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Weak Field Approx, algebra geodesic equation

  1. Mar 10, 2015 #1
    My book says in the slow arrow-10x10.png motion approx, so ## v << c ##, ##v=\frac{dx^{i}}{dt}=O(\epsilon) ##
    It then states:

    i) ##\frac{dx^{i}}{ds}=\frac{dt}{ds}\frac{dx^{i}}{dt}=O(\epsilon) ##
    ii) ## \frac{dx^{0}}{ds}=\frac{dt}{ds}=1+O(\epsilon) ##

    The geodesic equation reduces from ##\frac{d^{2}x^{a}}{dt^{2}}+\Gamma^{a}_{bc}\frac{dx^{b}}{ds}\frac{dx^{c}}{ds} =0 ## to ##\frac{d^{2}x^{i}}{dt^{2}}+\Gamma^{i}_{00}\frac{dx^{0}}{ds}\frac{dx^{0}}{ds}=O(\epsilon^{2})##

    Questions:
    1) I have no idea whatsoever where ii comes from.
    2) I see that in the slow arrow-10x10.png approx ##t## is approximately equal to ##s##, and so we can replace these in the first term, i.e ##\frac{dt}{ds}=1##, but isn't ##\frac{dx^{i}}{dt}=O(\epsilon) ## also negligible?

    3)From ii) I think I see why we neglect ##\frac{dx^{i}}{ds}\frac{dx^{i}}{ds}## compared to ##\frac{dx^{0}}{ds}\frac{dx^{0}}{ds}## so the the former gives ##(O(\epsilon))^{2}## and the latter ##(1+O(\epsilon))(1+O(\epsilon))=(1+2O(\epsilon)+(O(\epsilon))^{2}) ##, so as it has a non ## \epsilon ## term we don't neglect it.

    4) When we replace ##s## with ##t## for the fist term in the geodesic equation , why dont we do it for the second?


    5) Finally probably a stupid question, from a different source:http://www.mth.uct.ac.za/omei/gr/chap7/node3.html (just above eq 28), I'm confused why ##\frac{dx^{0}}{dt}=c##, I thought ##t## is coordinate time, and ##x^{0}## is also coordinate time, so this would equal ##1##? what is ##x^{0}## then?


    Your help is really appreciated thanks !
     
    Last edited: Mar 10, 2015
  2. jcsd
  3. Mar 10, 2015 #2

    PeterDonis

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    Which book?

    It comes from the fact that the 4-velocity is a unit 4-vector; its magnitude is 1 (or -1, depending on which metric signature convention you use). So we must have

    $$
    1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds}} = \sqrt{ \left( \frac{dx^0}{ds} \right)^2 - \Sigma \left( \frac{dx^i}{ds} \right)^2 }
    $$

    If you work it out you will see that it shows that i) implies ii).

    I think you mean ##d^2 x^a / ds^2## in the first term, correct? The geodesic equation starts out being written in terms of derivatives with respect to ##s##, not ##t##.

    Here I think the second term should just be ##\Gamma^i{}_{00}##, correct? That factor is already ##O(\epsilon^2)##, so you can set ##dx^0 / ds \approx 1## since the ##O(\epsilon)## term in it can be neglected.

    Yes.

    Essentially, we do; each factor of ##dx^0 / ds## becomes ##dx^0 / dt = dt / dt = 1## if we are only keeping terms to ##O(\epsilon^2)##. See above.
     
  4. Mar 10, 2015 #3

    PeterDonis

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    This is just using conventional units in which ##c## is in meters per second, instead of being just a dimensionless number with magnitude 1. In these units, ##x^0 = ct## instead of just ##t##.
     
  5. Mar 11, 2015 #4

    stevendaryl

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    Sort of an aside, it's not generally known that in the low-velocity, weak-gravity limit, the geodesics expressed in the appropriate coordinates is the same as the solutions of Newtonian gravity.

    In Newtonian gravity, you have an action

    [itex] A = \int dt (\frac{1}{2} m v^2 - m \Phi)[/itex]

    where [itex]\Phi[/itex] is the Newtonian gravitational potential. You minimize it to find the paths of a test particle. In GR, in the low-velocity, weak-field limit, you have a proper time:

    [itex] \tau = \int dt (1 - \frac{1}{2} \frac{v^2}{c^2} + \frac{\Phi}{c^2}) = t - A/mc^2[/itex]

    So minimizing [itex]A[/itex] is the same as maximizing [itex]\tau[/itex]

    The Newtonian potential [itex]\Phi[/itex] is related to the GR metric [itex]g_{\mu \nu}[/itex] through: [itex]g_{00} = 1 + \frac{2}{c^2} \Phi[/itex]
     
  6. Mar 11, 2015 #5
    Thank you very much !
    So I've done this and I've got ##\frac{dx^{0}}{ds}=(1+O(\epsilon))^{\frac{1}{2}}=1+\frac{O(\epsilon)}{2} ##
    Is this correct and then ignore the ##\frac{1}{2} ## ?
     
