In the following demonstration, there is an error, but I cannot find where. (I explicitely put the $c^2$ to keep track of units).

We consider a metric $g_{\mu\nu}$ with a signature $(-, +, +, +)$ :

$\boxed{g_{\mu\nu} = \begin{pmatrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}} \qquad\qquad \boxed{ g^{\mu\nu} = \begin{pmatrix} \frac{-1}{c^2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}} $

The tensor of a perfect fluid is: $ \boxed { \begin{aligned} T^{\mu\nu} &= \left(\rho c^2+P\right)u^{\mu}u^{\nu}+Pg^{\mu\nu}\\ T_{\nu}^{\mu} &= \left(\rho c^2+P\right)u^{\mu}u_{\nu}+P\delta_{\nu}^{\mu}\\ T_{\mu\nu} &= \left(\rho c^2+P\right)u_{\mu}u_{\nu}+Pg_{\mu\nu} \end{aligned} }$ where $\rho$ is the mass density.

We can easily check that $\rho c^2$ and $P$ have the same unit.

If we consider only the first component $00$, there is no pressure $P$ involved (I consider a cosmological case where the fluid is at rest in comoving coordinates). The only solution to have that is that $u^0u^0=-g^{00}$, so that the contravariant perfect fluid tensor become : $T^{00} = \left(\rho c^2+P\right)u^{0}u^{0}+Pg^{00}=\left(\rho c^2+P\right)\times{-g^{00}}+Pg^{00}=-\rho c^2 g^{00} = -\rho c^2 \times -\frac{1}{c^2}=\rho$ which is the expected result. So, to have the expected result, one should have $\boxed{\left(u^0\right)^2=+\frac{1}{c^2}}$.

Now come the weird part.

The metric can be written: $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=-c^2dt^2+dx^2+dy^2+dz^2=dt^2\left(-c^2+\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2}\right)\approx-c^2dt^2$ which is equivalent to :


which means : $\boxed{\left(u^0\right)^2=-\frac{1}{c^2}}$

So we clearly have a problem of sign here between the two expressions of $u^0$.

QUESTION: where is the mistake ?

  • $\begingroup$ $u^0 = \gamma c$, and not just $-g_{00}$, you don't cancel the pressure part on the calculation of $T^{00}$ $\endgroup$
    – Hydro Guy
    Commented May 5, 2013 at 14:01
  • $\begingroup$ If I consider a fluid at rest in comoving coordinates, then I think that $T^{00}=\rho$ $\endgroup$
    – Vincent
    Commented May 5, 2013 at 14:11
  • $\begingroup$ Even so, you can't do the reasoning that you are doing on the second part. You have $u^\mu=c\frac{dx^\mu}{ds}$, and you can't really set '$dx^2=dy^2=dz^2=0$ $\endgroup$
    – Hydro Guy
    Commented May 5, 2013 at 14:23
  • $\begingroup$ In the second part, as my fluid is at rest, I think that the terms $\frac{dx}{dt}$, $\frac{dy}{dt}$, $\frac{dz}{dt}$ are negligible compared to $c^2$, isn't it ? $\endgroup$
    – Vincent
    Commented May 5, 2013 at 14:31
  • $\begingroup$ I'm reviewing my calculations, and here's why I hate the $g_{\mu\nu}=diag(-1,1,1,1)$: you define $u^{\mu} = - c\frac{dx^\mu}{ds}$ because you have for time-like intervals in this signature $cd\tau = -ds$ and you always have $u^\mu=\frac{dx^\mu}{d\tau}$, and there you have your sign problem. $\endgroup$
    – Hydro Guy
    Commented May 5, 2013 at 14:41

1 Answer 1


A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, p, p)$ where $e$ is the local energy density, $p$ is the pressure in the local rest frame. So in general coordinates, the form of the energy-momentum tensor (as you have written) is

$$T^{\alpha\beta} = (e + p)u^{\alpha}u^{\beta} + pg^{\alpha\beta}.$$

Assuming units where $c = 1$. As you have correctly pointed out

$$T^{00} = (e + p)u^{0}u^{0} - p = e,$$

where you have correctly shown $u^{0}u^{0} = 1$.

