Energy-momentum density of point particle

[jostpuur]

I'm using a following notation. $(v^1,v^2,v^3)$ is the usual velocity vector, and

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

is the four velocity.

So a energy-momentum tensor of dust is

$$T^{\mu\nu} = \rho_0 u^{\mu} u^{\nu} = \frac{\rho_0}{1-|v|^2/c^2}\left[\begin{array}{cccc}<br /> c^2 & cv^1 & cv^2 & cv^3 \\<br /> cv^1 & v^1 v^1 & v^1 v^2 & v^1 v^3 \\<br /> cv^2 & v^2 v^1 & v^2 v^2 & v^2 v^3 \\<br /> cv^3 & v^3 v^1 & v^2 v^3 & v^3 v^3 \\<br /> \end{array}\right]$$

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

$$T^{\mu\nu} = m_0 \delta^3(x-x(t)) \sqrt{1-|v|^2/c^2} u^{\mu} u^{\nu}<br /> = \frac{m_0\delta^3(x-x(t))}{\sqrt{1-|v|^2/c^2}}\left[\begin{array}{cccc}<br /> c^2 & cv^1 & cv^2 & cv^3 \\<br /> cv^1 & v^1 v^1 & v^1 v^2 & v^1 v^3 \\<br /> cv^2 & v^2 v^1 & v^2 v^2 & v^2 v^3 \\<br /> cv^3 & v^3 v^1 & v^2 v^3 & v^3 v^3 \\<br /> \end{array}\right]$$

Is this correct?

 

[pervect]

I'm not sure whether or not there is a totally mathematically consistent formulation of the stress-energy tensor of a point particle or not. See for instance Stingray's comments (and more importantly his references) in the old thread

https://www.physicsforums.com/showthread.php?t=111148

If stingray is around, maybe he could comment in more detail.

 

[George Jones]

This looks the same as what I got.

In that post, I initially used inertial motion along the x-axix, Y means $\gamma$ and (x) means $\otimes$.

 

[jostpuur]

It seems there are some problems with point masses and metrics in general relativity, if I understood stingray's comments correctly. In special relativity that can probably be ignored.

My derivation went through the fact that

$$\delta^3(x-x(t))(c,v)^{\mu}$$

is a four vector, like a four current of a point charge, and for example

$$\delta^3(x-x(t)) u^{\mu}$$

is not. Good to see that George Jones got the same factors in a different way.