  7. Mar 11, 2015 #6

    PeterDonis

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    Basically, yes.
     
  8. Mar 14, 2015 #7
    Sorry just a side question, how does neglecting ##(O(\epsilon))^{2}## compare to neglecting ##O( \epsilon^{2} )##?
    Thanks.
     
  9. Mar 14, 2015 #8

    DrGreg

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    They are the same.

    To understand the notation better, see Big O notation.
     
  10. Mar 14, 2015 #9
    One more question sorry, on http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken] , going from (29) to (30), why the replacement of ## ,^{i} ##, so the partial derivatives, to ##\bigtriangledown ^{i} ##? , so this is the gradient function ? It has no significance right?
     
    Last edited by a moderator: May 7, 2017
  11. Mar 14, 2015 #10
    Sorry another, this may be a stupid question but, equation (41) ##R_{00}=...+O(\epsilon) ##, eq 42 ##R_{ab}=...+O(\epsilon) ## and then we sub (29) into (45), and obviusly don't neglect ##O(\epsilon)## in this substitution, but how is this consistent with neglectng it as done in (41) and (42)? Thanks.
     
  12. Mar 14, 2015 #11

    PeterDonis

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    He's using them both to mean the same thing. That's not really standard notation; in standard notation, ##\nabla## means the covariant derivative, not the partial derivative.

    No. What he means when he says those equations are correct to first order in ##\epsilon## is that any additional terms are ##O(\epsilon^2)##. The equations as he writes them contain all terms up to ##O(\epsilon)##; they are not being neglected. The reason no terms in ##\epsilon## appear in equation (41) is that the ##O(\epsilon)## terms cancel when you expand the 00 component of equation (40).
     
  13. Mar 27, 2015 #12
    So in eq (40) I have ##U^{\alpha}=(1,0,0,0)## and ##g_{00}=\eta_{00}+\epsilon h_{00}##, I can't see anything thats going to cancel
    Thanks.
     
  14. Mar 28, 2015 #13

    PeterDonis

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    No, you don't; the matter is slowly moving but you can't assume it's stationary. The article says (earlier on) that ##dx^i / dt## is ##O(\epsilon)##, and ##U^0## is the relativistic ##\gamma## factor so it depends on ##dx^i / dt##.
     
  15. Mar 29, 2015 #14
    So ##U^{0} U^{0}=\frac{c^{2}}{c^{2}-v^{2}}=\frac{1}{1-\frac{v^{2}}{c^{2}}}=1+\frac{v^{2}}{c^{2}}+...##
    I would conclude to neglect the 2nd term onward, I'm not seeing where the ##c^{2}## comes from, thanks.
     
  16. Mar 29, 2015 #15

    PeterDonis

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    Yes, but you need ##U_0 U_0## to expand equation (40). Lower the index on ##U^0## with ##g_{00}## and see what happens. Note that equation (32) gives you an explicit expression for ##g_{00}##.
     
  17. Mar 29, 2015 #16
    ##U_{0}U_{0}=\frac{1+\frac{2\Phi}{c^{2}}}{1-\frac{v^{2}}{c^{2}}}##, ##\Phi## the gravitational potential, unsure what to do next, is this gravitational potential a function of ##v## ?
     
  18. Mar 29, 2015 #17

    PeterDonis

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    Actually, you need two factors of ##g_{00}##, not one, because you're lowering two indexes (one for each ##U##).

    Also, I'm no longer sure there will be any terms of ##O(\epsilon)## in the expansion. ##v^2 / c^2## is ##O(\epsilon^2)##.

    Look at the equation for it earlier in the article and how that equation is derived.
     
  19. Mar 30, 2015 #18
    I've seen the derivation but I'm still unsure , theres equation 31, but this looks to me like to solve for ##\Phi## you need to integrate ##v^{2}## wrt space coordinates.

    We also have ##g_{00}=\eta_{00}+\epsilon h_{00}##, but we don't have an explicit expression for ##h_{00}##
     
  20. Mar 30, 2015 #19

    PeterDonis

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    The LHS of equation (31) is the acceleration ##dv / dt##, not ##v^2##. So it's telling you that the space derivative of ##\Phi## is equal to the time derivative of ##v##.

    However, you don't necessarily need to find an actual solution for ##\Phi##; you just need to determine whether it's ##O(\epsilon)##, ##O(\epsilon^2)##, etc.

    Yes, you do; comparing equations (30) and (31) gives you the relationship between ##h_{00}## and ##\Phi##. That relationship is expressed in equation (32); since ##g_{00} = \eta_{00} + \epsilon h_{00}##, you can read off ##h_{00}## from that equation.
     
  21. Mar 31, 2015 #20
    • Ah thanks, so ##\Phi## is of the order ##O(\epsilon^{2})## and so ##\frac{(1+2\Phi)^{2}}{1-\frac{v^{2}}{c^{2}}}=\frac{c^{4}}{c^{2}}+O(\epsilon^{2})=c^{2}+O(\epsilon^{2})##
     
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