# Calculating Energy-Momentum Tensor in GR

• Markus Kahn
In summary, the conversation discusses the attempt to rewrite the expression for the action, taking into account the factor of 1/2. The variation of the action is then performed, resulting in the expression for the energy-momentum tensor. It is mentioned that there may be a factor of 1/2 missing in the original expression and the signs should be double-checked. The final conclusion is that there is no further simplification possible for the energy-momentum tensor.
Markus Kahn
Homework Statement
During the derivation of the Einstein field equations via the variation principle we defined the energy momentum tensor to be
$$T_{\mu\nu} \equiv -2\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.$$
Assume now that
$$S_M = \int d^4x \sqrt{-g}(\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2),$$
where ##\phi## is the scalar field. Calculate ##T_{\mu\nu}##. Relevant Equations All given above My attempt was to first rewrite ##S_M## slightly to make it more clear where ##g_{\mu\nu}## appears $$S_M = \int d^4x \sqrt{-g} (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2).$$ Now we can apply the variation: \begin{align*} \delta S_M &= \int d^4x (\delta\sqrt{-g}) (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\ &= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\ &= \int d^4x \sqrt{-g}\left(- \frac{1}{2}g_{\mu\nu} [g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2] + \nabla_\mu\phi\nabla_\nu\phi \right)\delta g^{\mu\nu}\\ &= \int d^4x \frac{\sqrt{-g}}{2}\left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right)\delta g^{\mu\nu}. \end{align*} We then have $$\frac{\delta S_M}{\delta g^{\mu\nu}} = \frac{\sqrt{-g}}{2}\left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right) \quad \Rightarrow \quad T_{\mu\nu} = - \left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right)$$ Is this sensible? I'm not really sure if I performed the variation correctly, so I would appreciate if someone could give it a look. Markus Kahn said: Assume now that $$S_M = \int d^4x \sqrt{-g}(\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2),$$ where ##\phi## is the scalar field. $$Usually there is a factor of 1/2 multiplying ##\nabla_\mu\phi\nabla^\mu\phi##. I just wanted to verify that for your problem there is no such factor of 1/2.$$\begin{align*} \delta S_M &= \int d^4x (\delta\sqrt{-g}) (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\ &= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi \end{align*}$$In the second line you have too many ##\mu##, ##\nu## indices which is causing problems in your derivation. In the second line, I suggest writing ##(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2)## with different summation indices, e.g., as ##(g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2)## Last edited: JD_PM You are absolutely right, there should be a ##1/2## in front of the first term... I completely overlooked this. With this in mind we have$$ \begin{align*} \delta S_M &= \int d^4x (\delta\sqrt{-g}) (\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\ &= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi \\ &= \int d^4x \sqrt{-g}\left( -\frac{1}{2}g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2\right] + \nabla_\mu\phi\nabla_\nu\phi \right)\delta g^{\mu\nu} \end{align*} $$Which then would give$$T_{\mu\nu} = g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2\right] -2 \nabla_\mu\phi\nabla_\nu\phi.$$I don't see how this could be simplified further (if it can), maybe contracting one of the covariant derivatives with the metric, but other than that... Markus Kahn said:$$ \delta S_M = \int d^4x (\delta\sqrt{-g}) (\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\$
That missing factor of 1/2 should also appear in the second integral above.

Getting all the signs right is a headache. I haven't checked them carefully. Make sure you have all your signs correct in the original expression for the action.

I don't see any way to simplify the answer further.

Markus Kahn

## 1. How is energy-momentum tensor calculated in general relativity?

The energy-momentum tensor in general relativity is calculated using the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy in the universe. Specifically, the energy-momentum tensor is derived from the stress-energy tensor, which describes the density and flux of energy and momentum in a given region of spacetime.

## 2. What is the significance of the energy-momentum tensor in general relativity?

The energy-momentum tensor is significant because it represents the total energy and momentum present in a given region of spacetime. This is important in general relativity because the curvature of spacetime is determined by the distribution of energy and momentum, so the energy-momentum tensor is crucial for understanding the overall structure of the universe.

## 3. Can the energy-momentum tensor be calculated for all types of matter and energy?

Yes, the energy-momentum tensor can be calculated for all types of matter and energy, including both normal matter (such as atoms and molecules) and exotic matter (such as dark matter or dark energy). This is because the energy-momentum tensor is a mathematical representation of the distribution of energy and momentum, regardless of its physical nature.

## 4. How does the energy-momentum tensor relate to conservation laws?

The energy-momentum tensor is directly related to the conservation laws of energy and momentum. In general relativity, the energy-momentum tensor is conserved, meaning that the total energy and momentum in a given region of spacetime remains constant. This is a fundamental principle of the theory and is integral to understanding the behavior of matter and energy in the universe.

## 5. Are there any challenges or limitations in calculating the energy-momentum tensor in general relativity?

Yes, there are several challenges and limitations in calculating the energy-momentum tensor in general relativity. One major challenge is that the equations involved are highly complex and difficult to solve, often requiring advanced mathematical techniques. Additionally, the energy-momentum tensor may be difficult to calculate for certain types of matter or in extreme situations, such as near black holes or during the early stages of the universe's evolution.

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