# Continuity equation from Stress-Energy tensor

1. Sep 15, 2008

### Jonny_trigonometry

It is true that $$\frac{\partial}{\partial x^\beta} T^{0 \beta} = \gamma^2 c \left( \frac{\partial \rho}{\partial t} + \vec{\nabla} \bullet \left[ \rho \vec{v} \right] \right) = 0$$

but, how do we arrive at this point?

What is in $$T^{ \alpha \beta}$$

and how do we compute it for any alpha? I'm sorry if this is a no brainer. I missed some critical lectures.

2. Sep 16, 2008

### haushofer

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

$$T^{\mu\nu} = \rho_{0}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}$$

.

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 :)

3. Sep 17, 2008

### Jonny_trigonometry

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

4. Sep 18, 2008

### atyy

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