Pi and Einstein field equations

ignoramus1
I understand the difference betw mathematical and physical pi. I also understand that in non-Euclidean space the value of pi would differ depending on a surface's deviation from flatness.

But is there a different symbol for physical pi, to distinguish it from mathematical pi? Because I don't understand how pi works in the Einstein field equations (e.g., see: http://en.wikipedia.org/wiki/Einstein_field_equations)

does the pi in those equations have a variable value? in other words, if you were to replace pi with its mathematical value, would you get those equations wrong? ( or is there some part of the equations that makes pi change depending on curvature? or is there a different set of equations where pi variability is applicable??)

I am trying to figure out how to properly answer someone who claims: "But pi shows up in Einstein's field equations, so how can you say pi doesn't quite apply in relativity?"

I hope I have made my dilemma clear. I'd so very much appreciate help with this. I'm trying to explain these concepts to somebody (a case of the one-eyed leading the blind), and I don't want to misinform them.

Thank you.

Staff Emeritus
Gold Member
The symbol $\pi$ always stands for the mathematical pi. It would not make sense, for a couple of reasons, to define a notation for the ratio of circumference to diameter in curved spacetime. If someone were to define such a notation, it would have to be a tensor, not a scalar, because its value would depend on the orientation of the surface. In addition, it would depend on the size of the circle. Once you were done accounting for these issues, I think the tensor you'd end up defining would just be the Ricci curvature tensor. In flat spacetime, the Ricci curvature tensor equals zero, not $\pi$.

ignoramus1
The symbol $\pi$ always stands for the mathematical pi. It would not make sense, for a couple of reasons, to define a notation for the ratio of circumference to diameter in curved spacetime. If someone were to define such a notation, it would have to be a tensor, not a scalar, because its value would depend on the orientation of the surface. In addition, it would depend on the size of the circle. Once you were done accounting for these issues, I think the tensor you'd end up defining would just be the Ricci curvature tensor. In flat spacetime, the Ricci curvature tensor equals zero, not $\pi$.

I understand the gist of what you are saying, thank you. This answers the part of my question dealing with whether it makes sense to have 2 different symbols for pi.

I am still struggling to understand how $$\pi$$ works in the Einstein field equations, i.e., do those equations work in nonEuclidean space? "Of course, you birdbrain," you might say, but I need confirmation.

And BTW, though this is entirely not in the context of anyone's homework or schoolwork of any kind, I am wondering if I should have posted this question under a less specialized topic? Thanks.

Homework Helper
pi is the traditional 3.141592653... nothing more, nothing less. 8piG/3 is the constant of proportionality between the Einstein tensor and the the stress tensor. Nothing more, nothing less.

Yes the EFE work in non-Euclidean space.

IttyBittyBit
ignoramus1, I think part of your confusion stems from your associating pi with geometry - specifically, the ratio of the circumference of the circle to it's diameter. It is true that pi was first investigated in a geometrical context. However, in mathematics (and physics) pi is an extremely important constant which appears nearly everywhere and transcends it's usual geometric interpretation.

When you see pi in an equation, you might be tempted to think that there is some connection between that equation and the circle. This is not the case.

yuiop
Pi can be defined in ways that do not directly involve circles such as Pi = -sqrt(-1)*log(-1).

As others have mentioned, Pi is just a numerical constant.

Imagine a small creature on the surface of a sphere of sufficient size that the surface appears superficially flat to the creature. When the creature compares the circumference of a circle drawn on the surface of the sphere to the radius it discovers that the circumference is greater than 2*Pi*r. This does not mean the value of Pi has changed, but simply that the surface is not flat. A value of 2*Pi for the ratio of the circumference to the radius of a circle is only valid in Euclidean geometry. It is not the value Pi that has changed on the surface of a sphere but the geometry. If the same creature measured the inside angles of a triangle drawn on the surface of the sphere it would find the total is greater that 180 degrees and again this is a result of the non-Euclidean geometry and not due to a change in the value of a degree. In Relativity the circumference of a spinning flat disc is greater than 2*pi*r and this is due to the curvature of spacetime and again the failure of 2*Pi*r is not due to a change in the value of Pi but due to the non-Euclidean geometry of a relativistically spinning disc. In General Relativity gravity distorts space in a way that also results in non-Euclidean geometry.

The above is not meant to be a rigorous definition, but I hope it gives you the general idea.

nolanp2
You could also understand how the mathematical pi is still useful physically in a curved spacetime, for example when integrating over space in polar coordinates: We assume that the integration of an angle over a whole circle is given by 2 \pi, and you might wonder that since this wouldn't equate to the perimeter of a unit circle in curved space, we would need a new integration range.

However, since we know that we can always choose a locally flat coordinate system, we should also be able to argue that turning around by 2 \pi radians on a point still amounts to completing a full circle. Thus mathematical pi can still have a sensible physical interpretation in GR.