Differentiating determinants

  • I
  • Thread starter vahdaneh
  • Start date
6
0

Summary:

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

Main Question or Discussion Point

2.jpg

i'll be grateful for any advice I've already tried using
3.jpg

but it just gets massive and makes me get lost.
 

Answers and Replies

fresh_42
Mentor
Insights Author
2018 Award
12,120
8,475
I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero.
 
6
0
I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero.
4.jpg
 
fresh_42
Mentor
Insights Author
2018 Award
12,120
8,475
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.
 
samalkhaiat
Science Advisor
Insights Author
1,637
833
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)
[tex]- \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)[/tex] where [itex]g = \mbox{Det} (g_{\alpha \beta})[/itex]. Now, if that is your Lagrangian [itex]\mathcal{L}[/itex], then [tex]\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha}A_{\beta})} = 2 \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} ,[/tex] and the equation of motion is just [tex]\partial_{\alpha} \left( \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} \right) = 0.[/tex] So, from the identity (1), you find [tex]\partial_{\alpha} \left( F^{\alpha \beta} - \frac{1}{4} \left( F_{\mu\nu} ~^*\!F^{\mu\nu} \right) ~^*\!F^{\alpha\beta}\right) = 0,[/tex] where [itex]~^*\!F^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}[/itex].

You should have posted this in the Relativity forum.
 
Last edited:
6
0
Try looking up the literatures on the Born-Infield theory for the (highly non-trivial) proof of the following identity (in 4 dimensions)
[tex]- \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)[/tex] where [itex]g = \mbox{Det} (g_{\alpha \beta})[/itex]. Now, if that is your Lagrangian [itex]\mathcal{L}[/itex], then [tex]\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha}A_{\beta})} = 2 \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} ,[/tex] and the equation of motion is just [tex]\partial_{\alpha} \left( \frac{\partial \mathcal{L}}{\partial F_{\alpha \beta}} \right) = 0.[/tex] So, from the identity (1), you find [tex]\partial_{\alpha} \left( F^{\alpha \beta} - \frac{1}{4} \left( F_{\mu\nu} ~^*\!F^{\mu\nu} \right) ~^*\!F^{\alpha\beta}\right) = 0,[/tex] where [itex]~^*\!F^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}[/itex].

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
Science Advisor
Insights Author
1,637
833
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?
Which part of my post you did not follow?
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 [itex]\mathcal{L} = - \mbox{det} (g_{\mu\nu} + F_{\mu\nu})[/itex] describes a non-linear electrodynamics which tends to Maxwell’s theory in the weak field limit. As you can clearly see, when [itex]F[/itex] is small, you can drop the cubic term [itex]\big[(F \ \cdot ~^*\!F)~^*\!F \big][/itex] from the equation of motion and obtain the Maxwell’s equation [itex]\partial_{\alpha}F^{\alpha \beta} = 0[/itex].

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?
Well, you would get better answers in the relativity section and, I believe, you can ask to be moved. I only saw your post by accident.
 

Related Threads for: Differentiating determinants

Replies
1
Views
1K
Replies
1
Views
2K
Replies
0
Views
2K
Replies
1
Views
1K
Top