0
$\begingroup$

In Sean Carroll's GR book, pg. 89, there is this equation (2.93) involving the Jacobian of a general transformation:

$$\frac{\partial x^{\mu_1}}{\partial x^{\mu_1'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}dx^{\mu_1'}...dx^{\mu_n'}=\left|\frac{\partial x^{\nu}}{\partial x^{\nu'}}\right|dx^{\mu_1}...dx^{\mu_n}$$ where repeated indices are summed and $\left|\frac{\partial x^{\nu}}{\partial x^{\nu'}}\right|$ is the Jacobian of the transformation.

I have a hard time seeing why this is true. My understanding is that to calculate the Jacobian, we have to calculate the cofacters etc. Why does summing over the indices on the LHS gives $\left|\frac{\partial x^{\nu}}{\partial x^{\nu'}}\right|dx^{\mu_1}...dx^{\mu_n}$?

$\endgroup$
1
  • $\begingroup$ Do you know about differential forms and volume forms? $\endgroup$
    – ohneVal
    Commented Mar 19, 2021 at 8:13

1 Answer 1

1
$\begingroup$

It all comes from the fact that products of differentials are not just any kind of infinitesimals. They are elements of oriented hypersurface. E.g., $$ dx^{1}dx^{2}=-dx^{2}dx^{1} $$ $$ dx^{1}dx^{1}=-dx^{1}dx^{1}\Rightarrow dx^{1}dx^{1}=0 $$ This is best implemented with the formalism of differential forms and the wedge product. Try to convince yourself that these identities are true: $$ dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{n}}=\varepsilon^{\mu_{1}\cdots\mu_{n}}dx^{1}\wedge\cdots\wedge dx^{n} $$ $$ dx^{1}\wedge\cdots\wedge dx^{n}=\left( -1 \right)^s\frac{1}{n!}\varepsilon_{\mu_{1}\cdots\mu_{n}}dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{n}} $$ Then think about the fact that the contraction you've got there contains all possible permutations of the indices. And you've got the primes wrong on the RHS.

Note: The $\left( -1 \right)^s$ comes from raising the indices, $s$ being the number of negative eigenvalues of the metric. Thank you, @PraharMitra.

$\endgroup$
1
  • $\begingroup$ You're totally right. Thank you. I will edit. $\endgroup$
    – joigus
    Commented Mar 19, 2021 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.