Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I apologize for this simple question, but I lost a factor of 2 and can't find it anymore, so now I'm looking on the internet, perhaps one of you has some information about its whereabouts. :-)

Consider the electromagnetic field tensor $F_{\mu\nu}$ which corresponds to a differential form $F =\frac12 F_{\mu\nu} dx^\mu \wedge dx^\nu$. The wedge product of $F$ with itself can be expressed as

$$ F \wedge F = \frac14 F_{\mu\nu}\mathcal F^{\mu\nu} $$

where $\mathcal F^{\mu\nu} = \frac12 \epsilon^{\mu\nu\sigma\tau}F_{\sigma\tau}$ is the dual field tensor.

Writing $A = A_\mu dx^\mu$ for the vector potential, we have

$$ \int_V F \wedge F = \int_V (d A) \wedge F = \int_V d(A \wedge F) = \int_{\partial V} A \wedge F$$

But it appears to me that the latter integral is equal to

$$ \dots = \int_{\partial V} dx^\mu\ A_\nu \mathcal F^{\mu\nu} .$$

Yet, at the same time, we can write

$$ \int_V dx^\alpha \frac12 F_{\mu\nu}\mathcal F^{\mu\nu} = \int_V dx^\alpha (\partial_\mu A_\nu)\mathcal F^{\mu\nu} = \int_V dx^\alpha \partial_\mu (A_\nu\mathcal F^{\mu\nu}) = \int_{\partial V} dx^\mu\ A_\nu \mathcal F^{\mu\nu} .$$

Clearly, that cannot be.

I have lost a factor of two. Where is it?

share|improve this question
3  
For starters, your second equation equate a $4$-form and a $0$-form... –  Qmechanic Apr 4 '11 at 18:21
    
What at @Qmechanic said. Also, I'd check your definitions of wedge product and outer derivative and all other conventions you assumed. –  Marek Apr 4 '11 at 18:34
    
The last line is also very bizarre, why is the integration measure over the spacetime volume called $dx^\alpha$? Should it be just $d^4 x$? –  LuboŇ° Motl Apr 4 '11 at 18:58
2  
Just what everyone else said--it is very, very easy to double count factors of two depending on whether your convention absorbs the factorial factors into the antisymmetrization operator or the exterior derivative or whatever. –  Jerry Schirmer Apr 4 '11 at 23:10
1  
@Lubos: He is using nonstandard conventions, but the arguments are trivial to translate to normal notation. $dx^\alpha$ does mean $d^4x$, he just wanted to emphasize that there is no contraction of indices in this integral, it's all space, not over a surface. –  Ron Maimon Sep 4 '11 at 0:15
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.