Energy-momentum tensor and conservation of both energy and momentum

[Rory9]

Hi,

I believe you can use the "energy-momentum tensor" to express the conservation of both energy and momentum for fields ($$\partial_{\mu} T^{\mu \nu} = 0$$). But I'm wondering: why's a tensor needed, specifically, to describe this conservation of energy and momentum for fields? For particles, I believe a four-vector suffices (?). I'm not quite clear on this.

Thanks in advance for your wisdom. :-)

 

[Naty1]

I was reading Wikipedia recently on a related issue and although I can't answer your question this section might help:
http://en.wikipedia.org/wiki/Stress-energy_tensor#In_general_relativity_2

and note the line regarding partial and covariant derivatives...

I'd just love to know how and from where Einstein crammed everything used into his tensor formulation.

 

[Peeter]

I can't answer from a GR perspective, but we have a stress tensor in e&m too. There the time rate of change of energy and spatial divergence of the Poynting vector can be written as what I'd loosely call a "four-vector divergence" (not sure what the proper name is). That basically gives you the $T^{0k}$ part when put in tensor form.

Now I haven't worked through any calculations with the momentum part myself, but looking at:

http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor#Conservation_laws

it appears that considering the time rate of change of the poynting vector (which appears to be a momentum density) also has a corresponding conservation equation. Each of these looks like it gives a similar four-vector divergence equation so we have, in total, four such four-vector divergence equations for all the field energy and momentum density conservation conditions (ie: 4x4 second rank tensor equation).

 

[Stingray]

$\partial_\mu T^{\mu \nu} = 0$ implies the conservation of both linear and angular momentum. There are 10 independent PDEs there. These give you conservation laws for the 4 components of linear and 6 components of angular momentum (relativistically, angular momentum is a 2-form/antisymmetric rank-2 tensor).

 

[samalkhaiat]

$\partial_\mu T^{\mu \nu} = 0$ implies the conservation of both linear and angular momentum.

The invariance under spacetime translations (which leads to $\partial_{\nu}T^{\mu\nu} = 0$ ) DOES NOT imply Lorentz invariance ( which leads to the conserved angular momentum, $\partial_{\sigma}M^{\sigma \mu \nu}= 0$ ).
So it is not correct to say that $\partial_{\mu}T^{\mu\nu}=0$ implies the conservation of angular momentum.
Indeed, the angular momentum tensor is related to the canonical energy-momentum tensor by

$$M^{\sigma \mu \nu} = x^{\mu}T^{\sigma \nu} - x^{\nu}T^{\sigma \mu} + S^{\sigma \mu \nu}$$

where

$$S^{\sigma \mu \nu} = \frac{\partial \mathcal{L}}{\partial \partial_{\sigma}\phi} \Sigma^{\mu \nu} \phi$$

characterizes the spin angular momentum of the field.

Therefore $\partial_{\sigma} T^{\sigma \mu} = 0$ implies

$$\partial_{\sigma}M^{\sigma \mu \nu} = T^{\mu \nu} - T^{\nu \mu} + \partial_{\sigma}S^{\sigma \mu \nu} \neq 0$$

For this to vanish, we must impose Lorentz invariance! Indeed, we can show that Lorentz symmetry implies and is implied by

$$T^{\mu\nu} - T^{\nu\mu} + \partial_{\sigma}S^{\sigma\mu\nu} = 0$$

There are 10 independent PDEs there.

In a translation invariant theory and in the absence of fields at infinity,the FOUR equations$\partial_{\mu}T^{\mu\nu}=0$ leads to a TIME-INDEPENDENT energy-momentum 4-vector
$$P^{\nu} = \int d^{3}x \ T^{0 \nu}$$
Similarly, In a Lorentz invariant theory, the six equations $\partial_{\sigma}M^{\sigma\mu\nu}=0$ leads to a time independent angular momentum tensor

$$J^{\mu\nu} = \int d^{3}x \ M^{0 \mu \nu}$$

regards

sam

 

[Stingray]

Sam,
You're certainly right that there are 4 equations there rather than 10. I'm not sure what I was thinking at the time.

In any case, we were coming at this using different definitions and assumptions. I was looking at it as a special case of what's usually done in GR. There, you have a symmetric conserved stress-energy tensor derived from the functional derivative of the action with respect to the metric. $\nabla_a T^{ab} = 0$ comes from the diffeomorphism invariance of the action. It's then typical to define an angular momentum tensor about a point $z \in \Sigma$ by (going back to flat spacetime)
$$S_{ab} (\Sigma,z) = \int_\Sigma T^{c}{}_{[a} (x-z)_{b]} d S_c .$$
Stress-energy conservation does imply that this is conserved in an appropriate sense as long as you have the right fall-off conditions, hypersurfaces are chosen appropriately, etc.

