Defining pressure from the stress-energy tensor components Suppose I have a trial metric which when I plugged into the Einstein Field Equations produced a stress-energy-momentum tensor where the following components are non-zero:
$T_{tt}$, $T_{tr}=T_{rt}$, $T_{rr}$, $T_{\theta\theta}$ and $T_{\phi\phi}$, where $T_{rr} \neq T_{\theta\theta} \neq T_{\phi\phi}$. Can I write the "total" pressure $P$ as
$P = \sqrt{T_{rr}^{2}+T_{\theta\theta}^{2} + T_{\phi\phi}^{2}}$?
 A: $\let\a=\alpha \let\b=\beta \let\th=\theta$
In GR nothing can be said about physical significance of a tensor's components unless you know the metric. Unfortunately you didn't give that information. We might guess from coordinate names: $t$, $r$, $\th$,
$\phi$ suggest spherical space coordinates but it's a too vague clue for serious reasoning.
Just to play... A simple relationship between $T^{\a\b}$ and $p$ only exists if an orthonormal basis is used. E.g. assume
$$d\tau^2 = dt^2 - dr^2 - r^2 (d\th^2 + 
\sin^2\!\th\,d\phi^2)$$ 
(this is a Minkowsky spacetime in spherical coordinates). These coordinates are orthogonal but not orthonormal:
$$g_{tt} = 1 \quad g_{rr} = -1 \quad g_{\th\th} = -r^2 \quad g_{\phi\phi} = -r^2 \sin^2\!\th$$
Then for a still perfect fluid you'd write
$$T^{rr} = p \qquad T^{\th\th} = {1 \over r^2}\,p \qquad T^{\phi\phi} = 
{1 \over r^2 \sin^2\!\th}\,p.$$
The general form is
$$T^{\a\b} = (\rho + p)\,u^\a u^\b - p\,g^{\a\b}$$
where $u^\a$ is the 4-velocity of fluid.
A: The component $T^{ii}$ (not summed) of the stress-energy tensor represents the normal stress, which reduces to the pressure $p$ if it is independent of the direction.  
The formula you posted is not meaningful. To convince of this, if you think to a perfect fluid you have $T^{1 1} = T^{2 2} = T^{3 3} = p$. Your formula would give $p = \sqrt 3 p$ (?).
