Equations of motion from Born-Infeld Lagrangian

OhNoYaDidn't
Messages
25
Reaction score
0
We can write the Born-Infeld Lagrangian as:

[tex]L_{BI}=1 - \sqrt{ 1+\frac{1}{2}F_{\mu\nu }F^{\mu\nu}-\frac{1}{16}\left(F_{\mu\nu}\widetilde{F}^{\mu\nu} \right)^{2}}[/tex]

with [itex]G^{\mu\nu}=\frac{\partial L}{\partial F_{\mu\nu}}[/itex] how can we show that in empty space the equations of motion take the form [itex]\partial_{\mu}G^{\mu\nu}=0[/itex]
We should start with an Euler-Lagrange equation, but how can i write it for this Lagrangian?
 
Physics news on Phys.org
The EL equation for this case is
$$\partial_{\nu}\frac{\partial L}{\partial A_{\mu,\nu}}=0$$
where ##A_{\mu,\nu}=\partial_{\nu}A_{\mu}##. Using ##F_{\mu\nu}=A_{\nu,\mu}-A_{\mu,\nu}##, the rest should be straightforward. See also Jackson to see how covariant Maxwell equations are derived for ordinary ##F_{\mu\nu}F^{\mu\nu}## action. For other details about Born Infeld see Zwiebach - A First Course in String Theory.
 
Last edited:
  • Like
Likes   Reactions: vahdaneh and OhNoYaDidn't
Thank you, Demystifier.
I have never seen [itex]F_{\mu\nu}[/itex] written like that, but using that:
[itex]F_{\mu\nu}F^{\mu\nu}=(A_{\nu\mu}-A_{\mu\nu})(A^{\nu\mu}-A^{\mu\nu})=A_{\nu\mu}A^{\nu\mu}-A_{\nu\mu}A^{\mu\nu}-A_{\mu\nu}A^{\nu\mu}+A_{\mu\nu}A^{\mu\nu}[/itex]
[itex](F_{\mu\nu}\widetilde{F}^{\mu\nu})^{2}=((A_{\nu\mu}-A_{\mu\nu})\widetilde{F}^{\mu\nu})^{2}[/itex]

[itex]\frac{\partial L}{\partial A_{\mu\nu}} = \frac{-\frac{1}{4}({-A^{\nu\mu}+A^{\mu\nu}})+\frac{1}{16}A_{\mu\nu}F_{\mu\nu}(\widetilde{F}^{\mu\nu})^{2}}{\sqrt{ 1+\frac{1}{2}F_{\mu\nu }F^{\mu\nu}-\frac{1}{16}\left(F_{\mu\nu}\widetilde{F}^{\mu\nu} \right)^{2}}}[/itex]
What do i do with the [itex]\partial_{\nu}[/itex] now?
 
OhNoYaDidn't said:
I have never seen [itex]F_{\mu\nu}[/itex] written like that
Than you should first learn ordinary electrodynamics. See the Jackson's textbook.
 
OhNoYaDidn't said:
using that

You left out the commas. Look closely at what Demystifier posted; there are commas, so it's ##F_{\mu \nu} = A_{\nu , \mu} - A_{\mu , \nu}##. The commas are partial derivatives, so what he wrote is the same as ##F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu##.

As Demystifier said, you need a good background in ordinary electrodynamics for the topic under discussion.
 
  • Like
Likes   Reactions: OhNoYaDidn't

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K