# Help with gauge symmetry

1. Mar 8, 2015

### Mentz114

Refering to this paper "Theoretical Aspects of Massive Gravity" (http://arxiv.org/abs/1105.3735) about the spin-2 boson field and GR.

The author uses the Fierz-Pauli action ( I quote the massless part)

$-\frac{1}{2}\partial_\lambda h_{\mu\nu}\partial^\lambda h^{\mu\nu} + \partial_\mu h_{\nu\lambda}\partial^\nu h^{\mu\lambda}$

and states that these terms have the gauge symmetry

$\delta h_{\mu\nu} = \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$

for a spacetime dependent gauge parameter $\xi_\mu(x)$.

No problem there but I wonder if someone could show this explicitly ?

If I understand correctly this symmetry is diffeomorphism invariance, since Killings equations arise as the analog of conserved charges and currents.

Last edited: Mar 8, 2015
2. Mar 8, 2015

### Matterwave

You should plug in $h'_{\mu\nu}=h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$ into your equation and hopefully see that all the terms with $\xi$ cancel out so that you get the original equation back again.

3. Mar 8, 2015

### Mentz114

Thanks, I will try that.

Something that bothers me is that $h'_{\mu\nu}$ should be traceless, so shouldn't the gauge transformation be something like,

$h'_{\mu\nu}=h_{\mu\nu}+\partial_{(\mu}\xi_{\nu)}-\eta_{\mu\nu}\partial_\rho\xi^\rho$

4. Mar 9, 2015

### haushofer

Your h' "should not be traceless"; that's a specific gauge condition you can impose (see e.g. Zwiebach's stringtheory book). This gaugechoice is not preserved by a general gauge transformation as is often the case, but imposes a condition on the gauge parameter (vector field): it should be divergenceless, as you have noticed.

What you did is to introduce a compensating transformation, which pulls you back into your gauge choice again. You should then check if the Fierz Pauli action is actually invariant under such a transformation.

You may want to check similar cases, like the Donder gauge in treating grav. waves, or choosing the Minkowski metric for the worldsheet in string theory.

5. Mar 9, 2015

### haushofer

So, to be clear:

1) You have an action
2) This action is invariant under gct's (linearized)
3) To analyze e.g. the degrees of freedom of the field h you choose a specific gauge, e.g. one in which h is traceless
4) This gaugechoice breaks the gct's to a subgroup of transformations

6. Mar 9, 2015

### Mentz114

I just deleted a line because

2) is true because all tensorial contractions are invariant under gct's

So I suppose we add a condition (gauge choice) to get some physics.

Thank you for the response.

Last edited: Mar 9, 2015
7. Mar 9, 2015

### haushofer

No, that's not true. E.g., the divergence of a vector field using the partial derivative instead of the covariant one is not covariant wrt gct's. You can check this explicitly.

But yes, gauge degrees of freedom are just redundancies, so to extract physics you often have to gaugefix

8. Mar 9, 2015

### Mentz114

OK. So should I have said

2) is true because all covariant tensorial contractions are invariant under gct's.

Is the Fierz-Pauli action covariant ?

I'm having trouble relating this to all the material I'm reading on 'invariance under gauge transformations' which always use the EM field and the transformation of the 4-potential. This process is making a global symmetry (invariance) into a local one.

$\vec{A} \rightarrow \vec{A} + \nabla \Lambda$
$\phi \rightarrow \phi + \partial_t \Lambda$

If the transformed potential is plugged back into the EOMs they are unchanged.

I don't see how $h'_{\mu\nu}=h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$ is making a global symmetry into a local one.

9. Mar 9, 2015

### Ben Niehoff

A gauge transformation in EM acts on the 4-vector potential like this:

$$A'_\mu = A_\mu + \partial_\mu \zeta.$$
Surely you can see the resemblance to

$$h'_{\mu\nu} = h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu.$$
Although in this case you may have to think of a gauge symmetry as "a local redundancy in the degrees of freedom that simplifies the expression of the equations of motion" rather than "a global symmetry made into a local symmetry."

10. Mar 9, 2015

### Mentz114

Right.

So with $F_{\mu\nu} = \partial_\mu A_\nu-\partial_\nu A_\mu$ if we put in $A'_\mu$ the $\zeta$ terms cancel out.

What is the equivalent calculation in the F-P case ? I'll try the substitution again.

I hope I'm not looking argumentative - I need guidance.

