# A question about an identity in deriving Born-Infeld action

I have a question in David Tong's Example Sheet 4 Problem 5b, how to verify the last equation (*) on p.2? (There is a solution for example sheet 3, but seems to be no solution for example sheet 4.)

Problem 5b:

Show that the equations of motion arising from the Born-Infeld action are equivalent to the beta function condition for the open string,
$$\beta_\sigma\left(F\right)=\left( \frac{1} {1 -F^2} \right)^{\mu \rho}\partial_\mu F_{\rho\sigma }=0$$ Note: To do this, it will prove very useful if you can ﬁrst show the following results:
$$∂_μ\left[\operatorname{tr} \ln(1 − F^2)\right] = −4 ∂_\rho F_{μ\sigma}\left(\frac{F}{1-F^2}\right)^{\sigma \rho }$$

which requires use of the Bianchi identity for $F_{\mu \nu }$ and

\tag{*} \begin{align} \partial _\mu \left( \frac{ F}{1-F^2} \right)^{\mu\nu} &= \left( \frac{ F}{1-F^2} \right)^{\mu\rho} \partial_\mu F_{\rho\sigma} \left( \frac{ F}{1-F^2} \right)^{\sigma \nu} \\ &\qquad+ \left( \frac{1} {1 -F^2} \right)^{\mu \rho} \partial_\mu F_{\rho\sigma}\left( \frac{ 1 }{1-F^2} \right)^{\sigma \nu} \end{align}

In addition, as given in question 5a

$$F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$

My attempt to prove the problem:

LHS $$\partial_{\mu} \left( \frac{ F}{1-F^2} \right)^{\mu\nu} = \partial_{\mu} \left[ F^{\mu}_{\alpha} \left( \frac{1}{1-F^2} \right)^{\alpha \nu} \right] = \left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + F^{\mu}_{\alpha} \partial_{\mu} \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} \tag{1}$$

Using the formula in matrix cookbook for the derivative of inverse matrix, Eq. (53)

Eq. (1) becomes $$\left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2 F^{\mu}_{\alpha} \left[ \frac{1}{1 -F^2} \left( \partial_{\mu} F \right) F \frac{1}{1-F^2} \right]^{\alpha \nu}$$
$$= \left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2 \left( \frac{F}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} \tag{2}$$

The second term in Eq.(2) cancels the second term in the RHS in the problem sheet equation. We then need to show $$\left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + \left( \frac{F}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} - \left( \frac{1}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{1}{1-F^2}\right)^{\sigma \nu} =0 \tag{3}$$

then I didn't find a way to show Eq. (3) hold. I tried to combine the second and third term, and rearrange them, but didn't got a simple expression.

• Comment to the question (v1): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. Sep 1, 2013 at 7:02
• @DImension10AbhimanyuPS note the inclusion of a crucial omitted factor. Sep 1, 2013 at 11:18

$\partial_i \frac{F}{1-F^2}=-(\frac{F}{1-F^2})\partial_i (\frac{1-F^2}{F})(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})\partial_i(\frac{1}{F}-F)(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})(\partial_i(\frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})(- \frac{1}{F}\partial_i F \frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$

$=(\frac{1}{1-F^2})\partial_i F(\frac{1}{1-F^2}) + (\frac{F}{1-F^2})\partial_i F(\frac{F}{1-F^2})$

Eq. ($\ast$) in David Tong's notes follows from the more general eq. ($\ast\ast$)

\tag{**}\begin{align} \partial_{\mu}\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\nu} &=\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\rho} ~\partial_{\mu}F^{\rho}{}_{\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma}{}_{\nu} \\ &\qquad+\left(\frac{1}{1-F^2}\right)^{\lambda}{}_{\rho} ~\partial_{\mu} F^{\rho}{}_{\sigma} \left(\frac{1}{1-F^2}\right)^{\sigma}{}_{\nu} \end{align}

by putting the $\lambda$-index equal to the $\mu$-index, sum over $\mu$, and raise the $\nu$-index. Eq. ($\ast\ast$) is equivalent to eq. ($\ast\ast\ast$)

\tag{***}\begin{align} \partial_{\mu}\left(\frac{F}{1-F^2}\right) &=\left(\frac{F}{1-F^2}\right) \partial_{\mu}F \left(\frac{F}{1-F^2}\right) \\ &\qquad+\left(\frac{1}{1-F^2}\right) \partial_{\mu} F \left(\frac{1}{1-F^2}\right), \end{align}

if we implicitly imply that every matrix $F$ carries one upper and one lower index a la $F^{\lambda}{}_{\nu}$. (Note that that there is no ambiguity in writing the matrix expression $\frac{F}{1-F^2}$ as a fraction because the numerator and the denominator commute.)

Finally, eq. ($\ast\ast\ast$) follows by straightforward manipulations, which involve e.g. using the rule

$$\tag{****} \partial_{\mu}\frac{1}{M}~=~-\frac{1}{M}\partial_{\mu}M\frac{1}{M}$$

of how to differentiate an inverse matrix $M$. (In particular, the Bianchi identity that David Tong mentions is not used in the proof of eq. ($\ast\ast\ast$).)

• Thanks for your answer. Excuse me, I tried to use the matrix derivative in my attempt, as Eq. $(****)$ also in the matrix cookbook Eq. (53), and my Eq. (2) is not the same as (****). What is the reason for that? Sep 1, 2013 at 18:36
• It is solved in Trimok's answer, stupid me :( Sep 1, 2013 at 18:52
• Trimok's answer (v2) shows the manipulations that establishes eq. (***). Sep 1, 2013 at 21:26