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Energy-momentum density of point particle

  1. Jul 29, 2007 #1
    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?
     
  2. jcsd
  3. Jul 29, 2007 #2

    pervect

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    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.
     
  4. Jul 29, 2007 #3

    George Jones

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    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: Jul 30, 2007
  5. Jul 30, 2007 #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.
     
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