# Energy-momentum density of point particle

1. Jul 29, 2007

### 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

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

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

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

Is this correct?

2. Jul 29, 2007

### pervect

Staff Emeritus
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

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

3. Jul 29, 2007

### George Jones

Staff Emeritus
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 $\otimes$.

Last edited: Jul 30, 2007
4. Jul 30, 2007

### 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.