Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

MTW, Chap19, Stationary relativistic Vs Weakly gravitating source

  1. Mar 26, 2013 #1
    in MTW Ex 19.1, it was considered a Weakly gravitating body whereas in Ex 19.3, a Stationary relativistic source.
    What relativistic means here ? does it mean the source is very dense and massive ? or does it mean it is rotating at near the speed of light ?
    Can I specify the mass as the mass times Gamma ? where Gamma is 1/sqrt (1-vv)
    I need to find the non linear term which results from the Mass-energy which in this case no longer is the integral of the density , but rather a relativistic mass.
    Is this correct ? MTW mentions that we should consider a keplrian mass !! or did I misunderstood ?
    Please clarify
  2. jcsd
  3. Mar 26, 2013 #2


    Staff: Mentor

    Yes; these two exercises are considering two different kinds of objects.

    It means that the source's self-gravity is no longer negligible. Compare the opening sentences of the two sections:

    Section 19.1 (weakly gravitating source): "Consider an isolated system with gravity so weak that in calculating its structure and motion one can completely ignore self-gravitational effects." This means that the metric everywhere, including inside the source, can be expressed as the Minkowski metric plus a small perturbation: [itex]g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}[/itex].

    Section 19.3 (fully relativistic sources): "Abandon, now, the restriction to weakly gravitating sources. Consider an isolated, gravitating system inside which spacetime may...be highly curved...but refuse, for now to analyze the system's interior or the strong-field region near the system. Instead, restrict attention to the weak gravitational field far from the source..." This means that the metric inside or near the source can no longer be expressed as the Minkowski metric plus a small perturbation; but far from the source, it can.


    Not necessarily; a non-rotating source can still be relativistic.

    Where does the v come from? The source is stationary; that means we will work in a coordinate chart in which it is at rest.

    Why do you want to do this? What are you trying to figure out? You don't need to do this to work exercise 19.3.
  4. Mar 26, 2013 #3
    If you look at the following document ,


    you will see in section III, a multipole expansion. This answers question 19.1. However, right before equation 27 , the author mentions :
    Now the first integral is simply the (conserved) mass M of the system. (Technically this is the energy, but the
    difference only arises at the next order in velocity, which we have dropped

    This is why I'm considering this option otherwise I would not see how to get the non linear term proportional to M^2/R^2 ?
    A hint will be appreciated.
  5. Mar 26, 2013 #4
    But what does it mean a relativistic source - this is the term that got me confused...
  6. Mar 26, 2013 #5


    Staff: Mentor

    This is the next order in what this paper calls the "velocity scale" of the source, by which I think they mean how fast individual parts of the source are moving with respect to the source's center of mass. But you don't need to go to this order to obtain the M^2/R^2 term; that comes from nonlinear corrections to the static part of the metric, not from higher-order corrections in the non-static part (due to the motion of parts of the source). This statement on p. 452 of MTW may help:

    "Two types of nonlinearities turn out to be important far from the source: (1) nonlinearities in the static, Newtonian part of the metric, which generate metric corrections

    [tex]\delta g_{00} = - 2 M^2/r^2, \delta g_{jk} = \frac{3}{2} \left( M^2/r^2 \right) \delta_{jk}[/tex]

    (see exercise 19.3...)"
  7. Mar 26, 2013 #6


    Staff: Mentor

    A "relativistic" source here just means a source whose self-gravity is too strong to be neglected. But for an isolated source, the region of spacetime in which this is true will always be limited; once you get far enough away from the source, the corrections due to the source's self-gravity become too small to detect, so you can ignore them and use the linearized approximation.

    The reason MTW stresses this is that if a relativistic source is isolated, as above, you can still define a mass, linear momentum, angular momentum, etc. for it by looking at the metric far away from the source, where the linearized approximation can still be used. In other words, if you are far enough away from a relativistic source, you can ignore the fact that it is relativistic; you just measure its mass, momentum, angular momentum, etc. from far away the same as you would for any other source when you are far away from it. You only have to worry about the source's self-gravity if you need to know about the region near or inside of it; but for many purposes you don't need to know that.
  8. Mar 26, 2013 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    I think Peter already answered this. But let's restate.

    When you abandon the following assumption

    you say that the source is relativistic.

    Is that clear? I"m not sure how to clarify it any further, but if you're still confused you can still ask. Maybe explain what's confusing.
  9. Mar 28, 2013 #8
    thanks very much to you all. It is now clear.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook