# Killing tensor/vector very basic context question about consevation

1. Jan 6, 2015

### binbagsss

Timelike Killing vectors are associated with conservation of energy, and space-like Killing vectors with the conservation of momentum quantities.

But the energy-momentum tensor is always 'conserved' - well in GR, this goes to it's divergence being zero.

And , the FRW metric does not possess a time-like Killing vector - so energy is not conserved.
It possess space-like ones, and momentum is thus conserved.

But, how then, is the energy-momentum tensor conserved?

Depending on the geometry of the space, are there cases when momentum and energy are conserved seperately, and so energy-momentum is conserved, but also cases where the energy-momentum is only conserved as a single quantity. So a Killing vector could not suffice to give this conserved quantity. Instead would it be described by a Killing tensor?

Last edited by a moderator: Jan 6, 2015
2. Jan 7, 2015

### Staff: Mentor

Yes, but the energy and momentum that are conserved are global, not local, quantities.

Because the conservation law it obeys is local, not global. In a spacetime without a timelike Killing vector field, like FRW spacetime, there is no way to integrate the stress-energy tensor (that's the usual term for what you're calling the energy-momentum tensor) to get a globally conserved quantity. In a spacetime with a timelike KVF, there is. But in either case, the stress-energy tensor's covariant divergence is zero; that's the local conservation law, and it's always true.

3. Jan 8, 2015

### WannabeNewton

As Peter noted, given a matter distribution $T^{\mu\nu}$ the statement $\nabla_{\mu}T^{\mu\nu} = 0$ is local so it doesn't manifestly take into account the conservation laws for the gravitational field that the matter distribution interacts with i.e. it does not manifestly include the dynamics of the background space-time; this can be seen for example by transforming to local inertial coordinates wherein the above statement becomes $\partial_{\mu}T^{\mu\nu} = 0$ just as in flat space-time. This is of course because the dynamics of the background space-time is only gauge invariant at the level of curvature $R_{\mu\nu\gamma\delta}$, not at the level of the connection $\nabla_{\mu}$, due to the equivalence principle. In $\nabla_{\mu}T^{\mu\nu} = 0$ the dynamics of the gravitational field is incorporated in a complicated and non-gauge invariant way into $\nabla_{\mu}$.

Thus it isn't a surprise that $\nabla_{\mu}T^{\mu\nu} = 0$ holds in any space-time irrespective of the space-time symmetries, or lack thereof. Because a local conservation law cannot be construct from gauge invariant quantities that manifestly takes into account the dynamics of the gravitational field, again due to the equivalence principle, one must resort to quasi-local conservation laws. That is, one finds a non-gauge invariant quantity ("pseudotensor") $t^{\mu\nu}$ which is constructed only from $g_{\mu\nu}$ and $\partial_{\delta}g_{\mu\nu}$ while satisfying $\partial_{\mu}(T^{\mu\nu} + t^{\mu\nu}) = 0$, and as such can always be made to vanish in a local inertial coordinate system but which upon integration over a finite region of space-time ("quasi-local") yields gauge invariant conserved charges.

One such choice is the Landau-Lifshitz pseudotensor $t^{\mu\nu}_{LL}$, whose exact expression can be found here: http://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor#Metric_and_affine_connection_versions. So for example if we take any coordinate system $x^{\mu}$ on some region of this space-time and consider a space-like hypersurface $\Sigma$ in this coordinate system then we can define a conserved total 4-momentum $P^{\mu} = \int_{\Sigma}d^3 x(-g)(T^{\mu0}+ t^{\mu0}_{LL})$ that clearly includes the dynamics of both the matter distribution and the gravitational field it interacts with. It is important to note that as long as the Einstein equations are valid, $\nabla_{\mu}T^{\mu\nu} = 0$ is equivalent to $\partial_{\mu}(T^{\mu\nu} + t^{\mu\nu}_{LL}) = 0$; all the latter conservation law does is manifestly include the dynamics of the gravitational field by shifting it implicitly from $\nabla_{\mu}$ into $t^{\mu\nu}_{LL}$.

