Einstein, Curved Space, and Pi

Date: 10/09/1999 at 13:35:35
From: Steve
Subject: Pi and curved space...

I am in 8th grade and recently became interested in Einstein's General 
and Special Relativity theories. (Just so you know my question is NOT 
about Einstein's Relativity but how it relates to pi).

In Einstein's general relativity, space is described as being curved. 
If space is curved, then Euclidean geometry doesn't apply. And since 
one of the parts to Euclidean Geometry is that the relation between a 
circle's circumference and its diameter (pi) is always 
3.1415926535897923...  doesn't that mean that the value of pi changes 
when space is curved? Doesn't that also mean that sometimes the value 
of pi can actually be rational? Do you know of any methods to 
calculate pi in curved space?

Thanks for your help.

Date: 10/10/1999 at 21:49:27
From: Doctor Rick
Subject: Re: Pi and curved space...

Hi, Steve.

In general relativity, it is space-time (4-dimensional "space"), not 
just space, that is curved. But that concept is very hard to grasp 
(I'm not sure I can really picture it), so we get at the idea by 
considering curved space.

Even curved 3-dimensional space is hard to picture. It's easier to use 
a 2-dimensional example. The surface of a sphere is one good example; 
it has a constant curvature. But if you examine what happens to 
circles on the surface of a sphere, you may be surprised. What you'll 
find is that pi doesn't just have a different value, it actually has 
no meaning.

Pi is defined as the ratio of the circumference of a circle to its 
diameter. In flat 2-dimensional space (Euclidean geometry), it can be 
proved that this ratio is the same for every circle. Once you know 
this, you can calculate the value of the ratio - not exactly, but to 
any desired precision. You can also prove that pi is also the ratio of 
the area of a circle to the square of its radius, and so forth.

What happens on the surface of a sphere? Instead of straight lines, we 
have "geodesics" - the shortest distance between two points without 
leaving the surface. A geodesic on the surface of a sphere turns out 
to be a portion of a great circle - a circle, like the equator or a 
line of longitude on a globe, whose center is at the center of the 
sphere.

Look at a globe. It probably has a set of circles of different radii, 
all centered at the North Pole. These are lines of latitude. The line 
at 45 degrees of latitude, for instance, has a radius that is the 
portion of any line of longitude between 45 degrees and the North 
Pole. How long is this radius? It's half the distance from pole to 
equator, or 1/8 of the full line of longitude. The length of a line of 
longitude is the same as the length of the equator. Thus the circle's 
diameter is 1/4 of the length of the equator.

How long is the circumference of the circle? A little geometry will 
convince you that it is the length of the equator divided by the 
square root of 2.

Now, what is the ratio of the circumference of this circle to its 
diameter? It's 1/sqrt(2) : 1/4, or 2*sqrt(2) = 2.8284... That's 
noticeably less than pi.

But does this mean that "pi" on the surface of a sphere is 2.8284? No. 
Consider another circle on the globe, the equator. Its radius is 1/4 
the length of the equator. Its circumference is the length of the 
equator. So the ratio of the circumference of this circle to its 
diameter is exactly 2.

You might say that you were right, it is possible for "pi" to be 
rational in a curved space. But if "pi" depends on the radius of the 
circle, then the ratio of circumference to diameter is not fixed, and 
if the ratio isn't constant, there is no meaning in the definition of 
pi as THE ratio of circumference to diameter.

That's a pretty interesting result, even though it's a negative one. 
What it says is that we shouldn't take even the existence of pi for 
granted: there are geometries in which there is no such thing as pi.

I went a little fast at a few points. If you can't figure out what I 
am saying, please write back. Understanding curved space, even in 2 
dimensions, is a challenge.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/