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

\begin{equation} 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} \end{equation}

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{equation} \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} \end{equation}

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

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

\begin{equation} S_{einbein} = \frac{1}{2} \int d\tau \left( \frac{\dot{X}^2}{e} - e m^2 \right) \tag{3} \end{equation} 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?

  • $\begingroup$ 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) $\endgroup$ Jun 29, 2021 at 4:40
  • $\begingroup$ Does this answer your question? Variational principle for a point particle (massive or massless) in curved space $\endgroup$ Jul 1, 2021 at 2:53
  • $\begingroup$ 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. $\endgroup$
    – bolbteppa
    Sep 20, 2021 at 1:08


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