# Energy-momentum tensor for a scalar field (sign problem!)

1. Jun 23, 2008

### knobelc

Hi

I have a small subtle problem with the sign of the energy-momentum tensor for a scalar field as derived by varying the metric (s.b.). I would appreciate very much if somebody could help me on my specific issue. Let me describe the problem in more detail:

I conform to the sign convention $g_{\mu \nu} = (+,-,-,-)$. The Lagranagian for a real scalar field is

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

From Noether Theorem we find the energy-momentum tensor

$$T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \Phi)} \: \partial^\nu \Phi - \mathcal{L} g^{\mu \nu} = \partial^\mu \Phi \partial^\nu \Phi - \mathcal{L} g^{\mu \nu}.$$

Now I want to derive this via varying the action

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

in respect to $g_{\mu \nu}$. In particular it holds

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

$T_{\mu \nu}$ is defined so that varying the action derived from the total Lagrangian

$$\mathcal{L_{\rm tot}} = \frac{1}{16\pi G} R + \mathcal{L}$$

yields the Einstein field equations

$$G_{\mu \nu} = 8\pi G T_{\mu \nu}.$$

(Note that

$$\delta\int\frac{1}{16\pi G} R \sqrt{-g}\; dx^4 = \int G_{\mu \nu} \delta g^{\mu \nu}\sqrt{-g}\; dx^4,$$

therefore the - sign in the definition of $T_{\mu \nu}$.)

Now let's vary the lagrangian of the scalar field:

$$\delta \int \mathcal{L} \sqrt{-g}\; dx^4$$
$$= \int \delta(\mathcal{L}) \sqrt{-g} + \mathcal{L} \delta(\sqrt{-g})\; dx^4$$
$$= \int \delta \left( \frac{1}{2} g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi- V(\Phi ) \right) \sqrt{-g} + \mathcal{L} \left(-\frac{1}{2} g_{\mu \nu} \delta g^{\mu \nu}\right) \sqrt{-g}\; dx^4$$
$$= \frac{1}{2}\int \left( \delta g^{\mu \nu} \partial_\mu\Phi \;\partial_\nu\Phi - \mathcal{L} g_{\mu \nu} \delta g^{\mu \nu} \right) \sqrt{-g}\; dx^4$$
$$= \frac{1}{2}\int \left(\partial_\mu\Phi \;\partial_\nu\Phi - \mathcal{L} g_{\mu \nu} \right) \delta g^{\mu \nu} \sqrt{-g}\; dx^4.$$

Comparing this with the definition of the $T_{\mu \nu}$ yields

$$T_{\mu \nu} = -\partial_\mu \Phi \partial_\nu \Phi + \mathcal{L} g_{\mu \nu}$$

leading to the opposite sign as derived by the Noether Theorem.

I would appreciate very much if somebody could explain why I get the sign wrong. I know this is a subtle (and possibly unimportant) issue but getting the wrong sign without understanding why gives a bad feeling. Thank you for any help!

2. Jun 23, 2008

### George Jones

Staff Emeritus
According to Wald the Klein-Gordon energy-momentum tensor from Noether's theorem agrees with the Klein-Gordon energy-momentum tensor from varying the metric "up to a numerical factor." I do not know if the numerical factor is -1.

Wald says that in others cases, there is less agreement, and it is the energy-momentum arrived at by varying g that appears on the right of Einstein's equation.

If you have Wald, look near the bottom of page 457.

I first ran into differences between the canonical and symmetric energy-momentum tensors in section 12.10 of Jackson.

Last edited: Jun 23, 2008
3. Jun 23, 2008

### knobelc

I think, I got the reason for the wrong sign. Since I used the signature $g_{\mu \nu} = (+,-,-,-)$ my definitions of $T^{\mu \nu}$ and $\mathcal{L_{\rm tot}}$ are not correct. With my signature the correct expressions read as

$$\delta S = \delta\int \mathcal{L} \sqrt{-g}\; dx^4 = +\frac{1}{2}\int T_{\mu \nu} \delta g^{\mu\nu} \sqrt{-g}\; dx^4.$$

and

$$\mathcal{L_{\rm tot}} = -\frac{1}{16\pi G} R + \mathcal{L}.$$

With this I get everything right. :-)