The origin of the einbein

Posted by David Zaslavsky on August 14, 2010 1:56 AM — Edited August 15, 2010 7:34 PM

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You know that feeling you get when you’re reading a textbook and the authors start pulling equations out of thin air on the second page of the second chapter? No? Well then, my congratulations on having a life.

But seriously, I came across a gem of this sort in Superstring Theory volume 1 by Green, Schwartz, and Witten. Starting with the action for the relativistic point particle,

$$S = -m\int\udc s = -m\int\udc\tau\sqrt{-\dot{x}^2}$$

they introduce an “einbein” $e(\tau)$ to eliminate the problems that (1) the formula has a square root and (2) it doesn’t apply to massless particles. Fine, but WTF is an einbein anyway? And where do you come up with the equation in the very next sentence,

$$S = \frac{1}{2}\int\bigl(e^{-1}\dot{x}^2 - em^2\bigr)\udc\tau$$

I’ve been wondering about this for far too long. Here’s an explanation: we know experimentally that massless particles travel at the speed of light, along null geodesics. But the spacetime interval $\udc s$ along a null geodesic is zero. So when you try to compute the classical point-particle action $-m\int \udc s$ for a massless particle, it fails for two reasons: $m$ is zero and $\udc s$ is zero. The first of those seems maybe less serious, because it’s just a constant factored out of the action, but with $\udc s = 0$, there’s no way to even meaningfully do the integral.

The sensible way to work around this would be to try to get something inside the integral that can actually be integrated, i.e. something that isn’t equal to zero. Well, we’ve got two quantities that both go to zero in the massless limit; why not try to take a ratio and hope it’s well defined? With this approach, we can rewrite the action as

$$S = -m^2\int\frac{\udc s}{m}$$

Now, we can rewrite $\udc s$, using its definition $\udc s^2 = -g_{\mu\nu}\udc x^\mu\udc x^\nu$, in terms of $\dot{x}$,

$$\frac{\udc s^2}{\udc\tau} = -g_{\mu\nu}\ud{x^\mu}{\tau}\ud{x^\nu}{\tau} \equiv -\dot{x}^2$$

so the action becomes

$$S = -m^2\int\udc\tau\frac{\sqrt{-\dot{x}^2}}{m}$$

Then we can introduce the abbreviation $e\equiv\sqrt{-\dot{x}^2}/m$. This is the einbein. It generalizes the original integrand in a way that hopefully will not fall flat for a massless particle. From now on I’m going to treat $e(\tau)$ as an independent function, ignoring its definition in terms of $\dot{x}$ and $m$, and let’s see if we can recover that definition from the math.

So we’ve got the action

$$S = -m^2\int e\udc\tau$$

There’s just one problem, though: this leads to a useless (though at least accurate) equation of motion. Using the Euler-Lagrange equation to find the EOM for the one variable, $e$, gives

$$\ud{}{\tau}\pd{L}{\dot{e}} = \pd{L}{e} \quad\to\quad 0 = -m^2$$

($L = -m^2 e$) Well, what can we do about this? Perhaps try to eliminate the problematic factor of $m$? Using the definition of $e$, we can get

$$S = \int\udc\tau \dot{x}^2 e^{-1}$$

This is different: we have two variables, and thus two equations of motion. For $x$,

$$2(\ddot{x}e^{-1} - \dot{x}e^{-2}\dot{e}) = 0$$

and for $e$,

$$0 = -\dot{x}^2 e^{-2}$$

But unfortunately, neither of these is still specific enough. The EOM for $e$ tells you nothing except $\dot{x}/e = 0$, and the one for $\dot{x}$ can be reduced to $e = C\dot{x}$, where $C$ is an unknown constant. (From the originally chosen definition of $e$, you could tell that $C = i/m$, but we’re pretending that definition didn’t exist, remember?)

Evidently we can’t completely eliminate $m$ from the action, but we can’t leave it without $\dot{x}$ either. So the logical next thing to try is making a linear combination of the two forms of the action,

$$S = \int\udc\tau(A\dot{x}^2 e^{-1} - Bm^2 e)$$

where $A + B = 1$ for normalization. With this choice, the equation of motion for $e$ yields

$$0 = \dot{x}^2 + \frac{B}{A}e^2m^2$$

See, now that’s more sensible. This has the correct dependence on $m$, and it’s completely determined in terms of the quantities that enter into the action. Of course, $e$ still does depend on $\frac{B}{A}$, which could have any value, but it’s not a completely unknown quantity like $C$ was. So evidently there’s a whole spectrum of possible ways that the einbein $e$ could be defined, depending on exactly how we choose to define the action. This is actually related to the gauge invariance discussed in the next few lines of the GSW book.