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Interpreting microcanonical distribution

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jjr
#1
Sep3-14, 10:32 AM
P: 24
I'm trying to interpret the expression of a microcanonical distribution for energy [itex]E_0[/itex] of a particle of mass m moving about a fixed centre to which it is attracted by a Coulomb potential, [itex]Zr^{-1}[/itex], where [itex]Z[/itex] is negative. The function expression looks like this:

[itex]ρ_{E_0}(\textbf{r,p}) = \delta(E_0 - \frac{1}{2}m^{-1}p^2-Zr^{-1})[/itex].

Most of the stuff in the expression is understandable, but I am not sure what the delta signifies here. Any help?

Thanks!
J
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WannabeNewton
#2
Sep3-14, 10:35 AM
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In the microcanonical ensemble the system lies on a surface of constant energy in phase space so the probability distribution has to vanish off of the constant energy surface. The argument in the delta function just represents this surface.
jjr
#3
Sep3-14, 11:25 AM
P: 24
So [itex] E_0 ≠ \frac{1}{2}m^{-1}p^2+Zr^{-1} → ρ_{E_0}(\textbf{r,p}) = 0 [/itex] ?

Does the original expression actually say something about the distribution itself, or only about this property?

WannabeNewton
#4
Sep3-14, 11:38 AM
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Interpreting microcanonical distribution

Quote Quote by jjr View Post
So [itex] E_0 ≠ \frac{1}{2}m^{-1}p^2+Zr^{-1} → ρ_{E_0}(\textbf{r,p}) = 0 [/itex] ?
Yes.

Quote Quote by jjr View Post
Does the original expression actually say something about the distribution itself, or only about this property?
Well what you originally wrote down is not complete. It should be ##\rho = \frac{\delta(E - E_0)}{\Omega_{E_0}}## where ##\Omega_{E_0}## is the phase space volume accessible to the microstates. This simply says that at equilibrium the probability of the system being found in any of the accessible microstates is the same for all microstates.
jjr
#5
Sep3-14, 12:59 PM
P: 24
Great! Thanks for helping me out


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