QM + GR = black holes cannot exist

[brucep]

Brucep
Your post reminded me of this from the MIT site.
Please see pdf for equations, they come out screwy with my cut and paste quote here.





From… MIT

Pick pdf titled ‘How Gravitational Forces Arise from Curvature.’
Page 1 and 2.

I was looking for your post in the wrong thread. LOL The alert notified me but I guess my reading skills are getting suspect. LOL. Thanks for posting that. .

 

[RJBeery]

The real problem is nobody understands what I'm talking about because it's GR and very few people are familiar with how the theory works. You think you are but you don't have the scholarship since you haven't done the work to find out how it works [the actual physics]. Everything I've written about relativistic physics is based on my studies associated with Einsteins model of gravity. That's the way it will always be in a venue such as a public science forum. No problem but everytime you disagreed with my interpretation you're disagreeing with relativistic physics in general. You've made it clear many times that you don't trust the physics, predictions, experimental results, etc.. My underlying intent was to discuss the physics. All the other bullshit is the result of trying to discuss it where the cranks rule.

This is simply classic. Attention all of the little people here, brucep is the local GR expert! If you don't understand what he's saying then it's because he's speaking over your head...just trust his word as gospel.

According to general relativity, if you were to fall into a black hole, you shouldn’t notice anything strange when you cross the event horizon. Yes, you might feelstrong tidal forces, but you’d feel those outside the black hole as well.

They're are holidays of ignorance in this bloggers discussion of a predicted firewall. He calls it a Hawking firewall as if the prediction of a super hot bath of Hawking radiation impeding the progress of the falling observer was Hawkings idea. He also doesn't understand why the infalling observer won't be subject to tidal forces until spaghettifacation at r=0.

This is false. Tidal forces are dependent on mass of the black hole and distance from it. (Unless you're using an extremely odd definition of "spaghettification")

Then I provided another resource...from NASA

A tidal force is a difference in the strength of gravity between two points. The gravitational field of the moon produces a tidal force across the diameter of Earth, which causes the Earth to deform. It also raises tides of several meters in the solid Earth, and larger tides in the liquid oceans.

If the tidal force is stronger than a body's cohesiveness, the body will be disrupted. The minimum distance that a satellite comes to a planet before it is shattered this way is called its Roche Distance. The artistic image to the left shows what tidal disruption would be like for an unlucky moon. A human falling into a black hole will also experience tidal forces. In most cases these will be lethal! The difference in acceleration between the head and feet could be many thousands of Earth Gravities. A person would iterally be pulled apart! Some physicists have termed this process spaghettification!

I mean, I don't even BELIEVE in black holes but I know enough about the theory to know a poseur when I see one! How can you proclaim to be a local expert when you get something so basic as speghettification absolutely wrong?

 

[paddoboy]

I mean, I don't even BELIEVE in black holes but I know enough about the theory to know a poseur when I see one! How can you proclaim to be a local expert when you get something so basic as speghettification absolutely wrong?

I repeat.....
EVERY POSTER ignoring the mountain of evidence for BH's, is either willfully ignorant, or simply playing stupid for attention.
DISCLAIMER: I did partly pinch that sentence off another in the "Galileo" THREAD

 

[Billy T]

... It means the photon energy contributes to the local spacetime curvature as it passes along its natural path. In this sense it has a very small effect on the local spacetime geometry. ... I think this is enough for you to get answers to your query.

Then you seem to be clearly implying (or very clearly stating?) that two, side-by-side photons traveling initially along parallel lines spaced one nanometer apart for very long time, say from a distant star 100,000 light years way to enter my eyeball, will each make a tiny warping of local space (or in simple terms, mutually attract gravitationally) as they travel and eventually either (1) merge or (2) since they can pass thur each other * without noticeable interaction to "overshoot" and then attract each other again in an oscillation that repeatedly passes thru each other.

Is that what you intended to say?

