Use the Virial theorem to show the following...

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SUMMARY

The discussion focuses on applying the Virial theorem to derive the expected value of the canonical ensemble, specifically the expression ##\langle \cal H \rangle_k##. The Hamiltonian is defined with a scaling factor ##\lambda##, leading to the differentiation of the Hamiltonian with respect to ##\lambda##. The correct formulation involves taking the total derivative of the equation ##\lambda H(x_1,\ldots,x_N)=H(\lambda^{a_1} x_1,\ldots,\lambda^{a_N} x_N)##. Ultimately, the solution reveals that the expression ##\Bigl\langle x_k\frac{\partial H }{\partial x_k} \Bigr\rangle## equates to ##k_bT##.

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  • Understanding of Hamiltonian mechanics
  • Familiarity with the canonical ensemble in statistical mechanics
  • Knowledge of differentiation techniques in multivariable calculus
  • Concept of the Virial theorem in thermodynamics
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  • Study the derivation of the Virial theorem in statistical mechanics
  • Learn about Hamiltonian dynamics and its applications
  • Explore the concept of expectation values in the canonical ensemble
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Students and researchers in physics, particularly those studying statistical mechanics, thermodynamics, and Hamiltonian systems. This discussion is beneficial for anyone looking to deepen their understanding of the Virial theorem and its applications in canonical ensembles.

GravityX
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Homework Statement
I have to show the following with the virial theorem ##E=\langle \cal H \rangle_k## ##=k_bT \sum\limits_{i=1}^{N}a_i##
Relevant Equations
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The expression ##\langle \cal H \rangle_k## is the expected value of the canonical ensemble.

The Hamiltonian is defined as follows, with the scaling ##\lambda##

##\lambda \cal H ## : ##\lambda H(x_1, ...,x_N)=H(\lambda^{a_1}x_1,....,\lambda^{a_N}x_N)##

As a hint, I should differentiate the Hamiltonian with respect to ##\lambda##

Unfortunately, I don't know how exactly I should do this, i.e. form the derivative, I have now proceeded in such a way that I have formed the differential

$$H(\lambda^{a_1}x_1,...\lambda^{a_N}x_N)=\lambda^{a_1}x_1\frac{\partial H}{\partial x_1}+...\lambda^{a_N}x_N\frac{\partial H}{\partial x_N}$$

Then the derivative should look like this,

$$\frac{\partial}{\partial \lambda}H(\lambda^{a_1}x_1,...\lambda^{a_N}x_N)=a_1\lambda^{a_1-1}x_1\frac{\partial H}{\partial x_1}+...a_N\lambda^{a_N-1}x_N\frac{\partial H}{\partial x_N}$$

Is this correct
 
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Not entirely. Just take the total derivative of the equation
$$\lambda H(x_1,\ldots,x_N)=H(\lambda^{a_1} x_1,\ldots,\lambda^{a_N} x_N).$$
Your last equation is correct.

Next you have to think about, what
$$\left \langle x_k \frac{\partial H}{\partial x_k} \right \rangle$$
might be. For that start with the definition of how to take expectation values in the canonical ensemble. It's not too easy, and maybe you are allowed to use the result, which is the "virial theorem".
 
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Thank you vanhees71 for your help and sorry I'm only getting back to you now, I had two weeks Christmas break 🎅

I was able to solve the problem now, the expression ##\Bigl\langle x_k\frac{\partial H }{\partial x_k} \Bigr\rangle## we had stated in the lecture as ##k_bT##.
 
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