# Curvature at the origin of a space as described by a metric

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1. Feb 4, 2016

### Whitehole

1. The problem statement, all variables and given/known data
This is a problem from A. Zee's book EInstein Gravity in a Nutshell, problem I.5.5

Consider the metric $ds^2 = dr^2 + (rh(r))^2dθ^2$ with θ and θ + 2π identified. For h(r) = 1, this is flat space. Let h(0) = 1. Show that the curvature at the origin is positive or negative according to whether h(r) starts to turn downward or upward. Calculate the curvature for $h(r) = \frac{sin(r)}{r}$ and for $h(r) = \frac{sinh(r)}{r}$

2. Relevant equations

3. The attempt at a solution
$ds^2 = dr^2 + (rh(r))^2dθ^2 = dr^2 + (r^2h(r)^2)dθ^2$

If I let r → 0 then h(r) → 1 but (r^2h(r)^2)dθ^2 → 0
How can I tell what would be the behavior of h(r) if it will already be gone if I let r → 0.

For $h(r) = \frac{sin(r)}{r}$ and $h(r) = \frac{sinh(r)}{r}$ I know that when I let r → 0 from calculus the answer would be 1 and -1 respectively, but I don't know what to do with the "show" part. Can anyone give me hints?

2. Feb 4, 2016

### Orodruin

Staff Emeritus
You do not compute the curvature by simply looking at one component of the metric. You need to actually compute what the curvature is.

3. Feb 4, 2016

### Whitehole

I get your point but Zee did not show how to compute the curvature and stated in this section that I have to wait until some later chapters, that is why I'm confused why is that question in this chapter. I think maybe he wants me to compute the limit of h(r) as r → 0? In which h(r) are the givens above.

4. Feb 4, 2016

### George Jones

Staff Emeritus
Have you looked at the bottom of page 65 and top of page 66, and at appendix 1 for section 1.5?

5. Feb 5, 2016

### Whitehole

I have a question on appendix 1. He stated that the curvature is given by $R = \lim_{radius\rightarrow 0} {\frac{6}{radius^2}(1-\frac{circumference}{2πradius})}$.

Now from equation (4) in page 65, he gave the metric $ds^2 = dρ^2 + \sin^2(ρ)dθ^2$. To know if the space is curved or not. We apply the formula above.

The radius is $ρ$ so, $ρ = \int_0^ε\, dρ = ε$. I get that, but for the circumference, in flat space it should be $(radius)(dθ)$, what does $\sin(ρ)$ stand for here? It seems that it is representing the "radius" because he showed that, $circumference = \int_0^{2π}\, \sin(ε)dθ = 2π\sin(ε)$

Also, by applying the formula for curvature, $R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2π\sin(ε)}{2πε}) = \frac{6}{ε^2}(1-\frac{\sin(ε)}{ε}) = 1}$. How can it be 1 if $\lim_{radius\rightarrow 0} {\frac{\sin(ε)}{ε}} = 1$, so it should be 0.

6. Feb 5, 2016

### Orodruin

Staff Emeritus
It is not representing the radius, it represents how far you will move if you change $\theta$ by an amoun $d\theta$. Note that, since your space is curved you no longer have the direct connection between the radius and the circumference.

7. Feb 5, 2016

### Whitehole

Oh, then if I picture it in my mind, $ρ$ is the radius (0 to ε) which is a straight line? And $\sin(ε)$ is somehow the distance from $0$ to $ε$ but curved? So technically $\sin(ε)$ has a "longer" length than $ρ$?

8. Feb 5, 2016

### Whitehole

Oh, then if I picture it in my mind, $ρ$ is the radius (0 to ε) which is a straight line? And $\sin(ε)$ is somehow the distance from $0$ to $ε$ but curved? So technically $\sin(ε)$ has a "longer" length than $ρ$?

9. Feb 5, 2016

### Orodruin

Staff Emeritus
No, $\rho$ is the radius. From $\rho =0$ to $\epsilon$, the distance is $\epsilon$. The quantity $\sin(\epsilon)d\theta$ is the distance travelled along the circle of radius $\epsilon$ when you change the $\theta$ coordinate by $d\theta$. If you go around a full lap, you will therefore get a distance $\sin(\epsilon)2\pi$, not $\epsilon 2\pi$. This is the entire point.

10. Feb 5, 2016

### Whitehole

I know, but how can I picture $\sin(ε)$ only, without the $dθ$. For example, in flat space the circumference is $rdθ = r(2π)$, and I know $r$ is the distance from $0$ to $ε$. For curved space, the circumference is $\sin(r)dθ$, but where is this $\sin(r)$?

11. Feb 5, 2016

### Orodruin

Staff Emeritus
What do you mean by "where it is"? It is a property of the manifold. It is shorter around the circle than $2\pi r$. It is a property of the circle.

12. Feb 5, 2016

### Whitehole

What I mean is that, for a flat space the circumference is $r(2π)$, and $r$ is "the distance from the origin to some arbitrary point"". But for a curved space the circumference is $\sin(r)(2π)$, and $\sin(r)$ is...???

I know that this is a property of the manifold, etc. I'm just curious if there is a way to visualize this.

$R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2π\sin(ε)}{2πε}) = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{\sin(ε)}{ε})} = 1}$. How can it be 1 if $\lim_{radius\rightarrow 0} {\frac{\sin(ε)}{ε}} = 1$. Shouldn't it be $0$?

13. Feb 5, 2016

### Orodruin

Staff Emeritus
I suggest thinking of the surface of a sphere and how the radius and circumference will be related there.

No, you are completely neglecting the $1/\epsilon^2$ outside the parenthesis! If you struggle with this type of limits I suggest solidifying your calculus before attempting calculus on manifolds.

14. Feb 5, 2016

### Orodruin

Staff Emeritus
Just to clarify, it is not sin(r) for a general manifold. The function will be different for different manifolds. It may even be different for different points on the same manifold.

15. Feb 5, 2016

### Whitehole

Got it, L'hopital's rule, but how can I apply it to h(r) in post#1? h(r) is completely arbitrary.

$radius = \int_0^ε\, dr = ε$, $circumference = \int_0^{2π}\,εh(ε)dθ = 2πεh(ε)$

$R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2πεh(ε)}{2πε})} = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-h(ε))}$

Last edited: Feb 5, 2016
16. Feb 5, 2016

### Orodruin

Staff Emeritus
You got the circumference wrong.

17. Feb 5, 2016

### Whitehole

I've edited my post. Sorry for the carelessness. But still, h(r) is arbitrary, though it is given that h(0) = 1.

18. Feb 5, 2016

### Orodruin

Staff Emeritus
Yes, this is given, so what is the limit?

19. Feb 5, 2016

### Whitehole

By applying l'hospital's rule twice. $R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-h(ε))} = -3h''(ε)$

So depending whether $h''(ε)$ is positive or negative, it will tell me if the curvature is positive or negative.

20. Feb 5, 2016

### Orodruin

Staff Emeritus
Almost, a limit in which $\epsilon \to 0$ cannot depend on $\epsilon$ (it was what was approaching zero!). Replace $h''(\epsilon)$ by $h''(0)$. Also note that $h'(0)$ needs to be zero for this limit to be finite. If not you are not dealing with a smooth manifold.

Edit: Also note that if $h''(0) = 0$, then the manifold is flat at r = 0 and this does not necessarily imply that it is flat everywhere. An example of this would be $h(r) = 1 + r^4$.