# I Gravitation vs Curvilinear Coordinates

1. Jun 16, 2017

### davidge

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?

2. Jun 16, 2017

### Staff: Mentor

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 $X$

Are you sure? Try it!

No.

3. Jun 16, 2017

### davidge

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. Jun 16, 2017

### Staff: Mentor

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?

5. Jun 16, 2017

### davidge

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?
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.
Yes. I have computed all of the components a time ago and I remember that some of them were not zero.

6. Jun 16, 2017

### Staff: Mentor

But $r$ is not constant; it's one of the coordinates.

Try computing the Riemann tensor for the metric Weinberg gives with $r$ as one of the coordinates, instead of being just a constant.

7. Jun 16, 2017

### davidge

Ohhh, so this is the key thing
It would be great to do that and see the magic happening. I will perform the calculations later.

8. Jun 16, 2017

### Staff: Mentor

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.