- #36
CAF123
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So, for example, in the other equation I had when I rearranged dF = X dT + YdL for dL, I got $$\frac{\partial F}{\partial L} dL = - \frac{\partial F}{\partial L} dL,$$ where in this case L = L(T) so that this becomes $$\frac{\partial F(L,T)}{\partial L(T)} L'(T) dT = -\frac{\partial F(L,T)}{\partial T(L)} T'(L) dL$$ Provided L=L(T), this eqn then makes sense. Is this what you meant?voko said:Then the RHS is not an integral over a single path such as L = L(T).
Can you point me to a proof of this?Because ##dF## is a full differential, you can say that you can integrate over any path and the result will be the same, provided the endpoint are the same.
But this corresponds to $$ F_2 - F_1 = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{T_1}^{T_2} F_L(L_1, T) L'(T) dT + \int_{L_1}^{L_2} F_T(L, T_2) T'(L) dL + \int_{L_1}^{L_2} F_L(L, T_2) dL \\ = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{L_1}^{L_2} F_L(L, T_2) dL $$
Is your use of notation ##F_T## standard, ##F_T = \frac{\partial F}{\partial T}##? Is that top line, the general form of ##F_2 - F_1## and then by the choice of path, you were then able to say the middle terms were zero and so arrived at the bottom line?
Different in the sense that I said the value of T in the second term was T1 above?Observe this is different from what you had earlier.