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Value of energy-momentum tensor in weak field approximations

  1. Sep 30, 2011 #1
    My first question, so sorry if it's in the wrong forum.

    I'm trying to understand the Newtonian weak field approximations to general relativity. I can't see why, if the Schwarzschild metric (which can describe the gravitational field around the Sun) is a vacuum solution ([itex]T_{\mu\nu}=0[/itex] ) , do textbooks state that [itex]T=\rho c^{2}[/itex] when approximating Poisson's Equation (which also describes the gravitational field around the Sun) from the Einstein Field Equations?

    Thank you .
     
  2. jcsd
  3. Sep 30, 2011 #2

    Bill_K

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    In Newtonian gravitation the gravitational potential inside the sun obeys Poisson's equation, ∇2 = -4π Gρ where the mass density ρ ≠ 0. Outside the sun ρ = 0 and we have Laplace's equation ∇2 = 0. General Relativity works the same way except the source is the stress energy tensor Tμν, or in the case of the sun just T00 = ρc2.
     
  4. Oct 2, 2011 #3
    Sorry, I still don't see why if [itex]T^{00}=\rho c[/itex] there can be a Schwarzschild solution for [itex]T^{\mu\nu}=0[/itex] . Why aren't these two contradictory statements?
     
  5. Oct 2, 2011 #4

    pervect

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    I'm not positive I understand your difficulty, but I'll take a guess. Said guess being that you're cofused by how you can get gravity if T_uv = 0 everywhere.

    If you look at a Newtonian point charge, the charge density is zero everywhere except for the point, where the charge density is infinite.

    So except for one singular point, the charge density is 0.

    The Schwarzschild solution is vaguely similar in that you have an infinite singular stress-energy tensor at the center (the sigularity), and it's zero everywhere else, i.e. a vacuuum.
     
  6. Oct 2, 2011 #5
    Take a look at Bill's post again. In order to understand the answer to your question you have to understand the two different situations Possion, and Laplace's equations hold. Laplace's equation talks about a region of space that is void of sources, Possion's equation describes a region that has sources in that region.

    Take the Schwarzschild metric, it is a solution of
    [tex]
    R_{\alpha\beta}=0
    [/tex]
    Now take the weak field metric, and without any assumptions for now about the sources, to order in [itex]1/c^2[/itex], what does the Einstein tensor reduce to? Now what are the conditions we should put on [itex]T_{\alpha\beta}[/itex] to make our Newtonian limit make sense?

    Do you see how Laplace's equation is a special case of Possion's equation?
     
  7. Oct 3, 2011 #6
    So when we approximate Poisson's equation

    [itex]\nabla\cdot\nabla\phi=4\pi G\rho [/itex]

    by assuming [itex]T=\rho c^{2} [/itex]

    we are doing it for a point inside a gravitational source (eg the Sun).

    However, the Schwarzschild solution is for points outside a gravitational source, where [itex]T=0 [/itex]

    I think the fog may be clearing a little.

    But one new confusion. When I read my textbook I see that the derivation of the Newtonian limit is based on the assumption that we are dealing with a region filled with dust (ie no internal pressure). So we are assuming (still using the Sun as our example) that the Sun is “dust”. How is that a valid assumption? When I think of the Sun I think of this great seething mass of gas, with lots of pressure. So it's OK to assume that, even if for the Sun that
    [itex]pressure\ll density [/itex]

    ?

    Thanks for persevering.
     
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