Why do we impose de Donder gauge?

Question

In the field language, a massless particle corresponds to irreducible representations of the Lorentz group. In particular, given a spin-2 massless particle, we can embed the creation and annihilation operators into a symmetric traceless tensor $h_{\mu\nu}^T$ which satisfies $$ \square h_{\mu\nu}^T=0\,,\qquad\partial^\mu h_{\mu\nu}^T=0\,,\qquad \eta^{\mu\nu}h_{\mu\nu}^T=0.\,\qquad \qquad (1) $$

It is useful to consider an action for a massless spin-2 particle written in terms of a tracefull and symmetric tensor $h_{\mu\nu}$ which enjoys gauge symmetry $\delta h_{\mu\nu} = \partial_{(\mu}\xi_{\nu)}$ that is the linearised version of the Einstein-Hilbert action. In the following, I restrict to flat space-times.

In (almost) all text-books, it is used to fix the so called de Donder gauge

$$ \partial_\mu h^{\mu\nu}-\frac{1}{2}\partial^\nu h = 0. $$

Why do we use this gauge-fixing condition?

Why don't we impose $h=0$ and $\partial_{\mu}h^{\mu\nu}=0$ as gauge conditions separately, such that we reproduce Eq.(1) manifestly?

Answer 1

Actually this is not possible.

To impose the gauge $h=0$, I have to solve

$$ 2\partial\cdot \epsilon = -h\,,\qquad\qquad (1) $$

which always admits solutions.

This partial gauge fixing leaves a redundant gauge freedom $\delta h_{\mu\nu}=\partial_{(\mu}\epsilon^T_{\nu)}$ where $\partial\cdot \epsilon^T=0$.

If I want to remove $\partial_\mu h^{\mu\nu}=0$, then I have to solve

$$ \square \epsilon^\mu = \partial_\nu h^{\nu\mu}\,\qquad\qquad (2) $$ The quantity $\partial_\mu h^{\mu\nu}$ is a vector and we can decompose it into a transverse and longitudinal part

$$ \partial_\mu h^{\mu\nu} = A_\perp^\nu + \partial^\nu \pi\,\qquad\qquad (3) $$

Then, Eq.(2) implies

$$ \square \epsilon^\mu = A^{\mu}_\perp + \partial^\nu \pi\,\qquad\qquad (4) $$

As the parameter $\epsilon^\mu$ satisfies $\partial\cdot \epsilon=0$, the scalar field $\pi$ cannot be removed. Indeed, if I apply the divergence $\partial_\mu$ to both sides of Eq.(4) I get

$$ 0 = \square \pi $$

which is not true. Then, I am able to remove the transverse part of $\partial_\mu h^{\mu\nu}$, but not the longitudinal part.

Answer 2

It is pure nonsense to accept that there exists massless particle systems, premised on the belief that we don't know how to detect or measure the infinitesimal mass properties. The analytical analysis of differential geometry limits the fact-finding laboratory analysis of measuring these physical properties, and has no business in communicating quantum field theory with pure mathematical analysis. The nonsensical approach is to define transverse trace gauge invariant symmetric field equations to elementary particle systems belies the true understanding of quantum field theory. There are no known laws of symmetry conservation analysis coupled with definite calculations for doing quantum mechanics of physical systems that separates the WFE from the SFE. Even Herr Einstein's Field Equations being vacuum field equations possessing the properties of vacuum energy is not a coherent explanation of the properties of space-p-time.