# Homework Help: Least action principle for a free relativistic particle (Landau)

1. Dec 16, 2009

### provolus

Reading the Landau's "The classical theory of fields" (chapter 2, section 9 ) I have some doubts in explaining the steps in derivig the formula for the variation of the action for the relativistic free particle http://books.google.it/books?id=QIx...age&q="to set up the expression for"&f=false". Given the invariant element of measure:

$$ds=\sqrt{dx_idx^i}$$

where $$x^i$$ ( $$x_i$$ ) are the four contravariant (covariant) coordinates which parametrize the world line of the free particle, I have to vary respect $$x^i$$, that is I make the variation $$\delta x^i$$. So my doubts are about the second step of the formula before the 9.10, that is why:

$$\delta(ds)=\frac{d x_i \delta d x^i}{ds}$$

is obtained, instead of (IMH and erroneous O):

$$\delta(ds)=\frac{d x_i \delta d x^i}{2 \cdot ds}$$

???

Can someone be so kind to show me the steps?

Last edited by a moderator: Apr 24, 2017
2. Dec 16, 2009

### dextercioby

What is $\delta \left(x_{i}x^{i}\right) = ?$

3. Dec 16, 2009

### provolus

It should be:

$$\delta (x_i x^i) = \delta (c^2t^2-r^2) = 2 (c^2 t \delta x^0 - r \delta x^i$$)

but, sorry, I don't get the point... that is... should I calculate

$$\delta (dx_i dx^i)$$

?

4. Dec 16, 2009

### dextercioby

The "d" in the brackets is not important. That 2 you have obtained in front cancels the one in the denominator, thus giving you the final expression from Landau's book.

5. Dec 16, 2009

### provolus

thx for the moment. I hope to need no more help in covariant variation calculus... ;-)