Clearly $\frac{d P^{\mu}}{dx^0} = 0$ holds irrespective of whether or not the space-time possesses a Killing field so what does this have to do with them? Consider a space-time with a time-like Killing field $\xi^{\mu}$ and for simplicity assume it is asymptotically flat. One can always find an asymptotic Lorentz frame $(t,\vec{x})$ in which the space-time has no 3-momentum e.g. if one has a static black hole then this will be the asymptotic rest frame of the black hole. In this frame the space-time simply has a mass $M$ (called the ADM mass which in this case is equivalent to the Komar mass) that is constant, $\frac{dM}{dt} = 0$ i.e. the space-time has no dynamics; of course $\xi^{\mu} = \delta^{\mu}_t$ in these coordinates.

All this means is, instead of having to go through the formalism described above in order to make manifest the conservation laws due to the combined dynamics of the matter fields and the gravitational field they interact with, one can simply use $\xi^{\mu}$ to describe the dynamics of the matter fields alone since the space-time itself has no dynamics on account of $\xi^{\mu}$; essentially what one is doing is defining a static gravitational potential $\phi = \log(-\xi_{\mu}\xi^{\mu})^{1/2}$ to encode the dynamics of matter interacting with the static gravitational field. For example for a freely falling test particle one uses the conserved energy $E = -\xi_{\mu}p^{\mu}$ which, in a non-trivial way, includes both the gravitational potential energy and the kinetic energy of the particle (the split becomes manifest in the non-relativistic limit).

This is not unlike what one does in EM. In flat space-time we still have the local conservation law $\partial_{\mu}(T^{\mu\nu}_{\text{Matter}} + T^{\mu\nu}_{\text{EM}}) = 0$ for charged matter fields interacting with an EM field propagating in the flat space-time but if the EM field is static then it has no dynamics, just a constraint equation $\vec{\nabla}\cdot \vec{E} = 4\pi \rho$, and we simply describe the dynamics of the charged particle through an electrostatic potential $\phi$.

4. Jan 9, 2015

### binbagsss

And a spacelike killing vector field plays no role in this?

5. Jan 9, 2015

### Staff: Mentor

Not in what I was describing, no. A "conserved quantity" means a quantity that doesn't change with time. Integral curves of a spacelike KVF are (obviously) not timelike, so it makes no sense to say that a quantity that doesn't change along those integral curves doesn't change with time.

There is also another aspect that spacelike KVFs do play a role in: in a spacetime with any kind of KVF, the component of an object's 4-momentum that is parallel to that KVF will be a constant of the motion for free-fall motion. So in a spacetime with a timelike KVF, the "energy at infinity", which is the component of an object's 4-momentum along the timelike KVF, is conserved for free-fall motion. In a spacetime which has an axial KVF (i.e., a spacelike one with closed orbits), the "angular momentum", which is the component of 4-momentum along the axial KVF, is conserved for free-fall motion. And in a spacetime with a "linear" spacelike KVF (such as flat Minkowski spacetime), the linear momentum is conserved for free-fall motion. There is a sense in which these quantities are "global", not local, but it's not the same sense as we've been discussing; you still can't integrate these quantities to get a global conserved quantity.

6. Feb 2, 2015

### binbagsss

(Sorry to re-bump a old thread, but my question follows on so thought it's easiest)...So when we talk about s-l and t-l KVFs associated with momentum and energy conservation, we are talking about global conservation, and when we talk about the divergence of the energy-momentum tensor we are talking about local conservation, and this holds locally , regardless of whether a KVF exists. In a similar way, is it possible to just have local momentum conservation if no space-like KVF exists? (or local energy conservation if no T-L KVF exists?). Thanks

Also, just to confirm, without a time-like KVF global energy conservation is not possible?

7. Feb 2, 2015

### Staff: Mentor

There is only one local conservation law, the law that the divergence of the stress-energy tensor is zero. That always holds, regardless of what, if any, KVFs the spacetime has. When we talk about local conservation of energy or momentum, we are talking about particular components of the divergence of the stress-energy tensor in a particular local inertial frame (i.e., with a particular local split of spacetime into space and time); they're not different laws.

Not with the definition of "energy" we have been using. There are other possible definitions of energy, though. See this article from the Usenet Physics FAQ for a decent overview:

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html