Register to reply 
What symmetries are in the following action: 
Share this thread: 
#1
Jun2514, 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! 


#2
Jun2514, 03:10 PM

P: 895

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(D1)}{2} [/itex] free parameters. So for D=4, you have 6 dofs... 


#3
Jun2514, 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. 


#4
Jun2514, 04:10 PM

P: 895

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) 


#5
Jun2514, 04:16 PM

P: 895

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. 


#6
Jun2514, 04:34 PM

P: 895

For the H you were right.
[itex]H[/itex] is a p=3form, and a general pform in n dimensions has: [itex]\frac{n!}{(np)!p!}[/itex] ind. components. 


#7
Jun2614, 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. 


Register to reply 
Related Discussions  
EM Field Theory (Action Symmetries)  Advanced Physics Homework  0  
Internal (gauge) symmetries and spacetime symmetries  High Energy, Nuclear, Particle Physics  54  
Symmetries of NG action?  Beyond the Standard Model  5  
New action is key to Loop gains (the DittrichSpeziale action)  Beyond the Standard Model  6  
Action/reaction : if we do an action that requires energy of the whole universe  General Physics  2 