Gravitation vs Curvilinear Coordinates

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In Weinberg's book, it is said that a given metric ##g_{\mu \nu}## could be describing a true gravitational field or can be just the metric ##\eta_{\alpha \beta}## of special relativity written in curvilinear coordinates. Then it is said that in the latter case, there will be a set of Minkowskian coordinates ##\xi^\alpha (x)## such that
$$\eta^{\alpha \beta} = g^{\mu \nu}\frac{\partial \xi^{\alpha}(x)}{\partial x^\mu}\frac{\partial \xi^{\beta}(x)}{\partial x^\nu} \ (1)$$ everywhere, not just at every point ##X## and its infinitesimal neighborhood as the equivalence principle states.

Now what is the difference of everywhere to every point ##X##? Since if we pick up every point ##X## we will end up with all points that make the space we are considering?

Also, it is said in the book that for the metric which has coefficients
$$g_{rr} = 1; \ g_{\theta \theta} = r^2; \ g_{\varphi \varphi} = r^2 sin^2 \theta; \ g_{tt} = -1$$ it's possible to find a set of Minkowskian coordinates
$$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ that satisfies (1) above.

I don't understand, since by the above reasoning a sphere would'nt be a curved space, for it's possible to find that set of Minkowskian coordinates above that satisfy (1) at every point on the sphere. But we know that the sphere is a intrinsically curved space. Could the Riemann curvature tensor vanish for the sphere when we use spherical coordinates to evaluate it?
 

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  • #2
PeterDonis
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what is the difference of everywhere to every point ##X##?
If the spacetime is actually flat Minkowski spacetime, then you will be able to find one single coordinate chart in which the metric is ##\eta_{\alpha \beta}## everywhere.

If the spacetime is curved, then you will be able to find a coordinate chart in which the metric is ##\eta_{\alpha \beta}## at some chosen point ##X##, but if you then pick a different point ##X'##, the chart in which the metric is ##\eta_{\alpha \beta}## at ##X'## will be a different chart from the one in which the metric is ##\eta_{\alpha \beta}## at ##X##

it's possible to find that set of Minkowskian coordinates above that satisfy (1) at every point on the sphere
Are you sure? Try it!

Could the Riemann curvature tensor vanish for the sphere when we use spherical coordinates to evaluate it?
No.
 
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Are you sure? Try it!
I will try it. However, in Weinberg's book it is said that it's possible, by using $$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ Maybe the author made a mistake?

(Quote)
##"##For example, given the metric coefficients $$g_{rr} = 1; \ g_{\theta \theta} = r^2; \ g_{\varphi \varphi} = r^2 sin^2 \theta; \ g_{tt} = -1 \ \text{(6.4.2)}$$ we know that there is a set of ##\xi##s satisfying $$\eta^{\alpha \beta} = g^{\mu \nu}\frac{\partial \xi^{\alpha}(x)}{\partial x^\mu}\frac{\partial \xi^{\beta}(x)}{\partial x^\nu} \ $$ that is, $$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ but how could we have told that (6.4.2) was really equivalent to the Minkowski metric ##\eta_{\alpha \beta}##, if we weren't clever enough to have recognized it as simply ##\eta_{\alpha \beta}## in spherical polar coordinates? Or, on the other hand, if we change ##g_{rr}## to an arbritary function of r, how can we tell that this really represents a gravitational field, that is, how can we tell that Eqs. (6.4.2) now have no solution?##"##
 
  • #4
PeterDonis
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in Weinberg's book it is said that it's possible
It's possible in a space where the metric coefficients are as you gave them. But is a space where the metric coefficients are as you gave them a sphere? Just because the coordinates are "spherical coordinates" does not mean the geometry is that of a sphere. You can have spherical coordinates on flat Euclidean space.

To decide whether the metric is that of a sphere, or more generally whether it is even curved rather than flat, you need to compute its Riemann tensor. Have you tried that?
 
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Just because the coordinates are "spherical coordinates" does not mean the geometry is that of a sphere
But spherical coordinates automatically gives points that "live" on the surface of a sphere (for each r constant). So it automatically forms a sphere. How can we get a set of points forming a flat surface when using spherical coordinate system?
To decide whether the metric is that of a sphere, or more generally whether it is even curved rather than flat, you need to compute its Riemann tensor
I don't understand it. Since spherical coordinate system automatically gives points that form a sphere, it seems to me that the Riemann tensor will never vanish when we use spherical coordinates.
Have you tried that?
Yes. I have computed all of the components a time ago and I remember that some of them were not zero.
 
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PeterDonis
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spherical coordinates automatically gives points that "live" on the surface of a sphere (for each r constant)
But ##r## is not constant; it's one of the coordinates.

Since spherical coordinate system automatically gives points that form a sphere, it seems to me that the Riemann tensor will never vanish when we use spherical coordinates.
Try computing the Riemann tensor for the metric Weinberg gives with ##r## as one of the coordinates, instead of being just a constant.
 
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  • #7
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But ##r## is not constant; it's one of the coordinates.
Ohhh, so this is the key thing
Try computing the Riemann tensor for the metric Weinberg gives with rrr as one of the coordinates, instead of being just a constant.
It would be great to do that and see the magic happening. I will perform the calculations later.
 
  • #8
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But spherical coordinates automatically gives points that "live" on the surface of a sphere (for each r constant). So it automatically forms a sphere. How can we get a set of points forming a flat surface when using spherical coordinate system?
You might try dropping down one dimension and considering whether using polar coordinates (##r##, ##\theta## where ##x=r\cos\theta## and ##y=r\sin\theta##) does anything to change the Euclidean flatness of the two-dimensional x-y plane.
 
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