Least action principle for a free relativistic particle (Landau)

[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=QIxD3Z82AagC&pg=PA28&dq=%22to+set+up+the+expression+for%22&lr=&as_brr=3&as_pt=BOOKS&cd=3#v=onepage&q=%22to%20set%20up%20the%20expression%20for%22&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?

[dextercioby]

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

[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)

?

[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.

[provolus]

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