# Einstein-Hilbert action origin

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1. Oct 18, 2015

### Emilie.Jung

So, it always (at least in the books that I have read or scanned) that the Einstein-Hilbert action $$S=\int{\sqrt{g}d^4xR}$$ is directly posed without an explanation of its origin.
My question is how did it occur to Hilbert or Einstein to write down this specific form of action?

2. Oct 18, 2015

### JorisL

Carroll's book has a gentle introduction in chapter 4. (The online, older version might have them too)
Basically the argument is that this is the simplest possible action.
He then argues that Hilbert got to this form using similar arguments.
Finally he proves how to derive the EOM for this action from which the EFE follow.

3. Oct 18, 2015

### Staff: Mentor

As JorisL says, the basic argument is that it's the simplest possible action. But there are a couple of points worth expanding on.

First, what exactly does "simplest possible" mean? The idea is that the action should be a scalar that includes no higher than second derivatives of the metric. The only scalar derivable from the metric that meets that requirement is the Ricci scalar $R$. The $\sqrt{-g}$ comes in because it's needed to make the spacetime integration measure $d^4 x$ invariant under coordinate transformations, so the action itself is invariant.

Second, note that I said above "the only scalar derivable from the metric". There is also another scalar that could appear, namely a constant, not a function of anything. This constant is called the cosmological constant $\Lambda$, and the modern view is that it belongs in the Hilbert action--there is no a priori reason for not including it. Hilbert didn't include it originally because he simply didn't think of it.

4. Oct 19, 2015

### vanhees71

In addition one assumes that the equations of motion are 2nd order partial differential equations. Then $R$ and a comsomological constant are the only allowed Lagrange densities, because only $R$ is linear in the 2nd derivatives with coefficients not containing derivatives, which implies that by using the 4D Gauss integration theorem the action (integration by parts) can be rewritten with a Lagrangian with only 1st-order derivatives implying 2nd order field equations of motion. For a simple proof, see Landau, Lifshitz vol. II.

5. Oct 19, 2015

### Emilie.Jung

@PeterDonis Thank you for your answer. I have two questions though.

Why is that? Why not higher than third derivatives or even first derivatives of the metric?
Is there a proof that $\sqrt{-g}$ makes it invariant?

6. Oct 19, 2015

### vanhees71

You want only 1st-order derivatives under the integral defining the action. This can also be achieved by having a Lagrangian with 2nd order derivatives if these come only in linear form with coefficients that don't depend on derivatives, because then you can do an integration by parts, ending with an integral of an expression containing only $g_{\mu \nu}$ and its first (partial) derivatives.

To your second question, note that for an arbitrary diffeomorphism $q \mapsto q'$ you have
$$\mathrm{d}^4 q'=\mathrm{d}^4 q \mathrm{det} \left (\frac{\partial q'^{\mu}}{\partial q^{\nu}} \right ).$$
On the other hand the covariant components of the metric transform according to
$$g'_{\mu \nu}=\frac{\partial q^{\alpha}}{\partial q'^{\mu}} \frac{\partial q^{\beta}}{\partial q'^{\nu}} g_{\alpha \beta},$$
and thus
$$\mathrm{det} (g_{\mu \nu}')=\mathrm{det}^2 \left (\frac{\partial q^{\alpha}}{\partial q'^{\mu}} \right ) \mathrm{det} g_{\alpha \beta} = \mathrm{det}^{-2} \left (\frac{\partial q'^{\mu}}{\partial q^{\alpha}} \right ) \mathrm{det} g_{\alpha \beta}.$$
Thus the expression in question is a scalar, i.e.,
$$\sqrt{-g'} \mathrm{d}^4 q'=\sqrt{-g} \mathrm{d}^4 q,$$
where we have used $g=\mathrm{det} g_{\mu \nu}<0$.

7. Oct 19, 2015

### atyy

There is no reason, especially from the quantum point of view. Carroll comments on this in http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html.

"Indeed, in a world governed by quantum mechanics we expect all possible couplings between different fields (such as gravity and electromagnetism) that are consistent with the symmetries of the theory (in this case, gauge invariance). So why is it reasonable to set http://ned.ipac.caltech.edu/level5/March01/Carroll3/img596.gif = 0? The real reason is one of scales. ..."

The quantum point of view comes from Wilson, that gravity is an effective field theory: http://arxiv.org/abs/1209.3511 (see Eq 26 and the two paragraphs after it).

Last edited: Oct 19, 2015
8. Oct 19, 2015

### Staff: Mentor

First derivatives of the metric are present in $R$; they are lower order than second, which is the highest order present.

