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I'm having troubles in understanding a mathematical step in the derivation of the electromagnetic tensor. In Landau&Lifshitz's book I found that the action of a particle in an electromagnetic field is

$$ S = \int_a^b (-mc \, ds - \frac{e}{c}A_i \, dx^i).\tag{16.1} $$

Then they want to apply the least action principle and they immediately write

$$ \delta S = - \int_a^b \left(mc \frac{dx^i d\delta x_i}{ds} + \frac{e}{c} A_i d\delta x^i + \frac{e}{c} \delta A_i dx^i\right).\tag{23.1b}$$

I really can't understand this step. I have only basic knowledge on calculus of variation, but from what I know, if I have a functional

$$ I = \int f(x, y(x), y'(x)) \, dx $$

the "variation" of $I$ should be defined as

$$ \delta I = \int \frac{\partial f}{\partial y}\delta y \, dx $$

but in calculating $\delta \int mc \, ds$ I have a constant function which multiplies a differential which depends on the varying quantities, which is different.

I can't see how this definition of "variation" connects with $\delta S$ and I can't get what $d\delta x^k$ means... I'm not searching for a mathematical rigorous proof, but just an explanation of what's happening there or some more sub-steps. I think the key is to treat $ds$ as a function of $dt, dx, dy, dz$ but I tried and I couldn't get that expression out. I've also tried to search elsewhere but I found texts taking electromagnetic tensor as a definition, or reporting the same step without any clarifications.

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The action integral is understood to be a functional of the particle trajectory.

The variation means that instead of single trajectory $x^*(t)$ we consider all trajectories of the form $x^*(t)+\eta \delta x (t)$ where $\delta x(t)$ is any smooth function that is zero for boundary times.

The differential $ds$ can be expressed using functions $x_i'(t)$ and time differential $dt$. Then we can transform variation of the integral in the standard way Landau assumes.

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