# Invariance of Klein Gordon in General Relativity

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phyz2
TL;DR Summary
Confused bout KG Lagrangian in curved space
Hello!
I'm starting to study curved QFT and am slightly confused about the invariance of the Klein Gordon Lagrangian under a linear diffeomorphism.
This is $$L=\sqrt{-g}\left(g^{\mu\nu}\partial_\mu \phi \partial_\nu \phi-\frac{m^2}{2}\phi^2\right),$$
I don't see how ##g^{\mu\nu}\to g^{\mu\nu}+\partial^{(\mu}\xi^{\nu)}## should leave this invariant.. I get and extra piece which is ##\sqrt{-g} \partial^{(\mu}\xi^{\nu)} \partial_\mu \phi \partial_\nu \phi## and I don't see how this should be zero, even at the linear level.

Thanks for any help!

## Answers and Replies

Staff Emeritus
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What you wrote down as ##L## is not invariant because it includes the ##\sqrt g## from the volume element.

The expression in the parenthesis is manifestly a scalar.

Mentor
The expression in the parenthesis is manifestly a scalar.
For this to be true in curved spacetime (which is what I understand the OP to be asking about), wouldn't the derivatives need to be covariant derivatives instead of partial derivatives?

Staff Emeritus
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For this to be true in curved spacetime (which is what I understand the OP to be asking about), wouldn't the derivatives need to be covariant derivatives instead of partial derivatives?
For (first) derivatives of scalars there is no difference.

vanhees71
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One should emphasize that ##\sqrt{-g}\mathrm{d}^4 x## is generallt invariant too. So the action is generally invariant and thus the resulting field equations too. That makes the action principle so useful for finding adequate equations, that are covariant under a given symmetr.

Staff Emeritus
Homework Helper
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Agreed, I considered it subtextual that the volume element is invariant but I could have been more specific for clarity.

vanhees71
Summary:: Confused bout KG Lagrangian in curved space

Hello!
I'm starting to study curved QFT and am slightly confused about the invariance of the Klein Gordon Lagrangian under a linear diffeomorphism.
This is $$L=\sqrt{-g}\left(g^{\mu\nu}\partial_\mu \phi \partial_\nu \phi-\frac{m^2}{2}\phi^2\right),$$
I don't see how ##g^{\mu\nu}\to g^{\mu\nu}+\partial^{(\mu}\xi^{\nu)}## should leave this invariant.. I get and extra piece which is ##\sqrt{-g} \partial^{(\mu}\xi^{\nu)} \partial_\mu \phi \partial_\nu \phi## and I don't see how this should be zero, even at the linear level.

Thanks for any help!
The transformation for the metric you give is (a linearization of) the Lie derivative, an "active diffeomorphism". The Lie derivative of the Lagrangian is not zero.

phyz2
phyz2
The transformation for the metric you give is (a linearization of) the Lie derivative, an "active diffeomorphism". The Lie derivative of the Lagrangian is not zero.
Thanks! Could you elaborate a bit more? I thought the invariance under diffeos would be equivalent to the gauge invariance of the, say, Maxwell Lagrangian.. in that case the change of the action is zero.

Gold Member
2022 Award
The key is the action. If the action is invariant under general coordinate transformations then for sure the field-equations of motion derived from it are covariant equations. That implies that the Lagrangian ##\mathfrak{L}## should be a scalar density (!) or you write ##\mathfrak{L}=\sqrt{-g} L##, where ##L## is a scalar. Then
$$A=\int \mathrm{d}^4 q \mathfrak{L} = \int \mathrm{d}^4 q \sqrt{-g} L$$
is a scalar and thus the field equations of motion derived from Hamilton's principle using this action are generally covariant.

Thanks! Could you elaborate a bit more? I thought the invariance under diffeos would be equivalent to the gauge invariance of the, say, Maxwell Lagrangian.. in that case the change of the action is zero.
Sorry for the late reply. I understand your confusion. Let's take a scalar. Under a g.c.t. one has that phi(x) = phi'(x'). But the Lie derivative, what's commonly called an "infinitesimal g.c.t.", compares phi'(x) and phi(x) instead of phi'(x') and phi(x). Watch those primes like a hawk! And the Lie derivative of a scalar is in general non-zero, of course.

The Lagrangean density is a scalar density, so you should consider how such a density transforms under "infinitesimal g.c.t.'s". It's been a while, but with proper boundary conditions you should get a total derivative such that the action functional is invariant. See e.g. d'Inverno.

vanhees71
$$A=\int \mathrm{d}^4 q \mathfrak{L} = \int \mathrm{d}^4 q \sqrt{-g} L$$