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The Einstein-Hilbert Action:

$$S_{EH}=\frac{1}{2\kappa}\int \sqrt{-g}g^{ab} \left({\Gamma^c}_{ab,c} - {\Gamma^c}_{ac,b} + {\Gamma^d}_{ab}{\Gamma^c}_{cd} - {\Gamma^d}_{ac}{\Gamma^c}_{bd}\right) d^4x$$

Contains second derivatives of $g$ as the Christoffel symbols contain derivatives of $g$ but using integration by parts it can be brought into a form with only first derivatives of the metric tensor:

$$S_{EH'}=\frac{1}{2\kappa}\int \left( -{\Gamma^c}_{ab}\partial_c(\sqrt{-g}g^{ab}) + {\Gamma^c}_{ac}\partial_b(\sqrt{-g}g^{ab}) + \sqrt{-g}g^{ab}\left({\Gamma^d}_{ab}{\Gamma^c}_{cd} - {\Gamma^d}_{ac}{\Gamma^c}_{bd}\right) \right)d^4x.$$

Is this equivalent? I sometimes read about 'surface terms' but not sure what this means.

When you work it out it is:

$$S_{EH'}=\frac{1}{8\kappa}\int \sqrt{-g}\left( g^{ab}g^{de}g^{cf} +2 g^{ac}g^{bf}g^{de} +3 g^{ad}g^{be}g^{cf} -6 g^{ad}g^{bf}g^{ce} \right)\partial_c g_{ab}\partial_f g_{de} dx^4$$ (Although I might have got some of the constants wrong). I like this form because it is smilar to the Maxwell Action: $$S_M = \frac{1}{2}\int\sqrt{-g}(g^{ac}g^{bd}-g^{ad}g^{bc})\partial_a A_b \partial_c A_d dx^4$$

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  • $\begingroup$ The surface term they differ by involves both the metric and its derivative, and you cannot fix both the metric and its derivative on the boundary, so the two actions differ by a surface term that technically cannot be set to zero. If we consider the EH action with the GHY boundary term added, then the two are equivalent. From the point of view of formal variational calculus, total divergences are in the kernel of the Euler-Lagrange operator and the equations of motion from the two actions are the same. $\endgroup$ Mar 27 '20 at 17:10
  • $\begingroup$ @Bence If they are not equivalent then how do we know which one is "correct"? $\endgroup$
    – zooby
    Mar 27 '20 at 17:11
  • $\begingroup$ @zooby the gauge invariance of GR is such that you end up with a Hamiltonian that is everywhere zero except on the boundary, so when doing the Legendre transform, one cannot naîvely throw away boundary terms -- they contain nontrivial information. $\endgroup$ Mar 27 '20 at 17:13
  • $\begingroup$ 1) If we take the variational problem "literally" as opposed to "formally", then the EH action is only correct if we add the GHY term, which does make them equal. 2) If we only care about the classical EoMs, then they produce the same EoMs. $\endgroup$ Mar 27 '20 at 17:13
  • $\begingroup$ Why is GHY term needed? Isn't space infinite without a boundary? $\endgroup$
    – zooby
    Mar 27 '20 at 17:18
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  1. OP's 1st (2nd) action is (is not) covariant/geometric, respectively. OP's 2nd action transforms with a boundary term under general coordinate transformations.

  2. The reason for the GHY boundary term in the first place is explained e.g. in this Phys.SE post.

  3. OP asks in a comment:

    Why is GHY term needed? Isn't space infinite without a boundary?

    When deriving Euler-Lagrange (EL) equations from a variational principle we need to integrate by parts. We cannot do this unless we put boundary conditions (or fall-off conditions) at spatial infinity.

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