The action integral for the EM field in a dielectric medium

In summary, the action integral for the electromagnetic field in a matter with dielectric constant \epsilon is given by S_m=\frac{1}{8 \pi} \int dt d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right]where the standard expressions for the E field and B field E=-\frac{1}{c}\partial_t \bf A - \nabla \phi and \bf \nabla \times \bf A are used.
  • #1
MadMax
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0
Show that the action integral for the electromagnetic field in a matter with dielectric constant [tex]\epsilon[/tex] is given by [tex]S_m=\frac{1}{8 \pi} \int dt d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex]

where the standard expressions for the E field and B field [tex]E=-\frac{1}{c}\partial_t \bf A - \nabla \phi[/tex] and [tex]\bf \nabla \times \bf A[/tex] are used.

I'm familiar with the Lagrangian principle L = T - V, where T and V are kinetic and potential energy respectively, and I'm familar with the concept that minimising the action integral of a system leads us to finding the equations of motion for that system. In this case [tex]d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex] is the Lagrangian density of the system. I can see that the expression for Lagrangian density has units of energy per unit volume as one would expect.. But why does the Lagrangian density take that particular form?

Thank you for taking the time to read this; any help would be much appreciated.
 
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  • #2
For a system of particles, finding the Lagrangian is quite easy, since then you know that [tex] L = T - V [/tex] as you wrote, and the kinetic and potential energy are often known.

Now, however, you are working with a Lagrangian for the electromagnetic field. If I have understood things correctly, there is no general method of finding a Lagrangian.

To solve your problem, I would have started from the action and shown that the Euler-Lagrange equations then reproduces the Maxwell equations for fields in matter. This will then prove that the action [tex] S_m [/tex] reproduces the correct equations of motion and hence that it is the correct one.
 
  • #3
Thanks Jezuz. I think I'm happy with the problem now :smile:

Now for a related problem. I have

[tex]S_m=\frac{1}{8 \pi} \int dt d^3 \bf{x} \left[ \frac{\epsilon}{c^2} (\partial_t \bf{A})^2 - (\nabla \times \bf{A})^2 \right][/tex]

and apparently this can somehow be written as

[tex]S_m=\frac{1}{8 \pi} \int \frac{dw}{2 \pi} \int d^3r A_i(-w,r)[\epsilon(w,r) (w/c)^2 \delta_{ij} + \partial_k \partial_k \delta_{ij} - \partial_i \partial_j] A_j (w,r)[/tex]

where w is angular frequency I think.

to me this suggests using Fourier transformation and the expression for curl but damned if I know how.
 
  • #4
OK so I was correct in thinking I need to do a Fourier transform and expand the curl, in particular using the Levi-Civita tensor for the curl and Parseval's theorem for the transform.

First I should concentrate on the curl and use the following expression

[tex](\nabla \times A)_i = \varepsilon_{ijk} \partial_j A_k[/tex]

http://en.wikipedia.org/wiki/Levi-Civita_symbol

Anyway I still have lots of avenues to explore, just thought I'd update the thread a bit especially since it helps in terms of organising.

Please feel free to jump in at any time and make this a bit less painful and slow :P

Cheers.
 
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  • #5
Yepp, a Fourier transform should be used. I don't know if there is any differences between the variables [tex] x [/tex] and [tex] r [/tex], perhaps not.

There is a way to rewrite the levi-civita symbol in terms of delta's, but I can't remember it. This might simplify your calculations somewhat. I will post the expression if I find it.
 
  • #6
ah no worries jezuz got there in the end :) (should've said so earlier sorry)
 

1. What is the action integral for the EM field in a dielectric medium?

The action integral for the electromagnetic (EM) field in a dielectric medium is a mathematical expression that describes the total energy of the field in the medium. It takes into account both the electric and magnetic fields and their interactions with the medium.

2. How is the action integral related to Maxwell's equations?

The action integral is a direct consequence of Maxwell's equations, which are a set of fundamental equations that describe the behavior of electric and magnetic fields. The action integral is derived from these equations and provides a way to calculate the energy of the EM field in a dielectric medium.

3. What is the significance of the dielectric constant in the action integral?

The dielectric constant, also known as the permittivity, is a measure of how easily a material can be polarized by an electric field. It plays a crucial role in the action integral as it determines the strength of the electric field in the medium and, therefore, the total energy of the field.

4. How does the action integral change in the presence of a dielectric medium?

In the absence of a dielectric medium, the action integral reduces to the standard form for vacuum. However, in the presence of a dielectric medium, the action integral takes into account the effects of the medium on the electric and magnetic fields, resulting in a different expression for the total energy of the field.

5. What are the applications of the action integral for the EM field in a dielectric medium?

The action integral is a fundamental concept in electromagnetism and has applications in various fields, including optics, electronics, and telecommunications. It is used to calculate the energy of the EM field in dielectric materials and can also be used to study the propagation of electromagnetic waves through different mediums.

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