If this is true, then a photon falling in towards a tiny black Hole, BH, I think has a gravitational blue shift, but be that as it may if an enough do that the BH could be gaining mass as rapidly as it is losing it via Hawking radiation. In fact, if that is true and the photon flux is larger and long lasting, most of the BH's mass could be photons, which if the blue shift is also true begin to have sufficient energy (or EM field temperature) to destroy any matter - I.e. conditions like nano seconds after the Big Bang before it cooled enough for mater to "condense" out of the intense energy field. That was suggested here:

{in post 300} ... If the photons are bounded, as they would be in a black hole, they can even have inertia, the same way they do when they are bound in the energy states of electrons within atoms. In other words, black holes can exist without any matter in their cores or inner shells. It would behave exactly the same in terms of gravitation if all of the mass had been accreted to energy, and all of that mass converted to energy was orbiting in the region of the event horizon. It may not be able to form with this constitution, but almost certainly, it can end in this state. ...

* like two laser beams crossing thru each other in near perfect vacuum of deep space without scattering one each other.

 

[danshawen]

I believe in BHs, paddoboy. How could anyone not? Sagittarius A is on youtube, in all its glory, or at least, the stars buzzing around it. They are not doing so because the spot is just interesting.

Also, I'm not posting anything else on "Galileo" after RJBeery's expert witness testified; no way to top that. I'm satisfied with the conclusion stated in his first post there; just not the first post on this thread. If part of QT can be interpreted as ruling out black holes, and the observation persistss, guess which part needs to go, simply by weight of the scientific method? Sorry, turtles. No more holding up the world for you. A flat Earth really simplified time zones, but if it doesn't work, something's got to give.

 

[Motor Daddy]

Sorry, turtles. No more holding up the world for you.

What??? You mean it isn't turtles all the way down?

http://en.wikipedia.org/wiki/Turtles_all_the_way_down

A well-known scientist (some say it was Bertrand Russell) once gave a public lecture on astronomy. He described how the earth orbits around the sun and how the sun, in turn, orbits around the center of a vast collection of stars called our galaxy. At the end of the lecture, a little old lady at the back of the room got up and said: "What you have told us is rubbish. The world is really a flat plate supported on the back of a giant tortoise." The scientist gave a superior smile before replying, "What is the tortoise standing on?" "You're very clever, young man, very clever," said the old lady. "But it's turtles all the way down!"

—Hawking, 1988.

 

[brucep]

I don't believe I ever said anything like, "photons alone could not constitute the mass/energy of a black hole.". And in saying this I am not claiming that they could, or event that what we experience as EM radiation or photons, actually exists within an event horizon. I don't know anything about the physical reality or composition of what lies inside the event horizon, of what we call a black hole. This is where old maps used to say, "beyond here there be monsters". Today we temper that by saying beyond here {meaning an event horizon}, is the domain of theory, orr in the context of lay discussion very often imagination.

Ask a varied group of physists what E = mc^2 means and how it applies to physics and you will get many different and differing answers. For myself I will once again through out a paraphrased quote from a still up recovered source,

All mass can be thought of as an expression of energy, but not all energy can be thought of as mass.​

I don't believe that mass and energy are the same thing, though there is an obviously complex relationship between the two.

Then you seem to be clearly implying (or very clearly stating?) that two, side-by-side photons traveling initially along parallel lines spaced one nanometer apart for very long time, say from a distant star 100,000 light years way to enter my eyeball, will each make a tiny warping of local space (or in simple terms, mutually attract gravitationally) as they travel and eventually either (1) merge or (2) since they can pass thur each other * without noticeable interaction to "overshoot" and then attract each other again in an oscillation that repeatedly passes thru each other.

Is that what you intended to say?

If this is true, then a photon falling in towards a tiny black Hole, BH, I think has a gravitational blue shift, but be that as it may if an enough do that the BH could be gaining mass as rapidly as it is losing it via Hawking radiation. In fact, if that is true and the photon flux is larger and long lasting, most of the BH's mass could be photons, which if the blue shift is also true begin to have sufficient energy (or EM field temperature) to destroy any matter - I.e. conditions like nano seconds after the Big Bang before it cooled enough for mater to "condense" out of the intense energy field. That was suggested here:


* like two laser beams crossing thru each other in near perfect vacuum of deep space without scattering one each other.

