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provolus

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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:

[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?

[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?

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