# First variation of the action in relativistic notation - Landau & Lifshitz “Classical theory of fields”

In Landau & Lifshitz's book, Classical theory of fields, the action for a free particle is defined as:

$$\tag{8.1} S= \int ^b _a {-mc \ \text d s}=0,$$

where $$\text d s=c\,\text d t\sqrt{1-\frac {v^2}{c^2}}$$ is the the invariant space-time interval beetween points along the particles worldline. From the latter expression of the Lagrangian, it's easy to obtain the expressions of the momentum and the energy of the particle.

However, after a "classical" derivation, the author repeats the calculations with a different notation (I'll post the derivation from the "particle in a electromagnetic field" case omitting the 4-potential terms, because I think that my book contains an error in the free particle case):

Since $\text d s= \sqrt {\text dx^i \text dx_i}$: $$\delta S = -mc \int ^b _a\dfrac{\text dx_i \text d\delta x_i}{\text ds}.$$ Integrating by part, introducing the 4-velocity $u_i =dx_i/ds$, we get:

$$\tag{9.10}\delta S =-mcu_i\delta x^i|^b _a+\int_a ^bmc\dfrac{\text d u_i}{\text d s} \delta x^i \,\text ds.$$

I'm very puzzled about the meaning of these two lines; I have three questions:

1) What does the notation $\text d \delta x_i$ mean?

2) How does he obtain the $\delta S$ expression?

3) How does he pass from the first integral to the second?

If someone could explain in detail I'd be very grateful.

Note, I have no problem in getting the result, I can obtain it by replacing $\text d s =c\text d t \sqrt {...}$ and doing the variation on $v$.

## 2 Answers

Landau and Lifshitz apparently like to "cancel as many $ds$ in the numerator and the denominator as possible", perhaps for aesthetic reasons. However, to make sense of the integrals, it seems more pedagogical to express$^1$ everything as functions (and derivative of functions) of the parameter $s$, i.e., there should only appear a single $ds$ in an integral $\int \!ds (\ldots)$, and the integrand $(\ldots)$ should be a function of $s$. Manipulations between different integral expressions should be made with this understanding.

By the way, a similar issue often arises in the derivation of the Work-Energy theorem.

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$^1$ Instead of parameter $s$, one can also use the parameter $t$, as OP suggests.

• Hi @Qmechanic , after thinking about this for a while, I'm not sure it's a good idea to think about $s$ as a parameter when doing the variation of the trajectory. In fact, $s$ is a parameter defined by the trajectory itself, so thinking about it as a parameter could lead to absurd conclusions, like $\delta(\int ds)=\int \delta(1)ds=0$. – pppqqq Sep 14 '14 at 12:35
• @pppqqq : The variable $s$ plays a double role in L&L, but interpreted carefully, it still makes sense. In more detail: We are considering the proper time $\tilde{s}$ of a virtual path as a function of the parameter $s$, which happens to be the proper time $s$ of the classical path. – Qmechanic Sep 14 '14 at 12:58

The best explanation I found was here: http://fma.if.usp.br/~amsilva/Livros/Zwiebach/chapter5.pdf