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n1person
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I am currently reading and trying to solve most of the problems in Carroll's Geometry and Spacetime. I am generally okay at the math (I've done some mathy Riemannian Geometry type stuff), but am not overly good at some of the higher-level physics.
(Chapter 1, Question 13)
Consider adding to the Lagrangian for electromagnetism the additional term:
[tex]\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}[/tex]
a) Express this in terms of E and B
b) Show that including this term doesn't effect Maxwell's equations. Can you think of a deep reason for this?
Using -1,+1,+1,+1 flat metric
E&M Lagrangian:
[tex]\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_{\mu} J^{\mu}[/tex]
Levi-Civita Symbol:
[tex]\tilde{\epsilon}_{\mu\nu\rho\sigma}[/tex]
= -1 for odd permutations of 0123, +1 for even permutations, 0 otherwise
Electromagnetic Field Strength Tensor:
[tex]F_{\alpha\beta}=\partial_\alpha A_\beta - \partial_\beta A_\alpha [/tex]
(too lazy to type the whole matrix :P)Euler Lagrange Equations for a flat space-time field theory:
[tex] \frac{\partial\mathcal{L}}{\partial\Phi^i}- \partial_{\mu}(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi^i)} ) = 0[/tex]
a)
Just using some casework, I was able to get that it was equal to:
[tex]\mathcal{L}'=-\textbf{E} \cdot \textbf{B}[/tex]
b)
This one is less clear to me, following the procedure in the book:
[tex]\mathcal{L}_N=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + A_{\mu} J^{\mu}+\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma}= \mathcal{L} + \mathcal{L}'[/tex]
[tex]\Phi^i=A_\nu[/tex]
[tex] \frac{\partial\mathcal{L}'}{\partial A_\mu}= 0[/tex]
Index Lower Fun!
[tex] \mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}[/tex]
[tex]\frac{\partial}{\partial_\mu A_\nu} (\eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}) = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} (\frac{\partial}{\partial_\mu A_\nu} ( F_{\alpha\beta}) F_{\gamma\delta} + \frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) F_{\alpha\beta}) [/tex]
[tex] \frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) = \delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma [/tex]
So when we plug this in we get:
[tex] \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} ((\delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma) F_{\gamma\delta} + (\delta^\mu_\alpha \delta^\nu_\beta - \delta^\mu_\beta \delta^\mu_\alpha) F_{\alpha\beta}) [/tex]
At this point I think, well if whenever there are duplicates in [tex] \mu\nu\rho\sigma [/tex], the Levi-Civita Symbol is zero, and whenever there are aren't duplicates all the kroncher-deltas are zero, so this entire term is just zero.
Is this reasoning correct? I feel I am missing something cause I cannot think of a deep reason for this... I cannot think of an interesting physical interpretation...
Thanks for following along and reading till here :)
Homework Statement
(Chapter 1, Question 13)
Consider adding to the Lagrangian for electromagnetism the additional term:
[tex]\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}[/tex]
a) Express this in terms of E and B
b) Show that including this term doesn't effect Maxwell's equations. Can you think of a deep reason for this?
Homework Equations
Using -1,+1,+1,+1 flat metric
E&M Lagrangian:
[tex]\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_{\mu} J^{\mu}[/tex]
Levi-Civita Symbol:
[tex]\tilde{\epsilon}_{\mu\nu\rho\sigma}[/tex]
= -1 for odd permutations of 0123, +1 for even permutations, 0 otherwise
Electromagnetic Field Strength Tensor:
[tex]F_{\alpha\beta}=\partial_\alpha A_\beta - \partial_\beta A_\alpha [/tex]
(too lazy to type the whole matrix :P)Euler Lagrange Equations for a flat space-time field theory:
[tex] \frac{\partial\mathcal{L}}{\partial\Phi^i}- \partial_{\mu}(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi^i)} ) = 0[/tex]
The Attempt at a Solution
a)
Just using some casework, I was able to get that it was equal to:
[tex]\mathcal{L}'=-\textbf{E} \cdot \textbf{B}[/tex]
b)
This one is less clear to me, following the procedure in the book:
[tex]\mathcal{L}_N=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + A_{\mu} J^{\mu}+\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma}= \mathcal{L} + \mathcal{L}'[/tex]
[tex]\Phi^i=A_\nu[/tex]
[tex] \frac{\partial\mathcal{L}'}{\partial A_\mu}= 0[/tex]
Index Lower Fun!
[tex] \mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}[/tex]
[tex]\frac{\partial}{\partial_\mu A_\nu} (\eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}) = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} (\frac{\partial}{\partial_\mu A_\nu} ( F_{\alpha\beta}) F_{\gamma\delta} + \frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) F_{\alpha\beta}) [/tex]
[tex] \frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) = \delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma [/tex]
So when we plug this in we get:
[tex] \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} ((\delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma) F_{\gamma\delta} + (\delta^\mu_\alpha \delta^\nu_\beta - \delta^\mu_\beta \delta^\mu_\alpha) F_{\alpha\beta}) [/tex]
At this point I think, well if whenever there are duplicates in [tex] \mu\nu\rho\sigma [/tex], the Levi-Civita Symbol is zero, and whenever there are aren't duplicates all the kroncher-deltas are zero, so this entire term is just zero.
Is this reasoning correct? I feel I am missing something cause I cannot think of a deep reason for this... I cannot think of an interesting physical interpretation...
Thanks for following along and reading till here :)
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