Confirming Einstein, scientists find 'spacetime foam' not slowing down photons from faraway gamma-ray burst (Update)

March 16, 2015
This is the "South Pillar" region of the star-forming region called the Carina Nebula. Like cracking open a watermelon and finding its seeds, the infrared telescope "busted open" this murky cloud to reveal star embryos tucked inside finger-like pillars of thick dust. Credit: NASA

One hundred years after Albert Einstein formulated the general theory of relativity, an international team has proposed another experimental proof. In a paper published today in Nature Physics, researchers from the Hebrew University of Jerusalem, the Open University of Israel, Sapienza University of Rome, and University of Montpellier in France, describe a proof for one of the theory's basic assumptions: the idea that all light particles, or photons, propagate at exactly the same speed.

The researchers analyzed data, obtained by NASA's Fermi Gamma-ray Space Telescope, of the arrival times of from a distant gamma-ray burst. The data showed that photons traveling for billions of years from the distant burst toward Earth all arrived within a fraction of a second of each other.

This finding indicates that the photons all moved at the same speed, even though different photons had different energies. This is one of the best measurements ever of the independence of the speed of light from the energy of the light particles.

Beyond confirming the general theory of relativity, the observation rules out one of the interesting ideas concerning the unification of general relativity and quantum theory. While these two theories are the pillars of physics today, they are still inconsistent, and there is an intrinsic contradiction between the two that is partially based on Heisenberg's uncertainty principle that is at the heart of quantum theory.

One of the attempts to reconcile the two theories is the idea of "space-time foam." According to this concept, on a microscopic scale space is not continuous, and instead it has a foam-like structure. The size of these foam elements is so tiny that it is difficult to imagine and is at present impossible to measure directly. However that are traveling within this foam will be affected by the foamy structure, and this will cause them to propagate at slightly different speeds depending on their energy.

Yet this experiment shows otherwise. The fact that all the photons with different energies arrived with no time delay relative to each other indicates that such a foamy structure, if it exists at all, has a much smaller size than previously expected.

"When we began our analysis, we didn't expect to obtain such a precise measurement," said Prof. Tsvi Piran, the Schwartzmann University Chair at the Hebrew University's Racah Institute of Physics and a leader of the research. "This new limit is at the level expected from quantum gravity theories and can direct us how to combine and relativity."

Explore further: Gamma Ray Delay May Be Sign of 'New Physics'

More information: A Planck-scale limit on spacetime fuzziness and stochastic Lorentz invariance violation, Nature, 2015. DOI: 10.1038/nphys3270

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jscroft
4.5 / 5 (16) Mar 16, 2015
Ok is it just me or does the headline say the exact OPPOSITE of what the article says?

CONFIRMING EINSTEIN, SCIENTISTS FIND SPACETIME FOAM...

Yet this experiment shows otherwise. The fact that all the photons with different energies arrived with no time delay relative to each other indicates that such a foamy structure, if it exists at all, has a much smaller size than previously expected.


Sean_W
4.7 / 5 (15) Mar 16, 2015
Great article but it seems to invalidate the headline.
SoylentGrin
4.7 / 5 (14) Mar 16, 2015
I'm not sure how to reconcile the headline with the content of this article.

Headline:
Confirming Einstein, scientists find 'spacetime foam'


From the article:
The fact that all the photons with different energies arrived with no time delay relative to each other indicates that such a foamy structure, [b]if it exists at all[/b], has a much smaller size than previously expected.
Dethe
1 / 5 (15) Mar 16, 2015
does the headline say the exact OPPOSITE of what the article says
Yes, but what do you expect from PO headlines? They didn't find any evidence of space-time foam - or better to say, they were fooled with its properties as usually. IMO the trick is, the photons move with different speed across vacuum, but the same effect leads to their mutual attraction, so that they still arrive at the same moment. They just travel along different path, revolving each other..
Dethe
1 / 5 (16) Mar 16, 2015
The fact that all the photons with different energies arrived with no time delay .. indicates that such a foamy structure ....has a much smaller size than previously expected.
The inhomogeneities of space-time are actually quite large - they correspond the wavelength of CMBR, which is an analogy of Brownian noise at the water surface.. If we would have microwave eyes, we would see how the vacuum around us undulates like the hot air shivering above the camp air. But the same inhomogeneities have self-focussing effects for photons in similar way, like the foamy structure of hollow core waveguides. Apparently some parts of contemporary physics are more entertaining than others..
Shootist
4.6 / 5 (18) Mar 16, 2015
[ they correspond the wavelength of CMBR, which is an analogy of Brownian noise at the water surface..


horseshit on rye..
antialias_physorg
5 / 5 (12) Mar 16, 2015
There goes the LQG. Damn shame. Was a beautiful theory. But unless the foam is a lot smaller than envisioned there's probably no way to save it.

Oh well. Back to the drawing board.
Dethe
1 / 5 (12) Mar 16, 2015
new limit is at the level expected from quantum gravity theories
Everyone probably understands, why the ignorance of contemporary physics killed the dense aether model and scalar wave physics. But the ignorance is blind and it also delayed the understanding of mainstream formal theories (stringly & loopy theories), which are predicting some aspects of aether model, like the extradimensions and/or Lorentz symmetry violation. The basis of this misunderstanding is in belief, that the realm of quantum gravity theory exists at the esoteric distance scales somewhere around Planck length. But every theory reconciling general relativity and quantum mechanics should preferably operate at the distance scales BETWEEN these two theories, i.e. just at the human observer scale. The evidence of quantum gravity or extradimensions are all phenomena around us - the various forces, refractive phenomena, atoms, plants, bees and also human creatures. And of course the common photons and CMBR noise.
Dethe
1 / 5 (11) Mar 16, 2015
In addition, there is an apparent lack of will to consider the existence of phenomena, which are already recognized and described under different context as a confirmation of new theories. For example if we arrive at Earth and hit the atmosphere, then the gravity force following the inverse square law will get apparently violated with various short distance forces, which could be interpreted like the manifestation of extradimensions around Earth. But the mainstream physicists don't think in this way and they tend to ignore all existing phenomena just because they were observed and named before few centuries. It's sorta collective professional blindness: the physicists cannot think like the Martians, who are approaching the Earth while having no idea about existing concepts of physics - they can just see, that the gravity doesn't work normally at the proximity of Earth.
Doug_Huffman
2 / 5 (4) Mar 16, 2015
Ok is it just me or does the headline say the exact OPPOSITE of what the article says?
More click-bait. "Whaa'? They can't have."
Protoplasmix
5 / 5 (10) Mar 16, 2015
Ok is it just me or does the headline say the exact OPPOSITE of what the article says?
The quantum theories predict the foam but the results support the prediction of relativity that spacetime is continuous. As stated in the article: "The fact that all the photons with different energies arrived with no time delay relative to each other indicates that such a foamy structure, if it exists at all, has a much smaller size than previously expected." So it looks like the headline should say, " ... don't find 'spacetime foam'".
Dethe
1 / 5 (12) Mar 16, 2015
Actually just the physicists recently developed foamy structure of light waveguides, which is compensating the differences in refraction index for various wavelengths. So that one group of physicists doesn't know, what does this another one.. Interestingly enough, the same microtubule structure (1 D foam) is utilized with neuron fibers for dispersive-less propagation of neural signals. So we can say, that the human brain mimics the vacuum structure for the same reason - to enable the propagation of information along as long path as possible. Or better to say, we are adopted to observation of Universe in just the directions/ways, which enable the easiest propagation of light at distance.
arnold_townsend
4.8 / 5 (18) Mar 16, 2015
Everyone probably understands, why the ignorance of contemporary physics killed the dense aether model and scalar wave physics. But the ignorance is blind and it also delayed the understanding of mainstream formal theories


