# Energy momentum tensor of a scalar field by varying the metric

## Main Question or Discussion Point

Suppose you are given the Lagrangian of a scalar field $$\Phi(t)$$

$$\mathcal{L} = \frac{1}{2} \dot{\Phi}- \nabla \Phi - V(\Phi ).$$

By introducing covariant notation with $$\eta_{\mu \nu} = (1,-1,-1,-1)$$ this reads as

$$\mathcal{L} = \frac{1}{2} \eta^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ).$$

Let us now switch to GR, i.e. \$\eta^{\mu \nu} \rightarrow $$g^{\mu \nu}$$. The Lagrangian remains the same since covariant derivative of a scalar field is the same as normal derivative, i.e. $$\nabla_{\mu} \Phi = \partial_{\mu} \Phi$$. I derive the energy-momentum-Tensor $$T^{\mu \nu}$$ by varying $$g_{\mu\nu}$$ in the action

$$S = \int \mathcal{L} \sqrt{-g}\; dx^4,$$

i.e.

$$\delta S = \frac{1}{2}\int T^{\mu \nu} \delta g_{\mu\nu} \sqrt{-g}\; dx^4.$$

So we obtain with the variation $$\delta g_{\mu\nu}$$

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Sorry, I had some problems with the latex here. So I could not finish my first posted text. Here is the full version:

Suppose you are given the Lagrangian of a scalar field $\Phi(t)$

$$\mathcal{L} = \frac{1}{2} \dot{\Phi}- \nabla \Phi - V(\Phi ).$$

By introducing covariant notation with $\eta_{\mu \nu} = (1,-1,-1,-1)$ this reads as

$$\mathcal{L} = \frac{1}{2} \eta^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ).$$

Let us now switch to GR, i.e. $\eta^{\mu \nu} \rightarrow g^{\mu \nu}$. The Lagrangian remains the same since covariant derivative of a scalar field is the same as normal derivative, i.e. $\nabla_{\mu} \Phi = \partial_{\mu} \Phi.$ I derive the energy-momentum-Tensor $T^{\mu \nu}$ by varying $g_{\mu\nu}$ in the action

$$S = \int \mathcal{L} \sqrt{-g}\; dx^4,$$

i.e.

$$\delta S = \frac{1}{2}\int T^{\mu \nu} \delta g_{\mu\nu} \sqrt{-g}\; dx^4.$$

So we obtain with the variation $$\delta g_{\mu\nu}$$

$$\delta S = \delta \int \mathcal{L} \sqrt{-g} \;dx^4 = \int \left[\left( \frac{\partial \mathcal{L}}{\partial (\nabla_\mu \Phi)} \;\delta (\nabla_\mu \Phi) + \frac{\partial\mathcal{L}}{\partial g_{\mu\nu}} \;\delta g_{\mu\nu} \right) \sqrt{-g} + \mathcal{L} \;\delta (\sqrt{-g}) \right]dx^4.$$

Since our Lagrangian does not contain any covariant derivative $\delta (\nabla_\mu \Phi)=0.$ Furthermore it holds

$$\frac{\partial\mathcal{L}}{\partial g_{\mu\nu}} = \frac{\partial}{\partial g_{\mu\nu}} \left[\frac{1}{2} g_{\mu\nu} \partial^\mu\Phi \;\partial^\nu\Phi\right] = \frac{1}{2} \partial^\mu\Phi \;\partial^\nu\Phi$$

and it is well known in GR that

$$\delta(\sqrt{-g}) = \frac{1}{2}g^{\mu\nu} \sqrt{-g}\; \delta g_{\mu\nu}.$$

So we obtain finally

$$\delta S = \frac{1}{2} \int \left[ \partial^\mu\Phi \;\partial^\nu\Phi + g^{\mu\nu}\mathcal{L} \right]\delta g_{\mu\nu} \sqrt{-g}\; dx^4,$$

thus

$$T^{\mu \nu} = \partial^\mu\Phi \;\partial^\nu\Phi + g^{\mu\nu}\mathcal{L}.$$

My problem is now that the correct result should read as

$$T^{\mu \nu} = \partial^\mu\Phi \;\partial^\nu\Phi - g^{\mu\nu}\mathcal{L}.$$

Thus I get the sign in the second term wrong. Where is the error?? Sorry for this subtle problem but I have looked at it again and again and don't see it. So I would be happy if anybody could help me.

haushofer
Your variation isn't going correctly. First of all, your action contains derivatives of the scalar field, so you should include those. Covariant derivatives on scalar fields equal partial derivatives on scalar fields.

