- #1
Prathyush
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- TL;DR Summary
- Active diffeomorphism of a Schwarzschild metric under r ->c r
I am trying to understand active diffeomorphism by looking at Schwarzschild metric as an example.
Consider the Schwarzschild metric given by the metric
$$g(r,t) = (1-\frac{r_s}{r}) dt^2 - \frac{1}{(1-\frac{r_s}{r})} dr^2 - r^2 d\Omega^2 $$
We identify the metric new metric at r with the old metric at ##c r##
this gives
$$g(r,t) = (1-\frac{r_s}{ c r}) dt^2 - \frac{1}{(1-\frac{r_s}{c r})} dr^2 - c^2 r^2 d\Omega^2 $$
We can do a passive transformation to re-define the metric so that we have a Minkowski metric as ##r->\infty## We can define ##\tilde{r} = c r##
and we get the metric
$$g(r,t) = (1-\frac{\tilde{r_s}}{r}) dt^2 - \frac{1}{c^2 (1-\frac{\tilde{r_s}}{r})} d\tilde{r}^2 - \tilde{r}^2 d\Omega^2 $$
This does not look like the Schwarzschild metric and should not be a solution of GR.
This is not surprising because we simply dragged the metric numerically and hence introduced distortions to the original manifold.
Now I can define active diffeomorphism such that distance between ##(r, r+dr)## is actually the same as the distance between ##(c r, c r + cdr)## at fixed time then I won't introduce distortions(the ##c^2## in the denominator just cancels) but that is just an ordinary passive transformation. Maybe this is what duality between active and passive transformations means?
Please correct me if I am making a mistake or I have misunderstood something. I heard a claim that an active diffeomorphism is a symmetry in GR and generates new solutions of GR, I don't think it is correct.
Consider the Schwarzschild metric given by the metric
$$g(r,t) = (1-\frac{r_s}{r}) dt^2 - \frac{1}{(1-\frac{r_s}{r})} dr^2 - r^2 d\Omega^2 $$
We identify the metric new metric at r with the old metric at ##c r##
this gives
$$g(r,t) = (1-\frac{r_s}{ c r}) dt^2 - \frac{1}{(1-\frac{r_s}{c r})} dr^2 - c^2 r^2 d\Omega^2 $$
We can do a passive transformation to re-define the metric so that we have a Minkowski metric as ##r->\infty## We can define ##\tilde{r} = c r##
and we get the metric
$$g(r,t) = (1-\frac{\tilde{r_s}}{r}) dt^2 - \frac{1}{c^2 (1-\frac{\tilde{r_s}}{r})} d\tilde{r}^2 - \tilde{r}^2 d\Omega^2 $$
This does not look like the Schwarzschild metric and should not be a solution of GR.
This is not surprising because we simply dragged the metric numerically and hence introduced distortions to the original manifold.
Now I can define active diffeomorphism such that distance between ##(r, r+dr)## is actually the same as the distance between ##(c r, c r + cdr)## at fixed time then I won't introduce distortions(the ##c^2## in the denominator just cancels) but that is just an ordinary passive transformation. Maybe this is what duality between active and passive transformations means?
Please correct me if I am making a mistake or I have misunderstood something. I heard a claim that an active diffeomorphism is a symmetry in GR and generates new solutions of GR, I don't think it is correct.
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