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What symmetries are in the following action:

by bagherihan
Tags: action, symmetries
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bagherihan
#1
Jun25-14, 01:16 PM
P: 7
[tex] S=\int d^4x\frac{m}{12}A_μ ε^{μ \nu ρσ} H_{\nu ρσ} + \frac{1}{8} m^2A^μA_μ [/tex]
Where
[tex] H_{\nu ρσ} = \partial_\nu B_{ρσ} + \partial_ρ B_{σ\nu} + \partial_σ B_{\nu ρ} [/tex]

And [itex] B^{μ \nu} [/itex] is an antisymmetric tensor.

What are the global symmetries and what are the local symmetries?

p.s how many degrees of freedom does it have?

Thank you!
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ChrisVer
#2
Jun25-14, 03:10 PM
P: 914
Has [itex]A_{\mu}[/itex] anything to do with the [itex]B_{\mu \nu}[/itex]?

And what does it have dofs?
The Action is a (real) scalar quantity, so it has 1 dof.

if [itex]A_{\mu}[/itex] is a massive bosonic field, it should have 3 dofs.
and about [itex]B^{\mu \nu}[/itex] just by being an antisymmetric tensor (in Lorentz repr it is a 4x4 in your case matrix) will have:
[itex] \frac{D^{2}}{2}-D = \frac{D(D-1)}{2} [/itex]
free parameters. So for D=4, you have 6 dofs...
bagherihan
#3
Jun25-14, 03:36 PM
P: 7
Thanks ChrisVer,
[itex]A^\mu[/itex] has nothing to do with [itex]B_{\mu \nu}[/itex]
I meant the number dof of the thoery.
[itex]H_{\nu ρσ}[/itex] is antisymmetric, so it has only [itex]\binom{4}{3}=4[/itex] dof, doesn't it? thus in total it's 3X4=12 dof, isn't it?

And more important for me is to know the action symmetries, both the global and the local ones.

thanks.

ChrisVer
#4
Jun25-14, 04:10 PM
P: 914
What symmetries are in the following action:

For the symmetries you should apply the Noether's procedure ...
A global symmetry which I can see before hand is the Lorentz Symmetry (since you don't have any free indices flowing around)
ChrisVer
#5
Jun25-14, 04:16 PM
P: 914
Also I don't think you need the dofs of the strength field tensor anywhere, do you?
It gives the kinetic term of your field [itex]B_{\mu \nu}[/itex]
I am not sure though about the dofs now...you might be right.
ChrisVer
#6
Jun25-14, 04:34 PM
P: 914
For the H you were right.
[itex]H[/itex] is a p=3-form, and a general p-form in n dimensions has:
[itex]\frac{n!}{(n-p)!p!}[/itex] ind. components.
bagherihan
#7
Jun26-14, 03:31 AM
P: 7
You're probably right, it's the 6 dof of B that matters.
But apparently B has a gauge symmetry, so only 3 dof left.


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