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I am trying to expand the flat-space action

$$ S_{BI} = -T_p \int{d^{p+1}} \sigma \ \mathrm{Tr}\left( e^{-\phi} \sqrt{ -\det(\eta_{ab} + 4\pi^2\alpha^2 \partial_a\Phi^i\partial_b\Phi^i + 2\pi \alpha F_{ab}) \det(Q^{i}_{j}) } \right).\tag{1} $$

After some manipulation, I want to use the power series expansion of the natural log to expand the term

$$ \text{ln}[\delta^{c}_{b} + \lambda^2\eta^{cd} \partial_d\Phi^i\partial_b\Phi^i ]\tag{2} $$ to the fourth power in $\eta^{cd} \partial_d\Phi^i\partial_b\Phi^i.$

My problem is, I don't understand how to compute powers of $\eta^{cd} \partial_d\Phi^i\partial_b\Phi^i $, i.e. $(\eta^{cd} \partial_d\Phi^i\partial_b\Phi^i)^2$. How do I treat the indices?

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  • $\begingroup$ Does it help if you write out the summation explicitly? $\endgroup$ – d_b Sep 28 '20 at 5:28
  • $\begingroup$ @d_b I tried that at first but there were so many indices I kept confusing myself and I couldn't convince myself I was doing anything correctly $\endgroup$ – Hannah Sep 28 '20 at 5:50
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    $\begingroup$ maybe get some practice with a two-dimensional case and then you get the picture $\endgroup$ – Andrew Steane Sep 28 '20 at 8:00
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If you want to expand the Born-Infeld action$^1$ using $$\det(\mathbb{1}+M)~=~\exp({\rm tr}\ln(\mathbb{1}+M)),\tag{A} $$ then you need the logarithm of a whole matrix $$L~:=~\ln(\mathbb{1}+M)~=~-\sum_{n=1}^{\infty}\frac{1}{n} (-M)^n,\tag{B} $$ as opposed to the logarithm of a single matrix element (2). Then $$ L^a{}_b~=~M^a{}_b -\frac{1}{2}M^a{}_cM^c{}_b +{\cal O}(M^3),\tag{C}$$ and so forth. Here $a,b,c$ are world-volume indices.

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$^1$Additional complications arise from the (symmetrized) color-trace ${\rm Tr}$ in a non-abelian Dirac-Born-Infeld action (1). We will here assume for simplicity an abelian gauge group $U(1)$.

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  • $\begingroup$ This expression (B) is helpful to know, but I'm following the lead of my project adviser who got (2) by basically factoring out an eta so that the term would be in the form ln(1+x)... is that not valid or is it just a different way to get the same thing? $\endgroup$ – Hannah Sep 28 '20 at 21:13
  • $\begingroup$ Factoring out eta is correct, but it doesn't seem to lead to (2). $\endgroup$ – Qmechanic Sep 28 '20 at 21:19
  • $\begingroup$ I'm not seeing how it doesn't lead to (2)--I didn't show the steps but if I multiply everything in brackets by eta_(ac), I get back the first two terms in the first determinant in (1) $\endgroup$ – Hannah Sep 28 '20 at 23:29
  • $\begingroup$ Consider to include more details about your derivation of eq. (2) in the opening post. $\endgroup$ – Qmechanic Sep 29 '20 at 10:43

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