Proving Tensor Theorem: $\frac{\partial}{\partial t} \int {T^{0\alpha} d^3x} =0$

[Azrael84]

Hey,

Starting with the conversation law for the stress-energy tensor; T^{\alpha \beta}{}_{,\beta}=0. Does anyone know how I can prove:

\frac{\partial}{\partial t} \int {T^{0\alpha} d^3x} =0

for a bounded system (i.e. one for which T^{\alpha \beta}=0 outside a bounded region of space).

Seems really obvious intuivitvely that this is conservation of energy-momentum, but I just can't seem to get there mathematically.

My thoughts are maybe this involves the generalised Gauss's law so convert the divergence into integral form:

\int {T^{\alpha \beta}{}_{,\beta} d^4x} =\int{ T^{\alpha \beta}n_{\beta} d^3x

So given T^{\alpha \beta}{}_{,\beta}=0 we can say:

\int{ T^{\alpha \beta}n_{\beta} d^3x=0

Not sure where to go from there, if indeed this is the correct direction?

 

[clamtrox]

I think you have to be a bit more careful about integrating but this should give you a clue:

T^{\alpha \beta}{}_{,\beta}= \partial_t T^{\alpha 0} + \partial_i T^{\alpha i} = 0.

Then you just integrate in a spatial hypersurface over the area where T is non-zero. The spatial derivative gives you a boundary term which is zero, and you get the asked result.

 

[Azrael84]

I think you have to be a bit more careful about integrating

Do you mean because I set \int {T^{\alpha \beta}{}_{,\beta} d^4x} to zero, just because the integrand was zero. When I guess it could really integrate to be a constant?

<br /> T^{\alpha \beta}{}_{,\beta}= \partial_t T^{\alpha 0} - \partial_i T^{\alpha i} = 0 <br />

This is a typo that should be T^{\alpha \beta}{}_{,\beta}= \partial_t T^{\alpha 0} + \partial_i T^{\alpha i} = 0. Since we are contracting, not "dotting", right?

Assuming this. OK so integrate in spatial hypersurface over area where T is non zero.

We have from the above: \partial_t T^{\alpha 0} =- \partial_i T^{\alpha i}. Integrating this over d^3x leads to:

<br /> \int { \frac{\partial}{\partial t} T^{\alpha 0} d^3x} =-\int { \frac{\partial}{\partial x^i} T^{\alpha i} d^3x}<br />

Writing out the RHS explicitly:

<br /> \int { \frac{\partial}{\partial t} T^{\alpha 0} d^3x} =-\int { \frac{\partial}{\partial x} T^{\alpha x } dxdydz}-\int { \frac{\partial}{\partial y} T^{\alpha y } dxdydz}-\int { \frac{\partial}{\partial z} T^{\alpha z } dxdydz}<br />

Can I write this as in the following form?

<br /> \int { \frac{\partial}{\partial t} T^{\alpha 0} d^3x} =- T^{\alpha x }dydz-T^{\alpha y } dxdz}-T^{\alpha z } dxdy}<br />

Where do I go from here?

thanks again

 

[haushofer]

You should use Stokes on

<br /> <br /> \int { \frac{\partial}{\partial t} T^{\alpha 0} d^3x} =-\int { \frac{\partial}{\partial x^i} T^{\alpha i} d^3x}<br /> <br />

This RHS is equal to

<br /> \int \partial_{i}T^{\alpha i}d^3x = \int T^{\alpha i}[d^2x]_{i}<br />
Here the [dx] on the RHS is an oriented surface (formally defined with an epsilon tensor; see eg Wald) You shouldn't write this out explicitly, but just note that the fields which are used to construct T have compact support, and as such vanish at infinity. So the integral is 0.

 

[Azrael84]

You should use Stokes on

<br /> <br /> \int { \frac{\partial}{\partial t} T^{\alpha 0} d^3x} =-\int { \frac{\partial}{\partial x^i} T^{\alpha i} d^3x}<br /> <br />

This RHS is equal to

<br /> \int \partial_{i}T^{\alpha i}d^3x = \int T^{\alpha i}[d^2x]_{i}<br />
Here the [dx] on the RHS is an oriented surface (formally defined with an epsilon tensor; see eg Wald) You shouldn't write this out explicitly, but just note that the fields which are used to construct T have compact support, and as such vanish at infinity. So the integral is 0.

