The discussion focuses on deriving the effect of an external electric field (ε) on the heat capacity of perfect gas molecules with a permanent electrical dipole moment (u). Participants emphasize the importance of computing the partition function, specifically using the integral q = ∫₀²π ∫₀ᵖᶦ e^(-βuεcosΦ)sinΘ dΘ dΦ. The correct approach involves recognizing the need for the cosine of θ instead of φ in the exponential, as the external field points in a fixed direction, leading to results that include the hyperbolic sine function (sinh).
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No because you are in 3D and the angular measure you have to use isShadowz said: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.
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 said: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,
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.Shadowz said:Thanks, I got it. So we won't need r^2dr in the integral. I get the result that has sinh in it.