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I'm trying to derive the equation for the cosmological fluid:

$$\dot \rho + 3 \frac{\dot a}{a}(\rho +P)=0$$

by starting from the conservation of the stress-energy tensor:

$$\nabla^\mu T_{\mu \nu} = 0$$

with the stress-energy for a perfect fluid in its own frame being:

$$ T_{\mu \nu} = \text{diag} (\rho, a(t)^2 P,a(t)^2 P,a(t)^2 P) $$

in a spatially flat FLRW metric:

$$g_{\mu \nu} = \text{diag}(1,-a(t)^2,-a(t)^2,-a(t)^2)$$

But I keep getting a bogus answer! Consider the equation you get from $ \nabla^\mu T_{\mu \nu} = 0$ when $\nu =0$:

$$ \begin{align*} \nabla^\mu T_{\mu 0} &= 0 \\ g^{\mu \alpha}\nabla_\alpha T_{\mu 0} &= 0 \end{align*} $$

$T$ is diagonal, so $\mu$ must be zero, but $g$ is diagonal as well, so if $\mu$ is zero, then so is $\alpha$. This gives:

$$ \begin{align*} g^{0 0}\nabla_0 T_{0 0} &= 0 \\ \nabla_0 \rho &= 0 \\ \dot \rho &= 0 \end{align*} $$

Because $\rho$ is just a scalar, so the covariant derivative is the partial derivative. Except this answer is wrong.

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  • $\begingroup$ Any tensor component is a scalar. That doesn't mean the covariant derivative acts like the partial derivative component wise. Because really $T^{00} = x_\mu^0 x_\nu^0 T^{\mu\nu}$. Both these expressions are scalars, but to express the partial derivative of this scalar in terms of tensor quantities, you will need the Leibniz law and covariant derivatives. $\endgroup$ – Robin Ekman Mar 31 '14 at 15:24
  • $\begingroup$ Sometimes you see the notation $\nabla_\mu T_{\nu\rho}$. This does not mean the covariant derivative of the $\nu\rho$ component of $T$. It means the $\nu\rho\mu$ component of the covariant derivative of $T$. Note that the order of these operations matter! This is the case whenever your basis vectors are not parallel transported and you can see this in a simpler context if you consider the expression for curl in spherical or cylindrical coordinates. $\endgroup$ – Robin Ekman Mar 31 '14 at 15:29
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It is easier to make this calculation covariantly.

Given a time-like vector field $n^\mu$ representing the direction normal to the hyper-surfaces (in the Friedmann case the homogeneous and isotropic hyper-surfaces), you can always decompose the energy momentum tensor $T_{\mu\nu}$ as $$T_{\mu\nu} = \rho n_\mu n_\nu + 2n_{(\mu}q_{\nu)} + p \gamma_{\mu\nu}+ \Pi_{\mu\nu},$$ where $\rho \equiv T_{\mu\nu}n^\mu n^\nu$ is the energy density, $q_\alpha \equiv - T_{\mu\nu} n^\mu \gamma^\nu{}_\alpha$ the energy (heat) flow, $p \equiv T_{\mu\nu}\gamma^{\mu\nu} / 3$ the pressure and $\Pi_{\mu\nu} \equiv T_{\alpha\beta}\gamma^{\alpha}{}_\mu\gamma^{\beta}{}_{\nu} - p\gamma_{\mu\nu}$ all measured in the hyper-surfaces described by $n^\mu$, $\gamma_{\mu\nu} \equiv g_{\mu\nu} + n_\mu n_\nu$ is the spatial projector/metric and the signature is $(-1,1,1,1)$, i.e., $n_\mu n^\mu = -1$.

The energy momentum tensor "conservation" gives $\nabla^\mu T_{\mu\nu} = 0$. To project this result in the time direction (that is want you want when you put $\nu = 0$) you just multiply this expression by $n^\nu$, i.e., $$n^\nu\nabla^\mu T_{\mu\nu} = \nabla^\mu (T_{\mu\nu}n^\nu) - T_{\mu\nu}\nabla^\mu n^\nu = \nabla^\mu(-\rho n_\mu - q_\mu) - \mathcal{K}^{\mu\nu}T_{\mu\nu},$$ where $\mathcal{K}_{\mu\nu} \equiv \nabla_\mu n_\nu$ is the extrinsic curvature.

