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The following partial derivative mnemonic (with Jacobians) seems to work well in thermodynamics:

$\frac{\partial(A,B)}{\partial(C,B)}=\left(\frac{\partial A}{\partial C}\right)_B$

$\partial(A,B)=-\partial(B,A)$

Now it seems I can even treat the individual parts as single elements and get correct results without memorizing all sorts of partial derivative rules. For thermodynamics in particular all I need to know is $\partial(p,V)=\partial(S,T)$.

Can this mnemonic have a solid foundation in mathematics as it seems to work well?

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The first part of what you call a mnemonic is not a mnemonic but a fact:

$$ \def\jacob#1#2{\frac{\partial(#1)}{\partial(#2)}} \def\pderiv#1#2#3{\left(\frac{\partial #1}{\partial #2}\right)_{#3}} \jacob{A,B}{C,B} = \left| \begin{array}{cc} \pderiv ACB & \pderiv ABC\\ \pderiv BCB & \pderiv BBC \end{array} \right| = \left| \begin{array}{cc} \pderiv ACB & \pderiv ABC\\ 0 & 1 \end{array} \right| =\pderiv ACB\;. $$

The second part is only mnemonic in that the numerator and denominator of the Jacobian don't make sense individually, but its foundation in the antisymmetry of determinants is clear.

Then you just need the chain rule

$$ \jacob{x_1,x_2}{y_1,y_2}\jacob{y_1,y_2}{z_1,z_2}=\jacob{x_1,x_2}{z_1,z_2} $$

to justify treating Jacobians as fractions.

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I know that my particular results are correct. But the question was whether (in some framework) I can treat parts of the Jacobian individually without ever creating inconsistencies. In other words if it can be made a solid framework. Similarly to what non-standard calculus does for infinitesimals of ordinary differential calculus. –  Gerenuk Jan 23 '12 at 12:53
    
@Gerenuk: If you know it's correct, why do you call it a mnemonic? :-) I'm not sure what part of a solid framework is missing from my answer -- perhaps you could add an example of a relationship that you can derive using your approach but whose rigorous justification isn't covered by what I wrote? –  joriki Jan 23 '12 at 12:58
    
My worry is that if I always rely in a single $\partial(A,B)$ being a valid mathematical expression, I might end up having wrong results if I don't prove it's foundations. So my question is a multivariate-calculus version of math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio –  Gerenuk Jan 23 '12 at 13:04
    
@Gerenuk: Again, I'd find it much easier to respond to that if you gave a concrete example of a relationship you'd like to be able to justify. I believe the two relationships that you've stated so far are covered in my answer, and $\partial(p,V)=\partial(S,T)$, corresponding to $$\frac{\partial(p,V)}{\partial(S,T)}=1\;,$$ is a physical relationship that can't be justified mathematically. –  joriki Jan 23 '12 at 13:08
    
Everything I can think of I can readily justify. Yet, if you look at the other thread I gave in the comments, it seems far from trivial to treat even differential quotients as real fractions. It must be even harder for Jacobians. I'm asking if it is still possible. I don't have counter examples to my mnemonic, but I cannot prove that there are none. –  Gerenuk Jan 23 '12 at 13:16

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