As atyy said, third and higher order derivatives are expected to be present in principle according to the modern view. However, any such terms will have extra coefficients in front of them which will be very small, so small that at energies much less than the Planck energy (which means any energies we can expect to run experiments at in the foreseeable future), those terms are much too small to measure.

9. Oct 19, 2015

### samalkhaiat

I've derived the H-E action on these forums long time ago. Anyway, have a look at the derivation in the PDF below.

#### Attached Files:

• ###### Deriving the E-H Action.pdf
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10. Oct 19, 2015

### Emilie.Jung

@PeterDonis thank you.

So, is this enough explanation? Because I am not convinced how can this imply that? Why would having 1st derivatives of the metric in $R$ set a condition on the order of differentiation of the metric?Because in your previous answer you said:
That is to say, in the latter quote you said that given that the scalar should not include higher than 2nd order derivatives, SO Ricci is a good option because it fits the condition.
In the former quote, you say the opposite.
So, still I don't know how to explain this.

11. Oct 19, 2015

### Emilie.Jung

@vanhees71 thank you for the derivation and explanation. I understood the second part of the answer but I did not understand this
I still do not know how was the condition on the metric differentiation order set?

12. Oct 19, 2015

### Staff: Mentor

It doesn't. But you had asked:

Note the bolded part. I was simply pointing out that the bolded part of your question doesn't make sense, because first derivatives are included in "no higher than second derivatives". See below.

Yes. "No higher than second order derivatives" includes first derivatives. And the Ricci scalar $R$ includes first derivatives as well as second derivatives of the metric. But it doesn't include any higher order derivatives.

No, it doesn't. It just says that first derivatives are not higher order than second. See above.
The PDF that samalkhaiat linked to in post #9 goes into this. Basically, it's because we want the field equation that gets derived from the Lagrangian to be a differential equation of no higher than second order. The reason for that, as I understand it, is that in a pseudo-Riemannian spacetime, we don't know how to handle differential equations of higher than second order. But I'm not very familiar with the details.

13. Oct 19, 2015

### Emilie.Jung

14. Oct 20, 2015

### Emilie.Jung

@samalkhaiat Thank you a lottt for the pdf, very well structured. It was only now that I read it carefully with all the details. You said there," But, the principle of equivalence makes it impossible to have a non-trivial scalar function $L(g_{ab}; \partial_cg_{ab})$" Thus, you said this demands that there is also a second derivative of the metric present but with a certain condition. I did not understand this cause-effect relation. Would be grateful if you can explain this.

15. Oct 20, 2015

### samalkhaiat

I’m not sure I understand your confusion. So, I will only repeat what I have already done in the PDF.
In general, third and higher order partial differential equations (PDE) violate causality. For this reason, we assumed that our dynamical variable (the metric tensor $g$) satisfies second-order PDE’s: $$\partial^{2}g(x) = f(x) . \ \ \ \ \ (1)$$ Usually, second-order PDE’s (like (1)) follow from first-order (in derivative) Lagrangians (like $L_{(1)}=L_{(1)}(g,\partial g)$). But, the Principle of Equivalence (POE) says that the first-order Lagrangian $L_{(1)}(g,\partial g)$ is not good, i.e., it does not lead to non-trivial dynamical equations for $g$: POE lets you set $L_{(1)} = L_{(1)}(\eta , 0) = \mbox{const}$.
Since the POE does not allow you to set $\partial^{2}g = 0$, we look for a second-order Lagrangian $L_{(2)} = L_{(2)}(g,\partial g , \partial^{2}g )$. The equations of motion for this Lagrangian is given by
$$\frac{\partial L_{(2)}}{\partial g} - \partial \left( \frac{\partial L_{(2)}}{\partial (\partial g)} \right) + \partial^{2} \left( \frac{\partial L_{(2)}}{\partial (\partial^{2}g)} \right) = 0 . \ \ \ (2)$$ Now, according to our assumption (eq(1)), we want eq(2) to be second-order PDE in $g$. As you can easily see from eq(2), this will be the case provided that $$\frac{\partial L_{(2)}}{\partial (\partial^{2}g)} \propto A(g) .$$
That is to say, our second-order Lagrangian $L_{(2)}$ must depend linearly on $\partial^{2}g$:
$$L_{(2)}(g,\partial g , \partial^{2}g ) = A(g) \partial^{2}g + B(g , \partial g)$$

16. Oct 20, 2015

### Emilie.Jung

Why?

Why doesn't it let us to set $\partial^2g=0$ and if it doesn't why should we look for a second order lagrangian anyway?

