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Einstein's Equations in Dimension 4

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    In my general relativity class my professor mentioned that in dimension 4 there are only six statements in the Einstein equations and that this is exactly the number needed.

    2. Relevant equations

    3. The attempt at a solution
    Really I just want to understand why 6 statements are all that is needed (apparently exactly what is needed) in 4 dimensions. Any clarification would be greatly appreciated, thanks!
  2. jcsd
  3. Sep 16, 2012 #2


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    This is not about getting help on a specific homework or coursework exercise, so you're probably better off asking this in the "non-homework" relativity forum. I've moved this there.
  4. Sep 16, 2012 #3
    ok, thank you
  5. Sep 16, 2012 #4


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    Can you explain what you mean by <statement> ? I've never heard of <statements in the Einstein's equations> before.
  6. Sep 16, 2012 #5
    oh ok, by this my professor was speaking of the system of equations that is encompassed by the equation above. So originally there are 20 in dimension 4, but due to symmetry and divergence arguments there are only 6 independent solutions. At that point though he said this is exactly the number needed in dimension 4 and I am unsure as to why.
  7. Sep 16, 2012 #6


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    Hey, sorry I don't know the answer to your question, but I was wondering if you could explain how you get 20 equations in the first place, if each tensor has two indices and each index ranges over 4 possible values?

    EDIT: I'm rusty on this stuff, but having consulted Wikipedia, :tongue: I feel a little more confident commenting. To elaborate on what I said above, I would have expected 42 separate equations, not 20. Wikipedia then points out that these tensors are symmetric, and that each one therefore has 10 independent components, The article states that the EFEs are therefore a set of 10 equations, which is exactly what I would expect if you represent each tensor as a 4x4 matrix: the 4 diagonal elements are independent, and then due to symmetry, each of the 12 off-diagonal elements is equal to its mirror element reflected across the diagonal, so you would halve this number and end up with 4 + 6 independent components.

    It could be the "divergence properties" you mentioned above are what pare things down to 6, but I have no idea why this would be the ideal number of constraints for 4D. In any case there don't seem to be 20 equations to start with.

    EDIT 2: Aha! One need only read further:

    Last edited: Sep 16, 2012
  8. Sep 16, 2012 #7


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    A 4x4 matrix has 16 elements (not 20). Symmetry cuts this to 10 independent ones (4 diagonal, 6 off-diagonal). The field equations are 10 independent equations, which suffice to fix the 10 independent components of the metric.

    I don't understand the statement about 6, or about divergence. The field equations automatically give a divergence-free stress-energy tensor, so this is not an additional constraint. In fact, this is the standard argument for why we need the Einstein tensor in the field equations rather than the Ricci tensor; the Einstein tensor is constructed so that its divergence vanishes identically.

    The actual Einstein equations are 10 equations that determine the 10 components of the metric.

    If you used the Ricci tensor plus a separate equation for vanishing divergence, you'd have 10+4 equations, which would overdetermine the 10 components of the metric.

    The number 6 would seem to come from 10-4, but I can't understand why one would subtract.

    Maybe you misunderstood, and the statement intended by your professor was that the stress-energy tensor only had 6 independent components. (Symmetry makes it 10, and vanishing divergence would cut it to 6.)
  9. Sep 16, 2012 #8
    ya sorry I don't know where I got 20 from (I meant to write 16). Aside from that, my professor said that since G and g are divergence free (and so must 'T') this gives 4 constraints and so there are actually only 6 independent components in the Einstein equations at dimension n=4.
  10. Sep 16, 2012 #9


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    Did you see the quote I posted? It seems like the relevant question now is, how does having 6 independent constraints allow for the metric to have four "gauge-fixing degrees of freedom", and what does that mean? Maybe somebody here more well-versed in GR can comment.
  11. Sep 16, 2012 #10
    Wikipedia articles are horrible for learning anything, the terms they use are always taken out of context
  12. Sep 16, 2012 #11
    I actually didn't see that quotation, but I would also like to know what is meant by this because it may be just the answer I was looking for.
  13. Sep 16, 2012 #12


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    Look at for instance Carroll's notes, where this is very well treated.
  14. Sep 17, 2012 #13


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    p. 112, near equation 4.52

    This explains it. The 10-4 has the -4 because of coordinate invariance.
  15. Sep 17, 2012 #14
    You probably got the 20 from your professor telling you that in 4D, there are 20 linearly independent components to the Riemann tensor.
  16. Sep 19, 2012 #15
    Statements, did you perhaps mean how many solutions to the field equations there are?
  17. Sep 19, 2012 #16
    Yes indeed, you beat me to it.
  18. Sep 19, 2012 #17
    Thanks everyone, I appreciate the help!
  19. Sep 19, 2012 #18


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    The Einstein Field equations solve for a metric. The metric, in 4-D has 10 independent components due to symmetry. However, because one can get different metrics by changing the coordinate system (and thereby the basis vectors), one must not determine ALL 10 independent components of the metric, or else one is actually FORCING a coordinate system.

    One should be careful, though, that one is NOT solving even 6 components of the metric with the Einstein Field equations: one is putting 6 constraints on the 10 components of the metric.

