2
$\begingroup$

Consider the Born-Infeld Lagrangian, page 30 of Born-Infeld Action and Its Applications by Cong Wang.

$L_{BI} = \sqrt{\det (1+ F)}$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. I am trying to derive the EOM as done here. However I coudn't follow several steps. I am uncomfortable with the author treating $F_{\mu\nu}$ as a number without indices. If I follow the first few lines, it was shown that $L_{BI}$ becomes ${\det (1- F^2)}^{\frac{1}{4}}$ where $({F^2})^{\alpha\beta} = F^{\alpha\sigma}F^{\beta}_{\sigma}$

Following through the next steps,

$$\delta L_{BI} = \delta \exp(\frac{1}{4}\mathrm{tr} \ln(1-F^2)) =-\frac{1}{2} \exp(\frac{1}{4}\mathrm{tr} \ln(1-F^2)) (\frac{F}{1-F^2})^{\mu \nu} \delta F_{\nu \mu}$$

$$\delta \mathrm{tr} \ln(1-F^2) = (\frac{F}{1-F^2})^{\mu \nu} \delta F_{\nu \mu}$$

I am trying to understand this. Is the trace taken before taking the derivative? I would appreciate any help in proving this starting from $ \delta \mathrm{tr} \ln(1-(F^2)^{\alpha \beta})$. What matrix identies are to used?

$\endgroup$
0
$\begingroup$

For any matrix $A$: $$\delta \mathrm{tr}\, A = \mathrm{tr}\, \delta A$$ since $\mathrm{tr}$ is just a linear combination of matrix elements of its argument, and $\delta$ is linear. Writing the indices explicitly ($\delta_{ij}$ is Kronecker's delta): $$ \delta \, \mathrm{tr} A = \delta A_{ii} = \delta (\delta_{ij} A_{ij}) = \delta_{ij} \delta A_{ij} = \delta_{ij} (\delta A)_{ij} = \mathrm{tr} \, \delta A$$

Now let's take $A= \log (1-M^2) $ and go on: $$ \mathrm{tr}\,\delta \log (1-M^2) =\mathrm{tr}\,\left[ -(1-M^2)^{-1} \delta (M^2) \right]= -2 ({(1-M^2)^{-1}})_{ij} M_{jk} \delta M_{ki} $$ now writing $$ ({(1-M^2)^{-1}})_{ij} M_{jk} = \left(\frac{M}{1-M^2}\right)_{ik} $$ you have your result for any matrix $M$.

In the present case, things are slightly different because you have to treat Lorentz indices properly, and trace are computed by contracting with the inverse metric $\eta^{\mu\nu}$, so for instance $$ \mathrm{tr}\,\left[(1-M^2)^{-1} \delta (M^2) \right] = ((1-M^2)^{-1})^{\alpha\beta} \delta (M^2)_{\beta\alpha}, $$ but the key argument is the one expressed before.

$\endgroup$

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.