Effect of electrical field on heat capacity

[Shadowz]

Homework Statement

Given the perfect gas molecules with permanent electrical dipole moment u in the field \epsilon.
The potential energy is U = -u\epsilon\cos\Theta
Derive the additional effect of \epsilon on the heat capacity.

I need some hints, please help. Thanks.

 

[Zacku]

Hi,

You have to compute the partition function first...do you know how to do that?

 

[Shadowz]

Hi,

So my attempt was to compute q = \int_0^{2\pi} e^{-\beta U} d\Theta

But this gives me the Bessel function, so I am not sure if I am on the right track.

 

[Zacku]

Hi,

So my attempt was to compute q = \int_0^{2\pi} e^{-\beta U} d\Theta

But this gives me the Bessel function, so I am not sure if I am on the right track.

No because you are in 3D and the angular measure you have to use is
sin \theta d\phi d\theta and not just d\theta.

 

[Shadowz]

Hi,

Thank for your help. I agree.

So I tried

\int_0^{2\pi}\int_0^{\pi} e^{-\beta \mu \epsilon \cos\Phi}sin\Theta d\Theta d\Phi
but still gets the Bessel function.

Is my limit of integration wrong?

Should it be

\int_0^{2\pi}\int_0^{\pi}\int_0^\infty e^{-\beta \mu \epsilon \cos\Phi} r^2 drsin\Theta d\Theta d\Phi

Thanks,

 

[Zacku]

Hi,

Thank for your help. I agree.

So I tried

\int_0^{2\pi}\int_0^{\pi} e^{-\beta \mu \epsilon \cos\Phi}sin\Theta d\Theta d\Phi
but still gets the Bessel function.

Is my limit of integration wrong?

Should it be

\int_0^{2\pi}\int_0^{\pi}\int_0^\infty e^{-\beta \mu \epsilon \cos\Phi} r^2 drsin\Theta d\Theta d\Phi

Thanks,

Well the thing is that you should have \cos \theta instead of \cos \phi in the exponential. The reason for that is that you have an external field which point toward a non varying direction. In spherical basis and coordinates the only vector that fulfill this criteria is \hat{u}_z whose scalar product with \hat{u}_r gives \cos \theta.

 

[Shadowz]

Thanks, I got it. So we won't need r^2dr in the integral. I get the result that has sinh in it.

 

[Zacku]

Thanks, I got it. So we won't need r^2dr in the integral. I get the result that has sinh in it.

No you don't need the r^2dr part in the integral because your dipole has a fixed "length" and therefore its length shouldn't be taken as a degree of freedom.
I don't know exactly the result but a sinh sounds good to me.

 

[Shadowz]

Ya right. Thank you. The rest is easy, I think.