Now, the mistake comes from the way you have derived the four-velocity from the expression for the metric. The correct way to derive the four-velocity here is to first write the four-velocity as

$$u^{\alpha} = \frac{\mathrm{d}}{\mathrm{d}\tau}(x^{0}, x^{i})^{\mathsf{T}},$$

which gives

$$u^{0} = \frac{\mathrm{d}t}{\mathrm{d}\tau} = \gamma.$$

Then of course, in the fluid rest frame we get $(u^{0})^{2} = 1$ as before.

I hope this helps.

Edit. To address your comments:

For a particle with fixed spatial coordinates $x^{i}$, the interval elapsed as it moves forward in time is negative, $\mathrm{d}s^{2} = −\mathrm{d}t^{2} < 0$. This leads us to define the proper time via

$$\mathrm{d}\tau^{2} = -\mathrm{d}s^{2}.$$

The proper time elapsed along a trajectory through space-time will be the actual time measured by an observer on that trajectory. Some other observer, as we know, will measure a different time.

A path through space-time is specified by giving the four space-time coordinates as a function of some parameter, $x^{\mu}(\lambda)$, where typically for time-like paths the most convenient parameter to use is the proper time $\tau$. So we can write

$$\tau = \int \sqrt{-\mathrm{d}s^{2}} = \int \sqrt{-g_{\mu\nu}\frac{\mathrm{d}x^{\mu}}{\mathrm{d}\tau}\frac{\mathrm{d}x^{\nu}}{\mathrm{d}\tau}}\mathrm{d}\tau,$$

The tangent vectors $u^{\mu} = \mathrm{d}x^{\mu}/\mathrm{d}\tau$, are our four-velocities and these can be automatically normalised so we have

$$g_{\mu\nu}u^{\mu}u^{\nu} = u_{\mu}u^{\nu} = -1.$$

To convince your self of this you can consider the above expression in the fluid rest frame. Here, we have $\gamma = 1$ and $u^{i} = 0^{i}$. Thus, $u^{\mu} = (1, u^{i})^{\mathsf{T}}$. Now, again considering the rest frame of the fluid, we can clearly see

$$u^{0}u^{0} = 1.$$

Again, I hope this helps.

  • $\begingroup$ In your notations can you define what is $\tau$ in my notations (as a function of $s$ or/and $t$) $\endgroup$
    – Vincent
    Commented May 5, 2013 at 15:01
  • $\begingroup$ $\tau$ is the proper-time. Time measured in the rest frame of the fluid. It is defined as $\mathrm{d}\tau^{2} = \mathrm{d}S^{2}/c^{2}$. $\endgroup$
    – MoonKnight
    Commented May 5, 2013 at 15:07
  • $\begingroup$ Thanks. And what is the notation $(x^0, x^i)^T$ ? $\endgroup$
    – Vincent
    Commented May 5, 2013 at 15:08
  • 1
    $\begingroup$ The four-velocity as I have written it is contravariant. Contravaient vectors are usually written as column vectors. So, to make a vector represented by a row $(x, y, z)$ represent a contravariant/column vector, we take the transpose. This is very pedantic on my part, but something I always like to do when writing vectors in this way. For more about Covariance and contravariance see Wikipedia. $\endgroup$
    – MoonKnight
    Commented May 5, 2013 at 15:35
  • $\begingroup$ Oh, ok, I do not see this as pedantic but completely justified. But I still do not understand from where comes the minus between $\frac{dx^0}{d\tau}$ and $-\frac{dt}{d\tau}$ because for me $x^0=t$ (so when I write $g_{00}dx^{0}dx^{0}$ I get $-c^2dt^2$). $\endgroup$
    – Vincent
    Commented May 5, 2013 at 15:44

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