- #1
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- TL;DR Summary
- 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.
$$\delta S = \frac{1}{2}\int T_{\mu\nu} \delta g^{\mu\nu} \sqrt{g} d^4x$$
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.