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haushofer

Science Advisor

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I found an article about this by John Norton, called "Coordinates and covariance: Einstein's view of space-time and the modern view" ( foundations of physics, Vol.19, No 10, 1989 ). I will briefly outline the relevant parts of this article concerning my question.

He contends that general relativity has a geometric structure:

[tex]

M2 \equiv \ <M,\ g_{ab}, \ T_{ab}>

[/tex]

where I use the abstract index notation of Wald; a and b are NOT indices. This geometric structure has a coordinate representation as follows:

[tex]

M3 \equiv \ <A, \ g_{\mu\nu}, \ T_{\mu\nu}>

[/tex]

A is here an open subset of [itex]R^{4}[/itex], and [itex]g_{\mu\nu}[/itex] and [itex]T_{\mu\nu}[/itex] are the 4x4 matrices representing the metric and energy momentum tensor. Now to the duality between transformation on M2 and M3.

We have:

- a coordinate chart [itex]x^{\mu}: M \rightarrow A[/itex]

-an invertible function [itex]f: A \rightarrow A[/itex]

-a diffeomorphism [itex]h: M \rightarrow M[/itex]

The dual transformations are:

Active: h maps [itex]p \in M[/itex] with coordinate [itex]x^{\mu}(p)[/itex] to [itex]hp \in M[/itex] with coordinate [itex]x^{\mu}(hp)[/itex] ( because we are working in 1 chart ). This last coordinate can also be written as [itex]x^{\mu}(hp) = fx^{\mu}(p)[/itex].

Passive: the point p stays untouched; we simply go from one chart [itex]x^{\mu}(p)[/itex] to another [itex]x^{'\mu}(p) = (fx^{\mu})(p)[/itex]. If we apply this to our metric tensor, we have:

Active: h induces the carry-along map h* which maps geometric fields on the manifold to geometric fields on the manifold. For scalars we have for instance h*f(hp)=f(p), for vectors we have [itex]h*V^{a}(h*f)=V^{a}(f)[/itex] etc. So the metric is carried along, and we get [itex]h*g_{ab}[/itex].

Passive: f defines the transformation between the objects of geometric fields in the coordinate charts [itex]x^{\mu}[/itex] and [itex]x^{'\mu}[/itex]. For our tensor [itex]g_{ab}[/itex] this means that

[tex]

g^{'}_{\mu\nu} = \frac{\partial x^{\alpha}}{\partial x^{'\mu}}\frac{\partial x^{\beta}}{\partial x^{'\nu}}g_{\alpha\beta}

[/tex]

, the tensor coordinate transformation rule. Now he states that "the duality becomes most apparent with the aid of the notion of a carried-along coordinate chart". He defines this carried-along coordinate chart via

[tex]

hx^{\mu}(hp) \equiv x^{\mu}(p)

[/tex]

So this is what I don't understand. I understand that in the beginning we could state for an active transformation that on the manifold [itex]h \rightarrow hp[/itex] we induce a coordinate transformation [itex]x^{\mu}(p) \rightarrow x^{\mu}(hp) = fx^{\mu}(p)[/itex], but now he uses the diffeomorphism on the manifold to act on the coordinates! Why is this allowed? I don't see how one could possibly define this.

He then continues: the components of the carried-along object h*O at hp in the carried along coordinate chart hx equal numerically the components of the original object O at p in the coordinate chart x. For the metric this means that the carry-along is defined by

[tex]

(h*g)_{'\mu '\nu}(hp) = g_{\mu\nu}(p)

[/tex]

The statement then is that the matrix components of the LHS and RHS of this equation are defined by the tensor coordinate transformation rule, and so that this rule covers both passive and active transformations.

In my head moving the point on the manifold is something very different than switching your coordinatelabels. For instance, in the Schwarzschild case, going from Cartesian to polar coordinates is something very different than transforming actively from 100 kilometers from the singularity to 10 kilometers from the singularity.

Can someone clarify this? I'm sure more questions will follow (like: are the so-called "gauge transformations" related to relabeling your coordinates and using the tensor transformation law, or something else? ), but I would appreciate some help, because this is frustrating me already quite some time :D