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.