# Variational operator confusion

Let $$L=L(X, \dot X)$$ such that the first variation of $$L$$ is given by $$\delta L=\frac{\partial L}{\partial X}\delta X+\frac{\partial L}{\partial \dot X}\delta \dot X.\tag{1}$$ This is pretty standard as this is just the definition of our first variation. This also was of course a variation with respect to X(t). Now lets suppose we have an action $$S_0=\frac12 \int d \tau \left(e^{-1}(\tau) \dot X^2-m^2e(\tau)\right),\tag{2}$$ then its variation with respect to the field $$e(\tau)$$ will just follow as $$\delta S_0=\frac12\int d \tau\left(-\frac{1}{e^2} \dot X^2 \delta e - m^2 \delta e \right),\tag{3}$$ where our action takes the form $$\int d \tau L.\tag{4}$$ When dealing with variations, I am used to using the formula $$\delta f=\sum_i \frac{\partial f}{\partial x_i} \delta x_i.\tag{5}$$ But in these cases, we are using the phrase "with respect to" which leads me to think that instead of varying all the terms the Lagrangian depends on, we just vary the object we are varying with respect to. My question is, why are we not also varying the $$\dot X^2$$ term, and what is a useful formula for varying objects when taking the variation with respect to a single object in the Lagrangian? I have referenced other posts but they have not helped any.

One can in principle vary infinitesimally the $$S_0[X,e]$$ action (2) simultaneously wrt. both the $$e$$ and $$X$$ variables. The coefficient function in front of $$\delta X$$ will then give the Euler-Lagrange (EL) equation for $$X$$ in the same way that the coefficient function (3) in front of $$\delta e$$ is the EL equation for $$e$$. The point is that each coefficient function must vanish independently at a stationary field configuration.
• So if we vary wrt. something, like in the case of varying wrt. e, will we just hold the $X$ constant?
• Yes, "variation with respect to $e$" means taking the variations with $\delta X = 0$ May 3 at 13:06