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Slow motion approx, order of acceleration quick Q?

  1. Apr 1, 2015 #1
    I'm looking at equation (41) from equation (40) on http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken] and the second term, it says that these equations are given up to ##O(\epsilon)##.

    Looking at equation (32) for ##g_{00}## I see it is proportional to ##\Phi##, and from looking at (31) I see that ##\Phi ## is of the same order as ##d^{2}x^{i}/dt^{2}## .

    Looking at the 2nd term in (41) which has only been multiplied by a ##-1## from the ##g_{00}## I conclude that it is only possible to be given up to ##O(\epsilon)## if ##\Phi## , and so, ## d^{2}x^{i}/dt^{2}##, is of the order ##O(\epsilon^{2})##

    By definition of the slow approximation we have ##dx^{i}/dt=O(\epsilon)##. And time derivatives are neglected compared to space derivatives.

    It seems to me quite possible that ##d^{2}x^{i}/dt^{2}## being of the order ##O(\epsilon^{2})## can be justified.

    However I'm not sure how to show this properly/explicitly?

    Thanks.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 2, 2015 #2
  4. Apr 2, 2015 #3

    stevendaryl

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    Frankly, I don't understand the basis for their order [itex]\epsilon[/itex] analysis.

    Normally, I would think that to get Newtonian physics as an approximation to GR, we need something like this:

    1. Assume that [itex]g_{\mu \nu} = \eta_{\mu \nu} + O(\epsilon)[/itex]
    2. Assume that [itex]\frac{dx^j}{dt} = O(\sqrt{\epsilon})[/itex]
    3. Assume that [itex]\rho = O(\epsilon)[/itex]
    4. Assume that [itex]p/c^2 \ll \rho[/itex]
    5. Assume that [itex]\partial_t g_{\mu \nu}[/itex] is negligible compared with spatial derivatives.
    As for 4&5, I'm not sure how pressure and time-derivatives of the metric should rank in powers of [itex]\epsilon[/itex]

    But the big difference with what you've said is that I don't think that [itex]\Phi[/itex] should be [itex]O(\epsilon^2)[/itex]. Since [itex]g_{00} = 1 + \frac{2\Phi}{c^2}[/itex] it must be that [itex]\Phi = O(\epsilon)[/itex]
     
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