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Q & A: variable pi?

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Most recent answer: 05/16/2013
If gravity warps space-time then would the ratio between the diameter and circumference of a circle near a large mass be different to pi? Or, to put it another way, is pi a variable rather than a constant? Many thanks, Greg
- Greg Cowan (age 13)
Nelson, New Zealand

The answer to your first question is definitely yes. For the Earth, it comes out that the circumference is about an inch smaller than pi*diameter.

The answer to your second question is no. There's no need to make "pi" itself variable. It's well-defined as the ratio of the circumference to the diameter on a flat part of space. It just isn't equal to that ratio on other parts of space. Pi stays the same, but the circumference/diameter ratio depends on the location and size of the circle.

Mike W.

(published on 08/22/2008)

Follow-Up #1: pi in physics

Does that mean that one of the equations that use pi might suppouse to have the ratio of circumfirence to diameter instead and that the person who worked out the equation was wrong? For instance, the equation (uncertancy in position) * (uncertaincy in (mass * velosity)) = h/2pi might it mean istead the ratio of C and D. If it did, then might a large gravitational feild change the accuracy of measuring particles? Could there be apsolute cirtancy in black holes, but noone could communicate this information to the outside universe? Thankyou for your time and please try to answer me, Joe
- Joe Bishop (age 14)
No, there's no error involved in various mathematical uses of pi, which shows up in all sorts of places in mathematical relations used in physics. It just turns out that the original place where pi turned up in math, Euclidean geometry, is not the actual geometry of our universe.

As for the pi in that version of the Heisenberg uncertainty relation, it comes from  a property of sinusoidal waves. It is true that the relations between position and momentum become more complicated in regions where the deviations from Euclidean geometry are strong. In fact, we do not have an adequate theory of the regime where  quantum mechanics and gravity need to be integrated. Any connections between black hole physics and the uncertainty that arises from the 'measurement' processes in quantum mechanics are speculative.

Mike W.

(published on 05/16/2013)

Follow-Up #2: spin and geometry

The difference between the expected D/C is likely due to the Earth's angular momentum and not significantly affected by any gravitational pull.
- Jeff
Champaign, IL, USA
I'm not sure why you say that. The General Relativistic effects have been confirmed in detail in a wide variety of separate experiments on different gravitational sources. The General Relativistic effects of the Earth's spin ("frame dragging") and those of the spatial curvature were both recently confirmed, separately, by the Gravity Probe B experiment.

I suspect you're thinking this is measured by someone sending a signal to themselves around the Earth and forgetting that they're spinning. That effect would be huge compared to one inch (about 200 feet, for a signal going around the equator) and would change signs depending on which way the signal traveled. That's not the type of mistake people in this business make.

Sometimes we talk about sides of physics that are speculative, but this isn't one of them.

Mike W.

(published on 09/17/2009)

Follow-up on this answer.