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

  1. Dec 16, 2009 #1
    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?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Dec 16, 2009 #2

    dextercioby

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    What is [itex] \delta \left(x_{i}x^{i}\right) = ? [/itex]
     
  4. Dec 16, 2009 #3
    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]

    ?
     
  5. Dec 16, 2009 #4

    dextercioby

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    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.
     
  6. Dec 16, 2009 #5
    thx for the moment. I hope to need no more help in covariant variation calculus... ;-)
     
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