# Titor's Donut Shaped Singularity

Discussion in 'John Titor's Legacy' started by PaulaJedi, Mar 14, 2016.

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Regarding John Titor's Donut Shaped Singularity:

I am just thinking out loud. I am not claiming to be a scientist, mathematician, or even claiming to be correct. The following article is simply my own speculation. I could be very wrong. Make of it what you will. I do hope you find this interesting. Feel free to share your thoughts.

http://www.microsingularity.net/using-math-for-time-travel/

2. ### MartianSenior Member

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I'm not an expert on black holes, but it seems to me that they needn't be small, at least not when they're first formed. Also, they don't even need to be dense at first, if there's enough matter. The escape velocity is determined by equating kinetic energy to gravitational energy and solving for velocity:

½mv² = GMm/r
½v² = GM/r
v = sqr(2GM/r)

Then assume the escape velocity is the speed of light:

v = c = sqr(2GM/r)

We can then find the Schwarzschild radius by solving for r:

r = 2GM/c²

Next assume that we have a giant, homogeneous sphere of density p. We can define it as:

p = M/V

where the volume V is defined as:

V = (4/3)*pi*r³

If we choose a value for p, we can then solve for the radius at which it forms an event horizon. This would be a sphere that just barely forms a black hole.

p = M / ((4/3)*pi*r³)
M = ½rc²/G
p = ½rc² / ((4/3)*G*pi*r³)
p = (3/8)c² / (G*pi*r²)
r² = (3/8)c² / (p*G*pi)
r = sqr((3/8)c² / (p*G*pi))

It comes out to a huge number, but it's not infinite for any nonzero density. It makes me wonder how much interstellar hydrogen, etc, is required for a black hole to spontaneously form.

*shrug* I'm bored. lol

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All this math makes me realize what CERN could be doing -- creating artificial black holes via math. In other words, perhaps with math and physics we can create a singularity that doesn't suck us all into oblivion. But I really have no idea if that is possible or how to do it.

BTW, a sphere doesn't need to be giant to be dense, but you've already shown that above M/V -- mass doesn't indicate size.

Just thinking out loud again.

4. ### MartianSenior Member

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I was just rambling, as well. I do that sometimes.

At CERN, they'd most likely use relativistic mass increase which happens near light speed in order to create a black hole. The equation is:

m = m0 / sqr(1 - (v/c)²)

where m is the relativistic mass, m0 is the rest mass (which is constant), v is the velocity of the particle, and c is the speed of light.

The shape and size of the particle also change, but it should be possible to calculate the velocity at which it has the right mass and volume to form a black hole (I'd have to look up the other equations). This is because the mass approaches infinity as the particle velocity approaches the speed of light.

The problem they'd face after formation of a tiny black hole would be containment, and the likely solution would be to collide it with another black hole of equal mass.

This is all speculation though.

Last edited: Mar 15, 2016

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I really hope you`re right Martian, i always thought when 2 black holes collide they cant escape each others gravity, and so they merge to become one big black hole..OMG

6. ### MartianSenior Member

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Ok, Lorentz contraction is the name for the observed decrease in length of an object in the direction of motion near the speed of light. The length L is defined as:

L = L0 * sqr(1 - (v/c)²)

where L0 is the object's length at rest.

If we assume a particle is initially spherical, it will become ellipsoidal at relativistic velocities. The volume of an ellipsoid with semi-principal axes of lengths a, b, and c is:

V = (4/3)*pi*a*b*c

If we assume that the particle remains symmetric about one axis, it will have semi-principal axes of lengths a, a, and L. This means its volume will be:

V = (4/3)*pi*a²*L = (4/3)*pi*a²*L0*sqr(1 - (v/c)²)

The important thing to note here is that the volume decreases near light speed, so that should actually make it easier to form a black hole. However, this length contraction is only from the perspective of an outward observer, so it might be irrelevant.

7. ### MartianSenior Member

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If we assume that length contraction is irrelevant to the formation of a black hole, we can calculate the velocity required to form a black hole in a particle accelerator.

c² = 2*G*M/r
M = 0.5*r*c² / G
M = M0 / sqr(1 - (v/c)²)
sqr(1 - (v/c)²) = 2*M0*G / (r*c²)
1 - (v/c)² = 4*M0²*G² / (r²*c⁴)
c² - v² = 4*M0²*G² / (r²*c²)
v² = c² - 4*M0²*G² / (r²*c²)
v = sqr(c² - 4*M0²*G² / (r²*c²))

I'll need to look up the values for some well-known particles to see if this would work (the stuff inside the square root has to be positive for it to give a valid result).

8. ### MartianSenior Member

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It looks like it's simply not practical, at least based on the properties of protons and electrons. Oh well. lol

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So this means i can sleep easier at nights now?

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