Then you seem to be clearly implying (or very clearly stating?) that two, side-by-side photons traveling initially along parallel lines spaced one nanometer apart for very long time, say from a distant star 100,000 light years way to enter my eyeball, will each make a tiny warping of local space (or in simple terms, mutually attract gravitationally) as they travel and eventually either (1) merge or (2) since they can pass thur each other * without noticeable interaction to "overshoot" and then attract each other again in an oscillation that repeatedly passes thru each other.

Is that what you intended to say?
If this is true, then a photon falling in towards a tiny black Hole, BH, I think has a gravitational blue shift, but be that as it may if an enough do that the BH could be gaining mass as rapidly as it is losing it via Hawking radiation. In fact, if that is true and the photon flux is larger and long lasting, most of the BH's mass could be photons, which if the blue shift is also true begin to have sufficient energy (or EM field temperature) to destroy any matter - I.e. conditions like nano seconds after the Big Bang before it cooled enough for mater to "condense" out of the intense energy field. That was suggested here:


* like two laser beams crossing thru each other in near perfect vacuum of deep space without scattering one each other.

For the photons contributing to the local spacetime curvature the photons will follow the same geometry as directed by the spacetime curvature. To paraphrase an old saying matter tells spacetime to curve and the curvature determines the path. For the most part a photon doesn't interact with another photon. So they can't merge and exchange energy. My knowledge of quantum physics is pretty basic, IE weak. The redshift/blueshift is coordinate dependent phenomena. The local energy is a constant of the motion along the photons geodesics path so no local measurement will reveal a shift in energy. The reason the period of the light is frame dependent is because the tick ratio between the photons frame and the observers frame can vary.

 

[tashja]

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.

See linked img: http://i1373.photobucket.com/albums/ag380/rjbeery/golfball_black_holes_zps339d1899.png

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 would have witnessed the event horizon expand due to matter crossing it. This cannot happen. If the golf balls numbered 1 through (1.0 * 10^32 - 1) ever existed then we would theoretically be able to observe them, with perfect instrumentation, forever. But in this thought experiment the black hole is made of nothing but golf balls numbered 1 through (1.0*10^32)-1.This is a contradiction, therefore the event horizon cannot exist.''

Prof. Marolf:

Hi. The issue here is that there are many different notions of ³time² (or
even ³at this time²) for a black hole. There are some notions in which
golf balls never cross the horizon and black holes ³never form.² But
there are some in which they do. The discussion below appears to conflate
the two.

And BTW, although there are some definitions of time wrt which black holes
³never form,² what this means is that one cannot directly observe the
formation of an event horizon. But one can observer the formation of
something that is exponentially-close to having an event horizon. So very
quickly the distinction between whether it ³is now² or ³is not now²
actually a BH becomes entirely academic. It¹s like asking if a rubber
duck that a kid left in a bathtub an hour ago is ³still moving after he
pushed it² or whether it ³has really stopped.²

To reinforce this:

If the golf balls numbered 1 through (1.0 * 10^32 - 1)
ever existed then we would theoretically be able to observe them, with
perfect instrumentation, forever.

No, even with perfect instrumentation you cannot observe them forever.
There is a ³last photon² that they emit/reflect before crossing the
horizon. After that, there is no signal to detect. The same is true of a
star that collapses to form a BH. This is a good in-practice
observational definition of ³when the BH forms² .

Don Marolf

 

[brucep]

Brucep
Your post reminded me of this from the MIT site.
Please see pdf for equations, they come out screwy with my cut and paste quote here.





From… MIT

Pick pdf titled ‘How Gravitational Forces Arise from Curvature.’
Page 1 and 2.

