Variation of action for massive point particle (pp)

Question

So I'm pretty sure I'm missing something obvious, but for the life of me I cannot replicate the step between 1.2.2 and 1.2.3 in Polchinski Vol 1. Basically, I'm trying to find the variation of:

$$S_{pp} = - m \int d\tau \, (-\dot X^\mu \,\dot X_\mu )^{1/2} \tag{1.2.2}$$

Which ends up being:

$$\delta S_{pp} = -m \int d\tau \, \dot u_\mu \, \delta X^\mu.\tag{1.2.3}$$

My best attempt is as follows:

$\delta S_{pp} = -m \int d\tau \, \delta (-\dot X^\mu \,\dot X_\mu )^{1/2} $

$\delta S_{pp} = -m \int d\tau \, ^1/_2 (-\dot X^\mu \,\dot X_\mu )^{-1/2} $ $(- \delta \, \dot X^\mu \,\dot X_\mu)$

$\delta S_{pp} = -m \int d\tau \, ^1/_2 (-\dot X^\mu \,\dot X_\mu )^{-1/2} $ $(- \delta \, \dot X^\mu \,\dot X_\mu) + (-\dot X^\mu \, \delta \,\dot X_\mu)$

Which rearranging because of the symmetry of the last two terms is:

$\delta S_{pp} = -m \int d\tau \, (-\dot X^\mu \,\dot X_\mu )^{-1/2} $ $+ (-\dot X^\mu \, \delta \,\dot X_\mu)$

So here I substitute $$u^\mu = \dot X^\mu(-\dot X^\mu \dot X_\mu)^{-1/2}\tag{1.2.4}$$ to get:

$\delta S_{pp} = -m \int d\tau \space u_\mu \space \delta \dot X^\mu $

Here is where the problem really gets me (I'm sure I messed up some indices before but you know...) I try to integrate by parts and get:

$\delta S_{pp} = -m \int d\tau \, (u_\mu \dot X^\mu) - \int d\tau( u_\mu \delta \dot X^\mu)$

Once again by the power rule we have:

$\delta S_{pp} = -m \int d\tau (\delta u_\mu) \dot X^\mu + u_\mu (\delta \dot X^\mu) - \int d\tau( u_\mu \delta \dot X^\mu)$

Canceling the final two terms I get:

$\delta S_{pp} = -m \int d\tau (\delta u_\mu) \dot X^\mu$

Which, I mean, is sorta close? I can see that there is clearly something I'm missing or some rule that I applied incorrectly or something, but cannot for the life of me figure it out. I though about using the power rule (again) or integrating by parts (again) but I get nowhere.

Answer 1

You just have to remember that you can get rid of total derivatives in the integral. For this reason, in these kind of computations, you will often find the sentence "up to total derivatives". Also, you are missing a minus sign. \begin{equation} \delta S=-m\int d\tau \space\delta\bigg({\sqrt{-\dot{x}^\mu \dot{x}_\mu}}\bigg)=-m\int d\tau\space\frac{-\dot{x}_\mu\delta\dot{x}^\mu}{\sqrt{-\dot{x}^\mu \dot{x}_\mu}} \end{equation} Now, substituting \begin{equation} u^\mu=\frac{\dot{x}^\mu}{\sqrt{-\dot{x}^\mu \dot{x}_\mu}} \end{equation} we get \begin{equation} \delta S=-m\int d\tau\space (-u_\mu)\delta\dot{x}^\mu \end{equation} (which is like the expression you got except for a minus sign you forgot) and integrating by parts \begin{equation} \delta S=-m\int d\tau\space \frac{d}{d\tau}(-u_\mu\delta{x}^\mu) -m\int d\tau\space\dot{u}_\mu\delta{x}^\mu \end{equation} You can discard the first integral, being a total time derivative, thus you get \begin{equation} \delta S= -m\int d\tau\space\dot{u}_\mu\delta{x}^\mu \end{equation}

Edit

I forgot to mention that you can switch the variation and the derivative, i.e. \begin{equation} \delta \dot{x}^\mu=\delta \frac{d}{dt}x^\mu=\frac{d}{dt} \delta x^\mu \end{equation}