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Components of Einstein's Equations in 4 dimensions

  1. Sep 17, 2012 #1
    In this excerpt from the notes of Sean M. Carrol, he says:

    "Einstein’s equations may be thought of as second-order differential equations for the
    metric tensor field gμν. There are ten independent equations (since both sides are symmetric
    two-index tensors), which seems to be exactly right for the ten unknown functions of the
    metric components. However, the Bianchi identity ∇μGμν = 0 represents four constraints on
    the functions Rμν, so there are only six truly independent equations in (4.52). In fact this is
    appropriate, since if a metric is a solution to Einstein’s equation in one coordinate system
    xμ it should also be a solution in any other coordinate system xμ′ . This means that there are
    four unphysical degrees of freedom in gμν (represented by the four functions xμ′ (xμ)), and
    we should expect that Einstein’s equations only constrain the six coordinate-independent
    degrees of freedom."


    [itex]R_{\mu\nu}[/itex]-[itex]\frac{1}{2}[/itex]R[itex]g_{\mu\nu}[/itex]=8[itex]\pi[/itex]G[itex]T_{\mu\nu}[/itex] (Eq. 4.52)

    I am confused as to what the 6 independent equations represent. I'm getting lost on thinking that these equations represent 3 spatial coordinates and 1 time coordinate but then what do the other 2 components represent (obviously my understanding is flawed). Any clarification would be greatly appreciated, thanks.
     
  2. jcsd
  3. Sep 17, 2012 #2

    bcrowell

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    The indices mu and nu *each* take on 4 values, so it's 16 equations, not 4, until you start subtracting out the ones that aren't independent.
     
  4. Sep 17, 2012 #3
    Ok, so once you get down to the 6 which are independent, what do these represent physically?
     
  5. Sep 17, 2012 #4
    I've read that these 6 equations "propagate the 3-metric", if this is correct does this mean they describe time evolution of the spatial portions of the metric?
     
  6. Sep 17, 2012 #5

    dextercioby

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    Somehow related, IIRC the nr of DOF of GR is 2 and can be obtained in Lagrange formulation as 16 components of metric tensor - 6 because of symmetry of the tensor - 4 from the metricity condition ([itex] \nabla_{\mu} g^{\mu\nu} = 0 [/itex]) - 4 from the Noether identities ([itex] \nabla_{\mu} G^{\mu\nu} = 0 [/itex]) = 2, with G the Einstein tensor. In the Hamiltonian formulation, it's 16-6-8 first class constraints = 2.
    Going into quantum mechanics, this 2 are two possible eigenvalues of the 3rd component of the helicity operator, namely -2 and +2.
     
  7. Sep 17, 2012 #6
    thank you for the reply but I feel this is slightly more advanced than what I was looking for in my last question; I am more so looking for an understanding of what the 6 equations physically represent.
     
  8. Sep 17, 2012 #7

    dextercioby

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    Think about it, what do the Newton's equations represent ? It's the same for the EFE, they are PDE's whose solutions are the components of the metric tensor of the 4D spacetime in the canonical basis.
     
  9. Sep 17, 2012 #8

    bcrowell

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    When you count degrees of freedom, you don't necessarily get a unique interpretation. For example, I could define a point in space using (x,y,z), or by giving latitude, longitude, and height above sea level.

    So the best we can hope for here is at least one story that successfully accounts for why we have a certain number of d.f.

    For physical insight, maybe it would be helpful if we could work out the answer to the corresponding question for Maxwell's equations.
     
  10. Sep 17, 2012 #9
    I think I am starting to understand, thank you bcrowell and dextercioby.
     
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