You are so full of shit that if ever take a laxative you will disappear completely. Put down the bong and reacquaint yourself with reality.
version782
4.2 / 5 (5) Mar 16, 2015
Does this also invalidate Aether Wave Theory?
GoodElf
3 / 5 (2) Mar 16, 2015
Correct, there is no "quantum foam" structure in "empty spacetime" at the "Planck Length". Spacetime is "deformable" but at higher energies it becomes less and less "deformable" as probes like the LHC, at that scale, create their own "new" high energy particles/resonances. This is all about "resonances" at higher and higher energies as Feynman would have told us all.
It is "unfortunate" that so much effort has been placed into theories that "sound plausible" but have no experimental proof and is what happens when theory swamps experiment for so long, that it has become more "real" than the "reality". All "pop physics" talk in this stupid way and it is wrong. Dirac had his infinite "Quantum Sea" idea, swarms of "virtual particles" bobbing in and out of existence everywhere is an incorrect interpretation of quantum tunneling and are another fiction invented to solve these same problems and are like "epicycles" were to the Greeks because old researchers just could not accept Relativity.
Dethe
1 / 5 (10) Mar 16, 2015
It's complicated. The photons which are of shorter wavelength should propagate slower. But just these photons are also heavier and they also attract another lightweight photons, which are revolving them along longer path, which compensates their higher speed. So that as a whole the cluster of photons arrives in a single moment - but the heaviest photons should be localized around its center. The existing observations of gamma bursts really support this scenario.
In addition, the photons in gamma rays aren't always without delays. The closer gamma bursts (like the MKN501) often exhibit wider separation in arrival times of photons, than these more distant ones. IMO it's because the photons in nearby bursts had no time to form a dense cluster of photons bound by gravity. But the bursts from larger distance are homogenized already, because the photons of different speed already leaved the cluster and they were scattered.
El_Nose
3.7 / 5 (3) Mar 16, 2015
@Version

it one of a number on the quantity of the number of grains of sand you can hold in your hand the disproves AWT. Which as a model has some issues with predicting things you take for granted as predictable. but hey, i have some very good friends who are flat worlders so too each there own.
acronymous
3.9 / 5 (7) Mar 16, 2015
Ok is it just me or does the headline say the exact OPPOSITE of what the article says?

CONFIRMING EINSTEIN, SCIENTISTS FIND SPACETIME FOAM...

Yet this experiment shows otherwise. The fact that all the photons with different energies arrived with no time delay relative to each other indicates that such a foamy structure, if it exists at all, has a much smaller size than previously expected.



Yep. The headline got it 180 wrong.
rufusgwarren
2 / 5 (4) Mar 16, 2015
How is this a proof? Really?
rufusgwarren
2 / 5 (4) Mar 16, 2015
All we see is dual arrival, emission is yet to be confirmed. Again, how was this measured?
rufusgwarren
1 / 5 (4) Mar 16, 2015
Read Jung, synchronicity is usually explainable with statistics!
vic1248
5 / 5 (3) Mar 16, 2015
Einstein's Space-Time Is Fundamentally Smooth Theory vs. Space-Time Is Foamy Theory

http://www.space....oth.html
mikep608
1 / 5 (9) Mar 16, 2015
We exist in a solid electronic medium. Light is this medium having a reaction from protons and electrons. This reaction can only be pulled forward at the same speed. This is why light with different energy levels develope different wave length magnitudes.
here's my science page link
https://www.faceb...e?ref=hl
Benni
1 / 5 (5) Mar 16, 2015

@Ira, would you please explain this to us:

It's complicated. The photons which are of shorter wavelength should propagate slower. But just these photons are also heavier and they also attract another lightweight photons, which are revolving them along longer path, which compensates their higher speed. So that as a whole the cluster of photons arrives in a single moment - but the heaviest photons should be localized around its center. The existing observations of gamma bursts really support this scenario.
In addition, the photons in gamma rays aren't always without delays. The closer gamma bursts often exhibit wider separation in arrival times of photons, than these more distant ones. IMO it's because the photons in nearby bursts had no time to form a dense cluster of photons bound by gravity. But the bursts from larger distance are homogenized already, because the photons of different speed already leaved the cluster and they were scattered.


PhyOrgSux
2.6 / 5 (5) Mar 16, 2015
This "foam" seems to be same old aether just by another name;-P
vic1248
4.9 / 5 (14) Mar 16, 2015
The implications of this finding are of epic proportions.

First, they prove that light photons travel at a "constant speed" independent of their energies, a huge score for Einstein's "Theory of General Relativity." Second, they prove that space-time is "continuous" and not foamy as posited by quantum theoretical physicists, another huge score for Einstein's "Theory of General Relativity."

What all that means is, in the grand scheme of things, the concept of reconciling the Theory of General Relativity and the Quantum Field Theory is changing little by little to "either or," and the tides are favoring the Theory of General Relativity.

p.s. I saw some remarks regarding Aether Wave Theory, that has been ruled out a long time ago affirmatively.
PhysicsMatter
not rated yet Mar 16, 2015
The data showed that photons traveling for billions of years from the distant burst toward Earth all arrived within a fraction of a second of each other.

Did they say photons arrived at the same speeds of light so why there was fraction of a second delay among them if they came precisely from the same point of origin. Fraction of second could be huge, may be fraction of femtosecond they meant? So there is no foam, space time is continuous to infinity, back to classics.

I found interesting take on problems with relativity at:
https://questforn...ativity/
and problems with quanta at:
https://questforn...-quanta/

fred s
4.9 / 5 (9) Mar 16, 2015
"Ok is it just me or does the headline say the exact OPPOSITE of what the article says?

'CONFIRMING EINSTEIN, SCIENTISTS FIND SPACETIME FOAM...

Yet this experiment shows otherwise. . .' "

No, it doesn't; the headline is spot on.

What you missed is that Einstein's theory of General Relativity (GR) says that spacetime is smooth at all scales, no matter how tiny.

What quantum mechanics (QM) says, is that at the Planck scale, spacetime is topologically no longer a simple 4D manifold; it is foamy rather than smooth at such a scale.

So under GR, all photons emitted simultaneously, travelling through vacuum, would arrive exactly simultaneously, right down to the femtosecond over a 10-billion year journey.

QM, OTOH, says that photons with higher energies, thus shorter wavelengths, would be affected more by the quantum foam nature of spacetime, than would lower-energy, longer-wave photons. Thus, in this case, Einstein (GR) is confirmed, vs QM — as per the headline.
Protoplasmix
5 / 5 (1) Mar 16, 2015
Oops. Never mind...
mikep608
1 / 5 (4) Mar 16, 2015
We exist in a solid electronic medium. Light is this medium having a reaction from protons and electrons. This reaction can only be pulled forward at the same speed. This is why light with different energy levels develope different wave length magnitudes.
here's my science page link
https://www.faceb...e?ref=hl

This makes me think that they should study light of different energy levels. Why? Maybe the PARTICLE like property is an illusion that may be more attached to light with higher energy levels.
Uncle Ira
4.1 / 5 (9) Mar 16, 2015
@Ira, would you please explain this to us:

It's complicated. Blah, blah, blah,,,,


@ Bennie-Skippy I would if I could but I can't. I'm not the scientist like you aren't either. So Cher, you got to get somebody smarter than both of us to explain him to you. You might ask the Zephir-Skippy he is the one who wrote it to you..
TheGreatApe
1 / 5 (2) Mar 16, 2015
I came to crow about the genius of Einstein once again being confirmed, and was denied. Click bait. Wouldn't have read it if the headline read "Not confirming Einstein, scientists doesn't find 'spacetime foam' so it's not slowing down photons from faraway gamma-ray burst that they know of."

Protoplasmix
2.8 / 5 (4) Mar 16, 2015
Wait, my oops post was:
"Well, there's clearly an interaction (coupling?) that occurs between matter (energy) and spacetime, as gravitational lensing demonstrates. So the variable(s) for the foam remain -- hidden ? mm hmm..."

Edited to add: Because if there is no foam (or no direct interaction between energy and "empty" spacetime) then the force of gravity would have to propagate faster than light, correct?
David Aye
5 / 5 (1) Mar 16, 2015
Given that the speed of light can be altered by it passing through a solid, transparent substance, is it not possible that individual photons could be slowed down along their journey by passing through water ice (even microscopic) particles, or even nitrogen, methane, etc?
DarkLordKelvin
4.3 / 5 (6) Mar 16, 2015
Given that the speed of light can be altered by it passing through a solid, transparent substance, is it not possible that individual photons could be slowed down along their journey by passing through water ice (even microscopic) particles, or even nitrogen, methane, etc?


