The single particle density of states (Statistical physics)

Click For Summary
SUMMARY

The discussion focuses on calculating the single-particle density of states in statistical physics, specifically using the formula for the number of states in a six-dimensional phase space, given by (d³x d³p)/h³. Participants address three parts of a homework problem: integrating over spatial volume and momentum direction to find D(p)dp, applying the non-relativistic kinetic energy relationship ε = p²/2m to derive D(ε), and considering the relativistic energy-momentum relationship ε_{rel} = (p²c² + m²c⁴)¹/² to determine D(ε_{rel}). Key insights include the use of spherical coordinates for momentum integration and the conversion of variables for energy calculations.

PREREQUISITES
  • Understanding of statistical physics concepts, particularly density of states.
  • Familiarity with phase space and integration in multiple dimensions.
  • Knowledge of non-relativistic and relativistic kinetic energy equations.
  • Basic calculus skills, including variable substitution and integration techniques.
NEXT STEPS
  • Study the derivation of the density of states in statistical mechanics.
  • Learn about the application of spherical coordinates in momentum space integration.
  • Explore the implications of relativistic energy-momentum relationships in statistical physics.
  • Investigate the role of Planck's constant in quantum mechanics and statistical distributions.
USEFUL FOR

Students and researchers in physics, particularly those focusing on statistical mechanics, quantum mechanics, and thermodynamics, will benefit from this discussion.

SingBluSilver
Messages
1
Reaction score
0

Homework Statement


I'm having a little bit of trouble getting started with this problem. Can I get a little help?

Using: (number of states in the six-dimensional region d[tex]^{3}[/tex]x d[tex]^{3}[/tex]3p) = (d[tex]^{3}[/tex]x d[tex]^{3}[/tex]p)/h[tex]^{3}[/tex]
Which provides a convenient route to the single-particle density of masses.

a) Integrate over space (of volume V) and over the direction of the momentum p to determine D(p)dp, where D(p) denotes the number of states per unit interval of momentum magnitude.

b) Adopt the non-relativistic relationship between kinetic energy and momentum, ε = p[tex]^{2}[/tex] / 2m, and determine the number of states per unit energy interval, D(ε). Do you find agreement with our previous result?

c) Consider the relativistic relationship between the total energy and momentum, ε[tex]_{rel}[/tex] = (p[tex]^{2}[/tex]c[tex]^{2}[/tex]+ m[tex]^{2}[/tex]c[tex]^{4}[/tex])[tex]^{1/2}[/tex]. determine the number of states per unit interval of total energy, D(ε[tex]_{rel}[/tex])

(The rel after epsilon is supposed to be a subscript, not sure why it went superscript)

Homework Equations



(number of states in the six-dimensional region d[tex]^{3}[/tex]x d[tex]^{3}[/tex]3p) = (d[tex]^{3}[/tex]x d[tex]^{3}[/tex]p)/h[tex]^{3}[/tex]

The Attempt at a Solution



For part a I'm not exactly sure what I'm supposed to be integrating. Do I just convert d[tex]^{3}[/tex]x to a volume and use polar coordinates to convert d[tex]^{3}[/tex]p to 4pi*p^2dp and integrate that?
I'm also not understanding what parts b and c are asking. Can anyone push me in the right direction?

Any help would be appreciated.

Thank you
 
Last edited:
Physics news on Phys.org
This is probably too late now, but I think I can help you.

You should do just as you yourself suggested - integrate to get rid of all spatial dependence (assuming no x-, y-, or z-dependence the spatial part is just the volume, V) and then integrate to get rid of the angular part of the momentum (this is assuming spherical symmetry of course, so the integration over the solid angle just gives [tex]4\pi[/tex]). What is left is something like
[tex] 4 \pi * V * p^2 dp / h^3[/tex]
where h is Planck's constant.

For b) and c) of the problem. This is just a change of variables, like in ordinary calculus. Thus, you replace all p:s with p(E) and switch dp for dE, with the approprate "weight factor" ([tex]dE= (p/m) dp[/tex] for a free particle with [tex]E=p^2 /2m[/tex].
 

Similar threads

Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
27
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K