Is nonholonomic Xformation = Diffeomorphism

[friend]

Hagen Kleinert, in his book, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 4 Edition, generalizes the Equivalence principle with what he calls a nonholonomic transformation. He transforms equations in flat space into spaces with curvature and torsion using differential transformtions of the form

d{x^i} = \frac{{\partial {x^i}}}{{\partial {q^\mu }}}d{q^\mu }

where

{x^i} = {x^i}({q^\mu })

and where x is in flat space and q is in curved space. See equations 10.21 and 10.12 in the above link.

My question is whether this is a diffeomorphism as is required by diffeomorphism invariance of general relativity. Do all we need to do is make this differential transformation substitution in order to convert flat space physics to curved space physics? Thanks.

 

[atyy]

{x^i} = {x^i}({q^\mu })

That is the diffeomorphism, provided the change of coordinates is differentiable and invertible. The rest is just applying that change of coordinates.General relativity has two assumptions.

The first is the Einstein Field Equation which describes how energy curves spacetime.

The second is the equivalence principle which is the assumption that if you write the flat space laws of physics in general curvilinear coordinates, ie. in tensor form, then the generalization of the laws to curved space is to replace the flat spacetime metric with the curved spacetime metric. This is the same as saying that gravity can always be "locally" canceled out by an appropriate change of coordinates.

So it can be seen as all the laws of special relativity plus the Einstein Field Equation.

 

[friend]

That is the diffeomorphism, provided the change of coordinates is differentiable and invertible. The rest is just applying that change of coordinates.

I think Dr. Kleinert is proposing substitution in terms of the differential form of the transformation in order to avoid subtractions of the flat space coordinates themselves, which would be a distance in flat space but do not translate into distance in curved space. So if you replace subtractions in flat space by integrals of differential distance, then the curved space version can be achieve more easily with the differential transformations. Does this sound right to you?

General relativity has two assumptions.

The first is the Einstein Field Equation which describes how energy curves spacetime.

The second is the equivalence principle which is the assumption that if you write the flat space laws of physics in general curvilinear coordinates, ie. in tensor form, then the generalization of the laws to curved space is to replace the flat spacetime metric with the curved spacetime metric. This is the same as saying that gravity can always be "locally" canceled out by an appropriate change of coordinates.

So it can be seen as all the laws of special relativity plus the Einstein Field Equation.

This all suggests a strategy for deriving GR from first principles. Perhaps there is a programme that is doing this that you may have heard of. I'd appreciate your advice.

Suppose we start with the assumption that quantum theory is fundamental. This is suggested by the perspective that theories of larger phenomena are derived from theories of smaller phenomena - a reductionist view.

Let's say that there is a quantum mechanical formulation that is absolutely necessary, 1st quantization or 2nd quantization, I'm not sure which. We express this quantum theory in terms of a non-specific nonholonomic transformation in a form that makes the metric visible - maybe in terms of curvature. This makes the metric a dynamic field along with the rest of the quantum formulation. It's probably the case that the making the metric visible only occurs when the quantum theory is written in terms for finding the energy-momentum of the system. Maybe this allows solving for the energy-momentum in terms of metric variables such as curvature, at least if the metric has a certain signature and dimension. What problems would you see in this approach? Thanks.