Thank you in advance @samalkhaiat !

17. Oct 21, 2015

### samalkhaiat

Have you heard of “runaway” solutions? Systems described by higher (than second) order equations will develop unphysical non-causal solutions. A well-known case is the third order equation of Lorentz and Dirac that incorporates the effects of radiation reaction. It describes non-causal effects such as pre-acceleration of charges yet to be hit by radiation; infinitely self-accelerated electron and other diseases.
Newton’s equation is a second order equation. Maxwell’s equations and almost all equations of classical fields are second-order PDE’s. So, it is reasonable and natural to assume that $g_{\mu\nu}$ also satisfies second order equations.
This is the fundamental discovery of Riemann: if you can set all of the $\partial_{\mu}\partial_{\nu}g_{\alpha \beta}$ to zero at a point, then the space is flat. This is explained in many textbooks on GR. The idea is the following. Let the point in question be $x^{\mu}=0$. Choose new coordinates $\bar{x}^{\mu}$ defined by
$$x^{\mu} = A^{\mu}{}_{\nu} \ \bar{x}^{\nu} + B^{\mu}_{(\nu \rho)} \bar{x}^{\nu} \bar{x}^{\rho} + C^{\mu}_{(\nu \rho \sigma)} \bar{x}^{\nu} \bar{x}^{\rho} \bar{x}^{\sigma} .$$

Now, in D-dimensional space-time:

1) The symmetric metric tensor: $g_{\mu\nu}$ has $\frac{1}{2} D(D+1)$ components, and the matrix $A^{\mu}{}_{\nu}$ has $D^{2}$ elements. Since $D^{2} > D(D+1)/2$, we can choose $A^{\mu}{}_{\nu}$ to set $\bar{g}_{\mu\nu}(0) = \eta_{\mu\nu}$.

2) The first derivative of the metric: $\partial_{\rho}g_{\mu\nu}$ consists of $\frac{1}{2}D^{2}(D+1)$ numbers. This is equal to the number of parameters in $B^{\mu}_{(\alpha \beta)}$. Thus, there are just enough freedom in $B^{\mu}_{(\alpha \beta)}$ to set all $\bar{g}_{\mu\nu , \rho}(0) = 0$.

3) The second derivative of the metric: $\partial_{\mu}\partial_{\nu}g_{\alpha\beta}$ have $\frac{1}{4}D^{2}(D+1)^{2}$ components, while $C^{\mu}_{(\nu\rho\sigma)}$ have $\frac{1}{6}D^{2}(D+1)(D+2)$ parameters. Thus, $$\frac{1}{4}D^{2}(D+1)^{2} - \frac{1}{6}D^{2}(D+1)(D+2) = \frac{1}{12}D^{2}(D^{2}-1) ,$$ represent the number of components in the second derivative of $g_{\mu\nu}$ which cannot set zero by any coordinate transformations. This is exactly the number of independent components in the Riemann tensor $R_{\mu\nu\alpha\beta}$. In his PHD thesis, Riemann proved the following: “in the neighbourhood of any point, their exists a coordinate system (Riemann’s normal coordinates) such that $$g_{\mu\nu}(x) = \eta_{\mu\nu} - \frac{1}{3} R_{\mu\alpha\nu\beta} x^{\alpha}x^{\beta} + \mathcal{O}(x^{3}) ,$$ with $R_{\mu\nu\rho\sigma} = 0$, if and only if the space is flat”.
2 comes right after 1, is it not? You cannot obtain a second order equation from a third, fourth or seventh order Lagrangians, can you?

Last edited by a moderator: Oct 21, 2015
18. Oct 21, 2015

### Emilie.Jung

19. Oct 21, 2015

### Staff: Mentor

Just to add a brief note to this, since there are still some elements of $A^\mu{}_\nu$ left over--$D^2 - \frac{1}{2} D \left( D + 1 \right) = \frac{1}{2} D \left( D - 1 \right)$ of them to be exact--where do they go? The answer is that setting $\bar{g}_{\mu\nu}(0) = \eta_{\mu\nu}$ is not enough, by itself, to specify a unique coordinate chart $\bar{x}^\mu$. We also have to fix the exact orientation of the axes and the state of motion of an object at rest at the origin of the new chart. That requires $\frac{1}{2} D \left( D - 1 \right)$ parameters, exactly the number left over in $A^\mu{}_\nu$. (In 4-d spacetime, one way to think of this is that we are fixing the 6 parameters of a Lorentz transformation that orients the coordinate axes of the new chart in a particular way and sets the ordinary velocity of an object at rest in the chart to zero.)

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