If one has the action

$$L = \sqrt{g}{\cal L} = \sqrt{g}\Bigl(g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi -V(\phi)\Bigr)$$

I obtain when varying with respect to the metric ( varying with respect to the scalar field will give me the equations of motion, which are known I assume )

$$\delta L = \delta {\cal L} \sqrt{g} + {\cal L}\delta\sqrt{g} = ( \delta {\cal L} + \frac{1}{2}{\cal L}g^{\mu\nu}\delta g_{\mu\nu} )\sqrt{g}$$

which equals

$$(\frac{1}{2}\delta g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi + \frac{1}{2}g^{\mu\nu} {\cal L }\delta g_{\mu\nu})\sqrt{g}$$

Now remember that

$$\delta g^{\mu\nu} = -\delta g_{\mu\nu}$$

and you obtain

$$\delta L = \frac{1}{2} \sqrt{g}(g^{\mu\nu}{\cal L} - \partial_{\mu}\phi\partial_{\nu}\phi)\delta g_{\mu\nu}$$

From this you can easily obtain the energy momentum tensor. If I didn't make typos, that is.

haushofer
That was a quick answer from a long night drinking; ofcourse the last term should be

$$\delta L = \frac{1}{2} \sqrt{g}(g^{\mu\nu}{\cal L} - \partial^{\mu}\phi\partial^{\nu}\phi)\delta g_{\mu\nu}$$

First of all, the correct formula is

$$\delta g^{\mu\nu} = -g^{\mu \alpha}g^{\beta \nu}\delta g_{\alpha\beta}$$.

The whole paradox for me is then

$$\delta g_{\mu \nu}\;\partial^{\mu}\phi\partial^{\nu}\phi = \delta(g_{\mu \nu}\partial^{\mu}\phi\partial^{\nu}\phi)$$

$$= \delta(g^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi)$$

$$= \delta g^{\mu \nu}\;\partial_{\mu}\phi\partial_{\nu}\phi$$

$$= -g^{\mu \alpha}g^{\beta \nu}\delta g_{\alpha\beta}\;\partial_{\mu}\phi\partial_{\nu}\phi$$

$$= - \delta g_{\mu \nu}\;\partial^{\mu}\phi\partial^{\nu}\phi$$

This is a contradiction unless $\delta g_{\mu \nu} = 0$. This makes the difference between your and my solution. Can you resolve this?

haushofer
Maybe it's a little overdone, but my view on this is the following. Varying and raising indices don't commute. So,

$$[\delta, g_{\mu\nu}] \neq 0$$

This looks quite trivial, but it has consequences. When I wrote

$$\delta g_{\mu\nu} = -\delta g^{\mu\nu}$$

I ment explicitly

$$(\delta g)_{\mu\nu} = -(\delta g)^{\mu\nu}$$

You're used to freely change the position of the indices in a contraction, so eg for tensors of rank one,

$$X_{\alpha}Y^{\alpha} = X^{\alpha}Y_{\alpha}$$

However, this is not allowed when you consider variations which concern that object with which you raise and lower those indices: the metric itself! So I would say that

$$\delta(g^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi) \neq \delta(g_{\mu \nu}\partial^{\mu}\phi\partial^{\nu}\phi)$$

This can be traced back by the fact that

$$g_{\mu\nu}g^{\nu\rho} = \delta_{\mu}^{\rho} \rightarrow \delta\Bigl(g_{\mu\nu}g^{\nu\rho}\Bigr) = 0 \rightarrow \delta g_{\mu\nu} g^{\nu\rho} = - g^{\nu\rho}\delta g_{\mu\nu}$$

which creates that minus-sign. In fact, what you should have is

$$\delta(g^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi) = - \delta(g_{\mu \nu}\partial^{\mu}\phi\partial^{\nu}\phi)$$

It may look strange that lowering or raising indices on a tensor induces a minus-sign, but that's only when you look at the variation of the metric, which is a tensor. Ofcourse, this doesn't influence the transformation properties of the object, so it's still a tensor.

I agree that

$$\delta(g^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi) = - \delta(g_{\mu \nu}\partial^{\mu}\phi\partial^{\nu}\phi)$$,

but then problem arises that the way we write the Lagrangian determines the sign of the first term in the energy-momentum-tensor. How can we decide what way to write the Lagrangian, i.e. either

$$L = \sqrt{g}{\cal L} = \sqrt{g}\Bigl(g^{\mu\nu}\partial_{\mu}\phi\partial _{\nu}\phi -V(\phi)\Bigr)$$

or

$$L = \sqrt{g}{\cal L} = \sqrt{g}\Bigl(g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi -V(\phi)\Bigr)?$$