Almost certainly not .. any interaction like the ones you describe has a high probability of changing the direction of the photon. The chance that a photon coming from such a distant object that was initially on a trajectory that would intersect the Earth at the appropriate point in space (and time) would remain on that trajectory after a scattering or refraction event is negligibly small.
Protoplasmix
not rated yet Mar 16, 2015
3rd edit, sorry --

... or there could be 'gravitons' propagating at c, instead of foam... still QM tho'...
kamcoautomotive
Mar 16, 2015
This comment has been removed by a moderator.
Dethe
1 / 5 (7) Mar 16, 2015
Try to think about this: if the speed of photons would depend on their wavelength, then the photons would be massive. If the photons would be massive, they would attract each other and they would travel inside of gamma ray burst as a single body. After then their arrival times wouldn't differ anyway - and this is IMO what we are observing by now.
tonyv_414
1 / 5 (2) Mar 16, 2015
QM admits the difference between undisturbed dynamics versus the requirements of measurement (actual or abstract). GR does not.
GR spacetime requires continuum, but the measurement (observation) of spacetime, upon which all of mathematical accessibility and human sensorimotor functioning relies, requires discrete processing. Logical and mathematical processing, by definition, requires a "step-like" approach. That is what the mathematical "limit" is all about. To take a measurement is to take a "still photograph" of a point on the continuum. The still point is ALWAYS a fiction in a universe of continuous motion. This is what the "limit" in calculus is. Science will NEVER arrive at a singular meta-theory until it realizes that it needs the contradiction of discrete versus continuum in order to justify abstract modeling of any kind. Relativity is built on the "fictions" of the tensor calculus; the "limits" in the calculus are really no different than the fictitious foam required by QM
ursiny33
1 / 5 (3) Mar 16, 2015
If I was looking for a mechanical explanation of the different speeds between I high energy photons and low energy photons, it would assume a low energy photon has a smaller energy mass traveling at speed in a vacuum bubble generated around it by its speed and motion thru a dark matter environment pushing a path thru that dark matter environment, and the high energy has a larger electromagnetic mass that pushes thru more dark matter in its flight path with more dark matter particle resistance in its flight, but that's mechanical speculation
adam_russell_9615
5 / 5 (4) Mar 17, 2015
Given that the speed of light can be altered by it passing through a solid, transparent substance, is it not possible that individual photons could be slowed down along their journey by passing through water ice (even microscopic) particles, or even nitrogen, methane, etc?


Well, then there would be refraction, correct? And photons of differing energy would be refracted through different angles.
Dethe
1 / 5 (6) Mar 17, 2015
The different speed of light for different wavelengths follows from quantum field theory too. The gamma ray photons can materialize with photons of CMBR (actually the more, the higher energy they have) under temporal formation of massive particles, which would move with subluminal speed. These particles will decay immediately back into photons, but the speed of light as a whole will remain lowered. Such an explanation doesn't violate the relativity. In addition, there exist some limit for gamma ray photons, which would decay into massive particles (pions, etc.) in this way (so-called the GZK limit), but the long distance bursts seems to evade this limit. Therefore the situation, in which all photons in gamma ray burst arrive at the single moment is strange in this way or another - especially for long distance bursts.
mooster75
3.7 / 5 (3) Mar 17, 2015
[Everyone probably understands, why the ignorance of contemporary physics killed the dense aether model...
I thought that was you...
Dethe
1 / 5 (5) Mar 17, 2015
I can just agree with University of Cambridge...
mortoo
4.3 / 5 (3) Mar 17, 2015
My first thought was, why hadn't people looked at this earlier? Couldn't you see it in the wavelength dependence of pulsar timing?

So I looked at wikipedia about pulsars and it says, "Because of the dispersive nature of the interstellar plasma, lower-frequency radio waves travel through the medium slower than higher-frequency radio waves".

Is it because only radio waves not gamma rays are interacting?
Dethe
2.6 / 5 (5) Mar 17, 2015
Couldn't you see it in the wavelength dependence of pulsar timing
Why/how it should be observable? The scattering of radiowaves you cited above is caused with interstellar plasma, not with vacuum itself.
Navid010
5 / 5 (1) Mar 17, 2015
I really dont understand the topic
its opposite from the article
or am i wrong?
esophal
1 / 5 (1) Mar 17, 2015
This maybe true if space is the same. How about light travel thru freezing water?
thingumbobesquire
3.7 / 5 (3) Mar 17, 2015
I guess today we'll have to settle for the foam on green St Patrick's Day beer then.
Torbjorn_Larsson_OM
5 / 5 (2) Mar 17, 2015
Nice! This is consistent with earlier observations over cosmological distances of photon velocity and polarization (which also can be scattered by fluctuations).

Arkani-Hamed had a public presentation of today's physics posted on the web last year, and he noted that supersymmetry would control expected fluctuations to this degree.

[To wit, as I understood it: Supersymmetry relies on internal field 'space' degrees of freedom, and such freedoms would be small in string theory. They should couple with the large dimensions of space, and the result would be that only small fluctuations (relative space) would be observed.]
Torbjorn_Larsson_OM
5 / 5 (3) Mar 17, 2015
@fred: "What you missed is that Einstein's theory of General Relativity (GR) says that spacetime is smooth at all scales, no matter how tiny. What quantum mechanics (QM) says, is that at the Planck scale, spacetime is topologically no longer a simple 4D manifold; it is foamy rather than smooth at such a scale."

Not quite. GR predicts its own demise, because it breaks down at high energies, while QM doesn't need to. The idea of "quantum foam" is idiosyncratic, they are then attempting to apply QM directly on space. In reality, GR and QM plays nice, you can quantize GR and get the standard model graviton.

The problem comes then GR breaks down at high energies (so at small scales).

These types of experiments are the first that probe the Planck scale, but they do so weakly. You integrate out the small scale over the universe in order to say anything. If we can observe primordial gravity fluctuations and probe inflation, we may have another window.
Torbjorn_Larsson_OM
5 / 5 (2) Mar 17, 2015
@AAP: "There goes the LQG. Damn shame. Was a beautiful theory."

Well, it goes, but only because people still insist it is a "theory".

If it would be a theory of physics and not a construction of geometry, it would predict dynamics, energy & particles move. However it fails to do so, because it has no lower energy limit. Typically LQG mathematicians insert a toy model to emulate a harmonic oscillator (say taking a non-existent "limit" onto a HO) and then useful energy and particles to get anywhere, but it isn't predicted by the theory.

I thought it looked nice when I first heard about it, but after studying it, its shortcomings were obvious. It stopped being "beautiful" and turned ugly, in my eyes.
antialias_physorg
5 / 5 (2) Mar 17, 2015
LQG does (or did) make predictions in that there is a limit to the granularity and that granularity would have predicted that lower energy photons move a tiny bit faster than higher energy photons. Indirectly this would predict a maxium allowable energy for photons.
There had already been some indication that this was not so (an observed supernova a few years back where the uncertainty of measurement about the various wavelengths arriving was within a few seconds).

Well...there's plenty more theories out there.
El_Nose
5 / 5 (2) Mar 17, 2015
In other news the Headline is technically correct ... read more than the first 6 words in the sentence. It took me a moment too, cause I am used to headlines that are phrases... this one is a complete sentence and is meant to be read as such.... and yes it is still confusing.... but technically correct.
richardwenzel987
not rated yet Mar 17, 2015
Well, there's always the possibility that the universe is inconsistent. We've always assumed that the world we live in is logical and orderly, but in fact it might be illogical and random at the most fundamental levels. In that case we should be looking very closely at the Central Limit Theorem, from statistics. A collection of random variables can simulate order, at very coarse scales. Maybe it isn't even necessary to assume finite variance. In that case, the fun thing is, everything could fall apart at any moment. And so, eat, drink, and be merry...
Benni
1 / 5 (5) Mar 17, 2015
@Ira, would you please explain this to us:


@ Bennie-Skippy I would if I could but I can't. I'm not the scientist like you aren't either. So Cher, you got to get somebody smarter than both of us to explain him to you. You might ask the Zephir-Skippy he is the one who wrote it to you.


........well I just figured all you AWT guys have this thing whereby the AWT funny farm stuff is so innate to the way you think that you are equally capable of expositing it. You mean to tell me you are incapable of even following non-Differential Equation based science?
physicsbeany
5 / 5 (1) Mar 17, 2015
In other news the Headline is technically correct ... read more than the first 6 words in the sentence.


The headline is technically correct *now*; they edited it. Originally there were only those six words you alluded to.
antialias_physorg
5 / 5 (1) Mar 17, 2015
A collection of random variables can simulate order, at very coarse scales. Maybe it isn't even necessary to assume finite variance. In that case, the fun thing is, everything could fall apart at any moment.

Even with 'fundamental randomness' (which I find an appealing concept from a philosophical point of view as it would give an answer to "why THESE particular sets of physical laws") there doesn't exist full macroscopic randomness IF that fundamental randomness acts on a small enough scale entity independently.

