Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If one draws a circle on a sphere and measures the ratio of the diameter to the circumference, that value varies depending on the diameter of the circle compared to the diameter of the sphere it is drawn on (for a circle much smaller than the sphere it's drawn on, the ratio will converge to $\pi$, whereas for a circle the same size as the sphere, the ratio will be 2).

Does the same or similar hold for curved 4-dimensional space? As I approach a massive body, will the ratio of the circumference of a circle to its diameter change?

I found a number of sometimes conflicting statements online, including

http://www.physicsforums.com/archive/index.php/t-9869.html

http://mathforum.org/library/drmath/view/55198.html

http://www.last-word.com/content_handling/show_tree/tree_id/2339.html

share|improve this question
    
as the circle radius tend to zero, the ratio between circumference and radius will always tend to the planar value known as $\pi$, because any curved spacetime is always a minkowski plane in a small enough neighbourhood. But at a fixed radius of the circle, this ratio will change as you approach a curved spacetime region –  lurscher May 9 '12 at 16:22
    
I think Gauss suggested that curvature of space could be observed by measuring a very large triangle and checking to see if its angles added up to 180°. But it occurs to me now that gravitational lensing is a demonstration of exactly this effect. –  MJD May 9 '12 at 17:28

1 Answer 1

up vote 4 down vote accepted

The links you give seem to be arguing about whether the value of $\pi$ changes in a curved spacetime. It doesn't. $\pi$ is a constant defined as the ratio of a circle's circumference to diameter on a Euclidean plane.

However the ratio of a circle's circumference to diameter does change in a curved 4D spacetime. It will be greater than $\pi$ if the curvature is negative and less than $\pi$ if the curvature is positive (as in the example of your sphere).

As lurscher has commented, if you make the region of spacetime you're looking at very small the ratio will always tend to $\pi$, again just as in the example of your sphere.

share|improve this answer
    
What does it mean for space to have a negative curvature? If I envision the sphere (possibly too simple a model for 4D spacetime), drawing circles on the inside or outside of the sphere and then measuring the diameter (along the surface of the sphere) should result in the same ratio. –  Eric J. May 9 '12 at 19:05
    
@Eric Try this Wikipedia article. The short (and informal) summary is that if as you move away from a point in various directions, the surface always curves the same way, then the surface has positive curvature at that point; if it curves one way in some directions and the other way in other directions it has negative curvature. A surface may have different curvature at different points: a torus has some areas of positive curvature, some of negative curvature, and some flat areas. –  MJD May 9 '12 at 19:35
1  
An example of a negatively curved space is near the minimum of a saddle. The curvature changes elsewhere, though, you can embed a space of uniform negative curvature in Euclidean 3 space. –  Ron Maimon May 10 '12 at 2:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.