Gauge Invariance of Transverse Traceless Perturbation in Linearized Gravity

[PreposterousUniverse]

In linearized gravity we define the spatial traceless part of our perturbation $h^{TT}_{ij}$. For some reason this part of the perturbation should be gauge invariant under the transformation $$h^{TT}_{ij} \rightarrow h^{TT}_{ij} - \partial_{i}\xi_{j} - \partial_{j}\xi_{i}$$ Which means that the components of $h^{TT}_{ij}$ does not change under such a transformation. But I can't see why this should be true in general.

 

[haushofer]

It's been a while for me, and you don't give any references which makes it harder to answer your question, but isn't this whole business about fixing a gauge? I also don't see why your spatial traceless part is gauge invariant in general; it isn't. So I have a suspicion that this part is only gauge invariant after fixing part of the gauge. See e.g. "The basics of gravitational wave theory",

https://iopscience.iop.org/article/10.1088/1367-2630/7/1/204

eqns. of 2.37 and 2.44. Also, after eqn.2.20: "A metric perturbation that has been put into TT gauge..." So part of the gauge has already been fixed.

 

[vanhees71]

There is sometimes confusion about what "gauge fixing" means.

Let's start with the somewhat simpler case of the em. field, and I stick to special relativity for simplicity. You start with the homogeneous Maxwell equations and introduce the vector potential, leading to the conclusion that there must always be a vector field such that
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
Now it's immediately clear that any
$$A_{\mu}'=A_{\mu}+\partial_{\mu} \chi$$
describes the same physics, because it doesn't change the observable Field $F_{\mu \nu}$.

Now consider the inhomogeneous Maxwell equations (in Heaviside Lorentz natural units with $c=1$)
$$\partial_{\mu} F^{\mu \nu}=j^{\nu}.$$
Plugging in the potentials, you get
$$\Box A^{\mu} - \partial^{\mu} \partial_{\nu} A_{\nu}=j^{\mu}.$$
Since now $A^{\mu}$ is only determined up to a gauge transformation, we can impose one arbitrary condition on it, which is usually called a "choice of gauge" or "gauge fixing". In this relativistic notation it's almost irresistable to choose the Lorenz gauge condition,
$$\partial_{\mu} A^{\mu}=0,$$
because then, in this Lorentz components of the fields and the spacetime vector, everything decouples into four (apparently) independent wave equations
$$\Delta A^{\mu}=j^{\mu}.$$
Now the Lorenz-gauge condition doesn't fix the gauge completely. You can still use an arbitrary gauge field $\chi$ to put
$$A_{\mu}'=A_{\mu}+\partial_{\mu} \chi.$$
In order to keep the Lorenz gauge condition fulfilled (assuming $A_{\mu}$ does fulfill it), you get the constraint
$$\Box \chi=0.$$
So you can still, even staying within the class of Lorenz-gauge potentials, apply some restricted gauge transformation.

This doesn't help too much for the general case, because you have to solve the wave equation anyway for given sources $j^{\mu}$, for which you can use the retarded propagator "inverting" the d'Alembert operator, which is the usually needed solution in classical electromagnetics.

The issue, however changes for the free-field case, i.e., for $j^{\mu}=0$. Then you can impose as additional condition also $A^0=0$ without contradicting the wave equations for the potentials, and only with such an additional condition, you get the correct number of field-degrees of freedom, i.e., only then you completely "fix the gauge". Indeed, when you use a restricted gauge transformation to make both (!)
$$A^0=0, \quad \partial_{\mu} A^{\mu}=\vec{\nabla} \cdot \vec{A}=0,$$
you immediately see that the waves for the now "radiation-gauged" potential are transverse (solenoidal), as it shoud be, and any plane-wave mode has two polarization states (and not three as one might expect from a vector field, but that's a specialty of free massless relativistic fields, which all have two polarization states (except for a scalar field, which has of course only one polarization state), which can be choosen as the helicity, which for the em. field are the two circular polarized modes for each mode characterized by the wave vector $\vec{k}$, $\omega=|\vec{k}|$).

The same issue occurs for the 2nd-rank symmetric tensor field, and in the linearized version this describes a gravitational wave and can be treated as a special-relativistic field. One only needs to know that gauge invariance here is inherited from the general coordinate invariance (which is the gauge invariance principle underlying GR), which translates to the equivalence of the small deviations $h_{\mu \nu}$ of the metric from flat Minkowski-space metric, as in $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$. The linearized Einstein equation reads
$$\box h_{\mu \nu}+\partial_{\mu} \partial_{\nu} {h^{\rho}}_{\rho}-\partial_{\rho} \partial_{\nu} {h^{\rho}}_{\mu}-\partial_{\rho} \partial_{\mu} {h^{\rho}}_{\nu}=-16 \pi G \left (T_{\mu \nu}-\frac{T}{2} \eta_{\mu \nu} \right).$$
Now general covariance implies in linearized form that
$$h_{\mu \nu}'=h_{\mu \nu}-\partial_{\nu} \epsilon_{\mu} - \partial_{\mu} \epsilon_{\nu}$$
is equivalent with $h_{\mu \nu}$ for arbitrary vector fields $\epsilon_{\mu}$. It's also keeping the linearized Einstein equation exactly invariant, i.e., the left-hand side doesn't change under this "gauge transformation".

To decouple the equations for the components the most convenient gauge conditions are (note the arbitrary gauge field is now a vector field and thus we can impose 4 gauge conditions)
$$\partial_{\nu} {h^{\mu}}_{\mu}=2 \partial_{\mu} {h^{\mu}}_{\nu}.$$
This is the equivalent of the Lorenz gauge condition in E&M, because then the linearized field equation indeed decouples to
$$\Box h_{\mu \nu}=-16 \pi G \left (T_{\mu \nu}-\frac{T}{2} \eta_{\mu \nu} \right).$$
Now, as in E&M this gauge fixing condition doesn't fix the gauge completely either, because we can again make a gauge transformation
$$h_{\mu \nu}'=h_{\mu \nu} -\partial_{\mu} \epsilon_{\nu} -\partial_{\nu} \epsilon_{\mu}.$$
In order that $h_{\mu \nu}'$ still fulfills the above gauge condition the constraint on $\epsilon_{\mu}$ simply is
$$\Box \epsilon_{\mu}=0.$$
Now again for free fields, i.e.,
$$\Box h_{\mu \nu}=0$$
you have still four restricted gauge transformations to fix the gauge, and for a plane wave you can show that this can be indeed be used to show that for any $\vec{k}$ there are only two independent polarization degrees of freedom of a gravitational wave as is already known from the general theory of Poincare covariant field equations of free massless fields of any spin.