QM + GR = black holes cannot exist
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The waves on a body of water was an analogy.., waves on the ocean.... Not a direct example of EM wave and water.
Listen to rpenner on the anti parallel, definition.... I was confusing the definition of anti as not rather than against.., and not thinking it through.
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Billy T
I thought you'd be able to understand what I wrote. No, the photons don't pull on each other. They follow the natural path [the null geodesic] determined by the local spacetime geometry [curvature]. There's no pushing or pulling going on. Along the local inertial geodesic path spacetime is flat to some infinitesimal limit over some distance over the spacetime manifold. Tangent to every point on the manifold the spacetime is flat. No spacetime curvature no gravity. I'll add this: The only forces associated with GR are tidal forces [dg]. Since the local spacetime curvature is an infinitesimal [flat] over the local natural path it's first derivative is an infinitesimal leaving a uniform tidal field over the path. GR is a LOCAL theory of gravity. No action at a distance forces needed. Hope that works.
BTW assuming I wasn't answering your question is stupid. It's your problem.
I thought you'd be able to understand what I wrote. No, the photons don't pull on each other. They follow the natural path [the null geodesic] determined by the local spacetime geometry [curvature]. There's no pushing or pulling going on. Along the local inertial geodesic path spacetime is flat to some infinitesimal limit over some distance over the spacetime manifold. Tangent to every point on the manifold the spacetime is flat. No spacetime curvature no gravity. I'll add this: The only forces associated with GR are tidal forces [dg]. Since the local spacetime curvature is an infinitesimal [flat] over the local natural path it's first derivative is an infinitesimal leaving a uniform tidal field over the path. GR is a LOCAL theory of gravity. No action at a distance forces needed. Hope that works.
BTW assuming I wasn't answering your question is stupid. It's your problem.
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That's a pity but not unexpected. And btw 'anti-parallel' does *not* mean perpendicular - it means anti-parallel. And I see post #322 has elaborated on that. As for the rest of your thoughts there - best I say nothing.
I have never heard that one before. It sounds like that discussion wouldn't go very far if your firm in the belief that photons wouldn't interact with each other in any way. I was just thinking about waveguides when I was reading your previous post, so I assumed you where talking about EM waves traveling on the surface of water. I have never really heard about that type of experiment being done before. Then in a waveguide, EM waves do interfere with each other. They even interfere with themselves before they would have time to get back to a certain location. There is particle precognition with an action at a distance. EM waves that would reflect down a guide at a half wavelength will choose to not go down that path to begin with. There is little known about why this effect occurs, but it only happens down paths that are exactly one half wavelength. I think it has to do with it being in a state of quantum superposition, and they are able to cancel themselves out in that path, making them undetectable. There is no known reason why particles should be able to choose or prefer full wave paths that would result in reflection.
Dear layman; that's not what I said. Recall the immediate subject was *gravitational* interaction between photons. Conflating that with the well understood phenomenon of *classical* wave interference, either free-space or in waveguides, just gets one all in a tangle.
A full QED treatment is certainly beyond my capabilities (maybe e.g. Fednis48 could elaborate*) but there is absolutely no need of it. Classical EM covers the above admirably. *Which btw has become way off topic. Cheers.
Too late already to edit #329, but this is what such an edit would read:
To bring this somewhat back to being on-topic, it's occurred to me there is something interesting to consider in respect of interference in say a waveguide. Notwithstanding your idea that:
To bring this somewhat back to being on-topic, it's occurred to me there is something interesting to consider in respect of interference in say a waveguide. Notwithstanding your idea that:
That is not how either QED or classical EM treats it. The waves/coherent photons do go down - but phase is all-important and net energy density is zero. This does raise an interesting thought. How does one treat the assumed gravitational field of a propagating photon, given interference may null out, over an arbitrarily large path length, the collective energy of a coherent photon swarm? That is, a bystander adjacent to such guide section should not experience any net gravitational influence, despite each photon 'carrying' a gravitational field. Things start looking a bit strange when by classical SR, the 'electrogravitic' component of such a single photon gravitational field presumably should have a transverse, radial 'pancake' form infinitely thin (and infinitely intense!) delta function, the 'magnetogravitic' component being likewise transversely infinitely compressed. Makes one wonder a bit. Can't actually be so - no real infinities in nature. Maybe this is something for a separate thread. And I won't be starting one if so.
