QM + GR = black holes cannot exist
- Thread starter RJBeery
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A while back I went from posting here to just lurking for time management reasons, but I feel like RJBeery really deserves a friendly voice here. Newton's law of gravity says that in the Earth's gravitational field at a given height (or any fixed gravitational field for that matter), any object will experience the same gravitational acceleration. On the other hand, the Earth's acceleration toward the object will depend on the gravitational field produced by that object, which in turn depends on its mass. The closing acceleration between the Earth and the object is the sum of these two terms, so it also depends on the object's mass (very slightly).
And brucep: I would also like a citation for dr/dt = 2M/r^1/2. For one thing, it doesn't have the right units; I assume M here is standing in for some combination of the Earth's mass and the gravitational constant? Also, the equation as written gives velocity as a function of r independent of the height from which the object was dropped, which can't be correct.
It bothers me that this quote was not only posted, but quoted and liked multiple times. The whole point of Galileo's experiment was that the acceleration of an object under gravity is completely independent of its mass; in other words, gravitational mass is the same as inertial mass. The Earth pulls on any two masses equally, and the object's pull on the Earth must be added to that constant value.
Unless you're dealing with thermodynamics in the large-N limit, just about every function in nature is smooth if you look closely enough. It makes no sense to say that any objects less massive than the Earth fall at exactly the same rate, but as soon as they get more massive than the Earth they start to fall faster. Instead, you should be looking for a function that correctly predicts the small- and large-mass limits but interpolates smoothly between them. RJBeery's description meets this criterion.
And brucep: I would also like a citation for dr/dt = 2M/r^1/2. For one thing, it doesn't have the right units; I assume M here is standing in for some combination of the Earth's mass and the gravitational constant? Also, the equation as written gives velocity as a function of r independent of the height from which the object was dropped, which can't be correct.
No words needed. This may help folks with the conceptual problem.
What on earth do you believe that proves?
Yes, both masses accelerate toward a common center of mass. You seem to be assuming that because, in the bowling ball/earth relationship, the earth moves more toward the bowling ball than it would toward a base ball, that when comparring those two theoreticals.., bowling ball/earth and baseball/earth.., the bowling ball would hit the earth first or faster, by traveling a shorter distance... The problem you continue to completely ignore is that while it is true the earth accelerates more toward the bowling ball than the baseball, the baseball accelerates toward the earth more than the bowling ball. The two situations wind up being exactly the same as far as the total distance, time and closing speeds for the two drops.
This would be the same for all situations where the dropped mass is equal to or smaller than the earth's mass. The inertial resistance of both objects must be taken into account.., not just how the earth's inertial resistance is affected by the mass of the dropped object.
Though the difference between a baseball and bowling ball would be measureably insignificant, when viewed by an outside observer, the baseball would accelerate more toward the earth/baseball center of mass, than the bowling ball would toward the earth/bowling ball center of mass.., but the earth's acceleration in the two cases would also be different.............., by an amount that exactly offsets the difference in the accelerations of the baseball and bowling ball. The total distance in both cases is the same, unless you start saying you drop one from a second story window and the other from a first story window.
This is all about inertia, RJ. And you have to include how the inertia of all objects affects the end result. You cannot just realize that the earth moves more in one case and stop. You have to account for how each object accellerates.
Stop and think it through. You seem to have become stuck or vested in an incomplete view of what happens.
Stop saying this! Acceleration under Newtonian gravity is mass-independent.
Galileo, was using the earth as a fixed frame of reference.., which attributes all of the velocity of the dropped object to the smaller object. RJ introduced the fact that from an outside FoR both the dropped object and the earth accellerate toward their common center of mass, which is different when comparring dropped objects of different masses.
Galileo was wrong in one aspect it is not the earth pulling on the object, it is the earth and the object pulling on eachother...
If you are comparring the earth and a mass greater than that of the earth, then the earth becomes the dropped object, in Galilean terms. Once the distance between the two is such the the gravitational, potential of the larger mass exceeds that of the earth, at its surface, the two will begin to accelerate toward eachother faster than the terminal velocity associated with the earth's gravitational field.
