An alternative route to Einstein's equation is through the principle of least action, as we did previously to deduce the geodesic equation in curved spacetime in Geodesic equation from the principle of least action.

In this article, we will therfore go through the process of deriving the Einstein equations in vacuum and then in the presence of matter using **the variational approach**.

##### Action in vacuum

The derivation of the action from a set of equations of motion is very hard, not always possible, and there is no systematic way to do it. We therefore will begin by guessing the action and show that it gives the right answer.

So we will first seek an action S for gravitation that leads to the field equations of general relativity in the absence of matter and energy (in vacuum), that is, we will guess something like:

where L is a scalar Lagrange density and d^{4}V is the element of 4-volume. We thus need both a scalar and the 4-volume element.

The 4-volume element is easiest: we recall that in a locally Minkoskian coordinate system x^{α'}, the volume element is d4V=dx^{0'}dx^{1'}dx^{2'}dx^{3'}. If there is a positive-determinant Jacobian J^{α'}_{β} that transforms this to a general coordinate system x^{β}, we have:

It turns out however that the metric tensor in the general coordinate system is

so that if we define* g* to be the determinant of the 4x4 g_{γδ} matrix, we then have g=-(det J)2, so that if follows that det J = √-g. We thus see that the 4-volume element is

The simplest **Lagrangian L** that is a scalar function of the metric g_{αβ} and its derivatives is the **Ricci scalar R**, which can be obtained from the Riemann tensor, as we know from the previous article Bianchi identity and Ricci tensor

Our Lagrangian then is, **L=R **and we assume that the **Einstein-Hilbert action** could be epxressed as**:**

**Remark**: This integral is taken over the whole of space-time if it converges, and if not, S can still be made so by integrating over an arbitrarily large but compact region; this will still produce the field equations.

You can find an introduction to the Einstein-Hilbert Action at this end of the following Lenoard Susskind's video, starting from 1:11:00 to the end

By its own admission, Susskind has never been able to complete the entire derivation of the Einstein equation from this action, because it's too 'tedious'. Let's see how to do this.

##### Derivation of the Einstein equation from the Einstein Hilbert action

As we know from the principle of least action, the action variation of the action then requires **δS=0**

Knowing from the previous article Variation of the metric determinant that

we get

Setting δS=0, and given that δg^{μν} is totally arbitrary, we get the Einstein field equations in vacuo

if and only if we are able to demonstrate the second member drops off, i.e we have to show that:

Let us first remind us the expression of the Riemann tensor from our article Riemann curvature tensor part II: derivation from the geodesic deviation:

By contracting this tensor on the first and third indices (we set σ=α) we get the Ricci tensor R_{μν}:

Then varying it it gives,

The first two terms of this expression suggest that it could be the difference between **two covariant derivatives**. Let us prove that it is the case.

We know from our previous article Covariant differentiation exercise 2: calculation for the Euclidean metric tensor that the covariant derivatives are constructed from the following building blocks:

- - take the partial derivatives of the tensor
- - add a Γ
^{α}_{γβ}term for each upper index - - substract a Γ
^{γ}_{αβ}term for each lower index

So that we can write each covariant derivative as the sum of four terms (partial derivative + add one term for the upper index + substract two terms for the two lower indices)

and

we can thus verify that:

which is known as the **Palatini identity**.

**Remark**: a more formal way of demonstrating this identity is exposed in our article Palatini equation

So we can now write by replacing δR_{μν} by its expression

When considering the expression in brackets, we notice that the μ and ν indices cancel out, so that we are left with a tensor rank 1 tensor

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