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

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
Homework Helper
Messages
683
Reaction score
412
TL;DR
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.
 
Physics news on Phys.org
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.