1
$\begingroup$

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?

1 Answers 1

1

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.

  • 0
    @Gerenuk: OK, perhaps I should have asked more precisely for an *operation* that you'd like to perform that you can't justify. So far, we've covered three operations: replacing a Jacobian with a shared variable in the numerator and the denominator by a single partial derivative; swapping variables in the numerator and the denominator and simultaneously changing the sign; and cancelling "factors" that appear both in the numerator and in the denominator. These three operations I've rigorously justified. What else would you like to be able to do?2012-01-23