2
$\begingroup$

In Classical Mechanics a state of a system is either a pair $(q,p)$ or $(q,\dot{q})$ depending if we formulate the theory on the tangent or cotangent bundle of the configuration space.

The evolution of the system is thus a path of states $(q(t),p(t))$ or $(q(t),\dot{q}(t))$.

In Classical Field Theory a state of a system is again specified by the "coordinates" which are now the field values $\phi(\mathbf{x})$ and momenta $\pi(\mathbf{x})$.

The evolution of the state is $\phi(t,\mathbf{x})$ which is a path on the space of coordinates.

Now what about Classical General Relativity? It is tempting to say that the "states" are "Lorentzian geometries" so that a state is a pair $(M,g)$ where $M$ is a smooth manifold and $g$ is a Lorentzian metric tensor on $M$.

On the other hand, there is no evolution on that. Spacetime is given and it is not evolving with respect to anything. Actually the idea of "time evolution" is in some sense "inside of $(M,g)$" and not outside.

In that setting comparing to Classical Mechanics and Classical Field Theory, a state seems not to be a pair $(M,g)$.

In that case, what is a classical state in General Relativity and how does it compare to the Classical Mechanics and Classical Field Theory states?

$\endgroup$
3
$\begingroup$

There are multiple equivalent ways of formulating GR, some of them very exotic. In the standard, original formulation, there is no notion of a state. Don't have it, don't need it.

What you're asking for is more like what happens in the ADM formalism. The state is specified on an arbitrary spacelike surface, and consists of the metric and a set of momenta conjugate to the components of the metric.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.