[BUG] partial derivative product rule

[Martin Jackson]

Tevian and Matthias have pointed the way to some mathematically 
rigorous ways of thinking about the question Suda poses.  Here's a 
simple-minded take on the issue.

A physicist looking at this result might think in terms of 
dimensions.  If x,y,and z represent physical quantities (pressure, 
volume, temperature in the Boyle's law example Matthias mentions in 
his post) with specified physical dimensions, then this particular 
product of partial derivatives is dimensionless.  A physicist might 
expect the product to equal 1 since there is no reason to expect a 
pure number such 42 or pi to show up here.  That is, one might expect 
1 since the notation suggests that the changes in x,y and z all 
cancel. The fact that the product is -1 might surprise a physicist. 
On the other hand, the -1 here is a consequence of the fact that the 
problem involves an odd number of variables.  A  more general result 
is (in a spur of the moment notation)
partial(x_1)(x_2)*partial(x_2)(x_3)*...*partial(x_(n-1))(x_n)*partial(x_n)(x_1)=(-1)^n
In this light, the sign is just a consequence of how many variables 
are involved.  (To be even more general, we could use any permutation 
of {x_1,x_2,...x_n} in the "numerators" of the partial derivatives 
and any permutation in the "denominators."  These are connected to 
the faces of the octahedron in the approach Matthias gives.)

Martin







>  >>>>> Suda Kunyosying writes:
>
>     SK> Problem 54 of Stewart's section 15.5 can be proved easily enough. But
>     SK> is there a physical situation we can use to explain why the product:
>     SK> of partial derivatives (z wrt x, x wrt y, and y wrt z)= -1 ?
>
>One way to look at this involves the iterated coordinate transformations:
>	(x,y) |--> (x,z) |--> (y,z) |--> (x,z)
>and the stated identity follows from the fact that multiplying Jacobian
>matrices of 2 (or more) transformations yields the Jacobian matrix of the
>composition of the transformations; in this case, the product of the three
>matrices must be the identity.  (I think it is also necessary to use the
>composition property on two transformations in order to complete the proof.)
>
>A more geometric interpretation follows from the realization that the normal
>vector to the surface is just the (3-d) gradient of the function of
>3-variables which defines the surface, which can also be expressed in terms of
>the (2-d) gradients of the "graph" functions, so that the following vectors
>must be parallel:
>	grad(f)-k, grad(g)-j, grad(h)-i
>This again yields the desired result after a bit of algebra, but I don't see
>an obvious way to finish the argument geometrically.
>
>I see no physical context here -- and it's easy to get lost in the notation,
>since different variables are being held constant in each differentiation.
>
>Tevian
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