
#1
May1307, 06:10 PM

P: 19

The Problem;
Given H = U + PV and dU = TdS  PdV Find dH in terms of T, S, P, V My Solution; H = U + PV dH = dU + PdV + VdP dH = (TdS  PdV) + PdV + VdP dH = Tds + VdP My Question Am I missing a step between the first and second steps? I'm taking the derivative of both sides, but not specifying what the derivative is in respect to. (bad English, sorry) I learn this short hand from some physics guys, but I'm looking for the strict mathematical method. Any suggestions? Thanks, Bernie 



#2
May1307, 06:45 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

One way to make sense of what they're doing is via differential forms. If the sequence x_{i} are generalized coordinates on (a local patch of) the state space, so that any function f on the state space can be represented as a function of the x_{i}'s, then one property of differential forms is that
[tex] df = \sum_i \frac{\partial f}{\partial x_i} dx_i, [/tex] which, formally, looks just like the chain rule. Because of the formal similarity, differentials share many of the same properties as derivatives, such as d(fg) = f dg + g df.(Some formulations of differential forms take this property as part of the definition) (In differential geometry, it is customary to write i as a superscript, not a subscript. But I wrote it this way becuase I figured it was probably more familiar to you. In particular, so that it doesn't look like exponentiation) 



#3
May1307, 10:19 PM

P: 19

Thanks for your reply Hurkyl.
I haven't had DE yet, so I can't really comment on your reply. Bernie 


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