Variation of Metric and the Energy-Momentum Tensor: Where Am I Going Wrong?

In summary, the Energy-Momentum Tensor (EMT) is defined as the variation of the metric, and there are two different approaches to finding the EMT. The first approach involves varying ##g^{\mu\nu}##, while the second approach involves varying ##g_{\mu\nu}##. When using the first approach, the EMT has a negative sign, while using the second approach results in an EMT with a positive sign. The same issue arises when trying to compute the EMT for a scalar particle with a specific action. There is a conceptual error that needs to be addressed.
  • #1
Gaussian97
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I find two different expressions for the EM tensor for dust, and both derivations seem right to me.
Given the action ##S =-\sum m_q \int \sqrt{g_{\mu\nu}[x_q(\lambda)]\dot{x}^\mu_q(\lambda)\dot{x}^\nu_q(\lambda)} d\lambda## The Energy-Momentum Tensor (EMT) is defined by the variation of the metric
$$\delta S = \frac{1}{2}\int T_{\mu\nu} \delta g^{\mu\nu} \sqrt{g} d^4x$$
Then I use two different approaches, first one, because I want to vary ##g^{\mu\nu}## I find it better to write ##S =-\sum m_q \int \sqrt{g^{\mu\nu}[x_q(\lambda)]\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)} d\lambda##. Then
$$\delta S = -\sum m_q \int \frac{\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}{2\sqrt{g^{\mu\nu}[x_q(\lambda)]\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}} \delta g^{\mu\nu}d\lambda$$
And multiplying by ##1=\int \delta^{(4)}(x^\mu - x^{\mu}_q(\lambda))\frac{\sqrt{g}}{\sqrt{g}} d^4x##
$$\delta S = -\frac{1}{2}\sum m_q \int \frac{\delta^{(4)}(x^\mu - x^{\mu}_q(\lambda))\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}{\sqrt{g}\sqrt{g^{\mu\nu}[x_q(\lambda)]\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}} \delta g^{\mu\nu}d\lambda \sqrt{g}d^4x$$
Giving
$$T_{\mu\nu} = -\sum m_q \int \frac{\delta^{(4)}(x^\mu - x^{\mu}_q(\lambda))\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}{\sqrt{g}\sqrt{g^{\mu\nu}[x_q(\lambda)]\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}} d\lambda$$

The second approach, doing the variation to ##g_{\mu\nu}##, doing exactly the same I get
$$\delta S = -\frac{1}{2}\sum m_q \int \frac{\delta^{(4)}(x^\mu - x^{\mu}_q(\lambda))\dot{x}^\mu_{q}(\lambda)\dot{x}_{q}^\nu(\lambda)}{\sqrt{g}\sqrt{g_{\mu\nu}[x_q(\lambda)]\dot{x}_{q}^\mu(\lambda)\dot{x}_{q}^\nu(\lambda)}} \delta g_{\mu\nu}d\lambda \sqrt{g}d^4x$$
Now, because ##0=\delta(g_{\mu\nu}g^{\nu\lambda})## we must have ##\delta g_{\mu\nu} = -g_{\mu\alpha}g_{\nu\beta}\delta g^{\alpha\beta}## so I find
$$\delta S = \frac{1}{2}\sum m_q \int \frac{\delta^{(4)}(x^\mu - x^{\mu}_q(\lambda))\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}{\sqrt{g}\sqrt{g^{\mu\nu}[x_q(\lambda)]\dot{x}_{q\mu}(\lambda)\dot{x}_{q\nu}(\lambda)}} \delta g^{\mu\nu}d\lambda \sqrt{g}d^4x$$

Giving an EMT equal, but with a negative sign. The second one seems better because gives an energy density bounded for below, while the first one not, but I don't see any mistake.
 
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  • #2
Actually, I found the same problem trying to compute the Stress-energy tensor for a scalar particle with action
$$S = \int \left[\frac{1}{2}g_{\mu\nu}\partial^\mu\partial^\nu - V(\phi)\right]\sqrt{g}d^4x$$
Doing a variation ##\delta g_{\mu\nu}## I get
$$\delta S = \int \frac{1}{2}\delta g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi\sqrt{g}d^4x + \frac{1}{2}\int \left[\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-V(\phi)\right]g^{\alpha\beta}\delta g_{\alpha\beta}d^4x$$
which gives
$$T_{\mu\nu} = -\partial_\mu\phi \partial_\nu\phi - \frac{1}{2} g_{\mu\nu}\delta_\alpha \phi \delta^\alpha \phi + g_{\mu\nu} V(\phi)$$
While doing a variation ##\delta g^{\mu\nu}## I get
$$\delta S = \int \frac{1}{2}\delta g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\sqrt{g}d^4x - \frac{1}{2}\int \left[\frac{1}{2}g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi-V(\phi)\right]g_{\alpha\beta}\delta g^{\alpha\beta}d^4x$$
That gives an EMT
$$T_{\mu\nu} = \partial_\mu\phi \partial_\nu\phi - \frac{1}{2} g_{\mu\nu}\delta_\alpha \phi \delta^\alpha \phi + g_{\mu\nu} V(\phi)$$

So I don't know what I'm doing wrong but definitely, there's some conceptual error that I'm doing here.
 

1. What is the Stress-Energy Tensor for Dust?

The Stress-Energy Tensor for Dust is a mathematical representation used in the field of general relativity to describe the distribution of energy and momentum in a system of dust particles. It takes into account the effects of mass, energy, and momentum on the curvature of spacetime.

2. How is the Stress-Energy Tensor for Dust calculated?

The Stress-Energy Tensor for Dust is calculated by taking the energy-momentum tensor and setting the pressure to zero, as dust particles do not have pressure. This results in a diagonal tensor with the energy density as the only non-zero component.

3. What is the significance of the Stress-Energy Tensor for Dust in general relativity?

The Stress-Energy Tensor for Dust is significant because it allows us to model the behavior of systems with a large number of dust particles, such as galaxies or clusters of galaxies. It also plays a crucial role in understanding the dynamics of the universe on a large scale.

4. Can the Stress-Energy Tensor for Dust be used to describe other types of matter?

Yes, the Stress-Energy Tensor for Dust can be used to describe any type of matter that does not have pressure, such as non-relativistic particles. It can also be modified to include pressure and other properties for more complex systems.

5. How does the Stress-Energy Tensor for Dust relate to the Einstein field equations?

The Stress-Energy Tensor for Dust is one of the components in the Einstein field equations, which describe the relationship between the curvature of spacetime and the distribution of matter and energy. It represents the matter and energy content of a system and is used to solve for the metric tensor, which describes the geometry of spacetime.

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