Why isn't TT gauge over constrained?

[ergospherical]

In linearised theory, the polarisation tensor $A_{\mu \nu}$ (defined through $\bar{h}_{\mu \nu} = A_{\mu \nu} e^{ik_{\rho} x^{\rho}}$) transforms under a gauge shift $x \mapsto x + \xi$ with a harmonic function $\xi_{\mu} = X_{\mu} e^{ik_{\rho} x^{\rho}}$ like:
$$A'_{\mu \nu} = A_{\mu \nu} + i(k_{\mu} X_{\nu} + k_{\nu} X_{\mu} - (k \cdot X) \eta_{\mu \nu})$$If we form 4 equations $A'_{\mu 0} = 0$ (i.e. impose longitudinal gauge), then we have 4 equations in 4 unknowns $X_{\mu}$ that we can solve. So good so far...

If you also want to impose tracelessness, $A' = 0$, then since
$$A' = A - 2i(k \cdot X)$$you need to choose the $X_{\mu}$ such that $k \cdot X = -(i/2)A$. How come you can do this? Now you have 5 equations for 4 unknowns, so surely the system is over constrained compared to the 4 available degrees of freedom?

 

[PeterDonis]

the 4 available degrees of freedom?

How many degrees of freedom does a symmetric second rank tensor have? More than 4.

 

[ergospherical]

How many degrees of freedom does a symmetric second rank tensor have? More than 4.

I mean the 4 constraints $A'_{\mu 0} = 0$
Overall you have 10 - (Lorenz) - (Longitudinal) = 10 - 4 - 4 = 2, i.e. $A_{+}$ and $A_{\times}$

 

[PeterDonis]

I mean the 4 constraints $A'_{\mu 0} = 0$

Yes, but that has nothing to do with making $A$ traceless.

Overall you have 10 - (Lorenz) - (Longitudinal) = 10 - 4 - 4 = 2, i.e. $A_{+}$ and $A_{\times}$

2 is what what you have left after you have made all the gauge transformations you can make to simplify things. You start with 10 and then make gauge choices that eliminate a total of 8, leaving 2 physical degrees of freedom. Those gauge choices will involve a total of 8 equations that you can solve for 8 unknowns to eliminate 8 of 10 tensor components.

 

[PeterDonis]

we form 4 equations $A'_{\mu 0} = 0$ (i.e. impose longitudinal gauge)

Wouldn't you want to form a transverse gauge? Transverse traceless is the usual way of reducing to the 2 physical degrees of freedom (polarizations) in a gravitational wave.

 

[ergospherical]

The question is to show that you can make the polarisation satisfy $A_{\mu 0} =0$ and $A=0$. After already imposing Lorenz gauge you have residual gauge freedom in the form of harmonic functions $\xi$ of the form I specified, since these don’t affect the Lorenz gauge condition. But this residual gauge freedom gives you only 4 equations, one for each component of $X$, so it’s not obvious to me why it’s true that we can impose the 5 conditions…

 

[PeterDonis]

After already imposing Lorenz gauge

Which involves imposing 4 conditions, to remove 4 degrees of freedom.

you have residual gauge freedom in the form of harmonic functions $\xi$ of the form I specified, since these don’t affect the Lorenz gauge condition. But this residual gauge freedom gives you only 4 equations

Yes. One of them is tracelessness, and the other three are to make the tensor fully transverse.

But if you instead impose longitudinal gauge, which is what you said in the OP, then you can't also make the tensor transverse-traceless. You have to pick one or the other.

it’s not obvious to me why it’s true that we can impose the 5 conditions…

You can't. See above.

 

[PeterDonis]

If we form 4 equations $A'_{\mu 0} = 0$ (i.e. impose longitudinal gauge)

Btw, I'm not sure this constraint actually gives you "longitudinal gauge". The equation $A_{\mu 0} = 0$ is satisfied by tranverse-traceless gauge.

I'm also not sure this constraint is entirely independent of Lorenz gauge $\partial^\mu A_{\mu \nu} = 0$.