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 Exercise [19.02] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [19.02]
Here is exercise 19.2


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21 Jun 2010, 03:37

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.02]
On page 233 an expression for the exterior derivative operator is provided in coordinate terms, and involves antisymmetrization brackets. The expression for antisymmetrization brackets containing 3 terms is given explicitly on page 214. Combining these and applying them to dF (using the shorthand notation \partial_s\equiv\frac{\partial}{\partial x^s}) should yield:


\mathrm{d}F \equiv \partial_{[a}F_{bc]}\\ \\<br />= \frac{1}{3}\left(\partial_{a}F_{bc} + \partial_{b}F_{ca} + \partial_{c}F_{ab}\right)


The expression for d*F is similar.


The above factor of 1/3 seems to have been left out of Robin's solution.
In the derived equations, all instances of "4\pi" should be changed to "12\pi" to correct this.


(Note that there are 6 terms in the expression on p.214, but the negative ones are just doubles of the positive ones due to F being antisymmetric, i.e. F_{pq} = -F_{qp} ...hence the factor of 3 here, rather than 6.)


12 Feb 2012, 18:50

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [19.02]
Indeed the derivation by Robin apparently doesn't manage properly the anti-symmetrisation factor; however the final result is correct: the factor is 4\pi not 12\pi .
This because the anti-symmetrisation factor of exterior derivative simplifies with the factor generated by the contraction of F with Levi-Civita tensor in the Hodge star operator.
This is made explicit in my solution to exercise [19.3], in which these factors are worked out in detail; for completeness, I attach also my solution to [19.2], that I didn't post before because too much similar to Robin's one and because reuses the solution to [19.3].


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13 Feb 2012, 15:55

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.02]
Roberto, when I did this problem, I didn't use the Levi-Civita tensor in the Hodge star operator. Just like Robin, I simply applied the exterior derivative ("d") operator to the *F tensor which is explicitly given (in matrix form) on p.443.

Doing so introduces a factor of 1/3 on the LHS (which Robin left out), which doesn't cancel with anything: I end up with 12\pi on the RHS.

A quick check confirms that the explicit components of *F given on p.443 agree with the formula (also on that page):

*F_{ab} = \frac{1}{2}\varepsilon_{abcd}F^{cd} = \frac{1}{2}\varepsilon_{abcd}g^{cp}g^{dq}F_{pq}


...so I don't know where you get an additional factor of 3 from...?


13 Feb 2012, 19:45

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [19.02]
Yes, OK.
I realised that also in ex. 19.03, on which I relied to solve the 19.02, there is the same problem, and my solution to it contained a factor 3 error.

I reposted a corrected version of my solution to 19.03.


17 Feb 2012, 18:30

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.02]
I found the source of the problem.

The "4\pi" in the derived maxwell equations is correct, it's the initial expression that's missing a factor of 1/3.

On page 445, the second maxwell equation should read:


\mathrm{d} {^*}\!F = \tfrac{4\pi}{3} {^*}\!J


Note the fraction on the RHS - it's missing in the text, which is the error.

This is the same expression as in the caption to Fig. 19.1 (last line), though that's expressed in coordinate terms.


26 Feb 2012, 10:47

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [19.02]
Yes, I agree.

Using 4/3 in \mathrm{d} {^*}\!F = \tfrac{4\pi}{3} {^*}\!J would lead to correct Maxwell equations both in ex. 19.02 and in ex. 19.03.


26 Feb 2012, 18:38
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