Variational Principle for Relativistic Action

Question

I'm going through p. 27 in Landau & Lifshitz Classical Field Theory (vol 2), and I'm confused as to why only the contravariant part of the proper time is varied? They start with $$\delta S=-mc\delta\int_{a}^{b}ds=0 \tag{p. 27}$$ from which they then write $$\delta S=-mc\int_{a}^{b}\frac{dx_i\delta dx^i}{ds}=-mc\int_{a}^{b}u_id\delta x^i. \tag{p. 27}$$ My explanation is by comparing it to the concept of a dual space in linear algebra, where the covariant part of the proper time $dx_i$ acts as a function, so it would make sense to only vary the "vector" part of the proper time.
I'm also not entirely sure but I think he has defined $$u_i=\frac{dx_i}{ds}.$$

So in summation I have two questions

1. Only the contravariant part of the proper time is varied because analogous to the idea of an inner product in linear algebra we only vary the "vector" part of the inner product and keep the "function" part of the inner product constant.
2. Is $u_i=\frac{dx_i}{ds}$?

I know there's a similar question already on stackexchange but it doesn't help and he doesn't seem to consider the contra- and co-variant parts distinctly.

Answer 1

• L&L are considering Minkowski spacetime in SR (as opposed to a curved spacetime in GR), so the spacetime coordinates with lower indices $$x_i~:=~\eta_{ij}x^j$$ have been lowered by the constant Minkowski metric components $\eta_{ij}$. Hence we can lower the indices in $$u^i~=~\frac{dx^i}{d\tau}\tag{7.1}$$ and the differential $dx^i$ without worrying about derivatives of $\eta_{ij}$, cf. OP's 2nd question.

• Concerning OP's 1st question, L&L are actually varying both $dx^i$ and $dx_i$. They give the same contribution, so there is implicitly a factor $2$ (which is cancelled by a $1/2$ from a square root $ds=\sqrt{dx_idx^i}$).

• General warning: Be aware that a lot is implicitly implied in L&L.

References:

1. L.D. Landau & E.M. Lifshitz, Vol.2, The Classical Theory of Fields, $\S$9.