(I now use the same conventions) (I think the notations are clear enough if you are familiar with differential geometry. Further, I tagged this post as homework-and-excercises. What is the problem with this post? Why not simply leave it to others who may be interested if you do not like it?)

I am currently reading Andrew Strominger's lectures "Lectures on the infrared structure of gravity and gauge theory".

On page 9, it wrotes:

Maxwell's theory of electromagnetism is described by the action $$S=-\frac{1}{4e^2}\int d^4x\sqrt{-g}F_{\mu\nu}F^{\mu\nu}+S_M,$$ where $F=dA$. The equation of motion is $$d*F=e^2*j \implies \nabla^\mu F_{\mu\nu}=e^2 j_\nu,\tag{1}$$ where $*$ is the Hodge dual and the charge current is $$j^\nu=-\frac{\delta S_M}{\delta A_\nu}.$$

What I can not obtain is the symbol $\implies$ in Eq.(1). My proof of it is as following.

According to the definition of Hodge star, $$(*F)_{\mu\nu}=\frac{1}{2!}F^{\rho\sigma}\varepsilon_{\rho\sigma\mu\nu},$$ where $\varepsilon_{\mu\nu\rho\sigma}$ is the volume element $$\varepsilon_{\mu\nu\rho\sigma}=\sqrt{|g|}(dx^1)_\mu\wedge(dx^2)_\nu\wedge(dx^3)_\rho\wedge(dx^4)_\sigma.$$

According to the definition of exterior derivative, $$(d*F)_{\gamma\mu\nu}=(2+1)\frac{1}{2!}\nabla_{[\gamma}F^{\rho\sigma}\varepsilon_{|\rho\sigma|\mu\nu]}.$$ Taking the Hodge star again on the above equation, we have $$(*d*F)_\delta=\frac{1}{3!}\times\frac{3}{2}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta} =\frac{1}{4}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta} =\frac{1}{4}\nabla^{\gamma}F_{\rho\sigma}\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}.\tag{2}$$

Now using the properties for volume elements: $$\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}=\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{\mu\nu\gamma\delta}=(-1)^s 2!(4-2)!\delta^{[\rho}_{\ \ \gamma}\delta^{\sigma]}_{\ \ \delta},$$ where $s$ is the number of minus signs in the diagonalized metric tensor ($s$=1 or 3 depending on (-, +,+,+) or(+,-,-,-)). Thus $$(*d*F)_\delta=-\nabla^\rho F_{\rho\delta}$$

On the other hand, we have $$**j_\delta=(-1)^{s+1(4-1)}j_\delta=j_\delta.$$ Therefore, the first equation in Eq.(1) implies $$\nabla^\rho F_{\rho\delta}=-e^2 j_\delta.$$

My derivation arrives at a result which differs from Eq.(1) by a sign. Where is going wrong in the above derivation (since Andrew uses the second equation in Eq.(1) in several papers, I therefore expect something wrong in my derivation rather than a typo in his lecture notes)?


After comparing with Prahar's answer, I see the problem in my derivation. We used different definitions for the Hodge star. In my derivation, it is (take $*j$ as an example) $$(*j)_{\mu\nu\rho}=j^\sigma\varepsilon_{\sigma\mu\nu\rho}$$ while in Prahar's answer, he used (see Eq.(1) in the link) $$(*j)_{\mu\nu\rho}=j^\sigma\varepsilon_{\mu\nu\rho\sigma}$$ which differs by minus sigh from here.

If I had used Andi's definition, there will be no difference in most of the equations but Eq.(2) in the post will be $$(*d*F)_\delta=\frac{1}{3!}\times\frac{3}{2}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{\mu\nu]\rho\sigma}\varepsilon_{\delta\gamma\mu\nu} =-\frac{1}{4}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta} =-\frac{1}{4}\nabla^{\gamma}F_{\rho\sigma}\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}.$$

However, I would like to comment that, actually, the definition used here is more standard since it appears in Nakahara, John Baez and the text book of GR in my hand. I point it our for convenience of other readers.


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.