The point I was trying to make is that, if you consider phenomena like particle jumps and particle precognition with an action at a distance as being due to the mismatch of particle waves, under these two effects the mass of particles goes to zero when the particle waves are combined or reflected at 1/2 wavelength (the wavelengths of the particles would have to affect a change in mass in order for quantum gravity to accurately describe these two phenomena). Then quantum gravity would have to be described as particle waves in order to obtain that type of result that is what has been seen from quantum experiments. The mass of particles would have to be able to be shown to change combined at different wavelengths. The problem with that is though that quantum mechanics doesn't really seem to involve descriptions of combined particle waves, and it just assumes there is just a certain probability that a particle will be at any location. That would seem to be the reason why there cannot be a valid theory for quantum gravity. It would take describing hidden variables. It doesn't seem like quantum theory would even be equipped to approach the problem accurately if quantum gravity was dependent on wavelike interactions of virtual particles.
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layman you seem , reasonable
may I suggest a book that would expand on your thinking ?
just asking
river
What book would that be?
The Philosophers' Stone , by Joseph P. Farrell
I don't think this is really an issue. Form an SED approach Haisch, Reuda & Puthoff (and others) deal with the interaction between the ZPF and charged particles as a boundary conditions interaction. I think their work is a bit easier to understand in the context of inertia than gravity and both do still have unanswered issues, but dealing with the interaction bewptween ZPF EM wave or wavelike interactions is not the problem.
I thought that was what you were saying in your post 310, but this part of it left me in doubt:
"matter tells spacetime to curve and the curvature determines the path. For the most part a photon doesn't interact with another photon."
Especially the part I made bold. I think your "matter" is limited objects with non-zero rest mass - does not include the
M = E/ (c^2) mass. I certainly agree that M makes very very little warp of the local space time, but not sure it is zero.
So I restated you view and asked if that was what you were saying.
The actual comment was from John A. Wheeler.
"Spacetime tells matter how to move; matter tells spacetime how to curve." Wheeler's succinct summary of Einstein's theory of general relativity, in Geons, Black Holes, and Quantum Foam, p. 235.
That sums it up nicely. GR is a LOCAL theory of gravity. What determines the natural geodesic path, of an object, is the LOCAL spacetime curvature. The tensor I mentioned describes all the components that make up the matter that contribute to the LOCAL spacetime curvature. IE gravity GR style. When I mention "For the most part" I qualified this by stating I'm not an expert on quantum physics so I don't have to argue about whether photons interact under special circumstances with the cranks that frequent these threads. As far as GR is concerned photons don't interact with eachother. They don't merge as you asked. Gravity is a local theory while Newtons Law Of Gravitation is a theory where things tug at eachother globally. I'm sure you can figure out why the local theory makes predictions for natural phenomena which have been empirically verified and the other non local theory doesn't. For example Newton predicts the bending of lights path as it passes a large body of matter, like the Sun, is
dphi = 2M/R_sun
The correct prediction from a GR analysis is
dphi = 4M/R_sun
Newtons theory predicts the Sun is tugging on the photon while the correct local theory of gravity predicts the photon is following it's natural path, geodesic, through the LOCAL gravitational field.
The GR prediction of 4M/r is the basis for the lensing equations you!ve heard about. Over the entire path the local spacetime curvature is 0 at every point on the spacetime manifold and an infinitesimal when we give a set of points physical extent. So over the entire path, global perspective, the infinitesimals sum to
dphi = 4M/r
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I was playing around with wolfram alpha the other night, and it made me realize that maybe $$m = h/λc$$ could be a real possibility. It makes me wonder if the reason why $$E=mc^2$$ doesn't work out in a lot of equations for people is because it is already quantized. If it was already quantized than you could just substitute energy for the energy of the photon, $$E = hc/λ $$.
I decided to check my answer, and the units for the equation actually come out to units of mass. The Planck Constant is in Joule-seconds, $$ {kg*m^2}/s$$. The speed of light is in $$m/s$$. When you divide these units you end up getting $$ kg*m$$. Then when you divide that by the wavelength that is in meters, you are just left with kilograms.
I then took a look at this chart in the wiki.
