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8*pi in the Einstein field equations?

  1. Sep 15, 2013 #1
    A typical formulation of the Einstein equations is

    [tex]R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}[/tex]

    The [itex]\frac{G}{c^4}[/itex] make the units work out. What about the 8*pi? Why is this necessary?
     
  2. jcsd
  3. Sep 15, 2013 #2

    WannabeNewton

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    The field equations have to agree with Newtonian gravity in the Newtonian limit.
     
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