- #1

haushofer

Science Advisor

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In a desparate attempt, not for the first time, to understand some issues on coordinate transformations in GR

from a passive and active point of view, I opened this thread. I already read references of Wald, Rovelli and others,

(actually, I think I saw almost all articles and notes which try to explain this) and still don't get it, so I try to find my understanding here, hopefully with some help from you ;)

I will use scalar fields to keep things as simple as possible, and use the following notation:

\phi : scalar field

x^{\mu} : coordinates (in the text I simply call them x)

M : manifold

p,q : points on the manifold

O, O': charts (frames of reference)

A scalar field is characterized by the fact that under a coordinate transformation

[tex]

x^{\mu} \rightarrow x^{'\mu} \ \ \ \ (1)

[/tex]

one has

[tex]

\phi '(x') = \phi(x) \ \ \ \ (2)

[/tex]

With other words: the functional change of the field, [itex]\phi'[/itex], is such that this [itex]\phi'[/itex] with the new

coordinate x' as argument gives the same numerical value as [itex]\phi[/itex] at the old argument x.

Now I want to look at the variation

[tex]

\delta \phi (x) \equiv \phi '(x) - \phi(x) \ \ \ \ \ (3)

[/tex]

from both a passive and active interpretation of (1). First the active one, because that one I seem to understand.

The active interpretation of (1) means that we stay in one chart O, and that x and x'

refer to the different points p and q on the manifold M respectively. Then (3) has the interpretation of looking at the functional

change of the field, while staying in one point. Namely, first we perform (2) to go from

[itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex] (which gives us the transformed field evaluated in the point q!), and then reset the argument x' to the old value x. So (3) is evaluated in the point p. If [itex]\phi[/itex] would be the metric tensor instead, this is the way to find isometries: we look at the functional change of the metric. That active view makes perfect sense to me, and this is the way to derive the form of the Lie derivativeon fields if we write the coordinate transformation infinitesimally as

[tex]

x^{\mu} \rightarrow x^{'\mu} = x^{\mu} - \xi^{\mu}(x) \ \ \ \ (4)

[/tex]

giving that via a Taylor expansion (3) becomes

[tex]

\delta \phi (x) = - \xi^{\mu}\partial_{\mu} \phi(x)

[/tex]

But now the passive interpretation of the coordinate transformation, which should be easy (dark music is coming up now). Then (1) means we go from chart O to O'. Both x and x' label the same point p on the manifold M. But now we do the same thing as above. We perform the coordinate transformation (1) to go from [itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex], and then reset the argument x' in [itex]\phi'(x')[/itex] back to x. This gives us [itex]\phi'(x)[/itex]. But now in the chart O' the coordinate x indicates a different point q (namely, the point p was labeled by x'!), meaning that now in (3) we compare field values at different points! Or: both the arguments x at the RHS of (3) refer to different points, which of course is quite confusing. How is this to be reconciled with the passive point of view? What is the meaning of (3) from a passive point of view? And to what point does the argument x refer to on the LHS of (3)?

Can someone make this point (:D) clear? I'm sorry if people are fed up with this discussion, but I'd really like to understand this thoroughly.

In the end I would like to understand the transformation of the metric

[tex]

\delta g_{\mu\nu}(x) \equiv g_{\mu\nu}'(x) - g_{\mu\nu}(x)

[/tex]

and I can hardly believe that such an expression only makes sense in the active point of view (where it gives the Lie-derivative of the metric if one applies (4)).

from a passive and active point of view, I opened this thread. I already read references of Wald, Rovelli and others,

(actually, I think I saw almost all articles and notes which try to explain this) and still don't get it, so I try to find my understanding here, hopefully with some help from you ;)

I will use scalar fields to keep things as simple as possible, and use the following notation:

\phi : scalar field

x^{\mu} : coordinates (in the text I simply call them x)

M : manifold

p,q : points on the manifold

O, O': charts (frames of reference)

A scalar field is characterized by the fact that under a coordinate transformation

[tex]

x^{\mu} \rightarrow x^{'\mu} \ \ \ \ (1)

[/tex]

one has

[tex]

\phi '(x') = \phi(x) \ \ \ \ (2)

[/tex]

With other words: the functional change of the field, [itex]\phi'[/itex], is such that this [itex]\phi'[/itex] with the new

coordinate x' as argument gives the same numerical value as [itex]\phi[/itex] at the old argument x.

Now I want to look at the variation

[tex]

\delta \phi (x) \equiv \phi '(x) - \phi(x) \ \ \ \ \ (3)

[/tex]

from both a passive and active interpretation of (1). First the active one, because that one I seem to understand.

The active interpretation of (1) means that we stay in one chart O, and that x and x'

refer to the different points p and q on the manifold M respectively. Then (3) has the interpretation of looking at the functional

change of the field, while staying in one point. Namely, first we perform (2) to go from

[itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex] (which gives us the transformed field evaluated in the point q!), and then reset the argument x' to the old value x. So (3) is evaluated in the point p. If [itex]\phi[/itex] would be the metric tensor instead, this is the way to find isometries: we look at the functional change of the metric. That active view makes perfect sense to me, and this is the way to derive the form of the Lie derivativeon fields if we write the coordinate transformation infinitesimally as

[tex]

x^{\mu} \rightarrow x^{'\mu} = x^{\mu} - \xi^{\mu}(x) \ \ \ \ (4)

[/tex]

giving that via a Taylor expansion (3) becomes

[tex]

\delta \phi (x) = - \xi^{\mu}\partial_{\mu} \phi(x)

[/tex]

But now the passive interpretation of the coordinate transformation, which should be easy (dark music is coming up now). Then (1) means we go from chart O to O'. Both x and x' label the same point p on the manifold M. But now we do the same thing as above. We perform the coordinate transformation (1) to go from [itex]\phi(x)[/itex] to [itex]\phi'(x')[/itex], and then reset the argument x' in [itex]\phi'(x')[/itex] back to x. This gives us [itex]\phi'(x)[/itex]. But now in the chart O' the coordinate x indicates a different point q (namely, the point p was labeled by x'!), meaning that now in (3) we compare field values at different points! Or: both the arguments x at the RHS of (3) refer to different points, which of course is quite confusing. How is this to be reconciled with the passive point of view? What is the meaning of (3) from a passive point of view? And to what point does the argument x refer to on the LHS of (3)?

Can someone make this point (:D) clear? I'm sorry if people are fed up with this discussion, but I'd really like to understand this thoroughly.

In the end I would like to understand the transformation of the metric

[tex]

\delta g_{\mu\nu}(x) \equiv g_{\mu\nu}'(x) - g_{\mu\nu}(x)

[/tex]

and I can hardly believe that such an expression only makes sense in the active point of view (where it gives the Lie-derivative of the metric if one applies (4)).

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