I then put in the values for the lowest end of the spectrum of gamma rays at $$10^{-16}$$m. This gave me $$2.21 * 10^{-26}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-16 m)(2.998×10^8 meters per second))
Then at the other end of the spectrum there are long radio waves at $$10^8$$m. This gave me $$2.21 * 10^{-50}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^8 m)(2.998×10^8 meters per second))
I noticed that the long radio waves gave me a lot less comparisons than the gamma rays, but I assume that the mass of long radio waves would be so minute that it wouldn't have anything close to compare it too. It appears that according to my little equation that short wave lengths are more massive than long wave lengths of photons. This got me to check the other values of gamma rays.
At $$10^{-14}$$m I got an answer of $$2.21 * 10^{-28}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-14 m)(2.998×10^8 meters per second))
At $$ 10^{-12}$$m I got an answer of $$2.21 * 10^{-30}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-12 m)(2.998×10^8 meters per second))
Then what I noticed was that according to the corresponding quantities of the gamma rays that I input, the Compton Frequency matched the same units with the values seen on the chart from the wiki. It gave $$10^{24}$$,$$ 10^{22}$$, and $$10^{22}$$ respectively. Then it seems like if you found the mass in this manner, that you would be able to accurately find the frequency of the photon given the equation $$v={mc^2}/h$$.
I decided to check my answer, and the units for the equation actually come out to units of mass. The Planck Constant is in Joule-seconds, $$ {kg*m^2}/s$$. The speed of light is in $$m/s$$. When you divide these units you end up getting $$ kg*m$$. Then when you divide that by the wavelength that is in meters, you are just left with kilograms.
I then took a look at this chart in the wiki.
I then put in the values for the lowest end of the spectrum of gamma rays at $$10^{-16}$$m. This gave me $$2.21 * 10^{-26}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-16 m)(2.998×10^8 meters per second))
Then at the other end of the spectrum there are long radio waves at $$10^8$$m. This gave me $$2.21 * 10^{-50}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^8 m)(2.998×10^8 meters per second))
I noticed that the long radio waves gave me a lot less comparisons than the gamma rays, but I assume that the mass of long radio waves would be so minute that it wouldn't have anything close to compare it too. It appears that according to my little equation that short wave lengths are more massive than long wave lengths of photons. This got me to check the other values of gamma rays.
At $$10^{-14}$$m I got an answer of $$2.21 * 10^{-28}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-14 m)(2.998×10^8 meters per second))
At $$ 10^{-12}$$m I got an answer of $$2.21 * 10^{-30}$$kg.
http://www.wolframalpha.com/input/?i=(6.62607×10^-34 joule seconds)/((10^-12 m)(2.998×10^8 meters per second))
Then what I noticed was that according to the corresponding quantities of the gamma rays that I input, the Compton Frequency matched the same units with the values seen on the chart from the wiki. It gave $$10^{24}$$,$$ 10^{22}$$, and $$10^{22}$$ respectively. Then it seems like if you found the mass in this manner, that you would be able to accurately find the frequency of the photon given the equation $$v={mc^2}/h$$.
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Given a recent statement claiming photons do not gravitationally interact according to GR, looks like I need to repeat myself. According to GR, photons (strictly speaking, light-beams, as photon per se is a concept foreign to classical GR) *do* gravitationally interact in general. The *sole* exception being when such photons/'pencil-lead' light-beams are strictly parallel i.e. co-moving/co-propagating. That is stock standard GR, and afaik was first elaborated by Tolman et. al. back in 1931:
http://authors.library.caltech.edu/1544/1/TOLpr31a.pdf
If anyone wishes to continue disputing that is still the standard GR position and call it 'crank', it would be kind of ironic. Having reiterated my position and reasoning of #319, will take this occasion to firstly retract the following statement made in #330:
http://authors.library.caltech.edu/1544/1/TOLpr31a.pdf
If anyone wishes to continue disputing that is still the standard GR position and call it 'crank', it would be kind of ironic. Having reiterated my position and reasoning of #319, will take this occasion to firstly retract the following statement made in #330:
I was carelessly applying that to a bullet-like point-particle picture of photon. Some kind of fuzziness has to be assumed, whether owing to position indeterminacy for particle picture, or finite field distribution on a field picture. So all that can be said is that IF a freely propagating photon does have an active gravitational mass, the ensuing field must according to SR be everywhere strictly transverse to the propagation vector k. And among other things this raises some severe issues of which a bit I touched on in #330. So I have genuine doubts the stock standard GR position as above referenced is in fact correct. While there is no doubt about light possessing an effective passive gravitational mass, the more I think about consequences of an active gravitational mass for a freely propagating photon/light-beam, the less sensible it seems. But off-topic to elaborate here.