 

[Naty1]

You're certainly right that there are 4 equations there rather than 10. I'm not sure what I was thinking at the time.

Pure coincidence?:

In 4 dimensions, it takes 20 numbers to specify curvature at each point: 10 of these numbers are captured by the "Ricci tensor", while the remaining 10 are captured by the "Weyl tensor".

 

[Stingray]

Pure coincidence?:

In 4 dimensions, it takes 20 numbers to specify curvature at each point: 10 of these numbers are captured by the "Ricci tensor", while the remaining 10 are captured by the "Weyl tensor".

Yeah, I was probably thinking of the stress-energy tensor itself. It has 10 components. Well, the symmetric one does. Other definitions like those used by Sam are different.

 

[chwie]

nice discussion, but I think nobody answer the question. Why a tensor? If you use noether theorem you will find that symmetry over translation implies the energy-momentum tensor. In this case the equation of conservation is local (it means that the densities are not conserved in general inside a surface, but that the change of the densities is because of the amount that leaves the surface. some kind of continuity equation as the conservation of charge.) which is stronger than global conservation of energy.

Conversations can be express in terms of four vectors (as the conservation of charge in EM that is the four divergence of a four vector). If that is possible then what is the difference between charge conservation and energy-momentum conservation for a free particle? Well to describe local conservation of energy you need a a density of energy and a current density, the same for momentum. In that case we have four densities and 12 components of currents that is a total four equation with four terms each. If we want to express local conservation of energy and momentum in a covariant form we expect a tensor of rank two ( that is the case the diagonal terms are related to the densities and the off diagonals to the currents). This tensor at the moment can be reducible and each conservation can be expressed independently as a four vector. Now the argument is that for a free field (as for a free particle) the conversations of energy and momentum are not independent (that is the reason that Stress-energy tensor is symmetric, only ten independent components). In that case is not reducible and for that reason need to be express as a rank two tensor. The energy a momentum are not independent because I can write the energy as a function of momentum. I hope this can answer the question.

 

[bcrowell]

I believe you can use the "energy-momentum tensor" to express the conservation of both energy and momentum for fields ($$\partial_{\mu} T^{\mu \nu} = 0$$). But I'm wondering: why's a tensor needed, specifically, to describe this conservation of energy and momentum for fields? For particles, I believe a four-vector suffices (?). I'm not quite clear on this.

You've gotten lots of complicated and highbrow answers to what I think is really a much simpler question.

Energy-momentum vectors apply to both particles and fields. The stress-energy tensor applies to both particles and fields.

The energy-momentum vector tells you how much you have. The stress-energy tensor tells you the density and rate of transport of it.

Suppose you break up the stress-energy tensor into four columns, corresponding to t, x, y, and z. The y column, for example, tells you the rate at which energy and momentum are being transported in the y direction.

In the case of a point particle, the stress-energy tensor will blow up like a delta-function at a particular place.

-Ben

 

[pervect]

Somewhat a blast from the past. But anyway, the way I look at it is this. If you consider some region of space-time, one expects that region to contain some amount of energy E and some amount of momentum P due to the particles and fields in it.

But, the volume element doesn't transform as a scalar - it's not a So to make the description covariant, we need to figure out how a volume element transforms.

The end result as most textbooks explain is that one needs to sepcify the 4-velocity of an observer to get a covariant density. Hence, one feeds the 4-velocity into the stress-energy tensor to get the density of momentum and energy.

Alternatively, especially if one is dealing with time-orientable manifolds, one can think of a volume element as being a three-form, modulo some pesky sign issues. The textbooks don't spell this out, usually, I suspect it's because of said pesky sign issues - perhaps there's some deeper issue that I'm not aware of, but if that's the case I'm blissfully unaware of it :-).

A three-form has a dual that's a one-form, so a volume element is pretty much a vector, rather than a scalar, said vector being essentially the 4-velocity, i.e. a number proportional to the volume and pointing in the "time" direction, the direction that's orthogonal to the space the volume element occupies. (This gives the least problems with signs when one has a time-orientable manifold, in which case the vector always points to the future).

So the end conclusion is - energy momentum is naturally a 4-vector, and the density of energy momentum is naturally a rank two tensor. You feed said rank-2 tensor a vector that represents your volume element, and it spits out the total amount of energy and momentum contained in that volume.