QM + GR = black holes cannot exist

Once again, nothing that comes within 1.5 Schwarzchild radius of the EH, will never escape the BH's clutches, for the reasons given.
What interests you about me is nether here nor there, nor do I really care.
What I have stated is accepted mainstream cosmology.

 

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Hi Bruce. I've identified your problem.

By claiming that Galileo is "right" what you are maintaining is that fall times are independent of mass. In other words, from 10 meters above the Earth:
A golf ball would take 1.4 seconds
A bowling ball would take 1.4 seconds
A car would take 1.4 seconds
A building would take 1.4 seconds
The moon would take 1.4 seconds
The sun would take 1.4 seconds
A black hole would take 1.4 seconds

I'm confident that between my identification of your logical fallacy and the mathematical analysis I provided in the other thread you'll be able to reconcile this problem on your own.

Good luck!

 

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And of course I'm speaking of at and within the Schwarzchild radius, so why you happen to claim ,<c is beyond comprehension.
I am not and was not speaking of outside the Schwarzchild radius.

 

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False. You said

1.5 Schwarzschild radius is at "at and within the Schwarzschild radius". The area outside of R_s but less than 1.5*R_s can be escaped with a velocity < c.

 

Again, the original statement by RJBeery is totally wrong.
Nothing, within the Photon sphere [1.5 Schwarzchild radius] will ever escape the BH's clutches.

 

From you they're nonexistent. Cranks get to have it that way since they're never wrong.

 

This is false. You are confusing "no stable orbits" with "no escape". I won't point this out to you again.

 

Bruce, did you see my post where I identified your problem?

 

http://casa.colorado.edu/~ajsh/orbit.html

http://casa.colorado.edu/~ajsh/approach.html

The photon sphere

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1.5 Schwarzschild radii from the singularity.

This is the location of the innermost unstable circular orbit. To maintain circular orbit at this radius requires going at the speed of light. In particular light itself can orbit the black hole in circular orbits at this radius. For this reason, this location is sometimes called the photon sphere.

Inside 1.5 Schwarzschild radii, no circular orbits exist, stable or unstable: all free fall orbits fall into the black hole. To maintain orbit inside 1.5 Schwarzschild radii without falling in, it would be necessary to keep rockets constantly on burn.

Photons do not orbit in circles at the horizon, just skimming the surface. The place where photons orbit in circles is the photon sphere, at 1.5 Schwarzschild radii. Photons emitted at the horizon fall in; except that if a photon is emitted exactly vertically outward exactly at the horizon, then it will hover at the horizon, not moving at all.

http://casa.colorado.edu/~ajsh/singularity.html

 

News flash! Scientist recently rediscovered that there is a black hole at the center of almost every galaxy! The one in the Milky Way was discovered by having enough mass to be sufficient to be a suppermassive black hole by the orbits of other stars, and this massive object is not even visible being in our own galaxy!

 

Yes, you've identified the difference, but we aren't talking about free-falling. We're talking about theoretically escaping the black hole, under acceleration if needed, which is possible > R_s. What do you think the R_s signifies if the "actual" sphere-of-no-return is the photon sphere?

 

No, I'm saying nothing will ever escape a BH that wanders inside the photon sphere.
Now I'm not sure how much pedant or crank nonsense you may want to apply to gain credibility,

 

but the fact remains, unless something can obtain "c"it will never escape.

 

This is false. You apparently don't understand the difference between a free-fall and an object under acceleration. Ask one of your peers, I simply can't continue to tell you that you're wrong unless you consider the possibility that you are.

 

By definition the Schwarzschild radius is the distance from the center of a black hole where the escape velocity becomes greater than "c".

 

The EH signifies the parameter within which nothing including light escapes.
The photon sphere signifies the parameter at which light can orbit the BH.
Light can still escape within this parameter

 

So a rocketship able to travel at .99999c in a radial escape path from the black hole could not penetrate the photon sphere?

 

I understand the difference fairly well. If you can show me how something within the photon sphere [other then light] that can accelerate to "c" then I'll rethink the situation.

 

You don't need to accelerate to "c" to escape the photon sphere. You only need to accelerate to sqrt(2)c. You would need to accelerate to "c" at the event horizon.

 

Light emitted at just near but outside the EH, will arc back and secumb to the BH, unless emitted directly radially away....Even then it never escapes but appears to hover forever never getting away and never secumbing.

Back later, I must be off...
I'm listening.