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The problem is that in GR photons do not have a mass, but they follow the geodesics of spacetime curvature created by other massive bodies. In quantum mechanics, there is no spacetime geodesics to consider. If the photon has zero mass in quantum mechanics, it will not even be attracted to massive bodies. With no spacetime, the only way it could be affected by gravity is for it to have at least a very small amount of mass.
The reason why the photon is said to not have mass is because it was thought that an object with mass traveling the speed of light would then have infinite mass and wouldn't be able to move. More exactly, it would just have infinite mass relative to an observer saying that it is at rest according to GR. Meanwhile, it has recently started to become accepted in The Big Bang Theory that close to the moment of the Big Bang everything traveled FTL, and distant galaxies are immune to gravitational mass increase, not to mention all the circumstances the speed of light limit is broken in quantum mechanics. Then quantum mechanics just really wouldn't care if the photon had mass, only GR really would, and it wouldn't bring these kinds of infinities. Then GR and quantum mechanics are incompatible theories. Go figure...
In my post #338 the second to the last line came up in error. It was supposed to read, "It gave $$10^{24}$$,$$10^{22}$$, and $$10^{20}$$ respectively.", that actually was the decimal point the wavelengths are supposed to go to for gamma rays. The tex function seems to have gotten a bit sketchy since the changes to the website, and it was giving me troubles.
The reason why the photon is said to not have mass is because it was thought that an object with mass traveling the speed of light would then have infinite mass and wouldn't be able to move. More exactly, it would just have infinite mass relative to an observer saying that it is at rest according to GR. Meanwhile, it has recently started to become accepted in The Big Bang Theory that close to the moment of the Big Bang everything traveled FTL, and distant galaxies are immune to gravitational mass increase, not to mention all the circumstances the speed of light limit is broken in quantum mechanics. Then quantum mechanics just really wouldn't care if the photon had mass, only GR really would, and it wouldn't bring these kinds of infinities. Then GR and quantum mechanics are incompatible theories. Go figure...
In my post #338 the second to the last line came up in error. It was supposed to read, "It gave $$10^{24}$$,$$10^{22}$$, and $$10^{20}$$ respectively.", that actually was the decimal point the wavelengths are supposed to go to for gamma rays. The tex function seems to have gotten a bit sketchy since the changes to the website, and it was giving me troubles.
You really ought to try and read and understand that classic article I referenced earlier. In GR, light has no invariant/proper/rest mass, but it does have an effective active gravitational mass (with directional character in general case). A box full of incoherent radiation has in GR a net isotropic 'radiant mass' according to the famous mass energy equivalence E = mc^2. But for reasons touched on, I doubt it really is so. Photon-wall interactions add to the mix though and make it not at straightforward to analyze as for free-propagation-in-vacuum case.
danshawen
Valued Senior Member
And I don't understand why other posters here seem to resist the idea that although gravitation can curve the path of a photon, and EVEN A DROPPED GOLF BALL CAUSES THE WHOLE EARTH TO MOVE TO MEET IT, that the same gravitational interaction would have to mean that as the beam of light bends, the gravitating object ACTUALLY MOVES CLOSER TO THE BEAM.
E = mc^2 means EXACTLY what it says, not whatever you wish it to mean. It was classically derived from a consideration of the interaction of a photon with a solid mass. Bound energy can have inertia AND gravity. And it is also possible to (briefly) impart inertia to the energy of the vacuum also, and that is the real reason behind gravitation that appears to be "spooky action at a distance", and the 'curvature of space-time'.
E = mc^2 means EXACTLY what it says, not whatever you wish it to mean. It was classically derived from a consideration of the interaction of a photon with a solid mass. Bound energy can have inertia AND gravity. And it is also possible to (briefly) impart inertia to the energy of the vacuum also, and that is the real reason behind gravitation that appears to be "spooky action at a distance", and the 'curvature of space-time'.
Yes danshawen - the standard GR position - as per cited article.
I guess this was directed at me, but given no citation, that's a guess. That the inertial and active mass might be different may be 'heretical' but that doesn't automatically make it wrong. But to pursue it further here is wrong as this is all now way off OP topic. If you have some passion for this then think about starting a new thread - just on active mass of light/photon.
The latter stuff is clearly your own 'heretical' speculation, and should if you wish be taken up in a further separate thread.
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