4
$\begingroup$

I am working in a manifold diffeomorphic to $\mathbb{R} \times \mathbb{S}^2$ with metric $$g = dr^2 + f(r)^2\left(d\theta + \sin^2\theta~d\phi\right)$$ where $f(r)$ is always positive. Now, I posit an electric field $E = e(r)dr$ with $e(r)$ a function to be determined. Maxwell's electrostatics equations are as follow. $$ \begin{aligned} dE &= 0 \\ d\star E &= 0 \end{aligned} $$

The first equation is automatically satisfied by any $e(r)$. Now, at the moment I am obly able to take the Hodge dual of a form if I can express it in a orthonormal basis. So I move to a new basis:

$$dx^1 = dr,~~ dx^2 = f(r)d\theta,~~dx^3 = f(r)\sin(\theta)d\phi$$ and take the Hodge dual:

$$\star E = \star\left(e(r)dx^1\right) = e(r)~dx^2\wedge dx^3.$$

My problems start here. If I move back to the original frame to take the exterior derivative, I get an equation. If I take the derivative in this frame, I get a different one! Here are the two derivations.

Going back to the old frame:

$$\begin{align} d\star E &= d\left(e(r)f(r)^2\sin\theta~d\theta\wedge d\phi\right) \\ &=\partial_r\left(e(r)f(r)^2\right)\sin\theta~dr\wedge d\theta\wedge d\phi\\ &= 0 \\\\ &\leftrightarrow \partial_r\left(e(r)f(r)^2\right) = 0 \end{align}$$

Staying in the orthonormal frame: $$\begin{align} d\star E &= d\left(e(r)~dx^2\wedge dx^3\right) \\ &=\partial_1e(r) ~ dx^1 \wedge dx^2 \wedge dx^3\\ &= 0 \\\\ &\leftrightarrow \partial_1e(r) = 0 \end{align}$$

Now, to my understanding, these are different equations, since $\partial_r = \partial_1$ as per my change of basis. Something must be wrong, but it's escaping me at the moment. Most probably it's about how I take the exterior derivative, but I seem to be following the same rule in both coordinates.

I am probably making a silly mistake so thanks in advance for the patience.

$\endgroup$
1
+50
$\begingroup$

The mistake is writing the orthonormal basis as differentials. Note that $$d(dx^2) =d( f(r)d\theta\,)=f'(r)\,dr\wedge d\theta$$ which is not necessarily zero. Use a different name for the basis and everything becomes clear: $$e^1 = dr,~~ e^2 = f(r)d\theta,~~e^3 = f(r)\sin(\theta)d\phi$$ so that $$de^1 = 0\,\,\,\,\,\,\,\,\,de^2=\frac{f'}{f}e^1\wedge e^2\,\,\,\,\,\,\,\,\,de^3=\frac{f'}{f}\,e^1\wedge e^3+\frac{\cot{\theta}}{f}\,e^2\wedge e^3$$ Then in the orthonormal basis: \begin{align} d\star E &= d\left(e(r)\,e^2\wedge e^3\right) \\ &=\partial_1 e(r)\,e^1\wedge\,e^2\wedge e^3+e(r)\,de^2\wedge e^3-e(r)\,e^2\wedge de^3\\ &=\partial_1 e(r)\,e^1\wedge\,e^2\wedge e^3+e(r)\,\frac{f'}{f}e^1\wedge e^2 \wedge e^3-e(r)\frac{f'}{f}\,e^2\wedge \,e^1\wedge e^3\\ &=\left(\partial_r e(r)+2e(r)\frac{f'}{f}\right)\,e^1\wedge\,e^2\wedge e^3\\ &=\frac{1}{f^2}\partial_r\left(f^2 e(r)\right)\,e^1\wedge\,e^2\wedge e^3\\\\ &\leftrightarrow \partial_r\left(f^2 e(r)\right) = 0 \end{align} which is the same equation you got with the other method.

$\endgroup$
  • $\begingroup$ Perfect, thanks! A quick follow up. Would you say it is generally the case that it's easier to take Hodge duals in orthonormal bases, but easier to take exterior derivatives in coordinate bases? $\endgroup$ – Andrea Sep 6 '18 at 19:25
  • 1
    $\begingroup$ You're welcome! I would say that's marginally true. The Hodge dual formula is easier to remember in an orthonormal basis and the fact that $d^2=0$ simplifies derivatives in a coordinate basis. But it's often other considerations which tell you if you should be using one or the other, and it might be confusing to switch back and forth. In this case I'd have done like you did, hodge dual in the orthonormal basis and then switch back to the coordinate basis $\endgroup$ – John Donne Sep 6 '18 at 20:03
  • $\begingroup$ By the way, did you receive the bounty? $\endgroup$ – Andrea Sep 14 '18 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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