That will always be my favorite book on physics. It revealed so much for me. One of the most important things I learned is how the physics is conducted. The scientific method. It also made it possible for me to read scientific papers and have a clue of what's being discussed. LOL.

 

[paddoboy]

That will always be my favorite book on physics. It revealed so much for me. One of the most important things I learned is how the physics is conducted. The scientific method. It also made it possible for me to read scientific papers and have a clue of what's being discussed. LOL.

In the bloody absence of the LIKE button, may I add a LIKE to this post.
Thank you linesman, thank you ballboys!

 

[tashja]

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.

See linked img: http://i1373.photobucket.com/albums/ag380/rjbeery/golfball_black_holes_zps339d1899.png

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. Graham:

I am not really sure what you think is paradoxical about the situation,
but I will make a few comments.

1. Firstly, yes of course if one reversed the time in you situation you
would see golf balls coming out of the black hole (such a thing has a
name: its called a white hole, which is just the time reversal of a
black hole). This would only happen for a while, since presumably the
black hole formed by some other process.

2. There is nothing paradoxical about the "event horizon" (you are
really meaning the apparent horizon, which is essentially the event
horizon at a given time) increasing with time. In your situation you are
increasing the mass of the black hole by sending in golf balls, so if
you are increasing its mass the "event horizon" (R=2GM/c^2) will
increase with time. I use the term "event horizon" in quotation marks
deliberately, since the proper definition of an event horizon (which is
too complicated for me to enter into here) is teleological, and in
particular requires one to know the future evolution of the entire
spacetime. Since in you thought experiment you are feeding in an
infinite number of golf balls it would be correct to say it isn't really
well defined. The apparent horizon, which is often what people actually
mean in popular science discussions, can depend on time, and it only
coincides with the event horizon when the black hole is independent of
time (which it isn't here). I think this is the key point of confusion.

3. Finally, it is not really correct to speak of the black hole as "made
of golf balls", since once they fall into the black hole all information
about their structure is lost (bar their mass and angular momentum if
present). This is the content of the no-hair theorems, which assert
black holes are only characterised by their mass, angular momentum and
charge (properly speaking these results only apply to time-independent
black holes, but one would presume you could model the emission of golf
balls as a quasi-static process).

I hope these comments clear up any confusion. I should also add that if
you are at school it is great you are interested in these things, but I
wouldn't worry too much about any apparent paradoxes you find. The
problem is often what you read in the popular science literature or
online is often very imprecise, and at times downright inaccurate. I'm
sure I had similar thoughts at school, many years ago...

Kind regards
Alec Graham

 

[danshawen]

Unbound photons or energy cannot have inertia, but bound photons or energy can. A photon trapped in the event horizon of a black hole is in a bound state. So is a photon that is part of the energy of a bound electron in an atom. There are a series of careful thought experiments I have devised to show this, in preparation for SED (Stochastic Electrodynamics) making a comeback.

There was never anything wrong with the SED formulation of QT (which included classical inertia); just some math made cumbersome by not thinking carefully and avoiding the trap of absolute time and space. In the vacuum, absolute time and space do not exist. Neither will a correct formulation of gravitation in the context of interaction of vacuum energy with matter contain very much spatial geometry that is likewise a throwback to absolute space and time. No more using the centers of black holes (or anything else) for origins of coordinate systems. They are no more stationary or absolute than the baseball or golfball in the other thread.

 

[Billy T]

For the photons contributing to the local spacetime curvature the photons will follow the same geometry as directed by the spacetime curvature. To paraphrase an old saying matter tells spacetime to curve and the curvature determines the path. For the most part a photon doesn't interact with another photon. So they can't merge and exchange energy. My knowledge of quantum physics is pretty basic, IE weak. The redshift/blueshift is coordinate dependent phenomena. The local energy is a constant of the motion along the photons geodesics path so no local measurement will reveal a shift in energy. The reason the period of the light is frame dependent is because the tick ratio between the photons frame and the observers frame can vary.

Why not try to answer the question? Do the parallel traveling very closely spaced photons attract each other or not.