Yes if you have two earth's and drop them toward a common center of mass, they will pull harder on eachother than the earth pulls on a baseball, but they will also both have a far greater inertial resitance to accelleration than a baseball.., and the two will meet in exactly the same time it takes a baseball to fall to earth in the Galilean model where the earth is the FoR.
It is all about INERTIA, the force is accurately described by Newton's formula. The inertial resistance to that force is what keeps the results consitent.
Not entirely true. For this discussion, this is only true where the earth is the FoR, and all of the gravitational acceleration of a dropped object, is attributed to the smaller dropped object. The earth pulls harder on different masses, but those different masses resist that pull to different degrees, based on their mass. It is all about inertia!
Inertia is mass dependent and an object's inertial resistance to acceleration is mass dependent.
tashja
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Hi RJ. My apologies for misinterpreting your scenario. I did send the link to Prof. Ira Wasserman and here's what he said:
That's the way I see it...nicely put.
Excellent, Tashja, again 1000 thanks. If this doesn't resolve the discussion for everyone then nothing will.
You can't just sit there RJBeery in all your arrogance and pick and chose which bits of expertise will fit your general derision of accepted mainstream physics. They also told you BH's still exist, but you fail to recognise that.
Or do you now stand corrected on that issue?
Geometric units. The solution is dimensionless. The slight acceleration is nonsense. It's a local inertial free fall frame. The same frame for Newton First Law Of Motion. If the object in free fall is predicted to experience accelerations then the derivation would reflect that. It doesn't.
For Newton coordinates
dr/dt = [2M/r]^1/2
For local Schwarzschild coordinates
dr_shell/dt_shell = [2M/r]^1/2
For local rain coordinates
dr/dt_rain = [2M/r]^1/2
None of these derivations predict tidal accelerations effect the rate of free fall. With tidal accelerations you get the nonsense prediction that I exampled. That the rate at r=2M exceeds c. Newton doesn't predict it. Einstein doesn't predict it. What seems to confuse folks (sometimes) is thinking gravity is a force. That the mass of the object in free fall interacts with the mass M via a force. Over the natural path local spacetime curvature is an infinitesimal. Over a very long path the tidal accelerations may change the path slightly. For example the thought experiment where the dude is in the falling elevator with two particles and eventually can measure a slight deviation of the particle path with respect to his path. The tidal accelerations don't change the rate of free fall.
For some reason the brackets for the equations won't post. You know where the brackets go. Wrote the equations out again to see if I can get the brackets. Didn't work. ? ? What happens is the brackets appear for an instant then change to what you see now. When I call up edit the brackets are there. When I post they're gone. The brackets are gone from all the equations of motion I wrote down. At least on this IPAD. Could be a local problem. LOL. The brackets I used for the word 'sometimes' are gone also. Whatever. Maybe someone could tell me if the brackets are missing on their screen? Changed the bracket set and it worked. ??
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I asked 6 questions in post 234 here: http://www.sciforums.com/threads/qm-gr-black-holes-cannot-exist.142658/page-12#post-3230015
All questions related to Q1 which was: Does the photon make gravity {very little of course, but is it zero) and if it does: is the propagation away from the source at speed of light? To meaningfully ask this I considered an electron / positron pair orbiting their mass center initially with 2cm separation and showed that: (1) their gravitational field is not constant at say 1 meter from their mass center but varies [peak to least} with 1/4 their orbit period & (2) that as they are charged and accelerating they lose orbit KE to the EM field so spiral together - and eventual be come two 511Mev photons. I.e. the gravity field must get a growing hole at that time.
Only only me gave some discussion, but not any clear answer. One problem with his discussion suggesting that photons do "warp space time" {make gravity) is he speaks of the photon energy as the mass making this gravity. {but there is no unique "photon energy" E for all frames.)
