I'm doing the following homework question:

By invoking Stokes' theorem, according to which the integral of a vector field (which equals the field strength) over any two-dimension surface S that is bounded by that closed path (see 1.57), $$\Phi(x_1;x_1)=\exp\left[\oint-ieA_\mu(x)\mathrm d x^\mu\right]=\exp\left[\int_S-\tfrac{1}{2}ie F_{\mu\nu}\mathrm{ d} S^{\mu\nu}\right]$$ where $\int_S F_{\mu\nu}\mathrm d S^{\mu\nu}$ denotes twice the total electromagnetic flux through the surface $S$. The quantity $\Phi(x_1;x_2)$ is known as a non-integrable phase factor [305].

Equation 1.57 refers to the vector version of Stokes' theorem i.e. $\int_S \mathbf B\cdot \mathrm d \mathbf S=\oint_C\mathbf A\cdot \mathrm d \mathbf l$. Using this formula naively I'm able to find the answer but I'm not sure how to handle the fact that the dot product is using the Minkowski metric. I want to solve this using the theory of one-forms because it's also a nice practice. To do this I'm using the following definitions (which might be wrong)

\begin{align} A&=A_\mu\mathrm d x^\mu\\ F&=\mathrm d A\\ (\mathrm d A)_{\mu\nu}&=\partial_\mu A_\nu-\partial_\nu A_\mu \\ \mathrm d A&=(\mathrm d A)_{\mu\nu}\,\mathrm d x^\mu\wedge\mathrm d x^\nu \end{align}

Then I get the following \begin{align} \oint A_\mu\mathrm d x^\mu&=\int_{\partial S}A\\ &=\int_S\mathrm d A&\text{(Generalized Stokes)}\\ &=\int_S F\\ &=\int_S F_{\mu\nu}\,\mathrm d x^\mu\wedge\mathrm d x^\nu\\ &=\int_S F_{\mu\nu}\,\mathrm d S^{\mu\nu} \end{align} So I get an overall factor 1/2 wrong. Where does this factor come from?


1 Answer 1


If $A=A_\nu\, dx^{\nu}$ then $$ dA= \partial_{\mu} A_{\nu} \,dx^{\mu} \wedge dx^{\nu}\\ = \frac 12 (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})dx^{\mu} \wedge dx^{\nu}\\ = \frac 12 F_{\mu\nu} dx^{\mu} \wedge dx^{\nu} $$

  • $\begingroup$ Thanks that explains the discrepancy. On wikipedia's article for the exterior derivative en.wikipedia.org/wiki/… I found the definition $(\mathrm d\omega)_{\mu\nu}=\partial_\mu\omega_\nu-\partial_\nu\omega_\mu$. Does that definition agree with the one you're using? Or are those two different conventions? $\endgroup$ Commented Feb 25, 2021 at 12:51
  • 1
    $\begingroup$ It's not a notation that I would use precisely because of the factor of two. I Always write $\omega= \frac 1{p!} \omega_{i_1\ldots i_p} dx^{{i_1}}\cdots dx^{i_p}$, where the $p!$ eliminates the overcounting. $\endgroup$
    – mike stone
    Commented Feb 25, 2021 at 13:14

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