# Derivation of a metric

1. Apr 13, 2010

### Matterwave

1. The problem statement, all variables and given/known data

Show that the metric for a 3-sphere embedded in 4-space is:

$$ds^2=dr^2+R^2 sin^2(\frac{r}{R})(d\theta^2+sin^2\theta d\phi^2)$$

r is the distance from some "pole" and R is the radius of curvature of the 3-sphere.

My question:

I showed this by using the transformations (as suggested by the professor):

$$w=Rcos\chi$$
$$z=Rsin\chi cos\theta$$
$$x=Rsin\chi sin\theta cos\phi$$
$$y=Rsin\chi sin\theta sin\phi$$
$$r=R\chi$$

So, all I did was differentiate w, z, x, and y implicitly using all the product rules so that:

$$dw=-Rsin\chi d\chi$$
$$dz=R(cos\chi cos\theta d\chi - sin\chi sin\theta d\theta)$$
$$dx=R(cos\chi sin\theta cos\phi d\chi + sin\chi cos\theta cos\phi d\theta - sin\chi sin\theta sin\phi d\phi)$$
$$dy=R(cos\chi sin\theta sin\phi d\chi + sin\chi cos\theta sin\phi d\theta + sin\chi sin\theta cos\phi d\phi)$$

Ok. So, I squared each and set:
$$ds^2=dw^2+dx^2+dy^2+dz^2$$
After a lot of algebra, Lo, and behold I got exactly what my professor asked for!

I was overjoyed.

And then I got to thinking. Nowhere in my solution have I even invoked any condition that I am working on a 3-sphere. In fact, my metric before the coordinate transform is for flat space!

This leads me to think that all I have done is a coordinate transformation and NOT finding the metric of a 3-sphere. I tried the same method in just 2 dimensions going from Cartesian coordinates to Polar coordinates and I in fact got back the flat-space metric in polar coordinates.

So...my questions is...how the heck did I arrive at the metric for a 3-sphere embedded in 4-space WITHOUT even invoking the condition of a 3-sphere? How did I get to this metric by just doing a coordinate transformation!?

I refuse to believe that I just made a mistake somewhere and MIRACULOUSLY I got the right answer...There must be something I'm missing here.

2. Apr 13, 2010

### Fredrik

Staff Emeritus
Shouldn't the metric of $$\mathbb R^4$$ expressed in these coordinates contain a dR2 term? The last step is just to throw that term away, like you would throw away the dz2 from dx2+dy2+dz2 to get the metric for $$\mathbb R^2$$ from the metric for $$\mathbb R^3$$. (This isn't the most rigorous argument in the world, but at least in physics books, I think it's standard to just do these things without explanation).

When you latex sin and cos, write them as \sin and \cos. It will look better.

3. Apr 13, 2010

### Matterwave

lol, thinking this over, I think I get what happened...and it's actually quite obvious.

I assumed R was constant! LOL If I just wanted to do a coordinate transformation, I would have worried about dx/dR, dy/dR, etc terms. Therefore, I restricted myself to a 3-sphere. Also, on the infinitessimal level, the distance in a "straight" line is the same as the distance over a curved surface. Therefore, the derivation is valid.

I see that this is also what Fredrik was saying...but I didn't understand at the time.

I'll keep the latex hint in mind for future reference, thanks. =)