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GR0177 #5
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Problem
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This problem is still being typed. |
Thermodynamics }Degree of Freedom
Acording to the Equipartition Theorem, there is a $kt/2$ contribution to the energy from each degree of quadratic freedom in the Hamiltonian. In equation form, the average total energy is ) , where s is the degrees of freedom.
For a n-dimensional 1-particle system, the Hamiltonian is  , where  ( ) refer, respectively, to the  component of momentum (position).
Thus, for a 3-dimensional 1-particle system, one has 6 quadratic terms in the Hamiltonian, as in  .
Plugging in s=6, one finds that the average energy is 3kT.
(This revised solution is due to the GREPhysics.NET user kolndom.)
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Alternate Solutions |
kolndom
2005-11-09 20:38:28 | hi
In my opinion, we need only to consider the equipartition theorem, which reads for a system with single molucule energy E=Ap^2+Bq^2 each squared momentum or displacement term has same average kT/2. Since there are three translational degrees of freedom and 3 vibrational degrees of freedom and 3 rotational degrees of freedom, the total energy is (3+3)*kT/2=3kT.
daschaich
2005-11-09 21:24:22 |
There was a lot of discussion of this problem on the physicsgre yahoo group mailing list a few weeks back. Gbadebo Olaosebikan settled it with the following post. You can check out the earlier posts by poking around at
http://groups.yahoo.com/group/physicsgre/message/2378 />
// --------------------------
I think reibaretti's answer is the "right one" because
it uses the exact statement of the equipartition
theorem without any assumptions of the nature of the
oscillator.
The "n" in nKT/2 , we must remember strictly is
defined as the number of "squared" or quadratic terms
in the enrgy of a system. That this turns out to be
the number of degrees of freedom is only a corollary.
the energy of a three d oscillator which must be
assumed a particle UNLESS otherwise stated is
E = T + U
T = ((p_x)^2 + (p_y)^2 + (p_z)^2)/2m
U= mw^2(x^2 +y^2 +z^2)/2
There are definitely 6 "squared terms" in the energy E and so n=6 and the average energy becomes 6(kT/2)= 3kT.
Cheers
Debo
// --------------------------
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daschaich
2005-11-09 21:25:56 |
Which is exactly what kolndom wrote... sorry, didn't see your comment.
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NaijaProfessor
2005-12-05 22:15:51 |
A 3 dimensional harmonic oscillator has six degrees of freedom, 3 for KE and 3 for PE. Thus 6(1/2KT) = 3KT.
cheers.
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yosun
2005-12-06 19:15:48 |
kolndom, daschaich, NaijaProfessor --- thank you for flagging this problem; I have revised the solution above.
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Comments |
T
2005-11-10 18:56:23 | |  | kolndom
2005-11-09 20:38:28 | hi
In my opinion, we need only to consider the equipartition theorem, which reads for a system with single molucule energy E=Ap^2+Bq^2 each squared momentum or displacement term has same average kT/2. Since there are three translational degrees of freedom and 3 vibrational degrees of freedom and 3 rotational degrees of freedom, the total energy is (3+3)*kT/2=3kT.
daschaich
2005-11-09 21:24:22 |
There was a lot of discussion of this problem on the physicsgre yahoo group mailing list a few weeks back. Gbadebo Olaosebikan settled it with the following post. You can check out the earlier posts by poking around at
http://groups.yahoo.com/group/physicsgre/message/2378 />
// --------------------------
I think reibaretti's answer is the "right one" because
it uses the exact statement of the equipartition
theorem without any assumptions of the nature of the
oscillator.
The "n" in nKT/2 , we must remember strictly is
defined as the number of "squared" or quadratic terms
in the enrgy of a system. That this turns out to be
the number of degrees of freedom is only a corollary.
the energy of a three d oscillator which must be
assumed a particle UNLESS otherwise stated is
E = T + U
T = ((p_x)^2 + (p_y)^2 + (p_z)^2)/2m
U= mw^2(x^2 +y^2 +z^2)/2
There are definitely 6 "squared terms" in the energy E and so n=6 and the average energy becomes 6(kT/2)= 3kT.
Cheers
Debo
// --------------------------
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daschaich
2005-11-09 21:25:56 |
Which is exactly what kolndom wrote... sorry, didn't see your comment.
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NaijaProfessor
2005-12-05 22:15:51 |
A 3 dimensional harmonic oscillator has six degrees of freedom, 3 for KE and 3 for PE. Thus 6(1/2KT) = 3KT.
cheers.
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yosun
2005-12-06 19:15:48 |
kolndom, daschaich, NaijaProfessor --- thank you for flagging this problem; I have revised the solution above.
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physicsman
2010-01-18 22:27:50 |
It should be noted in the solution that the potential degrees of freedom come form being an harmonic oscillator. If this was a free particle only the translation terms would show up and the energy would be halved
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