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Deriving geodesic equation from energy-momentum conservation

  1. May 18, 2015 #1
    Hi all,

    I am trying to follow the calculation by samalkhaiat in this thread: https://www.physicsforums.com/threa...n-from-the-stress-energy-tensor.547502/page-2 (post number 36). I am having some difficulty getting the equation above equation (11) (it was an unnumbered equation) from the preceding equations:

    In particular, the equation states that:
    ## \frac{dX^c}{dx^0} \int d^3 x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_b (\sqrt{-g} T^{ab} + X^c \Gamma^a_{bd}\int d^3x \sqrt{-g} T^{bd} ) = \int d^3x \sqrt{-g} T^{ac} ## (*)

    I am confused to where the ## \frac{d X^c}{dx^0} ## comes from. I managed to massage equations so that I obtain equation (*), except that instead of

    ## \frac{dX^c}{dx^0} \int d^3 x \sqrt{-g} T^{a0} ## (**)

    I have instead

    ## \int d^3 x \partial_0 (\sqrt{-g} T^{a0} \delta x^c) ## (***)

    Help!!! Can someone help me see why the two equations above (equations (**) and (***) ) are equal to each other?
    Thank you so much for any help, this calculation is frying my brain!!! I can present my calculation up to this point if it is helpful at all.

    Any help will be much appreciated!!!
     
  2. jcsd
  3. May 20, 2015 #2

    samalkhaiat

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    You did not copy this correctly.
    The equation in question is [tex]\int d^{3} x \left[ \partial_{0} (\sqrt{- g} x^{c} T^{a 0}) + \partial_{j} (\sqrt{- g} x^{c} T^{a j}) + \sqrt{- g} x^{c} \Gamma^{a}_{b d} T^{b d} \right] = \int d^{3} x \sqrt{- g} T^{a c} . \ \ (R)[/tex]

    After that, I explained every step very carefully. Use Eq(4) and Eq(5), which are [tex]x^{c} = X^{c} + \delta x^{c} , \ \ \ \ (4)[/tex][tex]\Gamma^{a}_{b d} ( x ) = \Gamma^{a}_{b d} ( X ) + \delta x^{e} \partial_{e} \Gamma^{a}_{b d} . \ \ \ (5)[/tex] After the substitutions in (R), use Eq(7) and Eq(8), which are [tex]\int d^{3} x \ \sqrt{- g} \ \delta x^{c} \ T^{a b} = 0 , \ \ \ \ (7)[/tex] [tex]\int d^{3} x \ \partial_{j} ( \sqrt{- g} \ \delta x^{c} \ T^{a j} ) = 0 . \ \ \ (8)[/tex]

    Now, you do this and show me your working. Remember to do the [itex]\partial_{0}[/itex] differentiation in the first term of Eq(R). This to get you started: [tex] \frac{d}{ d x^{0}} \left( \int d^{3} x \ \sqrt{- g} \ \delta x^{c} \ T^{a 0} \right) + \frac{d}{ d x^{0}} \left( X^{c} \int d^{3} x \ \sqrt{- g} \ T^{a 0} \right) .[/tex] Now in this, the first term is zero because of (7), the second term give you [tex]\frac{d X^{c}}{d x^{0}} \int d^{3} x \ \sqrt{- g} \ T^{a 0} + X^{c} \int d^{3} x \ \partial_{0} ( \sqrt{- g} \ T^{a 0} ) .[/tex]


    Sam
     
  4. May 21, 2015 #3
    Dear Samalkhaiat,

    Thank you, both for your reply and your original post. I still have two questions. For your ease of reading, I am highlighting my 2 questions in boldfont.

    Starting from equation (R) of your post:
    ## \int d^3x [\partial_0 (\sqrt{-g} T^{a0} x^c) + \partial_j (\sqrt{-g} T^{aj} x^c) + \sqrt{-g}x^c \Gamma^a_{bd} T^{bd}] = \int d^3 x \sqrt{-g} T^{ac} ## (R)

    I believe I have the correct derivation for the third term. In particular, once I plug in the expansion of ##x^c## and ##\Gamma^a_{bc}## the third term gives me to first order:
    ## \int d^3x \sqrt{-g}x^c \Gamma^a_{bd} T^{bd}] = X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd} ##. (A)
    EDIT: Please see the end of my post!

    The first two terms, however, are still giving me problems. Proceeding from the first term of equation (R):

    ##\int d^3x \partial_0 (\sqrt{-g} T^{a0} x^c) = \int d^3x \partial_0 (\sqrt{-g} T^{a0} X^c) + \int d^3x \partial_0 (\sqrt{-g} T^{a0} \delta x^c) ##

    The second term is zero because of equation (7) in your post. The first term of equation (R) is therefore:

    ## \int d^3x \partial_0 (\sqrt{-g} T^{a0} X^c) = \frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_0 (\sqrt{-g} T^{a0}) ##. (B)

    The second term of equation (R) gives:
    ##\int d^3x \partial_j (\sqrt{-g} T^{aj} x^c) = \int d^3x \partial_j (\sqrt{-g} T^{aj} X^c) + \int d^3x \partial_j (\sqrt{-g} T^{aj} \delta x^c)## .

