Geodesic equations from action with auxiliary field

Question

A textbook says that the geodesic equations (for both massive and massless) can be derived from the following action:

$$ S = -\frac{1}{2} \int d\tau \:\eta \: (\eta^{-2} \dot{x}^\mu \dot{x}^\nu g_{\mu\nu}(x) + m^2) $$ where $\eta$ is an auxiliary field. The signature convention is here $(+,-,-,-)$. The e.o.m. derived are $$ \eta^{-1}\dot{x}^\mu \dot{x}^\nu \partial_\sigma g_{\mu\nu}(x) - \frac{d}{d\tau}( 2 \eta^{-1}g_{\sigma \mu}(x) \dot{x}^\mu ) = 0, \quad \quad \eta^{-2} \dot{x}^\mu \dot{x}^\nu g_{\mu\nu}(x) = m^2.$$

In the massive case, by redefining $\tau$ to be proper time, $\eta^{-1} = m$ and the geodesic equations come out.

1) I'm confused about the massless case: It seems that $\eta$ is undetermined as we have $\eta^{-2} \:0 =0$.

The text then goes to say that one can think of $S$ as the action of 4 fields $x^{\mu}$ living on a 1D space with metric $\eta(\tau)^2$. I understand that $ds^2 = d\tau^2 \eta(\tau)^2$ and therefore $d\tau \eta$ is the invariant measure.

2) I don't really understand what would motivate the factor of $\eta^{-2}$ next to $\dot{x}^\mu \dot{x}^\nu g_{\mu\nu}(x)$ when writing a theory for these 4 fields though. I realise $$\eta^{-2}\dot{x}^\mu \dot{x}^\nu g_{\mu\nu}(x) = \frac{dx^\mu}{ds}\frac{dx^\nu}{ds} g_{\mu\nu}(x).$$ I guess I don't get why one would want to write derivatives w.r.t. $s$ instead of $\tau$ when writing a theory of 4 fields on 1D space with coordinate $\tau$.

Answer 1

1. The einbein field $\eta(\tau)\neq 0$ is not a dynamical field because there is no $\dot{e}(\tau)$ present. It is a so-called auxiliary field.

2. The one-form $\omega:= \eta(\tau) \mathrm{d}\tau\in \Gamma(T^{\ast}I)$ on the 1-dimensional world-line (WL) manifold $I$ is invariant under WL reparametrizations $\tau\to\tau^{\prime}=f(\tau)$. WL reparametrization invariance is a gauge symmetry, cf. e.g. this Phys.SE post.

3. In the massive case $m>0$, the EL eq. for $\eta$ determines that $(m\eta)^2\approx g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}$ on-shell, cf. e.g. this Phys.SE post. It is possible to gauge-fix $\eta(\tau)= 1/m$, cf. e.g. this Phys.SE post.

4. In the massless case $m=0$, the reciprocal einbein field $\lambda(\tau)\equiv\frac{1}{\eta(\tau)}\neq 0$ is an undetermined Lagrange multiplier, cf. e.g. this Phys.SE post.

Answer 2

1) $\eta$ is always undetermined, there's nothing special about the massless case. You can see this by noting that you can write the EOM as

$$\ddot{x}^\mu + \Gamma^\mu{}_{\nu\rho}\dot{x}^\nu \dot{x}^\rho = \frac{\dot{\eta}}{\eta} \dot{x}^\mu,$$

so that the right hand side is completely arbitrary: just pick any nonzero $\eta(\tau)$, and you'll have a solution. Of course, changing $\eta$ and keeping the initial conditions fixed will just give you the same solution up to reparametrization.

2) This is just one way to understand why the action with $\eta$ works (beyond just "it gives the correct equations"), and how anyone would come up with it. We have a 1D metric $\gamma$ with just one component $\gamma_{\tau\tau} = \eta^2$. Now consider $x^\mu(\tau)$ for a fixed value of $\mu$: we're considering just one of the fields, not the whole vector. The derivative $dx^\mu/d\tau$ is the (single) component of a covector tangent to the worldline, so to get its square we have to form the combination $\gamma^{\tau\tau} \dot{x}^\mu \dot{x}^\mu$ (not summing over $\mu$), just like the square of a four gradient in spacetime is $g^{\mu\nu}\varphi_{,\mu}\varphi_{,\nu}$.

This explains the $\eta^{-2}$ factor: it's just the inverse metric. If you don't include it, you no longer have reparametrization invariance, which is just general coordinate invariance applied to our 1D worldline.