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.


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

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  • 6
    $\begingroup$ 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

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.

  • $\begingroup$ 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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