I had anriciapated this and in footnote of post 234 said:
" Also the photon's M = E/(c^2) mass is "frame dependent" so that would seem to be a problem with their "bending of space" view of gravity, I think. I do lean to the POV that all energy which is the same in all frames does make gravity. I.e. the gravity from a hot brick deceases slightly as it cools - Its temperature, random KE, is same in all frames. Any arguments on this POV? Before arguing , note that jar of half ice & half water has T = 0C in all frames. "
I.e. photons travel at same speed in all frames but don't have the same energy in all frames, thus if they "make gravity" via their "energy mass" = M = E /(c^2) it is a different amount in different frames, as E is frame dependent. This seems a problem to me. For example if photon passes near neutral hydrogen atom it would give a gravitational impulse to that H. That impulse accelerates the H towards the photon's point of closest approach. Many different frames all moving along the photon's path could measure the time required for the H to come to that "closest approach" point and as that motion is transverse to ALL of the frame' relative motion there should be no "time dilation" (at least in first order} Yet the E of the photon is different to first order in the different frames. {even linearly different in all frames whose relative speed difference is small compared to speed of light.]
Thus no one has been able to make even one sensible answer to my 6 questions - have a try at it if you think you understand any GR or QM as related to gravity.
All questions related to Q1 which was: Does the photon make gravity {very little of course, but is it zero) and if it does: is the propagation away from the source at speed of light? To meaningfully ask this I considered an electron / positron pair orbiting their mass center initially with 2cm separation and showed that: (1) their gravitational field is not constant at say 1 meter from their mass center but varies [peak to least} with 1/4 their orbit period & (2) that as they are charged and accelerating they lose orbit KE to the EM field so spiral together - and eventual be come two 511Mev photons. I.e. the gravity field must get a growing hole at that time.
Only only me gave some discussion, but not any clear answer. One problem with his discussion suggesting that photons do "warp space time" {make gravity) is he speaks of the photon energy as the mass making this gravity. {but there is no unique "photon energy" E for all frames.)
I had anriciapated this and in footnote of post 234 said:
" Also the photon's M = E/(c^2) mass is "frame dependent" so that would seem to be a problem with their "bending of space" view of gravity, I think. I do lean to the POV that all energy which is the same in all frames does make gravity. I.e. the gravity from a hot brick deceases slightly as it cools - Its temperature, random KE, is same in all frames. Any arguments on this POV? Before arguing , note that jar of half ice & half water has T = 0C in all frames. "
I.e. photons travel at same speed in all frames but don't have the same energy in all frames, thus if they "make gravity" via their "energy mass" = M = E /(c^2) it is a different amount in different frames, as E is frame dependent. This seems a problem to me. For example if photon passes near neutral hydrogen atom it would give a gravitational impulse to that H. That impulse accelerates the H towards the photon's point of closest approach. Many different frames all moving along the photon's path could measure the time required for the H to come to that "closest approach" point and as that motion is transverse to ALL of the frame' relative motion there should be no "time dilation" (at least in first order} Yet the E of the photon is different to first order in the different frames. {even linearly different in all frames whose relative speed difference is small compared to speed of light.]
Thus no one has been able to make even one sensible answer to my 6 questions - have a try at it if you think you understand any GR or QM as related to gravity.
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I certainly attempted to answer the question re a photon adding to gravity/spacetime curvature.
It does, but by an insignificant amount, solely due to its momentum.
I also raised the point re gravity making gravity due to the property of non linearity.
I gave you an answer in post 241. Over the path of the falling photon [that's right the photon is in free fall along the null geodesic] it's energy is a constant of the motion. The relativistic energy equation is derived from this. The energy and momentum are constants of the motion for all objects including light along the natural path.
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Ditto. Gravitational and inertial mass are equivalent. And the tidal acceleration has no effect on the rate the object falls. The problem folks are having is insisting they do. This occurs because of an erroneous derivation from Newtonian coordinates.
I see. So it's my arrogance rather than my veracity that's really bothering you?
They also told you BH's still exist, but you fail to recognise that.
Or do you now stand corrected on that issue?
You mean as in denying the existence of BH's in the face of overwhelming evidence, as compared to trickles of unknown certainties re quantum effects?
And of course "bothering me" is just a giant stretch of your Imagination.
see....
http://physicsworld.com/cws/article...black-hole-found-at-the-heart-of-dwarf-galaxy
I see you've given up the Galileo fight. I'll consider that a moral victory.
And no, I don't personally believe that black holes exist. The overwhelming evidence can be otherwise explained. You might have noticed that Tashja's expert opinions literally contradicted each other in certain aspects; that's a round-about way of saying that "we simply don't know yet".
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