I am working through the book Cosmology by Daniel Baumann, and in the subsection that covers the continuity equation (part of section 2.3.1 on perfect fluids) the author makes a claim that confuses me. He starts by stating that, in Minkowski space, energy and momentum are conserved, and therefore:

  • The energy density $\rho c^2$ satisfies the continuity equation, which means that the rate of change of the density equals the divergence of the energy flux: $$\dot{\rho}=-\partial_i\pi^i$$

  • The evolution of the momentum density satisfies the Euler equation: $$\dot{\pi}_i=\partial_iP$$

And here comes the claim:

These conservation laws can be combined into a four-component conservation equation for the energy-momentum tensor: $$\partial_\mu T^\mu_{\ \ \ \ \nu}=0$$

In the previous subsection, all the necessary information about the energy-momentum tensor was provided, including that the assumption of homogeneity and isotropy of the universe forces this tensor to take the form:

$$T_{00}\equiv\rho(t)c^2\qquad T_{i0}\equiv c\pi_i=0\qquad T_{ij}=P(t)g_{ij}(t,\vec{x})$$

So, what I am trying to do is to recover the equations

$$\dot{\rho}=-\partial_i\pi^i\qquad \text{and}\qquad \dot{\pi}_i=\partial_i P$$

from the four-component equation $\partial_\mu T^\mu_\nu=0$ (where I know we are using Einstein's summation convention for the index $\mu$ and that $\nu=0,...,3$).

For the $\nu=0$ component, this is my attempt:

$$\partial_\mu T^\mu_{\ \ \ \ 0}=0\ \Rightarrow\ \partial_0(-\rho c^2)+\partial_i(c\pi^i)=0\ \Rightarrow\ \dfrac{1}{c}\partial_t(-\rho c^2)+\partial_i(c\pi^i)=0\ \Rightarrow$$ $$\Rightarrow\ -c\dot{\rho}+c\partial_i\pi^i=0\ \Rightarrow\ \dot{\rho}=\partial_i\pi^i$$

where I have used that $\partial_0=c^{-1}\partial_t$ since $x^0=ct$, and where I have raised one of the indices in the energy-momentum tensor by using the metric tensor:

  • $T^0_{\ \ \ \ 0}=g^{0\beta}T_{\beta 0}=-T_00=-\rho c^2$
  • $T^i_{\ \ \ \ 0}=g^{i\beta}T_{\beta 0}=g^{i\beta}c\pi_\beta=c\pi^i$

So, instead of $\dot{\rho}=-\partial_i\pi^i$, I get $\dot{\rho}=\partial_i\pi^i$. I'm tempted to say this is an errata in the book but I suppose I'm probably wrong somewhere.

I have read that the energy-momentum tensor, written as $T^{\mu\nu}$, is symmetric and therefore $T^{\mu\nu}=T^{\nu\mu}$, but this means it's symmetric with both indices as upper indices, not that $T_{\mu\nu}$ is symmetric too, right? Because we need to use the metric to lower indices, and the product of two symmetric matrices isn't symmetric in general. So am I right in assuming that $T_{0j}=T_{j0}$, which I would need to deduce the second equation?

I'm thoroughly confused, any help would be greatly appreciated.

Edit: I'm not convinced that $T^\mu_{\ \ \ \ \nu}$ is also symmetric if I assume that $T_{\mu\nu}$ is symmetric. My calculations are as follows. I start with the components of $T_{\mu\nu}$, given by:

$$T_{00}\equiv\rho c^2\qquad T_{i0}\equiv c\pi_i\qquad T_{0j}=T_{j0}=c\pi_j \qquad T_{ij}=P g_{ij}$$

where the rest of the components equal zero. If I raise one of the indices, I get:

  • $T^0_{\ \ \ \ 0}=g^{0\beta}T_{\beta 0}=-T_{00}=-\rho c^2$
  • $T^i_{\ \ \ \ 0}=g^{i\beta}T_{\beta 0}=g^{i\beta}c\pi_\beta=c\pi^i$
  • $T^0_{\ \ \ \ j}=g^{0\beta}T_{\beta j}=-T_{0j}=-c\pi_j$
  • $T^i_{\ \ \ \ j}=g^{i\beta}T_{\beta j}=g^{ik}T_{kj}=Pg^{ik}g_{kj}=P\delta^i_j$

If $T^\mu_{\ \ \ \ \nu}$ is symmetric, then in particular $T^i_{\ \ \ \ 0}=T^0_{\ \ \ \ i}$, which means $\pi^i=-\pi_i$. But that isn't the case, since:


Then, where is my mistake in the calculations?


2 Answers 2


You are on the right track. Based on your calculations, your metric $g \equiv \eta = \mathrm{diag}(-1,+1,+1,+1)$ in cartesian co-ordinates. Here, I take $c=1$ as this makes calculations easier to write down. You can always re-insert $c$ at the end with some simple dimensional analysis. Two things to keep in mind now -

  1. The error in sign may well be due to an errata in the book. It ought to be $T_{i0}\equiv -\pi_i$ for the conventions used. This sign is positive for the (+---) signature used by Baumann in his lecture notes from which I presume, this book was developed.

  2. The symmetry is a property of the object in this case and not an artefact of the representation. In other words, it is the energy-momentum tensor that is symmetric here and not specifically its covariant form, but in general. Thus your assumption $T_{0j}=T_{j0}$ holds. It might be a good exercise to prove that the symmetry holds in general (for this case) for clarity.

Hope this helps :)

  • $\begingroup$ Thank you! But just to clarify, do you mean that if $T^{\mu\nu}$ is symmetric then $T^\mu_{\ \ \ \ \nu}$ and $T_{\mu\nu}$ are also symmetric, or do both indices need to be of the same type (both upper indices as in $T^{\mu\nu}$, or both lower as in $T_{\mu\nu}$)? $\endgroup$ Jan 30 at 10:15
  • $\begingroup$ @WildFeather the index type doesn't matter. It's the tensor $T$ that is symmetric here, regardless of the form. In other words, no matter how you raise or lower the indices of $T$ it will be symmetric. You can a proof here. $\endgroup$ Jan 30 at 13:52
  • $\begingroup$ I don't know if it's too late now to ask you this, but would you mind explaining why you said that for the (-+++) signature it ought to be $T_{i0}=-\pi_i$ instead of $T_{io}=\pi_i$? You were right, but I'm very curious to know the reasoning. Thank you! $\endgroup$ Mar 24 at 11:28

Just in case somebody else comes here and is confused by the same things I explained in the edit to my post, I will leave here the wonderful explanation I got about this in Math Stack Exchange:


To sum up, the symmetry of the tensor $T$ implies that $T^\mu_{\ \ \ \ \nu}=T^{\ \ \mu}_\nu$, not that $T^\mu_{\ \ \ \nu}=T^\nu_{\ \ \ \mu}$, hence my confusion.


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