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Continuity equation from Stress-Energy tensor

  1. Sep 15, 2008 #1
    It is true that [tex] \frac{\partial}{\partial x^\beta} T^{0 \beta} = \gamma^2 c \left( \frac{\partial \rho}{\partial t} + \vec{\nabla} \bullet \left[ \rho \vec{v} \right] \right) = 0 [/tex]

    but, how do we arrive at this point?

    What is in [tex] T^{ \alpha \beta} [/tex]

    and how do we compute it for any alpha? I'm sorry if this is a no brainer. I missed some critical lectures.
     
  2. jcsd
  3. Sep 16, 2008 #2

    haushofer

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    That T you've got there is the energy momentum tensor. In your case, it is the energy momentum tensor for non-interacting dus,

    [tex]
    T^{\mu\nu} = \rho_{0}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}
    [/tex]

    .

    Why this conservation is true, is another story. In the classical case, it can be derived from Noether's theorem. In the general relativistic case, the conservation is a consequence of something called diffemorphism invariance.

    I recommend you to take a look at the book of Inverno about general relativity, chapter 12 ( .1,2,3). There it is all explained :)
     
  4. Sep 17, 2008 #3
    thanks for the reply, but I can't find that book, is Inverno the author?
     
  5. Sep 18, 2008 #4

    atyy

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    haushofer probably means d'Inverno (Introducing Einstein's Relativity)
     
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