What is the relationship between Coulomb's gauge and Maxwell's Lagrangian?

In summary, the conversation discusses the process of finding the equation of motion of a Lagrangian by differentiating a determinant and references the Born-Infield theory and the Maxwell Lagrangian in relation to this process. The conversation also discusses the possibility of moving the discussion to a different forum for better answers.
  • #1
vahdaneh
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TL;DR Summary
i need to differentiate a determinant in my way to finding the equation of motion of my Lagrangian.
2.jpg

i'll be grateful for any advice I've already tried using
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but it just gets massive and makes me get lost.
 
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  • #2
I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero.
 
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  • #3
fresh_42 said:
I would go the troublesome way: write down the 24 terms of the determinant and differentiate. Many diagonals should be zero.
4.jpg
 
  • #4
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/threa...th-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.
 
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  • #5
vahdaneh said:
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.
 
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  • #6
samalkhaiat said:
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?
 
  • #7
vahdaneh said:
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.
 
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FAQ: What is the relationship between Coulomb's gauge and Maxwell's Lagrangian?

1. What is the definition of differentiating determinants?

Differentiating determinants refer to the process of finding the derivative of a function that involves determinants. It involves using the rules of differentiation to find the rate of change of a determinant with respect to its variables.

2. Why is differentiating determinants important?

Differentiating determinants is important because it allows us to find the rate of change of a determinant, which can be useful in solving problems in fields such as physics, economics, and engineering. It also helps us to better understand the behavior of a system described by a determinant function.

3. What are the rules for differentiating determinants?

The rules for differentiating determinants are similar to the rules for differentiating other functions. They include the power rule, product rule, quotient rule, and chain rule. However, there are also specific rules for differentiating determinants, such as the rule of differentiation by cofactors and the rule of differentiation by minors.

4. How do you differentiate determinants using the rule of differentiation by cofactors?

The rule of differentiation by cofactors states that the derivative of a determinant is equal to the sum of the products of the elements of a row or column of the determinant with their corresponding cofactors. This means that we can find the derivative of a determinant by multiplying each element in a row or column by its cofactor and then summing these products.

5. Are there any applications of differentiating determinants?

Yes, there are many applications of differentiating determinants, particularly in the fields of physics, economics, and engineering. For example, in physics, differentiating determinants can be used to find the velocity and acceleration of a particle in motion. In economics, it can be used to find the marginal cost and marginal revenue of a company. In engineering, it can be used to find the rate of change of a system's parameters.

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