Here is how that was asked with some comments and observations:

Then you seem to be clearly implying (or very clearly stating?) that two, side-by-side photons traveling initially along parallel lines spaced one nanometer apart for very long time, say from a distant star 100,000 light years way to enter my eyeball, will each make a tiny warping of local space (or in simple terms, mutually attract gravitationally) as they travel and eventually either (1) merge or (2) since they can pass thur each other * without noticeable interaction to "overshoot" and then attract each other again in an oscillation that repeatedly passes thru each other.

Is that what you intended to say?
If this is true, then a photon falling in towards a tiny black Hole, BH, I think has a gravitational blue shift, but be that as it may if an enough do that the BH could be gaining mass as rapidly as it is losing it via Hawking radiation. In fact, if that is true and the photon flux is larger and long lasting, most of the BH's mass could be photons, which if the blue shift is also true begin to have sufficient energy (or EM field temperature) to destroy any matter - I.e. conditions like nano seconds after the Big Bang before it cooled enough for mater to "condense" out of the intense energy field. That was suggested in earlier post by some one else.


* like two laser beams crossing thru each other in near perfect vacuum of deep space without scattering one each other.

 

[RajeshTrivedi]

Why not try to answer the question? Do the parallel traveling very closely spaced photons attract each other or not.

In Classical Newtonian Physics : No Gravity between travelling Photons.

In GR : Qualitatively Yes, because Photons have Energy Density, which causes curvature of space time, however minuscule.

The implication of Electromagnetism Vs nature of Gravity is not well understood. So I do not think that this question can be authoritatively answered as on date. In GR this question can be quantitatively attempted by those members, who have fairly decent idea about complex Energy/Space Tensor equations.

 

[Layman]

Why not try to answer the question? Do the parallel traveling very closely spaced photons attract each other or not.

Here is how that was asked with some comments and observations:

Then you seem to be clearly implying (or very clearly stating?) that two, side-by-side photons traveling initially along parallel lines spaced one nanometer apart for very long time, say from a distant star 100,000 light years way to enter my eyeball, will each make a tiny warping of local space (or in simple terms, mutually attract gravitationally) as they travel and eventually either (1) merge or (2) since they can pass thur each other * without noticeable interaction to "overshoot" and then attract each other again in an oscillation that repeatedly passes thru each other.

I think it has been proven that photons will red-shift and blue-shift in the presence of a gravitational field. Then I believe that the photon would have to have a very small amount of mass in order for the Higgs Field to ever accurately describe quantum gravitational affects. It is thought the Higgs Field creates drag on the waves on particles, and this causes the force of gravity on the quantum scale. Everything else in physics has already been unified, so it would seem like the photon would have to create some drag in the Higgs Field in order for gravity to be unified by this description of gravity as well.

Then I would guess that two photons would attract each other running in parallel. I think the affects of gravity would be more or less greater if the photons ran together in unison. That would be similar to the behavior of virtual particles. Then they are virtual particles, because they do not have the same mass as the individual particle would have alone. Then it could be possible that overlapping particle waves create a different value of mass for particles from them being dragged more or less through the Higgs Field. Take quantum jumping for instance, particle waves out of phase by half a wavelength will momentarily have zero mass during the course of a jump (like the electron orbit around the nucleus of an atom). Then particles in phase could create more drag in the Higgs Field making them more massive particles (then the electron would have to be seen to be made of virtual photons).

Then I think The Particle Zoo would hinder development of this type of theory. From the way I understand quantum theory, it seems like it could only work out if non-elementary particles was all just looked at as though they where made of the constituents of other fundamental virtual particles. Then these virtual particles where all actual particles; they just had a change in mass that was abnormal do to the combination of particles waves being in or out of phase creating different amounts of values of mass in the Higgs Field.

 

[Q-reeus]

Why not try to answer the question? Do the parallel traveling very closely spaced photons attract each other or not...

