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Relativistic Action: Mathematics

  1. May 22, 2014 #1
    Hello! I'm currently going through The Classical Theory of Fields - Landau, Lifshitz and I needed a bit of help on some of the math going on in a certain section.

    The book can be found here https://archive.org/details/TheClassicalTheoryOfFields

    On page 27, they give the action as:

    [itex]\delta S = -mc \delta \int\limits_a^b ds =0[/itex]

    With [itex]ds^2 = dx_i dx^i[/itex]. The definition of [itex]ds[/itex] does not bother me, but the steps they then take are very odd and I'm not quite sure how they obtain the equation before 9.10. If anyone could help that would be awesome!

    If anyone doesn't want to open that PDF here is the line I'm speaking of, which they basically jump right into without explanation (it is assumed knowledge):

    [itex]\delta S = -mc \int\limits_a^b \frac{dx_i \delta dx^i}{\sqrt{ds}}=-mc \int\limits_a^b u_i d\delta x^i[/itex]

    Where [itex]u_i = \frac{dx_i}{ds}[/itex].

    Thanks!

    EDIT: I guess this should have been posted in the Relativity subforum, whoops.
     
    Last edited: May 22, 2014
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  3. May 22, 2014 #2

    WannabeNewton

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    ##\delta ds = \delta (dx_i dx^i)^{1/2} = ds^{-1}dx_i \delta dx^i = u_i d \delta x^i## where I have used ##d \delta x^i = \delta dx^i##.
     
  4. May 22, 2014 #3
    I think you forgot a square root:

    [itex]\delta ds = \delta \sqrt{dx_i dx^i} = \frac{1}{2} dx_i \delta dx^i ds^{-1/2}[/itex]

    and therein lies my question, why is [itex]dx^i[/itex] effected by [itex]\delta[/itex] whereas [itex]dx_i[/itex] apparently not?

    I suppose we have:

    [itex]\sqrt{ds} = \frac{ds}{\sqrt{ds}}[/itex]

    Which would give the integrand as:

    [itex]\frac{dx_i \delta dx^i \sqrt{ds}}{2\bullet ds}[/itex]

    Which would yield:

    [itex]\frac{1}{2} u_i \delta dx^i \sqrt{ds}[/itex]


    I don't know if I'm using the [itex]\delta[/itex] operator correctly. I'm assuming it means find the differentials.
     
  5. May 22, 2014 #4

    WannabeNewton

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    I didn't forget a square root. ##\delta (dx_i dx^i)^{1/2} = \frac{1}{2}(dx_i dx^i)^{-1/2}\delta(dx_i dx^i) = ds^{-1}dx_i \delta dx^i = u_i d \delta x^i##.

    The line ##\frac{1}{2}(dx_i dx^i)^{-1/2}\delta(dx_i dx^i) = ds^{-1}dx_i \delta dx^i## makes use of ##ds = (dx_i dx^i)^{1/2}##
    and ##\delta (dx_i dx^i) = dx_i \delta dx^i + dx^i \delta dx_i = 2dx_i \delta dx^i##.
     
  6. May 22, 2014 #5
    Oh my I'm out of it apparently, thank you.

    Also thank you for the explanation of the rest, I really appreciate it. Working with differentials isn't my strong point (hopefully it will be soon!).

    Why is the last line true, I'm also new to four vector notation and the relations between them.
     
    Last edited: May 22, 2014
  7. May 22, 2014 #6

    WannabeNewton

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    ##dx_i \delta dx^i = dx^i \delta dx_i## (these are dummy indices so you can move them up and down at whim without changing the expression so long as Einstein summation is obeyed)

    More explicitly ##dx_i \delta dx^i = g_{ij}dx^i \delta dx^j = g_{ji}dx^j \delta dx^i = g_{ij}dx^j \delta dx^i = dx^j \delta dx_j = dx^i \delta dx_i## where ##g_{ij}## is the metric tensor and I have used the fact that it is symmetric in its indices as well as the fact that I can relabel dummy indices to whatever I want as long as Einstein summation is obeyed.
     
  8. May 22, 2014 #7
    Right, I remember reading in this text that:

    [itex]A^i = - A_i[/itex]

    so moving one up and the other down would cause no sign change.

    Thank you again!
     
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