So I recently pondered the following. Let's say I have an $2$ actions $S_1$ and $S_2$ which differ by a constant:

$$ S_1(\dot x_i, x_i) = S_2(\dot x_i, x_i) + \tilde c$$

Now their equations of motion will be identical in classical mechanics (without General Relativity) upon varying the coordinates $x_i \to x_i + \delta x_i$. Intuitively, I know this constant term will make a difference in general relativity. Is this hunch correct? What does the constant term $\tilde c$ look like in the form of Einstein Field Equations?

$$ G^{\mu \nu}+ \Lambda g^{\mu \nu}= \frac{8 \pi G}{c^4} T^{\mu \nu} $$

Or is there a better way to get the equations of motion in general relativity? Directly from the classical (without General Relativity) action?

  • $\begingroup$ The reason for the difference in the two calculations, is that in GR the metric is dynamical. Explicitly, the volume element, which contains the metric determinant $g$, means even a constant in the Lagrangian shows up in the field equations. That constant is the Lambda you wrote in the field equations. $\endgroup$
    – Eletie
    Dec 6, 2020 at 11:23
  • $\begingroup$ If you just wrote the Einstein Hilbert action and then the $c$, like in your first equation, it wouldn't contribute to the field equations. But that wouldn't make sense anyway, because in GR we're dealing with curved space so the volume form needs to be $\sqrt{-g} d^4 x$ in order to transform as a scalar. $\endgroup$
    – Eletie
    Dec 6, 2020 at 11:28
  • $\begingroup$ @eletie can you show the calculations? So I can accept your answer. Or comment including page showing the same? $\endgroup$ Dec 6, 2020 at 11:33
  • $\begingroup$ The calculation just goes about varying the metric field, in the standard variational way. Note this includes varying the metric determinant. It's in almost all GR textbooks but is a tedious calculation, so I'll link you to the wiki page en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action. It also includes the cosmological constant term too. Basically the variation of the metric determinant leads to a term proportional to $g_{ab}$ in the field equations, so the 'constant term' shows up in the field equations too. $\endgroup$
    – Eletie
    Dec 6, 2020 at 11:48

1 Answer 1

  1. An additive constant $\tilde{c}$ in the action functional cannot affect the Euler-Lagrange (EL) equations, i.e. in OP's case the EFE.

  2. Such constant $\tilde{c}$ renders the action functional non-local unless we can write it as an integral over spacetime.

  3. The cosmological constant term $\int\!d^4x~\sqrt{-g}\Lambda$ in the EH action is in contrast a local term that depends on the metric $g$, and affects the EFE.


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