After occasional lurking at SF for a long time, couldn't resist resurfacing here. In general photons do have an extremely weak gravitational interaction - strongest when their paths are anti-parallel, and zero when parallel. The latter is easy to figure if one just contemplates what photon frequency thus energy is 'seen' by a co-moving photon. If you don't figure it's zero - think again.

 

[Layman]

After occasional lurking at SF for a long time, couldn't resist resurfacing here. In general photons do have an extremely weak gravitational interaction - strongest when their paths are anti-parallel, and zero when parallel. The latter is easy to figure if one just contemplates what photon frequency thus energy is 'seen' by a co-moving photon. If you don't figure it's zero - think again.

I think they have a real word for "anti-parallel". It is called perpendicular. I really so no reasoning here to contemplate on. I tried thinking again, but nothing really comes to mind from this.

It should be obvious that any quantum gravity theory would have to come to zero for being at half wavelengths. At a half wavelength, two waves would be completely out of sync. They would in effect cancel each other out, eliminating the wave. Then particles traveling together at half wavelengths cannot travel that way indefinitely. The electromagnetic force would have to repel them, but the photon doesn't have charge.

That brings me to the point that the Higgs Boson can create two photons even though it doesn't have charge either. That is something that should be impossible in quantum physics, but that was what happened. The discovery was supposed to tell us what was wrong with quantum theory. Then we got something different, and everyone just ignores it (why am I not surprised?). There has to be an unexplained link between the photon and charge that we do not know about.

Charge does the job of keeping certain like particles from being able to come together. If electrons were bundles of virtual photons, then charge would basically be the limit of the amount of virtual particles that could take up one space. Universities have combined the amount of energy it would take to get the mass of an electron as according to $$E=mc^2$$. Basically, if you send enough energy of photons together that have the total combined energy needed to equal the mass of an electron, that is what you get.

 

[OnlyMe]

I think they have a real word for "anti-parallel". It is called perpendicular. I really so no reasoning here to contemplate on. I tried thinking again, but nothing really comes to mind from this.

Anti-parallel would include any path other than 0 and 180 degrees relative to eachother, not just perpendicular or 90 degrees. Which would expand the possibilities greatly in a 3D coordinate system.

It should be obvious that any quantum gravity theory would have to come to zero for being at half wavelengths. At a half wavelength, two waves would be completely out of sync. They would in effect cancel each other out, eliminating the wave. Then particles traveling together at half wavelengths cannot travel that way indefinitely. The electromagnetic force would have to repel them, but the photon doesn't have charge.

This assumes that individual EM waves/photons would interfere with eachother, something we don't see in any practical way, as far as this discussion is concerned. (I used wave/photon here only because it might be easier to think of EM waves as propagating independently, when thought of as photons.)... Two equivalent waves on different paths on the surface of a body of water, only apparently interfere where they are colocated. Before and as soon as they pass that apparent interference vanishes.

 

[rpenner]

No. For rays and vectors and directed line segments, anti-parallel means "in opposite directions."

http://en.wikipedia.org/wiki/Antiparallel_(mathematics)#Antiparallel_vectors

http://en.wikipedia.org/wiki/Antiparallel (Physics)

http://www.sophia.org/tutorials/parallel-and-antiparallel-vectors--3

 

[Layman]

This assumes that individual EM waves/photons would interfere with eachother, something we don't see in any practical way, as far as this discussion is concerned. (I used wave/photon here only because it might be easier to think of EM waves as propagating independently, when thought of as photons.)... Two equivalent waves on different paths on the surface of a body of water, only apparently interfere where they are colocated. Before and as soon as they pass that apparent interference vanishes.

Photons traveling inside of a body of water would be scattered and move around randomly, hence lowering the speed of light, but photons traveling on the surface of water could allow for them to move together as their particle waves, like EM waves in a waveguide. Then the EM waves mostly travel on the surface of the waveguide. The surface of water could be acting like a waveguide as well, and I think it would take them traveling together as a wave to interfere with each other. It seems like it could still fit in with what I said.