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MTW "Gravitation" page 263, equation 10.27?

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    (This is self-study.)

    In the equation just above 10.27 on page 263 of "Gravitation" by Misner, Thorne, and Wheeler, the first term is:
    [itex] \frac{\partial}{\partial x^{\beta}} (\frac{dx^{\alpha}}{d\lambda}) \frac{dx^{\beta}}{d\lambda} [/itex]
    which becomes the first term in (10.27):
    [itex] \frac{d^2 x^{\alpha}}{d\lambda^2} [/itex] ,
    where [itex] x^{\alpha}(\lambda) [/itex] is a geodesic curve ([itex] x^{\beta}(\lambda) [/itex] is the same curve, just an independent index). Note that the partial derivative in the first equation only applies to the quantity inside the parentheses.

    2. Relevant equations: See above.


    3. The attempt at a solution: The only thing I can see is that, somehow, the [itex] {\partial x^{\beta}} [/itex] in the bottom of the top equation gets exchanged with the [itex] {d\lambda} [/itex] of the [itex] \frac{dx^{\beta}}{d\lambda} [/itex] so that [itex] \frac{dx^{\beta}}{\partial x^{\beta}} [/itex] becomes 1 while the (newly formed) [itex] \frac{\partial}{d{\lambda}} (\frac{dx^{\alpha}}{d\lambda}) [/itex] becomes [itex] \frac{\partial d x^{\alpha}}{d\lambda d\lambda} [/itex], which becomes [itex] \frac{d^2 x^{\alpha}}{d\lambda^2} [/itex].

    To me, that seems like a lot of willy-nilly throwing around of d's and partials. The chain rule notwithstanding, I'm just not comfortable that that's right. BTW: I did not forget to change the partials to totals when I moved things around. I left them as they were until the last step to emphasis what distresses me about this approach.

    I've forgotten a lot of stuff about even elementary calculus, so perhaps I'm just forgetting how to handle the partial derivative of a total derivative in general. But, at the moment, I'm stumped. I would appreciate any insights anyone can provide to help me over this bump in the road.

    Thanks.
     
  2. jcsd
  3. Nov 16, 2014 #2

    TSny

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    Hello. Look at the unnumbered equation just above (10.21) on page 260. I believe this is what is being used in deriving (10.27).
     
  4. Nov 16, 2014 #3
    @TSny:

    Sometimes...my stupidity is just...stunning!

    Thank you so much for pointing that out to me. That fixed the problem and led me to understand what I was doing wrong with slinging the total and partial derivatives around. I've got it now. Thanks.

    It's late and I'm tired so I'm not going to try to offer anything tonight, but as soon as I can, I'll post the complete derivation using both the comma notation and the partial/total diff notation for anyone else who may be following along.

    Thanks again.
     
  5. Nov 17, 2014 #4
    Sorry. I hit "Post" instead of "Preview" before I was done. I've been unable to figure out how to delete the entire post or to edit it and get the LaTeX to work. I'll keep studying the forum docos to see what I'm doing wrong.

    Meanwhile, please see message #5 for the real reply.
     
    Last edited: Nov 17, 2014
  6. Nov 17, 2014 #5
    MTW p. 263 top (toward equation (10.27)):

    [itex] 0 = u^{\alpha}_{;\beta} {u^{\beta}} = (u^{\alpha}_{,\beta} + {\Gamma}^{\alpha}_{{\gamma}{\beta}}u^{\gamma}) u^{\beta} = u^{\alpha}_{,\beta} u^{\beta} + {\Gamma}^{\alpha}_{{\gamma}{\beta}}u^{\gamma} u^{\beta} [/itex]

    From this point on, we'll just focus on the first term, since that was the one giving me trouble.

    From the unnumbered equation just above equation (10.21) on page 260 (thanks TSny):
    [itex] f_{,\delta} u^{\delta} = {\partial}_{\mathbf{u}} f = \frac{df}{d{\lambda}} [/itex].

    Replacing f with [itex] u^{\alpha} [/itex] and [itex] {\delta} [/itex] with [itex] {\beta} [/itex] we get
    [itex] u^{\alpha}_{,\beta} u^{\beta} = {\partial}_{\mathbf{u}} u^{\alpha} = \frac{d}{d{\lambda}} u^{\alpha} [/itex].

    But, [itex] u^{\alpha} = \frac{dx^{\alpha}}{d{\lambda}} [/itex], so [itex] \frac{d}{d{\lambda}} (\frac{dx^{\alpha}}{d{\lambda}}) = \frac{{d^2}x^{\alpha}}{d{\lambda}^2} [/itex], which is the first term of (10.27).

    That completes the proof and we can go on our merry way, but I'd like to rework it now using the differential operators d & [itex] \partial [/itex].

    We start with the first term as given in MTW:
    [itex] \frac{\partial}{\partial x^{\beta}} \Big(\frac{dx^{\alpha}}{d{\lambda}}\Big)\frac{dx^{\beta}}{d{\lambda}}[/itex].

    Now, how to handle the partial derivative of a total derivative? This is actually elementary calculus that is very, very easy to understand, unless you're an old man like me and get mental blocks at crucial moments. To see how this works, let's replace the [itex] \Big(\frac{dx^{\alpha}}{d{\lambda}}\Big) [/itex] with an arbitrary function of the coordinates, [itex] f = f (x^\beta) [/itex]. This give us [itex] \frac{\partial f}{\partial x^{\beta}} \frac{dx^{\beta}}{d{\lambda}}[/itex]. From elementary calculus we know that [itex] \frac{\partial f}{\partial x^{\beta}} dx^{\beta} = df [/itex], so using [itex] u^\alpha [/itex] as our function we get [itex] \frac{\partial u^\alpha}{\partial x^{\beta}} dx^{\beta} = du^\alpha [/itex]. But what we really want is [itex]\frac{dx^\beta}{d\lambda}[/itex], so we get [itex] \frac{\partial u^\alpha}{\partial x^{\beta}} \frac{dx^{\beta}}{d\lambda} = \frac{du^\alpha}{d\lambda} = \frac{{d^2}x^{\alpha}}{d{\lambda}^2}[/itex] as done above.

    Thanks again to TSny for breaking my mental block.
     
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