# Geodesic equations from action with auxiliary field

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 $$\tau$$.

• @Qmechanic No. From Rubakov's 'Introduction to the theory of the early universe'. First volume. Appendix A. – Rudyard Dec 27 '18 at 19:48

## 2 Answers

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.

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.