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From my understanding when an object is spinning its inertial mass increases. By Einstein's equivalence principle if the inertial mass increases then also the gravitational mass increases. I have looked this up and this seems to be confirmed in neutron stars in the following paper(even though I don't understand most of the math) https://arxiv.org/abs/1003.5015. I wanted to ask if there is an angular velocity at which any sphere of mass M and radius R would turn into a black hole. Kind of an equivalent to the Schwarzschild radius but for angular velocity.

I tried giving it a go starting with:

$$ c^2 =\frac{2G}{r}*\frac{M_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

then by substituting

$$v=\omega*r$$

I got that the angular velocity for a sphere of mass M with rest mass $M_0$ and radius R to turn into a black hole is:

$$\omega=\frac{c}{R}\sqrt{1-\frac{4G^2M_0^2}{R^2c^4}}$$

Is this formula true in any way? Is it even possible for a sphere to rotate quickly enough for it to become a black hole?

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I've had great interest in this problem, and even decided to derive the equation myself. However, sadly, your equation is incorrect. If you check the units, you'll see they don't add up, not even closely. Here is the correct way to derive the equation..

First off, we take the Schwarzschild radius into consideration, let us start with this..

$$\large{\frac{mc^2}{\sqrt{\frac{\frac{2Gm}{c^2}}{r^2}}}=\frac{mc^2\sqrt{\frac{\frac{2Gm}{c^2}}{r^2}}}{\sqrt{\frac{\omega^2}{c^2}}}}$$

Where $m$ is mass, $c$ is speed of light, $G$ is gravitational constant, $\omega$ is the angular velocity, $\frac{2Gm}{c^2}$ is the Schwarzschild radius, and $r$ is the radius, we can single out $\omega$, and after doing the algebra, we come up with...

$$\omega = \frac{2mG}{cr^2}$$

Hence we have the equation, to calculate how fast something would have to move before it collapses into a black hole. And the dimensions are correct! $(\frac{1}{T})$

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  • $\begingroup$ I checked the units of my equation and they seem to add up, the part in the square root has no units, as it should have, and the multiplier in front is 1/T, which is correct, please let me know where you thought the mistake was so I can give it a triple check :). $\endgroup$ – Francesco Feb 18 '18 at 20:52
  • $\begingroup$ Also could you please explain for the motivation of your starting equation whcih I don't understand. Thank you $\endgroup$ – Francesco Feb 18 '18 at 20:53
  • $\begingroup$ The motivation comes from this problem, someone else had. I based them off of that, physics.stackexchange.com/questions/33916/… however this isn't 100% correct, they're missing the gravitational field inside of the equation, hence that is where I pull it out. symbolab.com/solver/algebra-calculator/… $\endgroup$ – Graham Best Feb 18 '18 at 21:05
  • $\begingroup$ That algebra calculator is equivalent to the work I did when algebraically finding the units. :) $\endgroup$ – Graham Best Feb 18 '18 at 21:05
  • $\begingroup$ If you feel you have learned what you need to know, I recommend marking this answer as the accepted answer!! :D $\endgroup$ – Graham Best Feb 18 '18 at 21:08

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