Now for a Friedmann metric the extrinsic curvature is simply $\mathcal{K}_{\mu\nu} = \Theta \gamma_{\mu\nu} / 3$ where $\Theta$ is the expansion factor and the heat flow $q_\mu$ is zero, hence, we obtain $$n^\nu\nabla^\mu T_{\mu\nu} = -\left[\dot{\rho}+\Theta(\rho + p) + \nabla_\mu q^\mu + \sigma^{\mu\nu}\Pi_{\mu\nu}\right],$$ where $\dot{\rho} \equiv n^\mu\nabla_\mu \rho$ and $\sigma^{\mu\nu}$ is the traceless symmetric part of the extrinsic curvature which is called shear tensor. At this point one can also introduces a scale factor satisfying $3\dot{a}/a = \Theta$.

The key ingredient to make this calculation correctly is the explicit introduction of the time vector $n^\mu$, note that it is its commutation with the covariant derivative and its derivative in terms of the expansion factor that introduces the missing terms in your result. If you want to make this calculation in the old fashion coordinate way you can just calculate everything in a specific coordinate system with all partial derivatives, Christoffel symbols, etc, and then project in the chosen direction.

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  • $\begingroup$ This doesn't seem to explain what went wrong with my original method. $\endgroup$ – user40381 Mar 1 '14 at 15:09
  • $\begingroup$ Did you read the last paragraph? $\endgroup$ – Sandro Vitenti Mar 1 '14 at 15:20
  • $\begingroup$ I don't understand why you said I should only project in the end. If $\nabla^\mu T_{\mu \nu} = 0$, then this is really four equations, one for each possible value of the free index $\nu$. The equation $\nabla^\mu T_{\mu 0} = 0$ should follow. $\endgroup$ – user40381 Mar 1 '14 at 15:44
  • $\begingroup$ That's why I wrote the problem in a coordinate free way, I showed what is the exact mean to the action of choosing the value of a free index. You can write the vector $n^\mu$ in a coordinate system in which $n^\mu \doteq (1,0,0,0)$, then, in this case you have T_{\mu\nu}n^\nu = T_{\mu0}, however the presence of a covariant derivative complicates things since it does not commute with the normal vector. $\endgroup$ – Sandro Vitenti Mar 1 '14 at 16:21
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Consider the (1,1)-tensor $T^{\mu}_{\phantom{\mu}\nu}=g^{\mu\lambda}T_{\lambda\nu}=\text{diag}(\rho, -P/c^2, -P/c^2, -P/c^2)$. Then $$ \nabla_\mu T^{\mu}_{\phantom{\mu}\nu} = \partial_\mu T^{\mu}_{\phantom{\mu}\nu} + \Gamma^{\mu}_{\mu\lambda}T^\lambda_{\phantom{\mu}\nu} - \Gamma^{\lambda}_{\mu\nu}T^\mu_{\phantom{\mu}\lambda} = 0. $$ For $\nu=0$, we find $$ \nabla_\mu T^{\mu}_{\phantom{\mu}0} = \partial_0 T^{0}_{\phantom{\mu}0} + \Gamma^{\mu}_{\mu 0}T^0_{\phantom{\mu}0} - \Gamma^{\lambda}_{\mu 0}T^\mu_{\phantom{\mu}\lambda} = 0. $$ Now, the only non-zero Christoffel symbols in this equation are $$ \Gamma^{1}_{10} = \frac{1}{2}g^{11}\frac{\partial g_{11}}{\partial x^{0}} = \frac{\dot{a}}{ca} = \Gamma^{2}_{20} = \Gamma^{3}_{30}, $$ so that $$ \nabla_\mu T^{\mu}_{\phantom{\mu}0} = \frac{\dot{\rho}}{c} + \frac{3\dot{a}}{ca}\rho + \frac{3\dot{a}}{ca}\frac{P}{c^2} = 0, $$ which is indeed the conservation law $$ \dot{\rho} + 3\frac{\dot{a}}{a}\left(\rho + \frac{P}{c^2}\right)=0. $$ You can also verify that $\nabla_\mu T^{\mu}_{\phantom{\mu}\nu} \equiv0$ for $\nu\neq0$.

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