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Energy-momentum density

(Redirected from Energy-momentum tensor)

The stress tensor or energy-momentum tensor is the corresponding conserved Noether current of any theory which is invariant under spacetime translations. In other words, any relativistic field theory must be invariant under (among other things) translations in space as well as translations in time, and corresponding to each of these symmetries will be a conserved current. These currents can be combined into a single second rank tensor whose divergence vanishes (this expresses the conservation of the current). For each conserved current there is a conserved charge. For the stress tensor, this charge is the energy-momentum of the field.

The Noether procedure gives the following formula for the stress tensor

T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^a)}\partial^\nu\phi^a-g^{\mu\nu}\mathcal{L}

where φa is the set of dynamical fields in the theory with Lagrangian L, and g is the metric tensor.

This formula, while easy to calculate, has one glaring problem: it need not be symmetric (for example, if φa is the electromagnetic vector potential of electrodynamics, this formula will not yield a symmetric tensor) . The stress tensor, in this form, cannot be coupled to gravity according to general relativity, because the Einstein tensor is symmetric; a symmetric tensor cannot be set equal to one that is not symmetric. There are two ways out of this problem.

Firstly, we note that since ∂μTμν=0 (i.e. T is a conserved quantity), then we may add to T the divergence of any totally antisymmetric third rank tensor: ∂μ(Tμν+∂ρχρμν)=∂μTμν+∂μρχρμν=0, where the first term vanishes by Noether's theorem, and the second term vanishes because the we are contracting a symmetric tensor (mixed partial derivatives commute) with an antisymmetric tensor χ. Thus we may choose an appropriate χ so that T becomes symmetric. The resulting symmetric form of the stress tensor is known as the Bellinfante tensor. The existence of such a tensor depends on the invariance of the theory under Lorentz transformations, so while any theory which is translation invariant will have a conserved stress tensor, only relativistic theories will have symmetric stress tensors and therefore only relativistic theories can be coupled to gravity. This is no surprise, of course, since general relativity is a relativistic theory.

An alternate route to the symmetric stress tensor is provided by the following formula

T^{\mu\nu}=\frac{\delta S}{\delta g^{\mu\nu}}

where S is the action of the theory. This formula has the advantage that it is always symmetric, but the disadvantage that it is harder to calculate despite being much more succinct to write down. The theory should be formulated on a spacetime with a generic metric, rather than Minkowski space, and then we simply take the functional derivative with respect to the metric. We can then set the metric equal to the Minkowski metric if we wish to work in Minkowski space.

If the theory is conformally invariant, as well as Lorentz invariant, then the stress tensor will be traceless, as well as symmetric. In fact, if the theory lives in more spacetime dimensions than 2, then one can prove that scale invariance along with Lorentz invariance imply that the theory is invariant under the full conformal group and therefore traceless. In 2 dimensions, while there is no such theorem, in all theories of interest, it is still true that scale invariant theories have traceless stress tensor. Note that the first formula above does not yield a traceless stress tensor. One must add another higher tensor to it to restore the tracelessness, analogous to the Bellinfante procedure. On the other hand, the tracelessness of the second form of the stress tensor follows immediately from conformal invariance.

Examples of energy-momentum tensors

For a perfect fluid with density ρ and pressure p, the energy-momentum tensor has components,

T^{ab}=(\rho+\frac{p}{c^2})u^au^b+pg^{ab},

where the ua denote the 4-velocity components of the fluid in some reference frame. For the special case when p = 0, we have the energy-momentum tensor for dust:

Tab = ρuaub

In Maxwell's theory of electromagnetism, the energy-momentum tensor may be written as:

T^{ab}=\epsilon_0 c^2(F^{a}{}_{c}F^{cb}+\frac{1}{4}g^{ab}F^{cd}F_{cd})

The energy-momentum density is a symmetric tensor and


10-26-2009 08:16:03
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