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The value of $\pi$, or the circumference divided by the diameter of a circle, is known with absurd precision, but I want it to be 3.

The circumference around a black hole outside the Schwarzschild radius is knowable. From the stationary frame outside of that radius, so is the diameter, but it is clear that we cannot traverse that diameter - in essence, it is infinite from the perspective of a traveler. In this instance, $\pi = 0$.

Can we do something like a line integral in GR to find the traversal diameter of a constant density sphere? If so, can we solve for the mass, $M$, and radius where $\pi' = 3$, $r_{3}$, in terms of normal things like density, $\rho$, and other universal constants? For the solution, assume that an instance of $\pi$ has the Euclidean value. The solution radius may be inside or outside the sphere.

In response to the comments, a 2D demonstration.

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closed as off-topic by Norbert Schuch, M. Enns, Cosmas Zachos, Dvij Mankad, Kyle Kanos Apr 13 at 20:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "We deal with mainstream physics here. Questions about the general correctness of unpublished personal theories are off topic, although specific questions evaluating new theories in the context of established science are usually allowed. For more information, see Is non mainstream physics appropriate for this site?." – Norbert Schuch, M. Enns, Cosmas Zachos, Dvij Mankad, Kyle Kanos
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    $\begingroup$ How about $3.2$ instead Indiana Pi Bill, Numberphile $\endgroup$ – MannyC Apr 13 at 15:38
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    $\begingroup$ $\pi$ is a dimensionless ratio of circumference to diameter. It is what it is, and it cannot be arbitrarily redefined. $\endgroup$ – David White Apr 13 at 16:03
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    $\begingroup$ What happened to old good 22/7 ? :-) $\endgroup$ – Poutnik Apr 13 at 18:22
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    $\begingroup$ Why the word "again" in the title? $\endgroup$ – Steeven Apr 13 at 18:37
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    $\begingroup$ Great catchphrase, I mean, title. $\endgroup$ – Avantgarde Apr 13 at 19:02
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If you see a π in a general relativity equation it is still the good old π. Ιn general relativity we do not use a different π, instead we say the circumference is no longer 2πr.

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  • $\begingroup$ Clearly, you did not read the question or chose to ignore its contents. $\endgroup$ – user121330 Apr 14 at 1:11
  • $\begingroup$ I gave the best answer I had, if you have a better one feel free to write your own $\endgroup$ – Yukterez May 11 at 1:10
  • $\begingroup$ I get it. Reading is hard. $\endgroup$ – user121330 May 11 at 22:42

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