Energy-momentum density of point particle

In summary, a energy-momentum tensor of dust is (u^0,u^1,u^2,u^3) = \frac{1}{\sqrt{1-|v|^2/c^2}}(c,v^1,v^2,v^3).
  • #1
jostpuur
2,116
19
I'm using a following notation. [itex](v^1,v^2,v^3)[/itex] is the usual velocity vector, and

[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?
 
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  • #2
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.
 
  • #3
This looks the same as what I got.

In that post, I initially used inertial motion along the x-axix, Y means [itex]\gamma[/itex] and (x) means [itex]\otimes[/itex].
 
Last edited:
  • #4
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

[tex]
\delta^3(x-x(t))(c,v)^{\mu}
[/tex]

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

[tex]
\delta^3(x-x(t)) u^{\mu}
[/tex]

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

1. What is energy-momentum density?

Energy-momentum density refers to the amount of energy and momentum contained within a given volume or point in space. It is a physical quantity used in physics to describe the distribution of energy and momentum within a system.

2. What is a point particle?

A point particle is a hypothetical object in physics that is used to represent a particle with no physical size or internal structure. It is often used in simplified models to describe the behavior of particles at a very small scale.

3. How is energy-momentum density calculated for a point particle?

The energy-momentum density of a point particle can be calculated by taking the product of its energy and momentum and dividing it by its volume. This results in a scalar quantity that represents the amount of energy and momentum per unit volume at a specific point in space.

4. What are the units of energy-momentum density?

The units of energy-momentum density depend on the system of measurement being used. In the SI system, the units are Joules per cubic meter (J/m³). In the cgs system, the units are ergs per cubic centimeter (erg/cm³).

5. How is energy-momentum density related to the energy-momentum tensor?

The energy-momentum density is a component of the energy-momentum tensor, which is a mathematical object used to describe the energy and momentum content of a physical system. The energy-momentum tensor includes additional components that account for the flow of energy and momentum in a system.

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