# A question about the Lagrangian for a massive or massless point particle

In my lecture I learned that the action that can be applied to the light ray is written like below: \begin{equation*} S[x;e]=(1/2)\int [(1/e)g_{ab}\dot x^a\dot x^b-m^2e]ds \tag{1} \end{equation*} where $s$ is the variable that replaces the role of time and $e(s)>0$ is said to be another auxiliary variable that is similar to the Lagrange multiplier. The dot symbol on $x$ means differentiation with respect to $s$ and $a$,$b$ ranges from $0$ to $3$.

Then, when $m>0$, the lecture note sets $$\frac{\delta S}{\delta e}=0\tag{2}$$ and derive $$(-g_{ab}\dot x^a\dot x^b)/e^2-m^2=0.\tag{3}$$

Here is my question. What on earth does $\frac{\delta S}{\delta e}$ mean? And why set it zero? Isn't it that usually that $\delta S=0$? I am very confused.

• That is the functional derivative of the action with respect to the tetrad ($e$) field. In general, $\delta S/\delta\phi=0$ is the equation of motion for some field $\phi$ appearing in your Lagrangian. This is equivalent to $\delta S=0$. Both equations tell you that you are in a critical point in configuration space. – Bob Knighton Mar 30 '18 at 15:07
• Using the chain rule, $\delta S=(\delta S/\delta e)\delta e+(\delta S/\delta x)\delta x+\cdots$, so that $\delta S=0$ implies both $\delta S/\delta e=0$ and $\delta S/\delta x=0$. – Bob Knighton Mar 30 '18 at 15:20

1. $e$ is a worldline einbein field. It is a non-dynamical auxiliary field in the sense that $\dot{e}$ is not present.
2. In the massive case $m>0$: If we integrate out the $e$-field, the action (1) takes the familiar square root form. Conversely, one may view the introduction of $e$ as a trick to get rid of the square root.
3. In the massless case $m=0$: $\delta S /\delta e\approx 0$ implements a light-cone constraint.