Continuity equation from Stress-Energy tensor

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Jonny_trigonometry
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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.
 
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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 :)
 
thanks for the reply, but I can't find that book, is Inverno the author?