    By equation (8) in your post, the second term is zero. The second term of equation (R) is therefore:

    ## \int d^3x \partial_j (\sqrt{-g} T^{aj} X^c) = X^c \int d^3x \partial_j (\sqrt{-g} T^{aj}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) ##. (C)

    Adding equations (A), (B), and (C) gives me that equation (R) becomes:

    ## \frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_0 (\sqrt{-g} T^{a0}) + X^c \int d^3x \partial_j (\sqrt{-g} T^{aj}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) + X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd} = \int d^3 x \sqrt{-g} T^{ac} ##.

    Combining the second and third terms, I obtain that equation (R) becomes:

    ## \frac{d X^c}{d t} \int d^3x \sqrt{-g} T^{a0} + X^c \int d^3x \partial_b (\sqrt{-g} T^{ab}) + \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) + X^c \Gamma^a_{bd} \int d^3x \sqrt{-g} T^{bd} = \int d^3 x \sqrt{-g} T^{ac} ##. (D)

    Which is equal to the equation above equation (11) in your original post if the following is true:

    ## \int d^4x (\partial_j X^c) (\sqrt{-g} T^{ai}) = 0 ## .

    So, my first question: is there a reason why said term is zero?


    Again, I really appreciate your help. Thank you so much for your answer!

    EDIT:
    Once I looked at my derivation of the third term of (R), I realized that I did not do it properly. The third term of (R) is
    ## \int d^3x \sqrt{-g} x^c \Gamma^a_{bd} T^{bd} = \int d^3x \sqrt{-g} T^{bd} (\Gamma^a_{bd} + \delta x^d \partial_d \Gamma^a_{bd})(X^c + \delta x^c) ##.

    To first order in ##\delta x## this is:

    ## \int d^3 x \sqrt{-g} T^{bd} (X^c \Gamma^a_{bd} + \delta x^c \Gamma^a_{bd} + X^c \delta x^d \partial_d \Gamma^a_{bd} ) ##.

    The second term is zero because we can write it as:

    ## \int d^3x \sqrt{-g} T^{bd} (\delta x^c \Gamma^a_{bd}) = \Gamma^a_{bd} \int d^3 x \sqrt{-g} T^{bd} \delta x^c ##,
    which is zero because of equation (7) of your post.

    So the third term of (R) is equal to:

    ## \int d^3 x \sqrt{-g} T^{bd} (X^c \Gamma^a_{bd} + X^c \delta x^d \partial_d \Gamma^a_{bd} ) = X^c \Gamma^a_{bd} \int d^3 x \sqrt{-g} T^{bd} + \int d^3 x \sqrt{-g} T^{bd} (X^c \delta x^d \partial_d \Gamma^a_{bd} ) ##.

    So my second question: is there a reason for the last term,

    ## \int d^3 x \sqrt{-g} T^{bd} (X^c \delta x^d \partial_d \Gamma^a_{bd} ) ##,

    to be equal to zero?
     
    Last edited: May 21, 2015
  5. May 21, 2015 #4

    samalkhaiat

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    For the same reason that allows us to pull [itex]X^{c}[/itex] out of the volume integral. [itex]X^{c}[/itex] is a function only of whatever we choose to parameterize the geodesic path. So, you could think of it as [itex]X^{c} ( s )[/itex].
    Notice that you used the same index contraction over [d] twice!! You need to be careful not to use the same index. Any way back to your question. Again for the same reason that lets you pull [itex]\Gamma[/itex] out of the integral. In the expansion of [itex]\Gamma ( x )[/itex], the term [itex]\partial \Gamma[/itex] is evaluated at [itex]x^{c} = X^{c}[/itex], so you can pull it out together with [itex]X[/itex] of that integral.
     
    Last edited: May 21, 2015
  6. May 21, 2015 #5
    Thank you again for your answer!
    Would you mind elaborating on the reason of why
    ## \partial_j X^c = 0 ## ?

    I understand that seeing ## X^c ## as a function of the geodesic parameter ##s## allows one to set ## \partial_j X^c = 0 ##, but isn't also true that
    ## \partial_j X^c = \partial_j (x^c - \delta x^c) = \delta^c_j ##,
    where ##\delta^c_j## is the delta function?

    Thanks!
     
  7. May 22, 2015 #6

    samalkhaiat

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    Well, that is all what you need. It is like writting [itex]x = 3 f(s) + \delta x[/itex].
    Why is that? First of all, you should not apply naive algebra on a deninition. Second, [itex]\partial_{j} x^{a} = \delta^{a}_{j}[/itex], and [itex]\delta x^{a}[/itex] is a function of [itex]x[/itex], so [itex]\partial_{j} ( \delta x^{a}) \neq 0[/itex].
    Look, as I said in the original post, one should think of [itex]X^{a} [/itex] as a spatially fixed point [itex]P[/itex] in the [itex]x[/itex]-coordinate system. Then, one tries to see what kind of curve does the point [itex]P[/itex] trace when the energy-momentum is covariantly conserved.
     
    Last edited: May 22, 2015
  8. May 22, 2015 #7
    This statement makes sense to me; I always got myself confused when dealing with partial vs. total derivatives. I think I understood the derivation now. Thank you for your help, and for your original post!
     
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