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Summary:: I find two different expressions for the EM tensor for dust, and both derivations seem right to me.

Given the action for a system of dust $$S =-\sum m_q \int \sqrt{g_{\mu\nu}[x_q(\lambda)]\dot{x}^\mu_q(\lambda)\dot{x}^\nu_q(\lambda)} d\lambda,$$ where I use the $(+,-,-,-)$ sign convention. The Energy-Momentum Tensor (EMT) is defined by the variation of the metric

$$\delta S = \frac{1}{2}\int T_{\mu\nu} \delta g^{\mu\nu} \sqrt{g} d^4x.$$

To compute that 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, is by doing the variation to $g_{\mu\nu}$, doing exactly the same steps 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. Furthermore, because the two derivations are so similar, I don't think an algebraic mistake can explain such difference, so the error must be a conceptual one.

Qmechanic
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Gaussian97
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1 Answers1

2

Possible conceptional mistakes:

  1. Note that the velocity $\dot{x}_{\mu}:= g_{\mu\nu}\dot{x}^{\nu}$ with lower index implicitly depends on the metric. In contrast the velocity $\dot{x}^{\nu}$ with upper index does not depend on the metric. This is important when we vary wrt. the metric.

  2. The stress-energy-momentum tensor depends on the sign convention for the metric, cf. this Phys.SE post.

Qmechanic
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  • Oh OK, so I should always consider that $x^\mu$ is constant (respect to the change of metric, i.e. $\delta x^\mu=0$?) 2. That means that if I have derivatives, for example for a scalar field, I should consider $\delta (\partial_\mu \phi)=0$ but $\delta (\partial^\mu \phi)\neq0$? 3. And in the case of, for example, a vector field $A^\mu$, should I consider $\delta A^\mu=0$ or $\delta A_\mu=0$?
    • – Gaussian97 Sep 25 '20 at 15:15
  • Yes. 2. Yes. 3. Well, that depends on the nature of the vector-field $A^{\mu}$. Perhaps the co-vector-field $A_{\mu}$ is more fundamental in that theory.
    • – Qmechanic Sep 25 '20 at 16:04