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Quick question of Einstein Field Equations

  1. Jul 16, 2012 #1
    I have seen and read a few different versions of the Einstein field equations (EFE). For example; R[itex]_{\mu\nu}[/itex] - [itex]\frac{1}{2}[/itex]g[itex]_{\mu\nu}[/itex]R = - 8[itex]\pi[/itex]GT[itex]_{\mu\nu}[/itex] , R[itex]_{\mu\nu}[/itex] - [itex]\frac{1}{2}[/itex]g[itex]_{\mu\nu}[/itex]R + g[itex]_{\mu\nu}[/itex][itex]\Lambda[/itex] = [itex]\frac{8 \pi G}{c^4}[/itex]T[itex]_{\mu\nu}[/itex] , and 8[itex]\pi[/itex]T[itex]_{\mu\nu}[/itex] = G[itex]_{\mu\nu}[/itex]

    So which one is the true equation?? And if you could explain why they're different that would be great!

    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 16, 2012 #2
    Yes, and?

    Also, LaTeX may be useful.

    [tex]R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = - 8 \pi G T_{\mu \nu}[/tex]

    Is written as:

    Code (Text):
    [tex]R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = - 8 \pi G T_{\mu \nu}[/tex]
    Though usually I think I've seen the right hand side as plus instead of minus.
  4. Jul 16, 2012 #3
    All three are correct. The first is the normal EFE. The second is the EFE with a cosmological constant. So, the first is the second with a vanishing cosmological constant. In the third, the left side is written as the Einstein tensor. The left side of the EFE collectively describes the curvature of space-time, so that's what the Einstein tensor does. It's just a briefer notation.

    EDIT: Also, in the last one, Newton's constant is set equal to 1 (the G on the right hand side in the other equations), to make the equation look nicer.
  5. Jul 16, 2012 #4
    Thank you Mark M your answer is extremely helpful. But why is the speed of light (c) included in the second EFE with the cosmological constant???
  6. Jul 16, 2012 #5
    That's the proper way to write it. Similar to the way the third equation set G = 1 so that it disappeared, the first and third equations set c = 1 so that it drops out of the equation.

    Also, I forgot to mention why the first has a minus sign on the right. That's a consequence of the sign signature. In relativity, since time is treated as another dimension in space-time, it must be distinguished from the three dimensions of space. Either you make the time coordinate of the metric negative, so that the metric reads [tex] ds^{2} = -dt^{2} + dx^{2} + dy^{2} + dz^{2} [/tex] or leave the time coordinate positive, and make spatial coordinates negative [tex] -ds^{2} = dt^{2} - dx^{2} - dy^{2} - dz^{2} [/tex] this first is called the mostly plus signature (-+++) and the latter is called mostly minus (+---). The second equation is using a different signature than the other two.
  7. Jul 16, 2012 #6
    That just means that that equation doesn't implicitly take [itex]c=1[/itex] or [itex]G=1[/itex]. It's common in relativity to do both just to reduce the number of constants running around for theoretical work, but putting them back in can be useful when an actual numerical calculation is being performed.
  8. Jul 17, 2012 #7
    [itex]R_{\mu\nu}-\dfrac{1}{2} g_{\mu \nu} R +\Lambda g_{\mu \nu} = \dfrac{8\pi G T_{\mu \nu}}{c^4}[/itex]
  9. Jul 17, 2012 #8
    Alright well thank you, that answers my questions! Thanks all!

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