[tex]

(u^0,u^1,u^2,u^3) = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v^1,v^2,v^3)

[/tex]

is the four velocity.

So a energy-momentum tensor of dust is

[tex]

T^{\mu\nu} = \rho_0 u^{\mu} u^{\nu} = \frac{\rho_0}{1-|v|^2/c^2}\left[\begin{array}{cccc}

c^2 & cv^1 & cv^2 & cv^3 \\

cv^1 & v^1 v^1 & v^1 v^2 & v^1 v^3 \\

cv^2 & v^2 v^1 & v^2 v^2 & v^2 v^3 \\

cv^3 & v^3 v^1 & v^2 v^3 & v^3 v^3 \\

\end{array}\right]

[/tex]

My first though was, that a energy-momentum tensor of a point particle would be this multiplied with a delta function, but after some struggling with the transformations of the delta function, I concluded that the result is, nontrivially

[tex]

T^{\mu\nu} = m_0 \delta^3(x-x(t)) \sqrt{1-|v|^2/c^2} u^{\mu} u^{\nu}

= \frac{m_0\delta^3(x-x(t))}{\sqrt{1-|v|^2/c^2}}\left[\begin{array}{cccc}

c^2 & cv^1 & cv^2 & cv^3 \\

cv^1 & v^1 v^1 & v^1 v^2 & v^1 v^3 \\

cv^2 & v^2 v^1 & v^2 v^2 & v^2 v^3 \\

cv^3 & v^3 v^1 & v^2 v^3 & v^3 v^3 \\

\end{array}\right]

[/tex]

Is this correct?