I tend to the answer that the first version is the more natural since it contains the "physical" derivation

$$\partial_{\mu} = \frac{\partial}{\partial x^\mu},$$

$$\partial^{\mu} = \frac{\partial}{\partial x_\mu}$$

which can only be defined by means of a metric, the object to be varied.

samalkhaiat
$$\delta(g^{\mu \nu}\partial_{\mu}\phi\partial_{\nu}\phi) = - \delta(g_{\mu \nu}\partial^{\mu}\phi\partial^{\nu}\phi)$$,
$$\delta_{g}\left( g^{ab} A_{a}B_{b}\right) = \delta g^{ab} A_{a}B_{b}$$

but then problem arises that the way we write the Lagrangian determines the sign of the first term in the energy-momentum-tensor. How can we decide what way to write the Lagrangian, i.e. either

$$L = \sqrt{g}{\cal L} = \sqrt{g}\Bigl(g^{\mu\nu}\partial_{\mu}\phi\partial _{\nu}\phi -V(\phi)\Bigr)$$

or

$$L = \sqrt{g}{\cal L} = \sqrt{g}\Bigl(g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi -V(\phi)\Bigr)?$$
there is no problem you can use any one.

Because $\mathcal{L}(g^{ab}) = \mathcal{L}(g_{ab})$, you need to decide about the sign so that you always have

$$T^{ab} \delta g_{ab} = - T_{ab} \delta g^{ab}$$

So when you take

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab}$$

you must then take

$$\delta \mathcal{L} = \frac{ \partial \mathcal{L} }{ \partial g^{ab} } \delta g^{ab} = - \frac{\partial \mathcal{L}}{\partial g_{ab} } \delta g_{ab}$$

the second equality follows from the identity
$$\left[ \delta g^{ab} = - g^{ac} g^{bd} \delta g_{cd} \ \right]$$

Thus

$$\delta S = \frac{1}{2} \int dx \sqrt{-g} \left( \mathcal{L} g^{ab} - 2 \frac{\partial \mathcal{L}}{\partial g_{ab}} \right) \delta g_{ab}$$

Then from

$$\delta S = \int dx \sqrt{-g} \frac{\delta S}{\delta g_{ab}} \delta g_{ab} \equiv - \frac{1}{2} \int dx \sqrt{-g} \ T^{ab} \delta g_{ab}$$

you find

$$T^{ab} = 2 \frac{ \partial \mathcal{L}}{\partial g_{ab}} - g^{ab} \mathcal{L}$$

*****************

But if you take

$$\delta \mathcal{L} = (+) \frac{\partial \mathcal{L}}{\partial g_{ab}} \delta g_{ab}$$

then you should write

$$\delta \sqrt{-g} = (-) \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab}$$

and define the T-tensor by

$$\delta S = \frac{1}{2} \int dx T^{ab} \delta g_{ab}$$

******************

Personally I use and stick to the followings

$$\delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{ab} \delta g^{ab}$$

$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial g^{ab}} \delta g^{ab}$$

$$\delta S = \frac{1}{2} \int dx \ \sqrt{-g} T_{ab} \delta g^{ab}$$

this again leads to the correct T-tensor

$$T_{ab} = 2 \frac{\partial \mathcal{L}}{\partial g^{ab}} - g_{ab} \mathcal{L}$$

regards

sam

So when you take

$$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab}$$

you must then take

$$\delta \mathcal{L} = \frac{ \partial \mathcal{L} }{ \partial g^{ab} } \delta g^{ab} = - \frac{\partial \mathcal{L}}{\partial g_{ab} } \delta g_{ab}$$
Why is that? Is this just a rule in order to get the "right" result? Without any further information I don't see a reason not to write

$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial g_{ab} } \delta g_{ab} = - \frac{ \partial \mathcal{L} }{ \partial g^{ab} } \delta g^{ab}$$

in this case. Or do I miss something?

Cheers, C.

samalkhaiat
Why is that? Is this just a rule in order to get the "right" result?
Exactly! you need your T-tensor,defined in GR to be

$$T^{ab} = -2 \frac{\delta S}{\delta g_{ab}} \ \ (1),$$

to agree with the (symmetric) canonical tensor, derived from Noether theorem,

$$T^{ab} = \frac{\partial \mathcal{L}}{\partial \partial_{a} \phi} \partial^{b} \phi - \mathcal{L} g^{ab}$$

Can you derive Eq(1) within the canonical formalism? No you can not. So you need a sigh manifesto.

Without any further information I don't see a reason not to write

$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial g_{ab} } \delta g_{ab} = - \frac{ \partial \mathcal{L} }{ \partial g^{ab} } \delta g^{ab}$$

in this case. Or do I miss something?
Yes, You would get an incorrect canonical energy-momentum tensor.

regards

sam

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