Understanding the Metric of a Curved Sphere: Exploring the Formula

[Niles]

Hi

This is not a homework question, but a question related to how to use the formula for the metric of a curved sphere (cylindrical coordinates):

<br /> ds^2 = dr^2 + R^2 \sin ^2 (r/R)d\theta ^2 <br />

The thing is - I haven't got the slightest idea of how to use it, and unfortunately, my astrophysics-book doesn't make a great effort out of explaining this. The Web hasn't got anything on this formula as well, so this is my last resort.

First, ds is the infinitesimal distance between two points (r, theta) and (r+dr, theta + d theta), right? If this is correct, then what is dr? Isn't that also the distance between the two points?

I really hope you can explain this to me.

 

[tiny-tim]

… the formula for the metric of a curved sphere (cylindrical coordinates):

ds^2 = dr^2 + R^2 \sin ^2 (r/R)d\theta ^2

Hi Niles!

That looks weird …

Spherical coordinates have:

ds^2 \,=\,dr^2\,+\,r^2d\theta^2\,+\,r^2sin^2\theta d\phi^2​

and cylindrical coordinates have:

ds^2\,=\,dr^2\,+\,r^2d\theta^2\,+\,dz^2​

Your formula doesn't look like either of them.

Also, how can r/R be an angle? And what's a "curved" sphere?

 

[Niles]

Ah man, this has nobody ever heard of this formula :-)?

Actually, I was wrong when saying a curved sphere. We are just finding the lengths between points on a sphere.

 

[nrqed]

Ah man, this has nobody ever heard of this formula :-)?

Actually, I was wrong when saying a curved sphere. We are just finding the lengths between points on a sphere.

But if you are on a sphere, dr should be equal to zero. What is the meaning of dr here??

 

[tiny-tim]

Hi Niles!

Is this something to do with black holes?

Is R the Schwarzschild radius?

 

[Niles]

Nono, not at all - R is the radius of the sphere, on which the points (r, theta) and (r+dr, theta + d theta) are.

Well, one of my questions regarding the equation is actually, why dr isn't 0 all the time. I thought this was a commenly known equation is cosmology?

 

[tiny-tim]

Hi Niles!

Got it!

R is the radius, r is "vertical arc-length".

So putting r = Rφ, where φ is a standard spherical coordinate, we have:

dR² = 0, dr² = R²dφ²,
​

and so the usual spherical coordinates:

ds^2 \,=\,dR^2\,+\,R^2d\phi^2\,+\,R^2sin^2\phi d\theta^2​

become:

<br /> ds^2\,=\,0\,+\,dr^2\,+\,R^2 \sin ^2 (r/R)d\theta ^2 <br /> ​

 

[Wiemster]

First, ds is the infinitesimal distance between two points (r, theta) and (r+dr, theta + d theta), right? If this is correct, then what is dr? Isn't that also the distance between the two points?

r, and theta are just two coordinates, sufficient to assign to each point a value of r and theta. Now how these coordinates relate to distances is given by ds. E.g. if you have a normal Cartesian coordinate system and you want to know the distance between the origin and a point (dx, dy, dz) infinitesimally far away then ds^2 = dx^2 +dy^2 +dz^2[/tex]. Now if you use angles as coordinates you get more elaborate relations between the coordinates and the distance, like the ones pointed out by tim. YOu can use almost any coordinates you like really. So that is what your relation gives, a distance function for you coordinates. To get more feeling for these coordinates you will probably have to look at their relation to coordinates you are used to (Cartesian, spherical...).<br /> <br /> Note dr is not the distance, ds is. dr is only the distance if dtheta is zero.

 

[jambaugh]

Let me add how you might use this equation.

Suppose you are tracking someone traveling on the Earth using radar. You have their distance from you r(t) and their angle theta(t) as functions of time.

If you want their speed you need to calculate ds/dt using this metric.
<br /> ds/dt =\sqrt{ \left(\frac{dr}{dt}\right)^2 + R^2 \sin ^2 (r/R)\left(\frac{d\theta}{dt}\right) ^2} <br />

Similarly if you wanted to find the total distance they traveled you would integrate the arclength using this metric in a path integral:

S = \oint_{p_1}^{p_2} ds

This is the geometric meaning of the metric. You may see how it would generalize to arbitrary coordinate systems and arbitrary metrics. The key to calculating the metrics in various coordinate systems is that these quantities (i.e. the geometry) should always be independent of your choice of coordinate systems. In general the squared differential ds^2 will be some homogenous quadratic function of the differentials in the coordinates. This is most succently expressed with a matrix (2-tensor) form.

 

[Niles]

Thanks guys. It's so good to know I can get help here.

 

[Niles]

I have a question regarding what Tiny-Tim wrote in his first post - in spherical coordinates we have:

<br /> ds^2 \,=\,dr^2\,+\,r^2d\theta^2\,+\,r^2sin^2\theta d\phi^2<br />

So dV in spherical coordinates come from ds^3?

 

[jambaugh]

I have a question regarding what Tiny-Tim wrote in his first post - in spherical coordinates we have:

<br /> ds^2 \,=\,dr^2\,+\,r^2d\theta^2\,+\,r^2sin^2\theta d\phi^2<br />

So dV in spherical coordinates come from ds^3?

No! ds is the differential of a path length or distance displaced when you displace the coordinates by dx dy and dz or dr, d\theta or what-ever.

One usually writes the square of ds since that is a quadratic in the differentials of the coordinates. But cubing it only gives you the cube of the arc-length element.

To get area or volume elements you need to write the metric as a matrix and take its determinant the area/volume will be the square root of this determinant times the product of the coordinate differentials. When it is diagonal like here then the determinant will be just the product of the diagonal terms. Hence (since we have a 2-surface) the differential area is:

dA =\sqrt{ab}

where ds^2 = a\cdot dr^2 + b\cdot d\theta^2

Hence: dA =\sqrt{ R^2\cdot \sin ^2 (r/R)}\cdot dr\cdot d\theta=R\cdot\sin(r/R)\cdot dr \cdot d\theta

Pick up a book introducing differential geometry if you get a chance. It may seem complicated at first but it is very systematic and once you get used to the notation you can parse general cases. The actual details can get a bit tedious.