1
$\begingroup$

In Weinberg's Gravitation,

the formula for the volume element in curviliniar coordinates is given by $$dV=h_1 h_2 h_3 dx^1 dx^2 dx^3.$$

The metric is given by $ds^2=h_1^2 dx_1^2+h_2^2 dx_2^2+h_3^2 dx_3^2.$

I am totally confused by this, I know that $\sqrt{g^{\prime}}=|\frac{\partial x}{\partial x^{\prime}}|\sqrt{g}$. If I take the initial metric to be euclidean, then clearly $\sqrt{g^{\prime}}=\det(J)$. The change in volume element is given by $dV=\det(J)^{-1}dV_{euclidean}$, which gives $dV=\frac{1}{h_1 h_2 h_3}dx_1 dx_2 dx_3.$

Where am I going wrong?

Also, the expression for curl, divergence etc have a factor of $\det(g)^{-\frac{1}{2}}$.

Why is this so? How do I prove it?

$\endgroup$

1 Answer 1

1
$\begingroup$

You're confusing yourself with your own notation. Instead of writing $dV$ all over the place, let's keep it explicit as $dx^{1} dx^{2} dx^{3}$ and such.

Let $dx^1$ and so on be in cartesian and $dx^{1'}$ be for some other curvilinear coordinates.

Then you get $$dx^{1'} dx^{2'} dx^{3'} = (\det J)^{-1} \, dx^1 dx^2 dx^3$$

Just move the determinant to the left and replace with $h$'s.

$$h_1 h_2 h_3 dx^{1'} dx^{2'} dx^{3'} = dx^1 dx^2 dx^3$$

And that is correct. There's some ambiguity in the notation because it's common in vector calculus to say $dV = \sqrt{|g|} dx^1 dx^2 dx^3$, to treat $dV$ as an "invariant" quantity that merely has equivalent descriptions in various coordinate systems. For some reason this isn't done in GR, so the meaning is slightly different. (Or perhaps this is why some prefer to say $d^4 x$ instead.)

$\endgroup$

Your Answer

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