- #1

haushofer

Science Advisor

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Dear all,

I have a question on the hole argument as presented here by John Norton,

http://plato.stanford.edu/entries/spacetime-holearg/

in particular section 3. The hole argument, as I understand it, can be used to show that GR is a gauge theory with the metric as gauge field and general coordinate transformations (gct's) as gauge transfo's (interpreted actively). Basically, one performs a gct and the metric then transforms as

[tex]

g_{\mu\nu} (x) \rightarrow g'_{\mu\nu} (x')

[/tex]

One can then evaluate the transformed metric ##g'_{\mu\nu}## in the old coordinate system ##\{x\}##,

[tex]

g'_{\mu\nu} (x)

[/tex]

which solves the Einstein equations if ##g_{\mu\nu} (x)## does. The difference between ##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ## is a gauge transformation (which is a Lie derivative of the metric wrt the vector field used for the gct). My question is about the following statement of Norton:

Here "spreadings" can be read as "##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ##". And I guess "invariant properties" can be read as scalars, right? But scalars are invariant when you don't reset the new coordinate to the old one (the good old scalar transformation law). Scalars are not

[tex]

ds^2 = g_{\mu\nu}(x) dx^{\mu} dx^{\nu}

[/tex]

and

[tex]

\tilde{ds}^2 = g_{\mu\nu}'(x) dx^{\mu} dx^{\nu}

[/tex]

respectively, which differ a gauge transformation (and hence are physically equal). Should I read "invariant" here then as "invariant up to a gauge transformation"? I understand that ##ds^2## equals

[tex]

ds'^2 = g'_{\mu\nu}(x') dx'^{\mu} dx'^{\nu}

[/tex]

because it's a scalar, but that's not how we define the "spreadings"; we compare ##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ## (and not ##g'_{\mu\nu} (x') ## and ##g_{\mu\nu} (x) ##!).

Thanks for any insights!

I have a question on the hole argument as presented here by John Norton,

http://plato.stanford.edu/entries/spacetime-holearg/

in particular section 3. The hole argument, as I understand it, can be used to show that GR is a gauge theory with the metric as gauge field and general coordinate transformations (gct's) as gauge transfo's (interpreted actively). Basically, one performs a gct and the metric then transforms as

[tex]

g_{\mu\nu} (x) \rightarrow g'_{\mu\nu} (x')

[/tex]

One can then evaluate the transformed metric ##g'_{\mu\nu}## in the old coordinate system ##\{x\}##,

[tex]

g'_{\mu\nu} (x)

[/tex]

which solves the Einstein equations if ##g_{\mu\nu} (x)## does. The difference between ##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ## is a gauge transformation (which is a Lie derivative of the metric wrt the vector field used for the gct). My question is about the following statement of Norton:

The two different spreadings share one vital characteristic upon which the hole argument depends: the two spreadings agree completely on all invariant properties.The two different spreadings share one vital characteristic upon which the hole argument depends: the two spreadings agree completely on all invariant properties.

Here "spreadings" can be read as "##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ##". And I guess "invariant properties" can be read as scalars, right? But scalars are invariant when you don't reset the new coordinate to the old one (the good old scalar transformation law). Scalars are not

*invariant*under a Lie derivative. E.g., the two 'spreadings' have intervals[tex]

ds^2 = g_{\mu\nu}(x) dx^{\mu} dx^{\nu}

[/tex]

and

[tex]

\tilde{ds}^2 = g_{\mu\nu}'(x) dx^{\mu} dx^{\nu}

[/tex]

respectively, which differ a gauge transformation (and hence are physically equal). Should I read "invariant" here then as "invariant up to a gauge transformation"? I understand that ##ds^2## equals

[tex]

ds'^2 = g'_{\mu\nu}(x') dx'^{\mu} dx'^{\nu}

[/tex]

because it's a scalar, but that's not how we define the "spreadings"; we compare ##g'_{\mu\nu} (x) ## and ##g_{\mu\nu} (x) ## (and not ##g'_{\mu\nu} (x') ## and ##g_{\mu\nu} (x) ##!).

Thanks for any insights!

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