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So the lecture notes to my thermodynamics course contain this relation: $$\left(\frac{\partial^2F}{\partial T^2}\right)_{V,N}=-\frac{1}{\frac{\partial^2U}{\partial S^2}}$$ With no further explanation given. I know that the Helmholtz Potential gives $\left(\frac{\partial F}{\partial T}\right)_{V,N}=S$ but how does one arrive at $\left(\frac{\partial S}{\partial T}\right)_{V,N}=-\frac{1}{\frac{\partial^2U}{\partial S^2}}$ ?

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$\left(\frac{\partial F}{\partial T}\right)_{V,N}=S$, differentiate this w.r.t Temp keeping constant V,N and then substitute from the first equation.

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  • $\begingroup$ I'm sorry, I can't quite follow $\endgroup$ Jan 17 '20 at 11:11
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From the differential expression for $dF$, we have: $ \left( \frac{\partial{F}}{\partial{T}}\right)_{V,N} = -S $. Taking another derivative w.r.t. $T$: $$ \left( \frac{\partial{F}^2}{\partial{T}^2}\right)_{V,N} = -\left( \frac{\partial{S}}{\partial{T}}\right)_{V,N}. $$

On the other side, from the differential expression for $dU$, we have: $ \left( \frac{\partial{U}}{\partial{S}}\right)_{V,N} = T. $ Taking another derivative w.r.t. $S$: $$ \left( \frac{\partial{U}^2}{\partial{S}^2}\right)_{V,N} = \left( \frac{\partial{T}}{\partial{S}}\right)_{V,N}. $$ We get the required result recalling that $\left( \frac{\partial{S}}{\partial{T}}\right)_{V,N} = \frac{1}{ \left( \frac{\partial{T}}{\partial{S}}\right)_{V,N} }$.

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