Exploring the Impact of Extreme Gravity on Pi

[ravisastry]

Hi, this thought came to my mind..not sure how correct this is. In extreme gravity, the space curves. Will the value of Pi change in this case, for a circle near such gravity ? pi = circumference/diameter.

 

[Mentz114]

If you draw a circle on the surface of a ball, is pi = circumference/diameter ?

Spatial curvature does change the ratio circumference/diameter and also the total angle in polygons.

 

[arkajad]

In extreme gravity you may even have problems with distinguishing space from time.

 

[D H]

Will the value of Pi change in this case, for a circle near such gravity ? pi = circumference/diameter.

In general the ratio of circumference to diameter for a circle varies with the size of the circle and the nature of the distance metric (which depends on geometry). In Euclidean space and using the Euclidean norm the ratio is constant and is equal to pi by definition. In any other space, or any other metric, the ratio of circumference to diameter may or may not be equal to pi. This does not mean that pi has a different value in that space / with that metric.

 

[bcrowell]

This does not mean that pi has a different value in that space / with that metric.

Except in Indiana, where it equals 3.

 

[ravisastry]

pi is a component of the fine structure constant and what i was thinking was...light follows curved path near high gravity. will the fine structure constant be different near such gravity because of pi ?

 

[Mentz114]

Except in Indiana, where it equals 3.

It is after all, very flat in Indiana.

Isn't the angle all the way around a circle always 2 pi ? And similarly with solid angles ?

 

[bcrowell]

pi is a component of the fine structure constant and what i was thinking was...light follows curved path near high gravity. will the fine structure constant be different near such gravity because of pi ?

If you study D H's #4, you'll see that there are two possible cases. (a) A pi appears in a particular formula because it's the ratio of the circumference of a circle to its diameter. (b) A pi appears there for some other reason.

In case a, it may be reasonable to speculate that the formula would be different in curved spacetime. In case b, not so much.

Do you think your example is case a, or b?

 

[bcrowell]

Isn't the angle all the way around a circle always 2 pi ? And similarly with solid angles ?

Yes. An angle is a local thing, and the local geometry of spacetime is always flat.

 

[pervect]

If you make sure you keep away from any singularities inside your circle, you'll find that for a small enough circle, the flatness of space-time will give you a value of pi for the ratio of circumference to diameter for a small enough circle as well.

If you draw a circle around a cosmic string, though, I'd expect that you'd find the ratio of circumference/diameter being non-pi even for a small circle - as long as the circle encloses the singularity (the cosmic string in this case).

 

[Passionflower]

An interesting, but perhaps no so trivial, question would be:

What is the circumference and area of a unit circle of radius (ruler distance) = 1 with its center at Schwarzschild coordinate r and a Schwarzschild radius of R₀ (we assume theta and phi is 0, so the circle is drawn 'flat')?

If we have a formula we can plot the diversion from pi in terms of the r coordinate and R₀.

Anyone?

Seems to me that to find the area we need to integrate 'slices' of the circle with constant r, the 'slices' are arcs of a circle of radius r.

 

[pervect]

By definition, we know the circumference of a circle at Schwarzschild coordinate R is 2*pi*R, and the area is 4*pi*R^2 - that's how the schwarzschild radial coordinate R is defined. What we don't know is the "radial distance to the center of the black hole". The question is probably basically meaningless. Among other issues, inside the black hole, r is not a spatial coordinate - i.e. if we consider two nearby points (r,t) and (r+dr,t), there is a timelike separation between these points, not a spacelike separation.

 

[Passionflower]

By definition, we know the circumference of a circle at Schwarzschild coordinate R is 2*pi*R, and the area is 4*pi*R^2 - that's how the schwarzschild radial coordinate R is defined. What we don't know is the "radial distance to the center of the black hole". The question is probably basically meaningless. Among other issues, inside the black hole, r is not a spatial coordinate - i.e. if we consider two nearby points (r,t) and (r+dr,t), there is a timelike separation between these points, not a spacelike separation.

Completely agree. I added a graph to show this. If we measure the ruler distance between R and R+1 for decreasing values of R we observe that the ruler distance increases. Also, observe the behavior of the radar distance (total roundtrip time) in an additional graph.

Just in case you wrote this in answer to my question then you completely misunderstood what I was asking, let me try it again but with slightly different words:

For a given Schwarzschild coordinate value R_c draw a unit circle of ruler length 1 around this point (we assume theta and phi is 0, so the circle is drawn 'flat'). What is the circumference and area given a Schwarzschild radius R₀. And obviously R_c > R₀.

But I assume you simply responded to the whole topic at hand.

 

[DrGreg]

(we assume theta and phi is 0, so the circle is drawn 'flat')

I've no idea what you mean by this. If [itex]\theta[/itex] and [itex]\phi[/itex] are both zero, you must be talking about a radial straight line, not a circle.

 

[Passionflower]

I've no idea what you mean by this. If [itex]\theta[/itex] and [itex]\phi[/itex] are both zero, you must be talking about a radial straight line, not a circle.

Aarg...I see my mistake.

I meant [itex]\theta=0[/itex], of course [itex]\phi[/itex] is not!

Ok, perhaps a picture would help.