So while everything could fall apart under such a premise the chances are so infinitesimal, that it is way more likely you would spontaneously walk through a wall by every atom in your body tunneling just at that moment (which is already so unlikley that you will need a googolplex of universes until that happens).
Dethe
1 / 5 (6) Mar 17, 2015
Whole the stuff is a demonstration of rather funny way, how the nature plays with scientists like the cat with mouse. The aether concept has been dismissed before one hundred years, because people believed, that that vacuum will scatter the light waves like the sparse gas. No one realized, that if light travels across vacuum in TRANSVERSE waves, it must propagate through it like the waves across foam, i.e. along membranes in similar way, like the ripples at the water surface - the waves in sparse gas are always LONGITUDINAL.

Now, when the smartest LQG physicists finally developed the foam model for vacuum with using of rather abstract and indirect methods, then the nature smiles again and it "says": "well, guys - you must try harder". There are essentially four options here.

At first, the foam model is BS and the light waves are propagating through empty vacuum in some miraculous way, which cannot be derived from solid state physics in apparent way. And we should live with it.
Dethe
1 / 5 (6) Mar 17, 2015
At second, the foam model is correct - but it applies to Planck scale or smaller, so it's essentially untestable today. This is also the outcome of the above article.

At third, the foam model is correct and its cells are quite large - but the photons do utilize some less trivial way, how to travel along this foam. I already pointed above, that in hollow core fibers or neurons the foam is actually used for elimination of scattering of light wave. This foam is one-dimensional, oriented in parallel to path of wave spreading and we can ask, how the normal foam will appear for particles, spreading with speed of light - well, it will also look one-dimensional, don't you think?

At fourth, we are simply fooled and the vacuum foam works rather normally - just the photons move together, but along differently curved spiral paths, so that the Lorentz symmetry is preserved globally, but still violated locally. This is IMO the most probable option with respect to massive character of photons.
Dethe
1 / 5 (6) Mar 17, 2015
The massive character of photons is also trivial consequence of LQG model, if we consider, that only massless particles can propagate through vacuum with speed of light. If their speed will depend on their wavelength/energy, then these particles cannot be massless anymore. After then we can ask, why not to consider the LQG consequently and consider the gravitational interaction of massive photons, which will keep them together. After then all photons in gamma ray burst can still arrive at target at the same time, they will just use slightly different paths during it. In another words, the LQG and foamy model of vacuum work well and in agreement with observations - we're just applying it non-consequentially.

Albert Einstein: "If the facts don't fit the theory, check the facts".
Benni
1 / 5 (6) Mar 17, 2015
The massive character of photons is also trivial consequence of LQG model, if we consider, that only massless particles can propagate through vacuum with speed of light. If their speed will depend on their wavelength/energy, then these particles cannot be massless anymore. After then we can ask, why not to consider the LQG consequently and consider the gravitational interaction of massive photons, which will keep them together. After then all photons in gamma ray burst can still arrive at target at the same time, they will just use slightly different paths during it. In another words, the LQG and foamy model of vacuum work well and in agreement with observations - we're just applying it non-consequentially.

Albert Einstein: "If the facts don't fit the theory, check the facts".


I'm curious & I'm sure Ira is too , have you ever seen a Differential Equation you could solve? You know, like maybe a few straight out of Einstein's GR?
Uncle Ira
4.1 / 5 (9) Mar 17, 2015
I'm curious & I'm sure Ira is too , have you ever seen a Differential Equation you could solve?


Alrighty Cher, now we are on the same frequency us together. This means you are going to do the differential equation for me so I won't be curious about whether you can do more than write differential equation six or five time on every thread. Choot, I can write down differential equation in every postum if I wanted too, but it does not mean I know how to cipher one.

Goody goody gee whiz, now I won't have to be curious any more.

Oh yeah, I almost forget. I don't mind, and I'm sure nobody else will mind much either if you need to post up more than a couple of postums to get it up there because of the letter limit thing. So Bennie-Skippy I will stand by to see you do the differential equation solving thing.
Dethe
1 / 5 (6) Mar 17, 2015
It would be a bit tricky, because these photons don't propagate along straight path, but along spirals - so we have to solve the quantum equations in 3D. It's rather job for computers. But we already know about analogy of massive photons from studies of freak waves at oceans together with their formal models. Under certain circumstances these solitons behave like the massive objects and they collect the energy of another waves from their neighborhood. I'm convinced that the photons must be massive, or the stars couldn't radiate their mass via photons. The massless light is matter of only Maxwell waves, which are transverse - but the gamma ray photons are rather pinpoint particles and they have 2-spin component, which gives them mass.. But it's difficult to prove it with observations, which indicate exactly the opposite. ;-)
Dethe
1 / 5 (6) Mar 17, 2015
The scientists are in position of teacher, who weights his pupils by their speed at racetrack. Now he faces the group of fatties, who are running slowly but in single group. So that he deduces, that these fat kids are probably running with the same speed like the thin sporty kids, because their speeds don't differ. But we actually didn't compare the gamma ray photons with the speed of these low energy ones, so that our deduction still holds water. Maybe the fat kids sport another hidden mechanism (solidarity), which keeps them in group.
Dethe
1.6 / 5 (7) Mar 17, 2015
The differential equations are nice and all - but until we will not realize, what actually happens there, then our formal models can only reproduce the insights, which we put into their derivations - at the best case. And because existing formal models really disfavor every idea of massive photon (not to say the photon exhibiting a gravity coupling at distance), the physicists really cannot derive them without deep rebuilding of existing theories.

The whole source of misunderstanding starts with belief, that when the light wave is formed with photons, then the photons must also follow the postulates of special relativity (which has been originally developed for light waves). But the light wave isn't equivalent to photon, so that the special relativity has actually nothing to say about photons as such (which are quantum objects by itself).
DarkLordKelvin
4.6 / 5 (9) Mar 17, 2015
I'm convinced that the photons must be massive, or the stars couldn't radiate their mass via photons.
What? Are you unfamiliar with that "most famous equation" from special relativity? Or the fact that stars are nuclear reactors? Or are you just mincing words in the same manner that the antiquated concept of "relativistic mass" used to? Photons have momentum, which is inversely proportional to their wavelength, but their intrinsic mass is zero. This is high-school level stuff now in the year 2015.

But it's difficult to prove it with observations, which indicate exactly the opposite. ;-)
Most scientists would take disagreement between observations and theoretical predictions as reasonably strong evidence that said theory is wrong, or at least severely flawed.
Dethe
1 / 5 (5) Mar 17, 2015
Are you unfamiliar with special relativity? Or the fact that stars are nuclear reactors
Even if the stars would burn the coal, it still wouldn't remove the problem, how they can radiate their matter into outside with massless photons. Do you have some smart proposal for it?
Most scientists would take disagreement between observations and theoretical predictions as reasonably strong evidence that said theory is wrong, or at least severely flawed
Yes, normal scientists really think so, but these smart ones - like the Albert Einstein - wrote: "If the facts don't fit the theory, check the facts".

Yes, we are really observing the group of gamma photons galloping together - this is a fact. Does this fact really imply, that these gamma photons are moving with speed of visible light? Maybe they just like each other.. ;-)
DarkLordKelvin
4.6 / 5 (9) Mar 17, 2015
Are you unfamiliar with special relativity? Or the fact that stars are nuclear reactors
Even if the stars would burn the coal, it still wouldn't remove the problem, how they can radiate their matter into outside with massless photons. Do you have some smart proposal for it?

It's not my proposal, but Einstein's, and it's kinda common knowledge. The nuclear reactions involved in stellar fusion are exothermic .. i.e. they release energy. Some of that energy is in the form of kinetic energy of massive particles, like nuclei and neutrons. Some of it is in the form of high-energy radiation, like gamma particles. Ultimately, much of that energy is released as EM radiation in one way or another, since the kinetic energy (heat) is radiated as visible or IR radiation as well.

In any case, the stars lose mass through conversion of mass to energy a la E=mc2 .. much of that released energy is then radiated away as EM radiation, by the mechanisms I mentioned above.
Dethe
1 / 5 (5) Mar 17, 2015
the stars lose mass through conversion of mass to energy a la E=mc2
Yes, this is what Einstein and "his" formula say. Now we should analyze the way, in which the matter transfers from place to place. Let say, we will surround the star with Dyson sphere, which will absorb all its energy. If the E=mc^2 equation is valid, then the star should gradually lose its mass and the mass of Dyson sphere should increase accordingly. Actually we already observed the similar effect with excited lead nuclei within mass spectrometer, so we can be rather sure, that this model is correct.

