- #1
robousy
- 332
- 1
Hey folks,
I have a pretty interesting Lagrangian setup:
\mathcal{L}=-\frac{1}{4}(\nabla_a u_b -\nabla_b u_a)(\nabla^a u^b -\nabla^b u^a)-\lambda(u_au^a-v^2)-\frac{1}{2}(\partial\phi^2)-\frac{1}{2}m^2\phi^2-\frac{1}{2\mu^2}u^au^b\partial_a\phi\partial_b\phi
My term u^a is a spacelike 5 vector that violates Lorentz Invariance in the 5th dimension only. The indices a,b run from 0 to 4.
My question:
In the paper I'm reading (http://arxiv.org/PS_cache/arxiv/pdf/...802.0521v1.pdf ), the above lagrangian is referred to have "the lowest order coupling". I'm guessing this comes from the last interaction term between u and \phi. Can anyone explain why the term has to be of the form \frac{1}{2\mu^2}u^au^b\partial_a\phi\partial_b\phi. Whats wrong with just \frac{1}{2\mu^2}u^au_a\phi^2 for example??
I have a pretty interesting Lagrangian setup:
\mathcal{L}=-\frac{1}{4}(\nabla_a u_b -\nabla_b u_a)(\nabla^a u^b -\nabla^b u^a)-\lambda(u_au^a-v^2)-\frac{1}{2}(\partial\phi^2)-\frac{1}{2}m^2\phi^2-\frac{1}{2\mu^2}u^au^b\partial_a\phi\partial_b\phi
My term u^a is a spacelike 5 vector that violates Lorentz Invariance in the 5th dimension only. The indices a,b run from 0 to 4.
My question:
In the paper I'm reading (http://arxiv.org/PS_cache/arxiv/pdf/...802.0521v1.pdf ), the above lagrangian is referred to have "the lowest order coupling". I'm guessing this comes from the last interaction term between u and \phi. Can anyone explain why the term has to be of the form \frac{1}{2\mu^2}u^au^b\partial_a\phi\partial_b\phi. Whats wrong with just \frac{1}{2\mu^2}u^au_a\phi^2 for example??
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