QM + GR = black holes cannot exist

[tashja]

RJ, here's Prof. Dolan's reply to your follow-up questions:

Prof. Dolan: I shall try to answer your questions:

RJBeery: If the event horizon has "formed", what is it composed of?

Prof Dolan: It is not composed of anything, it is just the boundary of a region of space from which light rays cannot escape.

RJBeery: If we can see the golf balls arbitrarily near R=0, and can do so for eternity, are we not receiving information, for eternity, from the region near R=0?

Prof. Dolan: This depends on where the observer is and how they are moving. Assuming "we" means static observers infinitely far away from the black hole then yes they can receive information for eternity. But the only information coming from R=0 left the origin before the black hole formed. Although they are receiving information for eternity, the amount of information per unit time decreases as time goes on, due to the redshift, and the total amount of information they receive is finite, even over an infinite period of time.

RJBeery:What about when the supposed event horizon has reached a thousand light-years in radius? We continue to see the golf balls near R=0 but their image has been pushed to the event horizon boundary?

Prof. Dolan: Again if "when" refers to time as measured by an observer at infinity then the golf balls at R=0 that can be seen are only ones that were there before the black hole formed, e.g. golf ball no 1 can be seen for all time, but the light is highly redshifted and the image is not pushed to the even horizon boundary, it stays at R=0 but it is the image of the golf ball before the black hole formed.

Instead of using golf balls, suppose we use Rolex watches, so that each image shows the time on the face of the watch. The image of Rolex no 1 at R=0 that is seen at infinity is that of the watch before the black hole formed, even for arbitrarily large times according to the observer. If the black hole forms at T_0 according to the clock of an observer at infinity, then Rolex no 1 can indeed be seen for ever, but the image will always show T_0, never T>T_0. If the black reaches a thousand light years in radius at time T_1000 (when Rolex no 1000 falls in, say) then the observer at infinity cannot see Rolex no 1 showing T_1000, because it is inside the event horizon when it's face shows T_1000. The observer will always see Rolex no 1000 frozen at the event horizon at 1000 light years radius, but the time on its face will also be frozen at T_1000, never later. At late times the observer at infinity sees Rolex no 1 forever, always at R=0 and always showing T_0 and Rolex no 1000 forever, always at R=1000 light years and always showing T_1000 (and both getting more and more red-shifted as the observer's time increases).

Hope this helps
Brian

​

 

[RJBeery]

If I was wrong you'd be writing

dr/dt = (2M/r)^1/2 (Newton, Einstein, and everybody else in physics)

As

dr/dt = (2M+m)^1/2

LOL

Regardless your juvenile analysis it's

dr/dt = (2M/r)^1/2

LOL

Your response is confusing. What are you talking about? Are you just blindly pointing to my "escape velocity" post and trying to apply it to the issue that has you so confused?

Listen dude, if you can accept that the MOON, the SUN and a BLACK HOLE would take less than 1.4 seconds to fall 10 meters to the surface of the Earth, but you somehow think that everyday objects like feathers, bowling balls and double-decker buses all behave according to a completely different set of "Galileo physics" such that they would all take the exact, same amount of time to fall (1.4 seconds)...then you have a cognitive disconnect with which I simply cannot help you. The more you continue to post, the more stubbornly ignorant you look.

 

[OnlyMe]

Your response is confusing. What are you talking about? Are you just blindly pointing to my "escape velocity" post and trying to apply it to the issue that has you so confused?

Listen dude, if you can accept that the MOON, the SUN and a BLACK HOLE would take less than 1.4 seconds to fall 10 meters to the surface of the Earth, but you somehow think that everyday objects like feathers, bowling balls and double-decker buses all behave according to a completely different set of "Galileo physics" such that they would all take the exact, same amount of time to fall (1.4 seconds)...then you have a cognitive disconnect with which I simply cannot help you. The more you continue to post, the more stubbornly ignorant you look.

RJ, the moon should fall toward the earth like any other object smaller than the earth would. Keeping in mind that by falling toward I mean the closing velocity, between the two objects. All objects of a smaller mass than the earth, starting from a stationary position relative to the earth, would reach the same relative terminal velocity, when dropped. Excluding atmospheric drag.

As soon as you step up to the sun or a black hole the result changes.., because the larger mass becomes the origin of the dominate gravitational field. The gravitational potential 1 meter above the surface of a star is far greater than it would be 1 meter above the earth. That greater potential changes the way the inertial resistance plays out. You start introducing greater terminal velocities. They would only fall toward each other , in the same manner as a smaller object falling toward the earth, until they reach a point where the gravitational potential is equivalent to that at the surface of the earth.

