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

[Whitehole]

Homework Statement

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 $h(r) = \frac{sinh(r)}{r}$

Homework Equations

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?

 

[Orodruin]

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

 

[Whitehole]

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

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.

 

[George Jones]

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

 

[Whitehole]

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

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.

 

[Orodruin]

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.

 

[Whitehole]

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.

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 $ρ$?

 

[Whitehole]

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.

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 $ρ$?

 

[Orodruin]

No, $\rho$ is the radius. From $\rho =0$ to $\epsilon$, the distance is $\epsilon$. The quantity $\sin(\epsilon)d\theta$ is the distance traveled 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.

 

[Whitehole]

No, $\rho$ is the radius. From $\rho =0$ to $\epsilon$, the distance is $\epsilon$. The quantity $\sin(\epsilon)d\theta$ is the distance traveled 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.

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 $\sin(r)$?

 

[Orodruin]

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.

 

[Whitehole]

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.

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.

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

 

[Orodruin]

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

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

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.

 

[Orodruin]

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...?

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.

 

[Whitehole]

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.

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

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

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

 

[Orodruin]

You got the circumference wrong.

 

[Whitehole]

You got the circumference wrong.

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

 

[Orodruin]

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

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

 

[Whitehole]

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

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.

 

[Orodruin]

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 $h(r) = 1 + r^4$.

 

[Whitehole]

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 $h(r) = 1 + r^4$.

Yes, I need to be pedagogic. Thank you very much for your help!