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The electromagnetic action in the language of differential geometry is given by

$$\displaystyle{S \sim \int F \wedge \star F},$$

where $A$ is the one-form potential and $F={\rm d}A$ is the two-form field strength.

At the extremum of the action $S$, $F$ is constrained by ${\rm d}F=0$ and ${\rm d}\star F=0$.


Now, generalise the above action to

$$\displaystyle{S \sim \int H \wedge \star H}$$

where $B$ is the two-form potential and $H={\rm d}B$ is the three-form field strength.

At the extremum of the action $S$, $H$ is constrained by ${\rm d}H=0$ and ${\rm d}\star H=0$.


Are there any qualitative differences between the two sets of equations in $d+1$-dimensions?

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    $\begingroup$ NB $\mathrm{d}F = 0$ is true for any $F$, not just at an extremum of the action. This is the Bianchi identity. $\endgroup$ – gj255 Oct 30 '16 at 23:57
  • $\begingroup$ Good point, my bad! $\endgroup$ – nightmarish Oct 31 '16 at 0:13

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