Now the question is, how the matter can travel from place to place in massless form?
DarkLordKelvin
4.6 / 5 (9) Mar 17, 2015
Let say, we will surround the star with Dyson sphere, which will absorb all its energy. If the E=mc^2 equation is valid, then the star should gradually lose its mass and the mass of Dyson sphere should increase accordingly
Yes, that's right ... Provided the sphere is perfectly absorbing at all wavelengths of EM radiation, as well as perfectly "sticky" so that all massive particles emitted are captured too. It would also need to have zero emissivity, so that none of the captured energy could be reradiated.
Now the question is, how the matter can travel from place to place in massless form?
As photons of course. What part of the "equals" in E=mc2 is giving you trouble?
Dethe
1 / 5 (6) Mar 17, 2015
None, in fact. If the photons transfer an energy, they must be also massive according to this equation.
Benni
1 / 5 (6) Mar 17, 2015
Let say, we will surround the star with Dyson sphere, which will absorb all its energy. If the E=mc^2 equation is valid, then the star should gradually lose its mass and the mass of Dyson sphere should increase accordingly


Yes, that's right ... Provided the sphere is perfectly absorbing at all wavelengths of EM radiation, as well as perfectly "sticky" so that all massive particles emitted are captured too. It would also need to have zero emissivity, so that none of the captured energy could be reradiated.
Alias: Mass/Energy Equivalence Principle formulated by Einstein in 1906, ten years before GR.

Protoplasmix
5 / 5 (2) Mar 17, 2015
the stars lose mass through conversion of mass to energy a la E=mc2

Yes, this is what Einstein and "his" formula say. Now we should analyze the way, in which the matter transfers from place to place
Use a Dyson sphere? Here, Zeph, try Solutions to Homework #4, all of Problem 2.
Benni
1.8 / 5 (5) Mar 17, 2015
If the photons transfer an energy


Photons ARE energy, they don't transfer energy. Photons transfer mass from place to place in the universe upon transformation to mass as in the case of electron pair production. If you don't understand electron pair production, type it into a Search engine & look at some Feynman diagrams under "Electron Pair Production", the transformation of gamma wavelength photons.
Dethe
2 / 5 (4) Mar 17, 2015
OK, if the photons are energy, they must be massive according to E=mc^2. From solution of Homework #4 follows, the Sun radiates 4.27x109kg of mass each second via photons. Can we agree with it?
Protoplasmix
5 / 5 (5) Mar 17, 2015
OK, if the photons are energy, they must be massive according to E=mc^2. From solution of Homework #4 follows, the Sun radiates 4.27x109kg of mass each second via photons. Can we agree with it?
Yup.
DarkLordKelvin
4.6 / 5 (9) Mar 17, 2015
OK, if the photons are energy, they must be massive according to E=mc^2.
No, that is not what E=mc^2 means. That describes the relationship between *intrinsic mass*, that is mass that is the same in any inertial reference frame, and energy. Photons have zero intrinsic mass, since they can never be at rest, and their energy is dependent on the relative velocity of an observer. You can define a "relativistic mass" for a photon, by just dividing its energy by c^2, but that is a deprecated concept. It's more useful to think about the momentum of a photon according to E=pc, from which we can see that because c is constant, the energy of a photon depends only on its momentum (and therefore its wavelength, since p=h/lambda, where h is Planck's constant).
Dethe
1 / 5 (5) Mar 17, 2015
Photons have zero intrinsic mass, since they can never be at rest
This is just the point. The zero "intrinsic" mass of photons is of zero physical relevance, just because the photons can never be at rest. It also implies, that the mass of photons is always nonzero.
energy of a photon depends only on its momentum and wavelenght
Why not, I do care only, if it can manifest itself as a mass - with gravity action at distance in particular.
Dethe
1 / 5 (5) Mar 17, 2015
Try to imagine, that the star inside of Dyson sphere explodes into cloud of photons, which will travel into outside. The gravity field of star will decrease and its excess will travel with speed of light toward Dyson sphere, where it will be absorbed. The gravity field manifest itself as a curvature of space-time - so we can also say, that the space-time around star will become more flat and portion of its curvature will travel outside with speed of light like the gravitational wave. This curvature will be located with actual position of photon cluster, which would exhibit its own gravity field in this way. Is such an interpretation acceptable for you?
Benni
2 / 5 (4) Mar 17, 2015
OK, if the photons are energy, they must be massive according to E=mc^2. From solution of Homework #4 follows, the Sun radiates 4.27x109kg of mass each second via photons. Can we agree with it?

Yup
rpaul_bauman
not rated yet Mar 17, 2015
Here is a one star comment, please do not read it. If space is a particle and its length is L and the time it takes a photon to jump from one space particle to the next is T then the speed the photon travles at is L/T which is C. Pretty simple isn't it !! But who would beleive that ?
DarkLordKelvin
4.5 / 5 (8) Mar 17, 2015
Try to imagine, that the star inside of Dyson sphere explodes into cloud of photons, which will travel into outside. The gravity field of star will decrease and its excess will travel with speed of light toward Dyson sphere, where it will be absorbed. The gravity field manifest itself as a curvature of space-time - so we can also say, that the space-time around star will become more flat and portion of its curvature will travel outside with speed of light like the gravitational wave. This curvature will be located with actual position of photon cluster, which would exhibit its own gravity field in this way.


To me that looks like a mishmash of physics "buzzwords", and doesn't make any sense at all. You start from a non-sensical premise (a star "exploding into photons") and it seems to go downhill from there. If you choose to eschew standard physical descriptions of basic concepts like photons and mass, it makes you quite difficult to understand.
Dethe
1 / 5 (5) Mar 17, 2015
Well, the stars sometimes explode and the spherical shell of photons gets released into free space during it. Now we are discussing bellow article about gamma ray bursts - so I wouldn't consider it downright a "non-sensical premise"..;-) My question was, what will happen with matter of star and it's gravitational curvature during it. Or what you're willing to accept about it at least.
DarkLordKelvin
4.3 / 5 (6) Mar 17, 2015
Well, the stars sometimes explode and the spherical shell of photons gets released into free space during it. Now we are discussing bellow article about gamma ray bursts - so I wouldn't consider it downright a "non-sensical premise"..;-) My question was, what will happen with matter of star and it's gravitational curvature during it. Or what you're willing to accept about it at least.
The question answers itself .. first of all, an actual star doesn't explode "into photons" .. some mass remains as particles - dust - after all that's how all elements heavier than iron are formed by stellar nucleosynthesis .. but more than that, there is usually a very dense, massive core remaining behind .. a dwarf star, neutron star or black hole, depending on original stellar mass. So, follow the mass distribution of the explosion in time, and you can determine how the gravitational curvature changes .. it will diminish (smoothly, I suppose) as mass-to-energy conversion proceeds.
Benni
2 / 5 (4) Mar 17, 2015
Einstein's view of "mass" & "energy" was simply that there were both the same. This is deduced from the gravity relationship between the two inverse products, all the characteristics of one carry over into the transformed product of the other. What I mean by this is best understood by example of a "light box".

If energy could trapped inside a perfectly mirrored box, the gravitational force that light box exerts within its' surroundings will increase. This is because the captured photons unable to escape do not lose inherent gravity upon transformation from mass, gravity is also conserved with either form of transformation. If you weigh the perfectly mirrored LB before it captured photons it would weigh less than before the capture of photons, hence the term Relativistic Mass often given to electromagnetism.

