- #1

zhouhao

- 35

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## Homework Statement

I am not sure whether the meaning of the equation ##(3)## which used for deriving momentum is as same as equation ##(4)##.I will make a detailed description below.

The lagrangian function for a free particle is ##L=-mc^2\sqrt{1-\frac{v^2}{c^2}} \quad (1)##

The action from point ##a## to point ##b## is :##S=\int_a^bLdt=-mc\int_a^bds##, because ##ds=cdt\sqrt{1-\frac{v^2}{c^2}}##

##S=-mc\int_a^b\frac{ds^2}{ds}=-mc\int_a^b\frac{dx_idx^i}{ds}=-mc\int_a^bu_idx^i##

In Landau's book:##\delta{S}=-mc\int_a^bu_id\delta{x^i} \quad (2)##

In my calculation,##\delta{S}=-mc[\int_a^b\delta{u_i}dx^i+\int_a^bu_id\delta{x^i}+\int_a^b\delta{u_i}d\delta{x^i}]##

##\int_a^b\delta{u_i}dx^i=0## is confirmed.But I can not confirm##\int_a^b\delta{u_i}d\delta{x^i}## equal to zero.I have no idea how Landau does for equation ##(2)##.

Let's still proceed in Landau's way:

Subsection integration for equation ##(2)##,we get ##\delta{S}=-mcu_i\delta{x^i}|_a^b+mc\int_a^b\delta{x^i}\frac{du_i}{ds}ds##

For moving path of the free particle ##\delta{S}=0##,##\quad (\delta{x^i})_a=(\delta{x^i})_b=0##, as a result,##\frac{du_i}{ds}=0##

Not fix ##\delta{x^i}_b## to zero,then ##\delta{S}=-mcu_i(\delta{x^i})_b \quad (3)##, according to definition of four dimension momentum,##p_i=mcu_i##.

Why equation ##(3)## equal to equation ##(4)##,I can not understand?

## Homework Equations

Definition of four dimension momentum : ##p_i=-\frac{\partial{S}}{\partial{x^i}} \quad (4)##

## The Attempt at a Solution

The meaning of Equation ##(3)## is just applying a variation to the

**real moving path**from ##a## to ##b## and not fixing ##\delta{x^i}_b## to zero.I can not figure out whether the physical meaning of equation ##(3)## equal to the one of equation ##(4)## or not.