many systems, that we can consider the density of systems, rho, such that rho is a function of space (x,y,z) and time (t).

Aah… maybe, in the OP this definition of rho was implied but never clarified, hence the confusion, Nametag.

Now, rho is defined as the limit of (number of particles in small area, DA)/DA as AD tend to zero. Provided that statisticaldensity fluctuations are unimportant.

Now, we need to prove that d(rho)/dt = 0 due to Hamilton’s equations of motion (I assume that you know what these are). So, dN(DA), the number of particles in DA is defined as:

dN=rhodpdq where dp and dq are the differentials of p and q.

Now consider a rectangle:

        4

| |
| |
| |
1| | 2
| |

 
        3

Consider now the time evolution of the flux of particles entering/leaving this rectangle.

At 1, dN

6

Drat! Accidentally hit submit! And edges 1 and 2 should be further apart!

Anyway, at 1, consider dN(1)=dtdp(rhoqdot)_q (qdot is rate of change of q, and ()_q means evaluate the quantity in brackets at q).

Then, the number leaving at 2 is

dN(2)= -dtdp(rhoqdot)_(q+dq), because we’ve moved dq through the phase space.

So the net change is

dN(12) = dtdp(rhoqdot)q - dtdp(rhoqdot)(q+dq)
i.e.
dN(12)= ((rhoqdot)_q - (rhoqdot)_(q+dq))dtdp

Now, dq is small, so we can Taylor expand the first term of dN(12) (does this help clarify things?), keeping only the first term, i.e.

dN(12) = - [d(rhoqdot)/dq]dtdqdp (hered denotes partial derivative).

Then apply the same reasoning to boundries 3 and 4 for the p co-ordinate to get

dN(34) = - [d(rhopdot)/dp]dtdqdp

Now, the density inside the rectangle will only change if ro is a function of time. So,

dN(inside)= drho/dtdtdq*dp

Add all these together to give you the continuity equation:

drho/dt + d(rhopdot)/dp + d(rhoqdot)/dq = 0

Now, you know that the systems follow Hamiltonian dynamics, so by applying that, and rearranging, you should get

drho/dt + rho{d(qdot)/dq + dpdot)/dp} + qdotdrho/dq + pdotdrho/dp = 0

The term in curly brackets vanishes (because of Hamilton’s version of Newton’s second law). The remainder of the terms are simply drho/dt using the chain rule of differentiation, and so,

**drho/dt = 0 **

Hope that’s helpful. If not, I’d reccomend you get a copy of Hand and Finch. Its very good. And, for what its worth, I am a physicist, so I do understand this.