    This is like having a system of 6 equations with 10 unknowns. With only 6 equations, the 10 unknowns are all undermined. There needs to be 4 additional equations, and those 4 additional equations come from the equations that define the coordinate system (x'=x'(x)).

    The reason the Einstein Field equations are actually 6 independent equations rather than 10 independent equations is a result of the Bianchi identities.
  20. Sep 19, 2012 #19


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    In [itex]D[/itex] dimensional space-time there are [itex](1/2)D(D + 1)[/itex] field equations
    [tex]G_{\mu \nu}= k T_{\mu \nu}. \ \ \ \ (1)[/tex]
    But we also have the identities,
    [tex]\nabla^{\mu}G_{\mu \nu} \equiv 0. \ \ \ (2)[/tex]
    These [itex]D[/itex] identities hold whether or not [itex]G_{\mu \nu}[/itex] themselves satisfy the field equations. Therefore, covariantly, the dynamics is described by only [itex](1/2)D(D - 1)[/itex] field equations (in 4 dimensions, only six out of ten equations are dynamical ones, and the remaining four ones are constraints). Indeed, the Einstein equation involves [itex]\ddot{g}_{ij}[/itex] but not [itex]\ddot{g}_{0 \mu}[/itex]. Thus, the invariance of the action under general coordinates transformation shows that the action principle and with it the field equations can not determined the metric tensor unless the coordinate system is specified in some non-covariant way.
    Of course, this failure is not dangerous. Similar situation arises in Maxwell’s equations
    [tex]\partial^{2}A^{\nu} - \partial^{\nu} \partial \cdot A = J^{\nu}. \ \ (3)[/tex]
    These four equations (for the four unknowns [itex]A_{\mu}[/itex]) do not determine [itex]A_{\mu}[/itex] uniquely, because the left-hand-sides of these equations are related by a single divergence identity analogous to eq(2),

    \partial_{\nu}( \partial^{\sigma}\partial_{\sigma}A^{\nu} - \partial^{\nu}\partial_{\sigma} A^{\sigma}) \equiv 0. \ \ (4)
    This identity reduces the number of functionally independent equations to three. In Maxwell’s case, the ambiguity in the solutions is removed by choosing a particular gauge. For example, given any solution [itex]B_{\mu}[/itex], one can always construct a new solution, [itex]A_{\mu}[/itex], such that
    [tex]\partial^{\mu}A_{\mu} = 0, \ \ \ (5)[/tex]
    by setting
    [tex]A_{\mu} = B_{\mu} + \partial_{\mu}\Omega ,[/tex]
    where [itex]\Omega[/itex] is given by
    [tex]\partial^{2}\Omega = - \partial \cdot B .[/tex]
    Thus eq(5) (regarded as a field equation) and eq(3), with appropriate boundary conditions, could generally determine [itex]A_{\mu}[/itex] uniquely.
    In similar way, in order to determine [itex]g_{\mu \nu}[/itex], it is necessary to introduce a coordinate condition, explicitly violating the invariance of the theory under general coordinates transformation. This procedure is often called “gauge fixing”. Of course this is an abuse of language and one must be careful about the difference between the choice of a coordinate system and gauge fixing. The former has physical meaning: inertial force (i.e. observable effects) actually depends on it. In fact, the properties of the solution to eq(1) depend on the choice of a coordinate system: in order to make definite predictions, one needs to add [itex]D[/itex] coordinate conditions (treated like field equations) to the Einstein equations.
    Mathematically, the most natural choice of the coordinate condition (invariant under [itex]GL(n)[/itex]) is that of de Donder
    [tex]g^{\mu \nu}\Gamma^{\sigma}{}_{\mu \nu} = 0. \ \ \ (6)[/tex]
    This is also called Harmonic Coordinate Condition. To understand why the term Harmonic is used, you need to use the fact
    [tex]\frac{1}{\sqrt{-g}}\partial_{\sigma}\left( \sqrt{-g} g^{\sigma \rho}\right) = - g^{\mu \nu}\Gamma^{\rho}{}_{\nu \mu}.[/tex]
    Now, if you insert [itex]\delta^{\rho}_{\mu} = \partial_{\mu}x^{\rho}[/itex] in the left hand side, you can rewrite the de Donder condition as
    [tex]\frac{1}{\sqrt{-g}}\partial_{\sigma}\left( \sqrt{-g}\ g^{\sigma \mu}\ \partial_{\mu} x^{\rho}\right) = 0.[/tex]
    Thus the term Harmonic coordinate condition makes sense because the operator [itex](1/ \sqrt{-g}) \partial_{\sigma}( \sqrt{-g}g^{\sigma \mu}\partial_{\mu})[/itex] is nothing but the curve space generalization of the d’Alembertian [itex]\partial^{2}[/itex].

    Ok, now I leave you with the following little exercise: Prove that one can always find a coordinate system such that [itex]g^{\mu \nu}\Gamma^{\rho}{}_{\mu \nu}=0[/itex]. Generalize the de Donder condition and show that there exists a coordinate system (called non-rotating) in which the energy-momentum tensor (of the matter field) is conserved globally.

    Last edited: Sep 19, 2012
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