# Deriving einbein action from Polyakov $p$-brane action

In this paper the author derives a "Polyakov style" $$p$$-brane action which is given by

$$$$S_{p}=-\frac{T_{p}}{2} \int d^{p+1} \xi \sqrt{-g}\left(g^{A B} h_{A B}-(p-1)\right) \tag{1}$$$$

where $$h_{AB} = \partial_AX \cdot \partial_B X$$ is the induced metric on the world volume of the $$p$$-brane, $$g_{AB}$$ is the intrinsic world volume metric, and $$T_p$$ is the $$p$$-brane tension. The sign convention is $$(-,+,\ldots,+)$$

A $$0$$-brane is a point particle with a non-negative mass $$m = T_0$$. So, substituting $$p=0$$ in the above equation we have

\begin{aligned} S_{PP} &= -\frac{m}{2} \int d\tau \sqrt{-g_{00}} \left( g^{00} \dot{X}^2 +1 \right), \ h_{00} =\dot{X}^2 \equiv \partial_0X \cdot \partial_0 X\\ &= \frac{1}{2} \int d\tau \left( \frac{m}{\sqrt{-g_{00}}} \dot{X}^2 - m \sqrt{-g_{00}} \right) \ \end{aligned} \tag{2}

where I have used $$g^{00} = 1/g_{00}$$.

This form is similar to the famous einbein action for a point particle

$$$$S_{einbein} = \frac{1}{2} \int d\tau \left( \frac{\dot{X}^2}{e} - e m^2 \right) \tag{3}$$$$ In order to make (2) have the same form of (3), I defined

$$e \equiv \frac{\sqrt{-g_{00}}}{m} \tag{4}$$

I'm not sure if this definition is consistent, because the einbein action is supposed to work for both massive and massless particles and my $$e$$ is ill defined for $$m=0$$. Where's my mistake?

• Looks like a duplicate of physics.stackexchange.com/q/137857, supplemented by the beginning of physics.stackexchange.com/q/240065 (in particular, Qmechanic's answers) Jun 29, 2021 at 4:40
• Does this answer your question? Variational principle for a point particle (massive or massless) in curved space Jul 1, 2021 at 2:53
• Coincidentally answered this also in the process of answering this. The point is that while $(4)$ is true for $m \neq 0$, if $m = 0$ then we even have $m = 0$ in $(2)$ so the derivation 'fails' in the $m = 0$ case. But clearly $(3)$ works even in the $m = 0$ case. This is exactly the same behavior that occurs in the standard derivation of $(3)$ as can be seen in my post above, it's actually a good sign that the $m = 0$ case fails to blindly transfer over, it's as if in $(3)$ we are taking an 'analytic continuation' to the $m = 0$ case. Sep 20, 2021 at 1:08