cristata
May24-10, 01:29 AM
I'm working from Feynman's definition of internal energy for the Debye theory of heat capacity. I'm trying to use that to derive the normal definition of heat capacity that I've seen. But I'm running into a problem. Note, in the following V_0 is frequency, whereas V is volume (that's how Feynman writes it).
U=\frac{3Vk_{B}^{4}T^{4}}{2\pi^{2}\hbar^{3}V_{0}^{ 3}}\int_{0}^{\Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
= \frac{12\pi V k_{B}T^{4}}{\Theta_{D}^{3}}\int_{0}^{\Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
= \frac{4\pi V}{3}\frac{9k_{B}T^{4}}{\Theta_{D}^{3}}\int_{0}^{\ Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
I get V = \frac{3N}{4 \pi} , in order to obtain
C_{V} =9Nk_{B}\left (\frac{T}{\Theta_{D}} \right)^{3}\int_{0}^{\Theta_{D}/T}\frac{x^{4}e^{x}}{\left (e^x - 1\right )^{2}} dx
Why would V be this amount? I don't understand why this must be the case, unless I'm making a mistake somewhere, but I can't see where. The only way I can get the normal definition of Debye heat capacity is if I set V equal to this.
http://books.google.com/books?id=Ou4ltPYiXPgC&pg=PA20&dq=feynman+%22is+the+debye+temperature%22&cd=1#v=onepage&q&f=false
U=\frac{3Vk_{B}^{4}T^{4}}{2\pi^{2}\hbar^{3}V_{0}^{ 3}}\int_{0}^{\Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
= \frac{12\pi V k_{B}T^{4}}{\Theta_{D}^{3}}\int_{0}^{\Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
= \frac{4\pi V}{3}\frac{9k_{B}T^{4}}{\Theta_{D}^{3}}\int_{0}^{\ Theta_{D}/T}\frac{x^{3}e^{x}}{\left (e^x - 1\right )} dx
I get V = \frac{3N}{4 \pi} , in order to obtain
C_{V} =9Nk_{B}\left (\frac{T}{\Theta_{D}} \right)^{3}\int_{0}^{\Theta_{D}/T}\frac{x^{4}e^{x}}{\left (e^x - 1\right )^{2}} dx
Why would V be this amount? I don't understand why this must be the case, unless I'm making a mistake somewhere, but I can't see where. The only way I can get the normal definition of Debye heat capacity is if I set V equal to this.
http://books.google.com/books?id=Ou4ltPYiXPgC&pg=PA20&dq=feynman+%22is+the+debye+temperature%22&cd=1#v=onepage&q&f=false