# Low energy quantum gravity

Hi everybody,

before I begin this exercise for myself I want to make sure I have a few things right. Would the Lagrangian for low energy quantum gravity be
$$\mathcal{L}=\frac{1}{2}\partial^\mu \bar{\psi}\partial_\mu \psi +\frac{1}{2}m^2\bar{\psi}\psi-\frac{1}{2}g_{\alpha\beta}R^{\alpha\beta}$$
or would it be
$$\mathcal{L}=\frac{1}{2}\nabla^\mu \bar{\psi}\nabla_\mu \psi +\frac{1}{2}m^2\bar{\psi}\psi-\frac{1}{2}g_{\alpha\beta}R^{\alpha\beta}$$
$$=\frac{1}{2}\left( \partial^\mu \bar{\psi}\partial_\mu \psi+\bar{\psi}\partial^\mu \partial_\mu \psi +\bar{\psi}g^{\alpha\beta}\Gamma^{\gamma}_{\alpha \beta}\partial_\gamma \psi \right)+\frac{1}{2}m^2\bar{\psi}\psi - \frac{1}{2}g_{\alpha\beta}R^{\alpha\beta}$$

That is, do the covariant derivatives act solely on the scalar functions or does the covariant derivative on the left act on both $\bar{\psi}$ and $\partial_\alpha \psi$?

If I use the first Lagrangian my equations of motion are incorrect , but I was also not under the impression that the covariant derivative was acting in the way $\nabla^\mu (\bar{\psi}\partial_\mu \psi )$. Or perhaps I am using the wrong Euler-Lagrange equations. Should the Euler-Lagrange equations have covariant derivatives in them or just normal partials like in the rest of QFT?

Thanks,

Ben Niehoff
Gold Member
You mean gravity coupled to a massive scalar? The action would be

$$S = \int d^4x \; \sqrt{-g} \Big(\frac12 \nabla_\mu \bar \phi \nabla^\mu \phi + \frac12 m^2 \bar \phi \phi + \frac{1}{16\pi G}\mathcal{R} \Big)$$

Don't forget the $\sqrt{-g}$ in the measure.

You mean gravity coupled to a massive scalar? The action would be

$$S = \int d^4x \; \sqrt{-g} \Big(\frac12 \nabla_\mu \bar \phi \nabla^\mu \phi + \frac12 m^2 \bar \phi \phi + \frac{1}{16\pi G}\mathcal{R} \Big)$$

Don't forget the $\sqrt{-g}$ in the measure.

That looks like what I got (except we have different units). But I'm still not sure about expanding the covariant derivatives. The covariant derivative reduces to a partial derivative for a scalar function, but with that substitution I don't get the correct Klein-Gordon equation. So I was wondering if the first covariant derivative acts on both $\bar{\psi}$ and $\partial_\mu \psi$?

Also so we are clear, this would be a low energy quantum general relativity, since the Klein-Gordon action is for all different spin particles and the Einstein Hilbert action is for coupling to the gravitational field. The result would be the field theory for massive particles (of all spins) couples to the gravitational field, correct?

Thanks,

Ben Niehoff
Gold Member
Yes, since the covariant derivative acts on a scalar field, it's really just an ordinary derivative. You can replace $\nabla_\mu \rightarrow \partial_\mu$ in both cases.

The important part is that you not forget the $\sqrt{-g}$.

Edit: To answer your other question, this Lagrangian will only be appropriate for a massive spin 0 field coupled to gravity.

Yes, since the covariant derivative acts on a scalar field, it's really just an ordinary derivative. You can replace $\nabla_\mu \rightarrow \partial_\mu$ in both cases.

The important part is that you not forget the $\sqrt{-g}$.

Edit: To answer your other question, this Lagrangian will only be appropriate for a massive spin 0 field coupled to gravity.

I'm missing something. The $\sqrt{-g}$ is important in the action, but I'm using Euler-Lagrange equations of motion. It doesn't even show up in them, that I'm aware. If I want to find the eqations of motion from that lagrangian, is there a different set of Euler-Lagrange equations that I should be using? If I only have the partials in the Lagrangian, then the equations of motion are not correct after applying the Euler-Lagrange equations.

Thanks,

EDIT: (in response to your edit) In another thread I recently read that the Klein-Gordon equation is correct for all spin particles, since every solution for spin particles is automatically a solution to the K-G equation.

here

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Would the correct Euler-Lagrange equations be

$$\nabla_\mu\frac{\partial\mathcal{L}}{ \partial (\partial_\mu \psi)}-\frac{\partial\mathcal{L}}{\partial \psi}=0$$

for this situation with a scalar field?

Ben Niehoff
Gold Member
The Euler-Lagrange equations are

$$\frac{\delta S}{\delta \phi(x^\mu)} = 0$$

If you are plugging things into some other equations, you need to stop and think here. Probably you are doing it wrong.

As for the other issue, it is true that particles of all spin satisfy the Klein-Gordon equation. However, the Klein-Gordon equation does not describe all of the physics of particles higher than spin 0.

The Euler-Lagrange equations are

$$\frac{\delta S}{\delta \phi(x^\mu)} = 0$$

If you are plugging things into some other equations, you need to stop and think here. Probably you are doing it wrong.

As for the other issue, it is true that particles of all spin satisfy the Klein-Gordon equation. However, the Klein-Gordon equation does not describe all of the physics of particles higher than spin 0.

This
$$\nabla_\mu\frac{\partial\mathcal{L}}{ \partial (\partial_\mu \psi)}-\frac{\partial\mathcal{L}}{\partial \psi}=0$$

gets the right equations of motion for the scalar function. As for the equations of motion for gravity, I know I can vary the Einstein-Hilbert action to get the field equations however, I was wondering if the following holds (I generated this purely by analogy)

Edit: I include the higher derivative parts
$$-\partial_\delta \partial_\gamma \frac{\partial R}{\partial (\partial_\delta \partial_\gamma g_{\alpha\beta})}+\partial_\gamma \frac{\partial R}{\partial(\partial_\gamma g_{\alpha\beta})}-\frac{\partial R}{\partial g_{\alpha\beta}}=R_{\alpha\beta}=0$$

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I realized as soon as I got on my bike that the above should also contain partials for the second derivatives of g in the scalar curvature. Please take that into account.

Ben Niehoff
Gold Member
This

$$\nabla_\mu\frac{\partial\mathcal{L}}{ \partial (\partial_\mu \psi)}-\frac{\partial\mathcal{L}}{\partial \psi}=0$$

gets the right equations of motion for the scalar function.

Then it may be right. You can check by varying the action with respect to the scalar field, and deriving the correct E-L equations.

As for the equations of motion for gravity, I know I can vary the Einstein-Hilbert action to get the field equations however, I was wondering if the following holds (I generated this purely by analogy)

Edit: I include the higher derivative parts
$$-\partial_\delta \partial_\gamma \frac{\partial R}{\partial (\partial_\delta \partial_\gamma g_{\alpha\beta})}+\partial_\gamma \frac{\partial R}{\partial(\partial_\gamma g_{\alpha\beta})}-\frac{\partial R}{\partial g_{\alpha\beta}}=R_{\alpha\beta}=0$$

This is most likely wrong.