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Manipulation within the Einstein Tensor in Einstein field equations

  1. Aug 3, 2014 #1
    Hello everybody. I was recently brainstorming ways to make the Einstein field equations a little easier to solve (as opposed to having to write out that monstrosity of equations that I started on some time ago) and I got an interesting idea in my mind.

    Here, we have the field equations:

    R[itex]\mu[/itex][itex]\nu[/itex] - [itex]\frac{1}{2}[/itex]g[itex]\mu[/itex][itex]\nu[/itex]R = ((8[itex]\pi[/itex]G)/c4)T[itex]\mu[/itex][itex]\nu[/itex]

    Now, notice that the curvature scalar R is apart of the equations. Now we know that:

    R = g[itex]\mu[/itex][itex]\nu[/itex]R[itex]\mu[/itex][itex]\nu[/itex]

    This means that we can express the equations as:

    R[itex]\mu[/itex][itex]\nu[/itex] - [itex]\frac{1}{2}[/itex]g[itex]\mu[/itex][itex]\nu[/itex]g[itex]\mu[/itex][itex]\nu[/itex]R[itex]\mu[/itex][itex]\nu[/itex] = ((8[itex]\pi[/itex]G)/c4)T[itex]\mu[/itex][itex]\nu[/itex]

    Now, the g[itex]\mu[/itex][itex]\nu[/itex] and the g[itex]\mu[/itex][itex]\nu[/itex] cancel each other out.

    This just leaves:

    R[itex]\mu[/itex][itex]\nu[/itex] - [itex]\frac{1}{2}[/itex]R[itex]\mu[/itex][itex]\nu[/itex] = ((8[itex]\pi[/itex]G)/c4)T[itex]\mu[/itex][itex]\nu[/itex]

    This reduces to:

    [itex]\frac{1}{2}[/itex]R[itex]\mu[/itex][itex]\nu[/itex] = ((8[itex]\pi[/itex]G)/c4)T[itex]\mu[/itex][itex]\nu[/itex]

    Finally, you should be able to then multiply both sides of the the equation by 2 to get:

    R[itex]\mu[/itex][itex]\nu[/itex] = ((16[itex]\pi[/itex]G)/c4)T[itex]\mu[/itex][itex]\nu[/itex]

    Now, you have the Ricci tensor derived and then you can derive the metric tensor from here (thus solving the equations).

    Would my logic in this process be accurate or is there some mathematical error within my logic that would render this invalid?
     
  2. jcsd
  3. Aug 3, 2014 #2

    Matterwave

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    You have made a mistake in doing your indices. You have overloaded the ##\mu,\nu## indices. The expression in terms of the Ricci tensor is:

    $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}g^{\rho\tau}R_{\rho\tau}=\frac{8\pi G}{c^4}T_{\mu\nu}$$

    There IS a trace reversed version of the equations, and they look like this:

    $$R_{\mu\nu}=\frac{8\pi G}{c^4}\left(T_{\mu\nu}-\frac{1}{2}Tg_{\mu\nu}\right)$$

    Where here similarly in the previous case ##T\equiv T_{\mu\nu}g^{\mu\nu}##.

    Basically you take the trace of the EFE's and then multiply it by ##g_{\mu\nu}/2## and add it back to the EFE's to get this equation. This may make things somewhat easier sometimes (especially useful in the vacuum case for when ##T_{\mu\nu}=0##), but you're really still just solving the same equation.
     
  4. Aug 3, 2014 #3

    WannabeNewton

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    Your indices are all messed up here. It should be ##g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta}## which won't give you anything useful beyond the canonical form of the EFEs.

    However there is a very slick and elegant way to calculate certain components of ##R_{\mu\nu}## if you have Killing field(s) available. If ##\xi^{\mu}## is a Killing field then it is very easy to show that ##R_{\mu\nu}\xi^{\nu} = \frac{1}{\sqrt{g}}\partial_{\nu}(\sqrt{g}\nabla_{\mu}\xi^{\nu})##. This lets you calculate the components of ##R_{\mu\nu}## along ##\xi^{\mu}## almost effortlessly. It is infinitely quicker to calculate this way if you have Killing fields available. It won't necessarily get you all the components of the Ricci tensor but it's better than nothing.
     
  5. Aug 3, 2014 #4

    Nugatory

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    Staff: Mentor

    Any time that your dummy indices are not paired, one up and one down and no other appearances, something has to be wrong. In this case, you'll see the problem if you try writing the summations out explicitly instead of using the summation convention.
     
  6. Aug 31, 2014 #5
    Friend

    You made ​​a mistake. The rules of algebra notational not allow to have more than two equal indices in a monomial. The rule of the sum of Einstein and determining which are the free indices can not be applied:

    Rμν - 1/2gμνgμνRμν = ((8πG)/c4)Tμν
     
  7. Sep 1, 2014 #6
    Please, it reads "notational algebra", read indicial algebra.
    Excuse me.

    VictorNeto
     
  8. Sep 1, 2014 #7
    Please, it reads "notational algebra", read indicial algebra.
    Excuse me.

    VictorNeto
     
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