# Differentiating determinants

• I

## Summary:

i need to differentiate a determinant in my way to finding the equation of motion of my Lagrangian.

## Main Question or Discussion Point  but it just gets massive and makes me get lost.

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fresh_42
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I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero.

• vahdaneh
I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero. fresh_42
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2018 Award
I'm no physicist so I considered it mathematically. The expression is basically
$$\dfrac{\partial}{\partial x_{\mu \vartheta}}\left(-\dfrac{1}{b^4}\cdot \det \begin{bmatrix}b&x_{12}-x_{21}&x_{13}-x_{31}&x_{14}-x_{41}\\x_{21}-x_{12}&-b&x_{23}-x_{32}&x_{24}-x_{42}\\x_{31}-x_{13}&x_{32}-x_{23}&-b&x_{34}-x_{43}\\x_{41}-x_{14}&x_{42}-x_{24}&x_{43}-x_{34}&-b\end{bmatrix} \right)$$
which is a big, but easy polynomial to differentiate. Maybe the matrix calculator here:
https://www.physicsforums.com/threads/list-of-online-calculators-for-math-physics-earth-and-other-curiosities.970262/#post-6164027will do it for you. Or you just consider the relevant of the 24 diagonals, i.e. those which have $x_{\mu \vartheta}$ in it, since the others become zero.

• vahdaneh
samalkhaiat
Summary: i need to differentiate a determinant in my way to finding the equation of motion of my Lagrangian.
i'll be grateful for any advice
Try looking up the literatures on the Born-Infield theory for the (highly non-trivial) proof of the following identity (in 4 dimensions)
$$- \mbox{Det} \left( g_{\mu\nu} + F_{\mu\nu}\right) = -g \left( 1 + \frac{1}{2} g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma} - ( \frac{1}{8} \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma})^2 \right) , \ \ \ (1)$$ where $g = \mbox{Det} (g_{\alpha \beta})$. Now, if that is your Lagrangian $\mathcal{L}$, then $$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha}A_{\beta})} = 2 \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} ,$$ and the equation of motion is just $$\partial_{\alpha} \left( \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} \right) = 0.$$ So, from the identity (1), you find $$\partial_{\alpha} \left( F^{\alpha \beta} - \frac{1}{4} \left( F_{\mu\nu} ~^*\!F^{\mu\nu} \right) ~^*\!F^{\alpha\beta}\right) = 0,$$ where $~^*\!F^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$.

You should have posted this in the Relativity forum.

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• vahdaneh and dextercioby
Try looking up the literatures on the Born-Infield theory for the (highly non-trivial) proof of the following identity (in 4 dimensions)
$$- \mbox{Det} \left( g_{\mu\nu} + F_{\mu\nu}\right) = -g \left( 1 + \frac{1}{2} g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma} - ( \frac{1}{8} \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma})^2 \right) , \ \ \ (1)$$ where $g = \mbox{Det} (g_{\alpha \beta})$. Now, if that is your Lagrangian $\mathcal{L}$, then $$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha}A_{\beta})} = 2 \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} ,$$ and the equation of motion is just $$\partial_{\alpha} \left( \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} \right) = 0.$$ So, from the identity (1), you find $$\partial_{\alpha} \left( F^{\alpha \beta} - \frac{1}{4} \left( F_{\mu\nu} ~^*\!F^{\mu\nu} \right) ~^*\!F^{\alpha\beta}\right) = 0,$$ where $~^*\!F^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$.

You should have posted this in the Relativity forum.
i don't know why i haven't tried differentiating it when its written in form of field tensor!
so if i'm not wrong with my calculations(cause i got the hint from your advice but i couldn't follow the rest of yours and tried writing it down for myself again) in the end, with considering coulomb's gauge should i have the same equation of motion as Maxwell's Lagrangian?

p.s. and sorry for posting it here i just posted one at physics homework and no one replied so i cut it short to determinant problem and brought it up here, should/can i move it to another forum?

samalkhaiat
I cannot say a lot about your way of solving the problem because you did not show me it. But in any case, you do not need to impose any gauge. The Lagrangian $\mathcal{L} = - \mbox{det} (g_{\mu\nu} + F_{\mu\nu})$ describes a non-linear electrodynamics which tends to Maxwell’s theory in the weak field limit. As you can clearly see, when $F$ is small, you can drop the cubic term $\big[(F \ \cdot ~^*\!F)~^*\!F \big]$ from the equation of motion and obtain the Maxwell’s equation $\partial_{\alpha}F^{\alpha \beta} = 0$.
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