Hey thanks haushofer. I can't claim to fully understand your reply, not quite at that level yet I don't think.

By [dx] being an oriented surface, do you mean for example on the x term on the RHS:

<br /> <br /> \int { \frac{\partial}{\partial x} T^{\alpha x } dxdydz}=\int { T^{\alpha x } (x^1) dydz-\int { T^{\alpha x } (x^2) dydz}<br /> <br />

...I'm thinking of the 4 volume, enclosed by a 3 hypersuface, of which two opposite x=const sides (at x1 and x2), have normal one forms, which are opposite signs?
Or maybe I'm way off.

Also have no idea what compact support means, I'm afraid, anyway to explain this in simpler terms?

Appreciate the reply though, thanks

 

[clamtrox]

Do you mean because I set \int {T^{\alpha \beta}{}_{,\beta} d^4x} to zero, just because the integrand was zero. When I guess it could really integrate to be a constant?

The integral of a function, which is zero everywhere (apart from a zero-measure set), is zero.

Since we are contracting, not "dotting", right?

T^{\alpha \beta}{}_{,\beta} means you take the dot product of a two-tensor and a one-form.

For the integral you need to use Stokes' theorem in some form. Here is where the caution is required. It certainly looks intuitively clear that you could turn a divergence into a boundary term without much problem, but Stokes' theorem is a bit more complicated in curved space. First, the integrand doesn't have a volume element but just a product of the coordinates. Secondly it has a partial derivative over the tensor field, not a covariant derivative.

 

[Azrael84]

<br /> T^{\alpha \beta}{}_{,\beta}<br /> means you take the dot product of a two-tensor and a one-form. [/tex]

I think you have made an error here, this is the contraction of a two-tensor and one form. So no minus signs creep in on the spatial elements. As a simpler case consider conservation of particles,

N^{\alpha}{}_{, \alpha}=0 which has the expanded form of \frac{\partial}{\partial t} N^{0}+\frac{\partial}{\partial t} N^{x}+\frac{\partial}{\partial t} N^{y}+\frac{\partial}{\partial t} N^{z}=0

Not..

\frac{\partial}{\partial t} N^{0}-\frac{\partial}{\partial t} N^{x}-\frac{\partial}{\partial t} N^{y}-\frac{\partial}{\partial t} N^{z} =0

As many gen rel texts will verify. Similarly for energy conservation, you are contracting not dotting with this "4D divergence".

 

[Azrael84]

Also, is it definitely Stokes I need not Gauss?

 

[clamtrox]

Gauss' theorem is a special case of the more general Stokes' - you can call it whatever you want, usually on curved manifolds you only talk about Stokes.

 

[Azrael84]

Gauss' theorem is a special case of the more general Stokes' - you can call it whatever you want, usually on curved manifolds you only talk about Stokes.

OK. I see, I've never actually studied that, don't suppose you know where I can learn about it online? or any specific book references?

Or if anyone could show me explicitley where to go after the point I finished at in my prev post, would be most appreciative.


Thank you both for your help

 

[pervect]

Hey,

My thoughts are maybe this involves the generalised Gauss's law so convert the divergence into integral form:

\int {T^{\alpha \beta}{}_{,\beta} d^4x} =\int{ T^{\alpha \beta}n_{\beta} d^3x

So given T^{\alpha \beta}{}_{,\beta}=0 we can say:

\int{ T^{\alpha \beta}n_{\beta} d^3x=0

Not sure where to go from there, if indeed this is the correct direction?

Pick a particular hypersurface bounded below by t0 and above by t0+h as two sides of the hypersurface enclosing the hypervolume.

Pick the other hypersurfaces so that they are at spatial infinity. Then they won't contribute to the surface integral because the stress-energy tensor is bounded.

Now perform the surface integrals, and set it equal to the volume integral of the divergence. The later must be zero, because the divergence is zero.

 

[haushofer]

OK. I see, I've never actually studied that, don't suppose you know where I can learn about it online? or any specific book references?

If you're a physicist willing to learn differential geometry and topology, I recommend Nakahara's book (diff.geometry and topology in physics). For learning GR and differential geometry and tensor analysis on its way I would recommend Sean Carroll's notes (which are online or as a book) or Wald. Wald is quite formal for a first encounter to GR, but one of the most complete books on the subject :)