Dethe
1 / 5 (5) Mar 17, 2015
The question answers itself .. first of all, an actual star doesn't explode "into photons" .. some mass remains as particles - dust
Dust will remain at place - but what will happen with matter traveling across space in form of photons? Apparently some sort of people is scarred just with thinking about trivial consequences of relativity and mass energy equivalence, if it could ruin their religion..
DarkLordKelvin
4.4 / 5 (7) Mar 17, 2015
Dust will remain at place - but what will happen with matter traveling across space in form of photons?
As I understand it (we are coming to the limit of my understanding of GR), only intrinsic mass (matter) can actually cause the sort of "gravitational curvature" you referred to. EM radiation will *respond* to that curvature, but cannot itself induce it. Therefore, any changes in "curvature" will occur during the mass-to-energy conversion (as I said already). Now I suppose, for your "explosion" *specifically*, it might appear to an observer that the "curvature change" was associated with the photon front, since I think both will propagate at the speed of light, and both will have a magnitude/intensity that falls off as 1/(distance)^2. However, I don't think that applies generally, since m->E conversion doesn't REQUIRE a simultaneous photon flux; for example in alpha or beta decay, the energy released goes into KE of emitted massive particles.
Dethe
1 / 5 (6) Mar 17, 2015
only intrinsic mass (matter) can actually cause the sort of "gravitational curvature" you referred to
So that when the star will release cloud of photons, then the portion of its space-time curvature will simply disappear and it will not emerge, until the photons will not get absorbed somewhere else? After then it will suddenly reappear at the target?
DarkLordKelvin
4.2 / 5 (5) Mar 18, 2015
only intrinsic mass (matter) can actually cause the sort of "gravitational curvature" you referred to
So that when the star will release cloud of photons, then the portion of its space-time curvature will simply disappear and it will not emerge, until the photons will not get absorbed somewhere else? After then it will suddenly reappear at the target?


No, that's not what I said; there is no "disappearing". For each (small) decrease of intrinsic mass associated with photon emission, there's speed of light propagation of the associated "curvature change". The inverse happens at the other end, assuming that the radiation is absorbed in a way that it changes the intrinsic mass of the absorbing entity. The information that the mass of the "absorber" has increased (i.e. your "gravitational curvature") propagates outwards at the speed of light. You seem insistent on making "simplifications" that are not consistent with established physical theories .. why?
Dethe
1 / 5 (6) Mar 18, 2015
So that when the star releases cloud of photons, then the portion of its space-time curvature will drop to lower value. This lower value will propagate with speed of light together with photons at the target like the gravitational wave. When the photons will arrive at target, then the curvature of space-time will suddenly drop (because the gravitational wave will arrive at the same moment) and after then it suddenly rise (because of mass transferred with photons)? I'm not doing any simplifications, I'm just asking you for your interpretation of this process.
antialias_physorg
5 / 5 (7) Mar 18, 2015
Even if the stars would burn the coal, it still wouldn't remove the problem, how they can radiate their matter into outside with massless photons. Do you have some smart proposal for it?

There is a law for the conservation of energy - not one for mass. Mass is just one form of energy (of many). A photon can transmit energy. While it has no mass it does have a momentum.
Note: You could convert all of a star into massless state (e.g. by hitting it with an identical star made of anti-matter). But that conversion would still satisfy conservation of energy (and conservation of momentum).
DarkLordKelvin
4.2 / 5 (5) Mar 18, 2015
So that when the star releases cloud of photons, then the portion of its space-time curvature will drop to lower value. This lower value will propagate with speed of light together with photons at the target like the gravitational wave. When the photons will arrive at target, then the curvature of space-time will suddenly drop (because the gravitational wave will arrive at the same moment) and after then it suddenly rise (because of mass transferred with photons)?


I don't like the term "suddenly" because it's fuzzily defined and tends to imply "non-physical" interpretations, but other than that, yes, that's my understanding of the process, as. I already wrote.

But I am not really sure how important such "localized" interpretations really are. Consider that if you are observing from outside the Dyson sphere inside which all of that was occurring, you would not detect any changes at all.
Fleetfoot
4.6 / 5 (9) Mar 18, 2015
what will happen with matter traveling across space in form of photons?
.. only intrinsic mass (matter) can actually cause the sort of "gravitational curvature" you referred to.


The source term in GR is the stress-energy tensor:

http://en.wikiped...y_tensor

That includes all forms of energy and momentum so the energy of the cloud of emitted photons measured in the rest frame of the star does contribute. The shell theorem means only the total of the star, photons and sphere matters, and that is constant outside.

Note also that E=mc^2 is a simplification only valid for particles with mass which can be brought to rest. The full equation which is valid for all particles including photons is

E^2 = (pc)^2 + (mc^2)^2

where p is the magnitude of the momentum. Since a photon can't be brought to rest, you have to use that equation, and since E=pc for a photon, m=0.
DarkLordKelvin
4 / 5 (4) Mar 18, 2015
what will happen with matter traveling across space in form of photons?
.. only intrinsic mass (matter) can actually cause the sort of "gravitational curvature" you referred to.
The source term in GR is the stress-energy tensor:

[link]

That includes all forms of energy and momentum so the energy of the cloud of emitted photons measured in the rest frame of the star does contribute. The shell theorem means only the total of the star, photons and sphere matters, and that is constant outside.
Ok, but the stress-energy tensor for a field can be defined in a "curvature-free" region of spacetime, correct? So wasn't my statement that only the intrinsic mass contributes to curvature essentially right? Otherwise you'd have to account for the "self-curvature" induced by the photon-field, which is what Dethe seems to be claiming. In other words, is it possible for EM fields (of arbitrary intensity) to induce spacetime curvature in GR? I thought not, but I may be mistaken.
Protoplasmix
5 / 5 (6) Mar 18, 2015
... which is what Dethe seems to be claiming. In other words, is it possible for EM fields (of arbitrary intensity) to induce spacetime curvature in GR? I thought not, but I may be mistaken.
Seems more like Dethe's example would result in a termination of spacetime curvature propagating outwards at c, since that's what a nearby test particle would experience.

Note that a combined pseudotensor can be constructed (e.g, Landau-Lifshitz) to extend the stress-energy tensor to provide a way for showing what portion curves spacetime and what portion propagates along the curvature.
indio007
1 / 5 (3) Mar 19, 2015
Too bad the inventors of tensors calculus proved long ago that pseudo-tensors like Einstein's don't exist.
DarkLordKelvin
4.2 / 5 (5) Mar 19, 2015
Too bad the inventors of tensors calculus proved long ago that pseudo-tensors like Einstein's don't exist.


Ummm .. not sure what you mean there, the Landau-Lifshitz and Einstein pseudo tensors certainly "exist" in the sense that they can be meaningfully defined, and they seem to do a pretty good job at explaining observable phenomena, so they also appear relevant to "real" physical phenomena. Is it possible that you just mean that the inventors of tensor calculus proved long ago the pseudo-tensors like the ones used in GR aren't really tensors?
indio007
1 / 5 (3) Mar 19, 2015
Too bad the inventors of tensors calculus proved long ago that pseudo-tensors like Einstein's don't exist.

Is it possible that you just mean that the inventors of tensor calculus proved long ago the pseudo-tensors like the ones used in GR aren't really tensors?


Do not exist is a quote.
See for yourself.

Mechanics. - On the analytic expression that must be given to
the gravitational tensor in Einstein's theory
T. Levi-Civita
http://arxiv.org/...04v1.pdf
"Now it is well known that differential invariants of the 1^0 order which are intrinsic
i.e.,like G*, exclusively formed with the coefficients of ds^2 and with their first derivatives, do
not exist.This is enough to render, at least in general, not admissible
the form of the gravitational tensor taken by Einstein. "
Protoplasmix
5 / 5 (5) Mar 19, 2015
Quoting the 2nd paragraph from Landau-Lifshitz link:

"Some people object to this derivation on the grounds that pseudotensors are inappropriate objects in general relativity, but the conservation law only requires the use of the 4-divergence of a pseudotensor which is, in this case, a tensor (which also vanishes). Also, most pseudotensors are sections of jet bundles, which are perfectly valid objects in GR."
indio007
1 / 5 (2) Mar 19, 2015
Quoting the 2nd paragraph from Landau-Lifshitz link:

"Some people object to this derivation on the grounds that pseudotensors are inappropriate objects in general relativity, but the conservation law only requires the use of the 4-divergence of a pseudotensor which is, in this case, a tensor (which also vanishes). Also, most pseudotensors are sections of jet bundles, which are perfectly valid objects in GR."


Another sourceless unattributable comment wedged into an article.... wikipedia is full of them.
Dethe
1 / 5 (4) Mar 19, 2015
Note that a combined pseudotensor can be constructed (e.g, Landau-Lifshitz) to extend the stress-energy tensor to provide a way for showing what portion curves spacetime and what portion propagates along the curvature
The pseudotensors do violate the dimensionality of 4D space-time metric of general relativity by implying, there are another additional dimensions, in which these pseudotensors can be constructed. Which is also why the pseudotensors don't exist in spherical coordinates, only cylindrical ones. The problem with temporal phenomena in general relativity is actually quite simple: at the moment, when the only time dimension serves as a coordinate of space-time metric, then everything becomes frozen in it. There is no other additional space or time dimension, in which the changes of space-time metric could propagate.