 

[RJBeery]

As soon as you step up to the sun or a black hole the result changes.., because the larger mass becomes the origin of the dominate gravitational field. The gravitational potential 1 meter above the surface of a star is far greater than it would be 1 meter above the earth. That greater potential changes the way the inertial resistance plays out. You start introducing greater terminal velocities. They would only fall toward each other , in the same manner as a smaller object falling toward the earth, until they reach a point where the gravitational potential is equivalent to that at the surface of the earth.

OnlyMe, I can only try to teach so many people at one time (and to what end? I'm not getting paid for this). I spent a LOT of time putting together my other thread on this subject using, you know, MATH rather than phrases like "changes the way the inertial resistance plays out". If you want to continue on this subject please move it over to the other thread. I would also ask that you actually READ the opening post to that thread and you will perhaps appreciate why you guys are mistaken on this point. Thanks.

 

[tashja]

More replies from the experts to RJ's golf balls thought experiment:

RJBeery: Start with an existing black hole and an event horizon radius R at time T. Say the black hole is being "fed" an infinite series of golf balls, one after the other, which are all stamped numerically such that the current golf ball external to the event horizon is 1.0 * 10^32.

Now, starting at time T, run the clock backwards to T_past until R_past = R/2. What does the scene look like? Do golf balls with numbers less than 1.0 * 10^32 appear? If they do then there is a time T_crossover such that T_past < T_crossover < T where we could have witnessed the event horizon expand due to matter crossing it. In my understanding of GR, this cannot happen because golf balls external to the event horizon remain theoretically observable (with perfect instrumentation) forever. But in this thought experiment the black hole at time T is made of nothing but golf balls numbered 1 through (1.0*10^32)-1. I find it difficult not to view this as a contradiction, so what is the resolution?

Prof. Booth:
Black hole horizons can certainly expand and matter can certainly cross the horizon and no longer be visible afterwards. There are several technical issues about exactly how the expansion happens and exactly what one means by horizon (due to its definition the event horizon can evolve in strange ways) but this expansion happens generically as matter falls in.

On the other hand an event horizon is, by definition, a null surface. As such you can never see the horizon directly - by construction light cannot escape from it to be seen. So you would never actually see the horizon itself expand but could certainly monitor the effects of that expansion.

I think perhaps your confusion comes from this statement:

''In my understanding of GR, this cannot happen because golf balls external to the event horizon remain theoretically observable (with perfect instrumentation) forever.''

Now, for a stationary observer at infinity in a stationary spacetime (ie one where the horizon isn't growing), it's true that the observer never sees anything cross the horizon - it just gets red-shifted out. However, in this case everything is dynamical. Things can certainly be seen to happen - for example when numerical relativists run black hole merger simulations you can see (the effects of) black holes merging. The gravitational wave observatories have been built precisely to measure the gravitational waves emitted in this kind of situation. Similarly if LISA (the space observatory) is ever built, one of the signals that it will be looking for are the gravitational wave emitted as stars fall into the central galactic black hole - not so different from the golf balls in your example.
Does this answer your question? I could get into the technicalities of how things expand, but
I don't think that you are looking for that are you?

Best,

Ivan Booth
Department of Mathematics and Statistics
Memorial University
St. John's, NL A1C 5S7
Canada

 

[paddoboy]

I have a few questions (sort of on topic but not much related to BH.):
First is: Do photons have a very tiny gravitational field?
They have no rest mass so I guess the answer is no* (and at end of post support that with simple observation). Here is the problem that makes me have some doubt:

Photons/Light do create a minimal very tiny amount of spacetime curvature due to momentum.
By the same token, gravity or spacetime curvature also creates more gravity and spacetime curvature, due to the property of nonlinearity.

 

[brucep]

Your response is confusing. What are you talking about? Are you just blindly pointing to my "escape velocity" post and trying to apply it to the issue that has you so confused?

Listen dude, if you can accept that the MOON, the SUN and a BLACK HOLE would take less than 1.4 seconds to fall 10 meters to the surface of the Earth, but you somehow think that everyday objects like feathers, bowling balls and double-decker buses all behave according to a completely different set of "Galileo physics" such that they would all take the exact, same amount of time to fall (1.4 seconds)...then you have a cognitive disconnect with which I simply cannot help you. The more you continue to post, the more stubbornly ignorant you look.

It's confusing to you since you don't even understand your juvenile analysis and what the consequences would be if you were right.

