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There are some points in this wikipedia chapter. Main equation is:

$$ T^{\alpha \beta} \, = \left(\rho + {p \over c^2}\right)u^{\alpha}u^{\beta} + p g^{\alpha \beta} $$

where $c$ is explicit.

The one for the trace is:

$$T = 3p - \rho c^2$$

that seems contradictory with:

$$T^{\alpha\beta} = \left( \begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$$

with trace $3p+\rho$ (difference in sign and value of last term).

The expression for the four-velocity:

$$u^{\alpha} = (1, 0, 0, 0)$$

is not the usual one $(c, 0, 0, 0)$.

Finally, the metric:

$$g^{\alpha\beta} \, = \left( \begin{matrix} - c^{-2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)$$

also with explicit $c$, it is also not the usual:

$$\left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)$$

Are the wikipedia equations in this chapter using a coherent notation ? If yes, how to explain the previous points ?

pasaba por aqui
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    Thinking of tensors as "matrices" always ended up confusing me. In fact, I believe you can't strictly take a trace of an object like $T^{\mu\nu}$, since this quantity will turn out to be basis dependent. I find it better to think of the "trace" for such objects to be $g_{\mu\nu} T^{\mu\nu}$. See this answer here. – Philip Aug 21 '20 at 21:32

1 Answers1

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The trace is ${T^\mu}_\mu$ not $T^{\mu\mu}$, so the minus sign from lowering lowering the $\mu=0$ index accounts for the sign difference. Also $c=1$ for most people, so the he wiki may not be consistent, but getting right is easy.

Looking at the Wiki, it is $g^{\mu\nu}$nthat has the $c^{-2}$, and to lower the index you need $g_{\mu\nu}$ which has the $-c^2$, so the wiki article is consistent.

mike stone
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  • I looked the wiki article it appears consistten, You did not say that your metric matrix ``$g$'' was actually the inverse metric. I amended my answer. – mike stone Aug 21 '20 at 21:46
  • Sorry, question edited. However, still surprised by $u^{\alpha}=(1,0,0,0)$ instead of $u^{\alpha}=(c,0,0,0)$ – pasaba por aqui Aug 21 '20 at 21:53