# Question on the Einstein-Hilbert action

Does it make sense to write that the Einstein-Hilbert action as

$$$$S=\int\mathcal{L}\left(g^{\mu\nu},\partial^{\lambda}g^{\mu\nu}\right)\sqrt{-g}\,\mathrm{d}^4x=\frac{1}{2\kappa}\int R\,\sqrt{-g}\,\mathrm{d}^4x\tag{1}$$$$

where $$\kappa\equiv 8\pi G/c^4$$ and $$x\equiv\left(t,\vec{x}\right)$$? If not, what is the correct way of writing it?

I wrote the above because the Maxwell action can be written as

$$$$S=\int\mathcal{L}\left(A_{\nu},\partial_{\mu}A_{\nu}\right)\,\mathrm{d}^4x\tag{2}$$$$

but I am not entirely sure if what I wrote for the Einstein-Hilbert action is correct.

• Yes it is correct, btw it should also contain the double derivative of metric. Metric is the choice of field in EH action. This expression is also true for higher derivative gravity
– KP99
Jul 28 '21 at 14:44

Morally, yes, but in detail I would suggest a few edits to what you wrote.

First, I would tend to write the fundamental variables with lower indices: the metric $$g_{\mu\nu}$$ (rather than the inverse metric $$g^{\mu\nu}$$) and the derivative of the metric $$\partial_\mu g_{\nu\lambda}$$ (rather than the inverse metric times the first deritvative of the inverse metric, $$\partial^\mu g^{\nu\lambda}=g^{\mu\alpha} \partial_\alpha g^{\nu\lambda}$$). The reason is mostly aesthetic, although it is important to realize that the true partial derivative operator has a downstairs index $$\partial_\mu$$, and the partial derivative with an upstairs index is just shorthand for a matrix product of the inverse metric with the partial derivative.

Second, in the normal, covariant way it is written, $$$$S = \int {\rm d}^4 x \sqrt{-g} R,$$$$ the Einstein-Hilbert action actually involves second derivatives of the metric. You can see this since, schematically, the Riemann curvature scalar $$R$$ is related to the Christoffel symbols $$\Gamma$$ via $$R \sim \partial \Gamma + \Gamma \Gamma$$, and since $$\Gamma\sim g^{-1} \partial g$$ depends on derivatives of the metric, keeping only the highest derivative terms we expect that $$R \sim \partial \Gamma \sim \partial \partial g$$ (and an explicit calculation confirms that the second derivative terms we expect based on this schematic argument really are there and do not cancel). The equations of motion still turn out to be second order (and not third order like you might naively expect based on the appearance of second derivatives in the action), because when you vary the action you find the third derivative terms take the form of a total derivative, which we can assume vanishes on the boundary (or, even better, we can add the Gibbons-Hawking-York boundary term to cancel these third derivative terms in the variation of the action).

Finally, since $$\sqrt{-g}$$ is itself a function of the metric, I would consider it optional whether or not you include it explicitly in the Lagrangian.

Putting this together, I would re-express the form in your question as $$$$S = \int {\rm d}^4 x \sqrt{-g} \mathcal{L}(g_{\mu\nu}, \partial_\mu g_{\nu\lambda}, \partial_\mu \partial_\nu g_{\lambda \sigma})$$$$

• Thanks for your insights! I have edited my original question as I have left out a part of my expression. In the edited expression, is what I have defined for $x$ correct? Jul 28 '21 at 15:18
• @Thomas Yes, $x$ combines 3 space and 1 time coordinates. Jul 28 '21 at 15:20
• "not third order like you might naively expect" Or fourth, even.
– J.G.
Jul 28 '21 at 15:34
• @J.G. The EH action is linear in second derivatives (since it is linear in $R$ which contains terms like $\partial \partial g$ but not $(\partial \partial g)^n$ for $n\geq 2$), so I think the highest order terms that could show up are third order. It's been a while since I explicitly checked though so I could be wrong :) Jul 28 '21 at 16:11