Your juvenile analysis includes a component where you sum the gravitational relationship between M and m to your prospective equation of motion. So the correct equation of motion

dr/dt = (2M/r)^1/2.

Becomes completely bogus

dr/dt = (2M+m/r)^1/2

when r=2M the Schwarzschild event horizon


dr/dt = (2M+m/2m)^1/2 = (1+m/2M)^1/2. (c=1)

LOL^2

Since nobody really wants to see you get hurt pushing the report button I feel obligated to give you a vote of overconfidence

*PLONK*

 

[OnlyMe]

OnlyMe, I can only try to teach so many people at one time (and to what end? I'm not getting paid for this). I spent a LOT of time putting together my other thread on this subject using, you know, MATH rather than phrases like "changes the way the inertial resistance plays out". If you want to continue on this subject please move it over to the other thread. I would also ask that you actually READ the opening post to that thread and you will perhaps appreciate why you guys are mistaken on this point. Thanks.

RJ, I am responding to your posts in this thread. I use common lay oriented language for two reasons; the math would go over the heads of many of those reading posts on this forum, and working out the math would be more work for me than I am willing to put into it. (I haven't used the math involved for at least 30 years.)

I have obviously not been following your other thread whatever it may be.., but if it is consistent with what you have posted here, you are wrong...

Yes, a large object pulls on the earth with more force than a small one does, making the earth move further toward it as it falls toward the earth... But that is not the end of things.., because the earth pulls harder on the smaller object making it fall faster toward the earth, that moves less toward it. But the net result in either case is the same. As long as you are talking about only two objects with no other external forces, even Newton's field equation holds true. There is no difference in closing velocities, unless you start talking about gravitational potentials greater than the earth's. And even then if you drop a feather and the earth toward a neutron star, there will be no difference in their terminal velocities.

You seem to be ignoring the inertia and interial resistance of all objects involved.

 

[paddoboy]

RJ, that is taking only the effect of the different masses of the falling objects, on the earth's mass and inertial resistance, into account. You are correct in stating that the larger mass will "pull" the earth toward it more than the smaller mass would, but what is being neglected is that the earth "pulls" the smaller mass toward it more than it does the larger mass. You have to account for the inertial resistance of all objects.

The net result is that if you begin with the same separtion distance and assume both objects are at rest relative to oneanother, it does not matter what the mass of each object is, the total mass and distance between the masses, is all that matters. They, a small and large object dropped toward a larger object, fall toward the larger object at the same rate. This will hold true as long as you are comparring how two objects with masses smaller than the third object, interact with the larger mass gravitationally.

Totally in agreement:

I'm with Dr Toad on this.
In normal conditions on earth, air resistance plays a part and we see the hammer hitting the ground first.
In an airless environment, like on the Moon, and the results achieved with the NASA Apollo XV experiment, there is no air resistance to affect the results and they hit at the same time.
Although the hammer has more mass, it also has more inertia compared to the feather, and this extra mass and more inertia, cancels each other out, and they end up hitting at the same time.
That's the way I see it, but I'm willing to be shown different.

With BH's, there is far more evidnce to support them, then any theoretical quantum mathematical evidence invalidating them.
BH's exist, almost certainly...The only question is type and the nature of the Singularity and whether it does exist or not.

 

[paddoboy]

RJ, here's Prof. Dolan's reply to your follow-up questions:
RJBeery: If the event horizon has "formed", what is it composed of?
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

I find it really hard to believe that someone out to rewrite 20th/21st century cosmology/physics, could ask such an inane question, with such a logical answer.

​

 

[brucep]

RJ, I am responding to your posts in this thread. I use common lay oriented language for two reasons; the math would go over the heads of many of those reading posts on this forum, and working out the math would be more work for me than I am willing to put into it. (I haven't used the math involved for at least 30 years.)

I have obviously not been following your other thread whatever it may be.., but if it is consistent with what you have posted here, you are wrong...

Yes, a large object pulls on the earth with more force than a small one does, making the earth move further toward it as it falls toward the earth... But that is not the end of things.., because the earth pulls harder on the smaller object making it fall faster toward the earth, that moves less toward it. But the net result in either case is the same. As long as you are talking about only two objects with no other external forces, even Newton's field equation holds true. There is no difference in closing velocities, unless you start talking about gravitational potentials greater than the earth's. And even then if you drop a feather and the earth toward a neutron star, there will be no difference in their terminal velocities.

You seem to be ignoring the inertia and interial resistance of all objects involved.

