# Why is a gauge transformation more than just a coordinate transformation?

In linearized gravity, we may make a coordinate transformation $x'^\mu=x^\mu+\epsilon\xi^\mu$. This coordinate transformation leads to the perturbation of the metric (away from Minkowski) as $h'_{\mu\nu}=h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$.

Solving for the connections, Riemann, Ricci tensors, we arrive at \begin{equation} R_{\mu\nu}=\frac{1}{2}\left[ -\Box h_{\mu\nu}+\partial_\mu V_\nu +\partial_\nu V_\mu\right] \end{equation} (where the Ricci tensor written above is first order in small perturbations). At this step, it is often written that if we choose $V_\mu=0=\partial_\kappa h^\kappa_\mu-\frac{1}{2}\partial_\mu h$, then we may the equation above becomes $R_{\mu\nu}=\Box h_{\mu\nu}=0$ (assuming we are in vacuum, $T_{\mu\nu}=0$).

1. I get the impression that a gauge transformation is more than just a change of coordinates, but I have a hard time visualizing this. Is there an intuitive way to understand how a gauge transformation is more than just a change of coordinates?

2. Choosing $V_\mu=\partial_\kappa h^\kappa_\mu-\frac{1}{2}\partial_\mu h =0$ will make $R_{\mu\nu}=\Box h_{\mu\nu}=0$, but why are we allowed to make this choice? It looks like we are just choosing to put a restriction on the form of $h_{\mu\nu}$ (I realize that this is not really what's going on, but this is what it looks like to me).

3. I read through this question, but am still somewhat uncomfortable with the topic. What are the allowed/restricted forms of $V_{\mu}$ that we may choose (and why)?

## 2 Answers

I get the impression that a gauge transformation is more than just a change of coordinates, but I have a hard time visualizing this. Is there an intuitive way to understand how a gauge transformation is more than just a change of coordinates?

It is basically a coordinate transformation. But it is better to use the "active" perspective to treat this matter.

Consider a spacetime $(M^0,T^0)$, where $M^0$ is the spacetime manifold and $T^0$ is a preferred tensor field, or a collection of preferred tensor field.

When we do linear perturbation of the spacetime, we take a whole family of such spacetimes $(M^\epsilon,T^\epsilon)$ with $\epsilon$ being a real parameter that sort of signals how "close" $M^\epsilon$ is to $M^0$.

Our "perturbed" quantities live in $M^\epsilon$, our "unperturbed" quantities live in $M^0$, however we'd like to work on $M^0$ all the time, even with perturbed quantities.

For the perturbation scheme to be realized, you need $M^0$ and $M^\epsilon$ to be diffeomorphic, so you assume there is a smooth family of diffeomorphisms $$\psi_\epsilon:M^0\rightarrow M^\epsilon,$$ where this family satisfies $\psi_0=\text{Id}_M$ (it is the identity map for $\epsilon=0$).

We pull back $T^\epsilon$ to $M^0$ as $$T=(\psi_\epsilon)^*T^\epsilon\approx T^0+\epsilon T^{(1)},$$ where at the last equality we assumed that the perturbation is linear eg. $M^\epsilon$ differs from $M^0$ only to first order.

The diffeomorphism $\psi_\epsilon$ is not unique however, so consider another one, $\phi_\epsilon$. With this, we can define $$T'=(\phi_\epsilon)^*T^\epsilon\approx T^0+\epsilon T^{(1)}{'}.$$

The question is how are the primed and unprimed perturbations related. To answer this, we note that the difference between the two diffeos can be described on $M^0$ as $$\phi_\epsilon=\psi_\epsilon\circ\chi_\epsilon,$$ where $$\chi_\epsilon=\psi^{-1}_{\epsilon}\circ\phi_\epsilon.$$

This satisfies $\chi_0=\text{Id}_M$, so there is a vector field $X$ corresponding to it: $$X=\frac{d}{d\epsilon}\chi_\epsilon |_{\epsilon=0}.$$ Expressing the primed $T$ with $\chi$: $$T'=(\phi_\epsilon)^*T^\epsilon=(\psi_\epsilon\circ\chi_\epsilon)^*T^\epsilon=(\chi_\epsilon)^*(\psi_\epsilon)^*T^\epsilon\approx(\chi_\epsilon)^*(T^0+\epsilon T^{(1)}).$$ Now we apply the $O(\epsilon)$-approximation to $\chi_\epsilon$ too. For an arbitrary tensor field $Q$, we have $$\mathcal L_XQ=\frac{d}{d\epsilon}(\chi_\epsilon)^*Q |_{\epsilon=0},$$ where $\mathcal L$ is the Lie derivative, so we have $$T'\approx T^0+\epsilon T^{(1)}+\epsilon\mathcal L_X(T^0+\epsilon T^{(1)})\approx T^{0}+\epsilon(T^{(1)}+\mathcal L_XT^0)+O(\epsilon^2),$$ so the perturbing field itself transforms as $$T^{(1)}{'}=T^{(1)}+\mathcal L_X T^0.$$

For your situation, $T^0=\eta$, $X=\xi$, $T^{(1)}=h$, and we have $$\mathcal L_\xi \eta_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu,$$ so we get the transformation rule $$h'_{\mu\nu}=h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu.$$

So basically, the conclusion is, from a more geometrical point of view, the gauge transformation exists because there is a certain ambiguity in identifying the perturbed spacetime with the background spacetime. The ambiguity is expressed with the gauge transformation.

If you change the metric perturbation by $\partial_{\mu}V_{\nu} + \partial_{\nu}V_{\mu}$, then a direct computation will show that the linearized versions of the Riemann tensor and the Ricci tensor remain unchanged. This is akin to Electromagnetism, where you have a covector field $A$ and the Maxwell tensor $F = dA$, and adding any covector $A^{\prime}$ to $A$ leaves $F$ unchanged as long as $dA^{\prime} = 0$. So you have:

-) A background spacetime $M$ (Minkowski).

-) A "potential", 2-covariant tensor field $h_{\mu\nu}$ defined on $M$ (akin to the potential $A_{\mu}$ of EM)

-) The Riemann tensor $R_{\alpha\beta\mu\nu}$ (akin to the Maxwell tensor $F_{\mu\nu}$).

-) A set of rules that say how you can change the potential without changing the "active" quantity $R_{\alpha\beta\mu\nu}$.

So for any physical situation of interest, you are free to chose between different versions of the potential (you would want to choose the one that simplifies the description of the system under consideration). Similar to a change of coordinates, really...