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As discussed here and within, the Lagrangian for a massless particle, using the $(-,+,+,+)$ metric signature, is

$$L = \frac{\dot{x}_\mu \dot{x}^\mu}{2e} - V,\tag{1}$$

where $\dot{x}^\mu := \frac{dx^\mu}{d\lambda}$ is the velocity, $\lambda$ is some worldline parameter, $e$ is the auxiliary einbein and $V$ is the potential.

The EL equations give us the EOMs

$$\dot{x}_\mu \dot{x}^\mu = 0,\tag{2}$$ $$\ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho - \frac{\dot{e}\dot{x}^\mu}{e} + e\partial^\mu V = 0,\tag{3}$$

where $\Gamma^\mu_{\sigma\rho}$ are the Christoffel symbols of the metric $\eta_{\mu\nu}$ for some choice of coordinates. After this, I'm not sure how to proceed, for the following reasons.

In the $V=\text{constant}$ case, the system is underdetermined, and we are free to choose some $e$, such as setting $e=1$. We then get the consistent EOMs

$$\dot{x}_\mu \dot{x}^\mu = 0,\tag{4}$$ $$\ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho = 0.\tag{5}$$

In the general case, however, we seem to end up with the two EOMs being inconsistent.

For example, suppose we use Cartesian coordinates and a $z$-direction potential such as $$V = z,\tag{6}$$ choosing initial conditions satisfying the null velocity condition such as $\dot{x}^\mu \left(0\right) = \left(1,0,0,1\right)$. The EOM for the coordinates becomes

$$\ddot{x}^\mu = \left(0,0,0,-1\right),\tag{7}$$

yielding $\dot{x}^\mu \left(\lambda\right) = \left(1,0,0,1-\lambda\right)$, which fails to satisfy the null velocity condition for $\lambda \neq 0$.

Qmechanic
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tomdodd4598
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2 Answers2

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You are free to choose $e$. Your freedom to choose $e$ corresponds to your freedom to choose how to parameterize the curve. You can think of reparameterization invariance as a gauge invariance on the world line (see Section 1.1.2 of Tong's string theory notes). In particular, if you chose $e=1$, your first equation is the condition that the particle is massless, and the second equation becomes the standard form of the geodesic equation with a force term equal to $\partial^\mu V$.

Andrew
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  • But if I just choose $e=1$ after getting the EOMs, isn't the system overdetermined? On the other hand, if I set $e=1$ before getting the EOMs, then surely the condition that $\dot{x}_\mu \dot{x}^\mu = 0$ isn't enforced? – tomdodd4598 Sep 09 '22 at 03:08
  • @turbodiesel4598 The system is actually underdetermined if you try to solve for $e$ as well as $x$. (See https://en.wikipedia.org/wiki/Gauge_theory). However, you are right it is important that you do not set $e=1$ until after you derive the equations of motion. This is because the operations of differentiating (in this case, by varying the action) and plugging in a value for the variable you are differentiating do not commute. – Andrew Sep 09 '22 at 12:03
  • Indeed, I've seen that solving for $e$ and $x$ is not possible in the case of no potential. However, with a potential, if I get the EOMs, then fix $e$, I now have the null four-velocity condition as well as the $x$ EOM, which seems to me like an overdetermined system. On the other hand, if I just ignore the null condition and solve the $x$ EOM, then the null condition isn't necessarily enforced for all potentials $V$. I've updated the question to reflect the doubts I have. – tomdodd4598 Sep 09 '22 at 15:33
  • @turbodiesel4598 it's not overdetermined. If the particle had mass, you would have a condition for a timelike four velocity instead of a null one. Does that bother you? – Javier Sep 09 '22 at 18:23
  • @Javier no, because in that case I'm not fixing $e$, so I have one more equation, solving for one more unknown than in the massless, fixed $e$ case. – tomdodd4598 Sep 09 '22 at 18:46
  • @turbodiesel4598 You shouldn't be able to solve for $e$ even when there's a potential. If you can, then something is wrong. I am not sure offhand where the problem is in the equations you presented. – Andrew Sep 09 '22 at 18:54
  • The problem was that the numerator in my expression for $e$, $\dot{x}\mu \left( \ddot{x}^\mu + \Gamma^\mu{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho \right)$ is equal to $2 \dot x_\mu \nabla_{\dot x} \dot x^\mu$, which is zero. This means we instead have the interesting condition $e\dot x^\mu \partial_\mu V = 0$. I have updated the question again to make it a little clearer. – tomdodd4598 Sep 10 '22 at 15:56
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It is usually implicitly assumed that OP's Lagrangian (1) for the massless relativistic point particle exhibits world-line (WL) reparametrization covariance, cf. point 4 in my related Phys.SE answer here. This puts strong conditions on the potential $V$, i.e. the 1-form $V\mathrm{d}\lambda$ should be a WL invariant. One allowed potential is an E&M background $V=-qA_{\mu}(x)\dot{x}^{\mu}$. OP's example (6) is not allowed.

Qmechanic
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