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=QIxD...for%22&f=false. Given the invariant element of measure:

[tex]ds=\sqrt{dx_idx^i}[/tex]

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

[tex]\delta(ds)=\frac{d x_i \delta d x^i}{ds}[/tex]

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

[tex]\delta(ds)=\frac{d x_i \delta d x^i}{2 \cdot ds}[/tex]

???

Can someone be so kind to show me the steps?

dextercioby

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

provolus

It should be:

[tex]\delta (x_i x^i) = \delta (c^2t^2-r^2) = 2 (c^2 t \delta x^0 - r \delta x^i[/tex])

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

[tex]\delta (dx_i dx^i)[/tex]

?

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... ;-)