[later]
I may have made a bad mistake. In the paper the phrase '.. surviving two derivative terms ...' is used in connection with the action. Later the author refers to 'two-derivative terms'. So I could have dropped 2 terms from the action if I mis-interpreted the 'two derivative' bit.

I think in view of this ghastly ambiguity I'm dropping this paper in protest.

I need to go to the original paper ( or maybe Carroll has something I should read).

Thanks to all respondees.

Last edited: Mar 9, 2015
11. Mar 9, 2015

### Mentz114

12. Mar 9, 2015

### dextercioby

If you leave aside 2 terms of the Lagrangian action, you can't show the gauge symmetry anymore.

13. Mar 9, 2015

### Mentz114

Yes, I dropped two terms. But, I was right in that the two terms I used are invariant under the 'traceless gauge' I proposed. See equations (5), (6), (7) and (8) in this

http://arxiv.org/pdf/hep-th/0606019v2.pdf

However I can see now that the full Lagrangian is invariant under the transformation $h'_{\mu\nu} = h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$ which is diffeomorphism.

So all my original doubts are settled. Progress at last.

Thanks again.

14. Mar 9, 2015

### dextercioby

Yes, the reduction from 4 to 2 terms in the presence of a subsequent tracing condition reminds one of the reduction of the standard E-m Lagrangian density to the so-called "Fermi form', if the Lorenz gauge is imposed from the start.

15. Mar 9, 2015

### Mentz114

This is from the first paper I cited

The real underlying principle of GR has nothing to do with coordinate invariance
or equivalence principles or geometry, rather it is the statement: general relativity is the
theory of a non-trivially interacting massless helicity 2 particle. The other properties are
consequences of this statement, and the implication cannot be reversed.

(my bold emphasis)

Is this graviton-graviton interaction ?

If this is the case, are there interaction terms in the Lagrangian ?

If so, I take that to mean an interchange of something between gravitons, or possibly gravitons changing into other gravitons.

To get sourced gravitons do we need a source term in the Lagrangian ?

16. Mar 9, 2015

### dextercioby

Yes, GR is perturbatively an infinite sum of graviton-graviton interaction terms. Yes, gravitons can self-couple, but there is no Yang-Mills version of GR, as shown by Henneaux et al. here: N. Boulanger, T. Damour, L. Gualtieri, M. Henneaux, Inconsistency of
interacting, multi-graviton theories, Nucl. Phys. B597 (2001) 127–171

17. Mar 10, 2015

### haushofer

If you see a tensor, you should always ask the question "tensor under WHAT transformations?". In this case you have linearized gct's; linearized, because there is only a partial derivative in the transformation of the metric perturbation h, and not a covariant one. In this case: Yes, the FP-action is covariant wrt these linearized gct's, but you should check that explicitly by plugging the transformation into the action and see that you get at most total derivatives in the Lagrangian.

Another example: the Newtonian potential is often called a scalar, but it is only a scalar under the Galilei group. Most certainly NOT under gct's! Otherwise we wouldn't have a Newtonian form of the equivalence principle. A related example is the connection, which is not a tensor under gct's, but it is a tensor under the Poincare group.

18. Mar 10, 2015

### Mentz114

Interesting.

It seems the summing of a series is not required, as Deser says

We now derive the full Einstein equations, on the basis of the same self-coupling requirement,
but with the advantages that the full theory emerges in closed form with just one added (cubic)
term, rather than as an infinite series,

I don't know if this is important but it makes it easier for me to understand the theory.

19. Mar 10, 2015

### Staff: Mentor

There's another wrinkle here, too, though. Suppose I start with a spin-2 field theory on flat Minkowski spacetime, and compute this perturbative sum. As mentioned here, Deser showed that this infinite sum actually converges to the Einstein-Hilbert action. But many spacetimes described by this action do not have the same topology as flat Minkowski spacetime; for example, Schwarzschild spacetime has topology $R^2 \times S^2$, not $R^4$. So viewing GR this way requires one to believe that summing the infinite perturbation series can change the topology of the underlying spacetime in which the fields are defined, which doesn't seem right; or, alternatively, one must believe that this view of GR is only valid for solutions that have the same topology as the underlying spacetime, which leaves out many important solutions.

Last edited: Mar 11, 2015
20. Mar 10, 2015

### dextercioby

But perturbative methods are not simple approximations. They neglect the full topology, they are only 1st order variations of the flat spacetime metric. That way the full nonlinear theory becomes linear. The result of Henneaux et al. merely states that the only way a spin 2 field can self-couple is the perturbative Hilbert-Einstein action, plus the important fact that there's only 1 type of gravitons.