That is to say, I don't doubt, that these changes may exist and propagate, but they wouldn't follow 4D general relativity, but actually violate it instead.
Dethe
1 / 5 (4) Mar 19, 2015
The consequential logical thinking and formal math is not so free legue as the contemporary theorists - who are adding various parameters to existing theories rather freely - believe, because these terms usually violate (increase) the dimensionality of existing models on background. The adding of pseudotensors into general relativity has made a hyperdimensional model from the original rather consistent and clean 4D model. What the scientists are trying to describe with it as a gravitational waves belongs into hyperdimensional dark matter effects in this way. I can't say, that these effects cannot physically exist in this moment - but to consider them as a consequence of vanilla 4D general relativity in its original form is rather brave. Do we really search for consequences of general relativity - or rather violations of it? This is what matters here.
indio007
1 / 5 (2) Mar 19, 2015
Time as a dimension has a big problem. It's a point. The past is gone and the future hasn't happened. What's left? The point like present. spacetime does not represent reality.
Protoplasmix
5 / 5 (2) Mar 19, 2015
"Some people object to this derivation on the grounds that pseudotensors are inappropriate objects in general relativity ... Also, most pseudotensors are sections of jet bundles, which are perfectly valid objects in GR."
Another sourceless unattributable comment wedged into an article.... wikipedia is full of them.
Trying to say a pseudotensor doesn't exist, to me, is like saying sqrt(-1) doesn't exist.

As far as I know, pseudotensors are the only way to comply with an exact conservation law when defining an integral energy-momentum and generalizing it to curved spacetime. See http://web.mit.ed.../gr7.pdf , and this may be helpful too: http://math.ucr.e..._gr.html - the important thing is not to confuse artifacts in a particular coordinate system with real physical effects.
Fleetfoot
5 / 5 (6) Mar 19, 2015
the stress-energy tensor for a field can be defined in a "curvature-free" region of spacetime, correct?


It defines the amount of 'stuff' in a region, whether that region has curvature or not is a global question.

So wasn't my statement that only the intrinsic mass contributes to curvature essentially right?


No.

Otherwise you'd have to account for the "self-curvature" induced by the photon-field


You have to include the energy density of the photons.

is it possible for EM fields (of arbitrary intensity) to induce spacetime curvature in GR?


Definitely! Consider the fact that for the 48000 years or so, the universe was dominated by the energy in the radiation we now see as the microwave background, it had more effect than all the matter in the universe:

http://en.wikiped...ated_era

which is what Dethe seems to be claiming


That's another sock account for the crackpot previously known as "Zephyr".
Dethe
1 / 5 (4) Mar 19, 2015
Trying to say a pseudotensor doesn't exist, to me, is like saying sqrt(-1) doesn't exist
This is just the difference between formally thinking mathematicians and physicists. Of course, we can say, that the waterfall can be described with hyperbola, because the hyperbole is also valid curve - but the parabola is the only correct curve in its physical model. The pseudotensor introduces additional dimensions into 4D general relativity, so it cannot be postulated within 4D space-time metric. If you need more exact reasoning, http://www.jstor..../2371768]here is[/url]. The pseudotensor exist within relativity, but its value must be always zero.
Dethe
1 / 5 (4) Mar 19, 2015
Correct link. Compare also T. Levi-Civita's derivation of the divergence of pseudo-tensor.
Protoplasmix
5 / 5 (4) Mar 19, 2015
Time as a dimension has a big problem. It's a point. The past is gone and the future hasn't happened. What's left? The point like present. spacetime does not represent reality.
infinite spacetime - in two finite dimensions, we have diagrams: Carter-Penrose diagrams
Protoplasmix
5 / 5 (3) Mar 19, 2015
Trying to say a pseudotensor doesn't exist, to me, is like saying sqrt(-1) doesn't exist.
This is just the difference between formally thinking mathematicians and physicists...
Zeph, to me, mathematics is the physics of numbers :)
richardwenzel987
5 / 5 (2) Mar 19, 2015
Another possible solution would be to have a quantum gravity theory allow for a dualism, in which space would appear continuous or quantized, depending on the sorts of measurement being performed on, or within, that space. In this case, it would probably only allow granularity in local measurements in a local frame. Photons traveling across vast distances are not interacting with any sort of experimental device anywhere along that long path, although there may be interactions unknown to us. But they are not "measured" until they are detected and their arrival times are determined. That measurement would not require that the intervening space be granular, or reveal any granularity. Quantized space only appears when you look at it in a particular way.
Dethe
1 / 5 (3) Mar 19, 2015
solution would be to have a quantum gravity theory allow for a dualism
Above I'm explaining how this dualism works for cluster of photons: it travels smoothly across space as a single large particle, whereas the photons inside of it are moving chaotically. It's dual process to the conduction of electrons inside of superconductor - the individual electrons are moving wildly, but their charge waves (collective excitations) propagate smoothly. The observable reality just rhymes at different scales. The electrons or photons just do the same like the communities of people - they're forming pairs or clusters for to propagate through random life easier. The gamma ray clusters traveling across Universe unscattered in this way can also serve as a trivial example of selection at the physical level.
Dethe
1 / 5 (3) Mar 19, 2015
BTW The photons within gamma ray cluster must be also entangled by their gravity at distance for being able to travel collectively. This mutual connections helps them to overcome obstacles - the quantum fluctuations of space-time - in similar way, like the formation of Cooper pairs in Type-I superconductors or even larger clusters in Type-II superconductors from electrons. IMO the cluster of photons in gamma ray burst fits all characteristics of single giant photon including its vortex shape which is known from boson condensates - it just propagates itself in slightly subluminal speed as a whole.
DarkLordKelvin
3.7 / 5 (3) Mar 20, 2015
solution would be to have a quantum gravity theory allow for a dualism
Above I'm explaining how this dualism works for cluster of photons: it travels smoothly across space as a single large particle, whereas the photons inside of it are moving chaotically. It's dual process to the conduction of electrons inside of superconductor - the individual electrons are moving wildly, but their charge waves (collective excitations) propagate smoothly. The observable reality just rhymes at different scales. The electrons or photons just do the same like the communities of people - they're forming pairs or clusters for to propagate through random life easier. The gamma ray clusters traveling across Universe unscattered in this way can also serve as a trivial example of selection at the physical level.


This just seems like a mess of words ... can you show how these effects arise out of the mathematical formalism of GR? Can you derive the force binding the photons?
Dethe
1 / 5 (4) Mar 20, 2015
The photons have energy E = h f = m c^2, i.e. their mass equals m = h f / c^2, which you can use for calculation of their binding force according to Newton gravity law.
Fleetfoot
5 / 5 (7) Mar 20, 2015
The photons have energy E = h f = m c^2, i.e. their mass equals m = h f / c^2, which you can use for calculation of their binding force according to Newton gravity law.


They also have momentum of p=hf/c and since (mc^2)^2 = E^2 - (pc)^2 clearly their mass is m=0.

The equation E=mc^2 is a simplified version that you get by assuming p=0. That is fine for massive particles, it's just the energy measurement in their rest frame, but you can't use that for photons because they don't have a rest frame.
Fleetfoot
5 / 5 (2) Mar 20, 2015
This just seems like a mess of words


An accurate assessment, it's just gibberish.
Dethe
1 / 5 (4) Mar 20, 2015
since (mc^2)^2 = E^2 - (pc)^2
Nope, this is just ad-hoced modification of mass-energy equivalence in an effort to make the photons massless. At the moment, when the position and speed of photons can be defined, they have rest frame in the same way, like another particles. What did you say applies to light wave, which really has no rest frame defined - but the photons is different object. Most of people here don't understand, just the existence of photons represents a violation of relativity and Maxwell's theory with quantum mechanics.
DarkLordKelvin
4 / 5 (4) Mar 20, 2015
since (mc^2)^2 = E^2 - (pc)^2
Nope, this is just ad-hoced modification of mass-energy equivalence in an effort to make the photons massless.
What are you talking about? Momentum of light has been known since before QM or relativity, and the rest of that is the *definition* of total energy in SR. The massless-ness of photons is a consequence of them traveling at c, which brings us back to the result highlighted in this article, which provides one of the most stringent tests of that.
At the moment, when the position and speed of photons can be defined, they have rest frame in the same way, like another particles.
Prove it. Establish a *valid* transformation within SR that shows photons with v=0.
Dethe
1 / 5 (3) Mar 20, 2015
The massless-ness of photons is a consequence of them traveling at c
This is just my point - the photons are massive, so that they cannot travel with speed of light. This animation illustrates it.
Establish a *valid* transformation within SR that shows photons with v=0
Special relativity was developed and derived for Maxwell waves, never for photons and it can say nothing about photons.
indio007
1 / 5 (1) Mar 20, 2015
Trying to say a pseudotensor doesn't exist, to me, is like saying sqrt(-1) doesn't exist.