RJ, I am responding to your posts in this thread. I use common lay oriented language for two reasons; the math would go over the heads of many of those reading posts on this forum, and working out the math would be more work for me than I am willing to put into it. (I haven't used the math involved for at least 30 years.)

I have obviously not been following your other thread whatever it may be.., but if it is consistent with what you have posted here, you are wrong...

Yes, a large object pulls on the earth with more force than a small one does, making the earth move further toward it as it falls toward the earth... But that is not the end of things.., because the earth pulls harder on the smaller object making it fall faster toward the earth, that moves less toward it. But the net result in either case is the same. As long as you are talking about only two objects with no other external forces, even Newton's field equation holds true. There is no difference in closing velocities, unless you start talking about gravitational potentials greater than the earth's. And even then if you drop a feather and the earth toward a neutron star, there will be no difference in their terminal velocities.

You seem to be ignoring the inertia and interial resistance of all objects involved.

If I was wrong you'd be writing

dr/dt = (2M/r)^1/2 (Newton, Einstein, and everybody else in physics)

As

dr/dt = (2M+m)^1/2

LOL

Regardless your juvenile analysis it's

dr/dt = (2M/r)^1/2

LOL

Correction line 2
dr/dt = (2M+m/r)^1/2 (not really LOL)

 

[brucep]

Correction line 2
dr/dt = (2M+m/r)^1/2 (not really LOL)

Just in case somebody doesn't get it

dr/dt = (2M+m/2M)^1/2 = (1+m/2M)^1/2 is > c.

For example

m_earth = .00444meter in geometric units and 2M for a solar mass = 2954meter

So

dr/dt = (1+.00444m/2954m)^1/2 = 1.001225988057 (c=1)

Complete nonsense. It predicts the greater the mass of m the greater it exceeds c at r=2M. So not only does RJBerry predict Galileo is wrong but that Einstein is wrong also. That Newton is wrong and every other gravitational physicist working in the field is wrong. The extra speed is the consequence of RJBerry including irrelevant bullshit in his analysis. So Tashja what do you think of that? Do you understand what I've been writing down?

 

[brucep]

I find it really hard to believe that someone out to rewrite 20th/21st century cosmology/physics, could ask such an inane question, with such a logical answer.

​

I concur. LOL^2

 

[RJBeery]

Just in case somebody doesn't get it

dr/dt = (2M+m/2M)^1/2 = (1+m/2M)^1/2 is > c.

For example

m_earth = .00444meter in geometric units and 2M for a solar mass = 2954meter

So

dr/dt = (1+.00444m/2954m)^1/2 = 1.001225988057 (c=1)

Complete nonsense. It predicts the greater the mass of m the greater it exceeds c at r=2M. So not only does RJBerry predict Galileo is wrong but that Einstein is wrong also. That Newton is wrong and every other gravitational physicist working in the field is wrong. The extra speed is the consequence of RJBerry including irrelevant bullshit in his analysis. So Tashja what do you think of that? Do you understand what I've been writing down?

Brucep, where did I write that equation? Stop typing it and please post a quote, or a link, or a post # reference.

 

[brucep]

Brucep, where did I write that equation? Stop typing it and please post a quote, or a link, or a post # reference.

It's the equation of motion for your goofball analysis. It's simple

This is the equation of motion that describes the rate an object falls in the gravitational field. Derived by Newton and Einstein (and everybody else who has a clue). Galileo derives it experimentally. He concludes that the rate an object falls has no association with the objects mass.

dr/dt = (2M/r)^1/2

You conclude that isn't true and claim the gravitational interaction between M and m must be included.

That's what the +m is derived from. Get it? The +m results in the unphysical prediction at r=2M. It's unphysical because your analysis doesn't have anything to do with reality (physics). You've had this explained to you to no avail.

 

[RJBeery]

It's the equation of motion for your goofball analysis.

I asked for a link to the post. I don't recognize that equation.

 

[RJBeery]

Also, I'm calling out RUSS_WATTERS for bringing up this Godforsaken subject. Why aren't you defending the very point you felt you were being so clever to make in the first place? Because I'm the only person who knew the right answer?

 

[brucep]

G

I asked for a link to the post. I don't recognize that equation.[/

Also, I'm calling out RUSS_WATTERS for bringing up this Godforsaken subject. Why aren't you defending the very point you felt you were being so clever to make in the first place? Because I'm the only person who knew the right answer?

Your answer isn't even wrong it's unphysical. Delusional.

 

[Trippy]

I asked for a link to the post. I don't recognize that equation.

You've been given a perfectly reasonable explanation.

 

[RJBeery]

No words needed. This may help folks with the conceptual problem.