I'm not saying psuedotensors in general don't exist. I'm saying Einstein's pseudo tensor doesn't exist.
DarkLordKelvin
4 / 5 (4) Mar 20, 2015
the stress-energy tensor for a field can be defined in a "curvature-free" region of spacetime, correct?


It defines the amount of 'stuff' in a region, whether that region has curvature or not is a global question.


Well, ok, I guess this gets exactly at the point of my confusion, I guess. If EM fields can cause spacetime curvature, then doesn't that mean that there is a non-zero gravitational attraction between a pair of photons, like Dethe seems to be claiming? I had always thought that was strictly ruled out by GR, but I never could manage to make it through the tensor math by myself, so I always had to be content with a qualitative understanding of the principles.

[ctd]
DarkLordKelvin
3.7 / 5 (3) Mar 20, 2015
@Fleetfoot

Perhaps the "box of photons" example will serve to illustrate my confusion better. A box that is perfectly reflective on the inside has an intrinsic mass of m when empty. Now someone makes an unspecified change to the box so that it's measured mass (T00 component of stress-energy tensor) becomes m+dm.

1) is there any way for an external observer to tell whether the mass change dm arises from matter being placed in the box, or from trapping an equivalent amount of EM radiation inside it? (I am almost certain the answer is "no", I just want to make sure.)

2) Assuming the answer to 1 is "no", then doesn't that mean that "trapped" EM radiation is somehow different than "free" EM radiation, because we can clearly define a reference frame where the box is at rest, but we cannot define a frame where an equivalent density of EM radiation propagating through free space would be at rest, right?

3) Does the apparent diff in 2 have to do with momentum conservation?
DarkLordKelvin
3.7 / 5 (3) Mar 20, 2015
The massless-ness of photons is a consequence of them traveling at c
This is just my point - the photons are massive, so that they cannot travel with speed of light.
Prove it (that gif demonstrates nothing). All available experimental evidence indicates your claim is wrong.
Establish a *valid* transformation within SR that shows photons with v=0
Special relativity was developed and derived for Maxwell waves, never for photons and it can say nothing about photons.
I don't think that's correct. Photons are just quantized fields in the framework of QED, so I don't see how the difference between continuous energy density of the fields vs. quantized "occupation number" based energies of the fields should make a difference.
richardwenzel987
not rated yet Mar 20, 2015
One more crazy idea: it would be nice to have a symmetry of the quantum gravity duality with the well-known wave/particle duality. If you had enough photons from a very distant source to pass them through a double slit arrangement, you might then find a departure from the expected diffraction pattern, which might be interpreted in terms of space quantization. A very, very, sensitive detector would be required. In other words, arrival times might not be the measurement to perform to detect space quantization, but rather diffraction pattern in a double-slit experiment.
Fleetfoot
5 / 5 (4) Mar 20, 2015
1) is there any way for an external observer to tell whether the mass change dm arises from matter being placed in the box, or from trapping an equivalent amount of EM radiation inside it?


No.

2) Assuming the answer to 1 is "no", then doesn't that mean that "trapped" EM radiation is somehow different than "free" EM radiation, because we can clearly define a reference frame where the box is at rest, but we cannot define a frame where an equivalent density of EM radiation propagating through free space would be at rest, right?


The difference is between one photon and more than one. We cannot define a rest frame for a single photon but you can define a frame in which the total momentum for a group of photons (minimum 2) is zero because momentum is a vector. Unless the photons are moving in exactly the same direction, you can always find a frame in which they are moving in opposite directions and Doppler shift allows you to make their frequencies equal.
Fleetfoot
5 / 5 (3) Mar 20, 2015
If you plot the energy versus the momentum for a single particle with non-zero mass over a range of observer frames around the particle's rest frame, you get a hyperbola (a.k.a. the mass shell).

If you do the same for the scalar sum of the energies versus the vector sum of the momenta for 2 or more photons, you also get a hyperbola. The aggregate of the photons is indistinguishable from a single particle having the energy in the zero momentum frame.
Captain Stumpy
5 / 5 (2) Mar 20, 2015
ATTENTION AND WARNING!
to anyone using ZEPHIR/dethe links

IF it links back to his own site, as does his pseudoscience aether link above, then he can gain a LOT of information about anyone who uses the link
especially if he is the admin or moderator of the site linked

DO NOT EVER CLICK ANY LINKS THAT ARE HIDDEN, SHORTENED or go to his aether site unless you are a phenomenal hacker!!!!

This animation illustrates it
@the IDIOT TROLL ZEPHIR
your animation short link goes to a known PSEUDOSCIENCE SITE and is NOT supported by ANY SCIENTIFIC EVIDENCE

therefore you are simply (again) pushing your aw/daw stupidity and trying to gain more acolytes for your religion
do you get paid per head or what?
reported
Dethe
1 / 5 (3) Mar 20, 2015
I don't see how the difference between continuous energy density of the fields vs. quantized "occupation number" based energies of the fields should make a difference
Photons contain 2-spin component (gravitational wave) which gives them mass. In linearized, Einstein–Maxwell theory on flat spacetime, an oscillating electric dipole is the source of a spin-2 field. The mass of photons depends on their wavelength and it gets zero for microwave photons, the longer wavelength photons are even unstable tachyons.

Protoplasmix
5 / 5 (3) Mar 20, 2015
@CS, for what it's worth, I checked on Zeph's  "divergence of pseudo-tensor" link which goes to a pdf file at sjcrothers.plasmaresources.com - various assertions of Crothers on alleged flaws with relativity have already been debunked--
here: http://www.youtub...7-rUyW3I
here: http://www.youtub...HLJoe7pA
and here: http://www.youtub...I6jECJxk

Anything to add, Zeph?
indio007
1 / 5 (1) Mar 20, 2015
@CS, for what it's worth, I checked on Zeph's  "divergence of pseudo-tensor" link which goes to a pdf file at sjcrothers.plasmaresources.com - various assertions of Crothers on alleged flaws with relativity have already been debunked--
here: http://www.youtub...7-rUyW3I

Anything to add, Zeph?

debunked my ass. You can't debunk ric=0
Captain Stumpy
5 / 5 (2) Mar 21, 2015
@CS, for what it's worth, I checked on Zeph's  "divergence of pseudo-tensor" link which goes to a pdf file at sjcrothers.plasmaresources.com - various assertions of Crothers on alleged flaws with relativity have already been debunked--
@Protoplasmix
thanks for that update -
i was specifically talking about his embedded link to here: http://www.aether...tons.gif
it is embedded in the quote
This is just my point - the photons are massive, so that they cannot travel with speed of light. This animation illustrates it.
linked as "illustrates it."

This is known pseudoscience. and if you visit his site he can collect info OR do worse

i also don't ever open crothers etc
they are a favourite of the EU crowd... and EU is debunked in so many places it is not even funny
it makes one wonder how ANYONE can still believe in it... but then you meet idiots like CD, zephir and jvk

THANKS for the links
appreciate it

viko_mx
1 / 5 (2) Mar 21, 2015
Gravity is local programed behavior of structure of vacuum of space wich defines some behavior and interaction of elementary particles. Gravity is not force or bended space wich is very untenable idea idea.
Fleetfoot
5 / 5 (4) Mar 21, 2015
Photons contain 2-spin component (gravitational wave) which gives them mass.


Photons are massless and have spin 1.

Gravitons would be massless and have spin 2 (but at present they are hypothetical).

Gravitational waves are macroscopic effects of GR, not particles at all.
Protoplasmix
5 / 5 (5) Mar 21, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here: http://www.staff....ons.html
DarkLordKelvin
4.3 / 5 (6) Mar 21, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here: http://www.staff....ons.html


@Protoplasmix That link is simply spectacular! Everyone who has had interactions with those passionate deniers of science who afflict the forums at PO and elsewhere should read it, even if they don't know much about (or care much about) relativity. Old Gerard, who is a good writer to begin with, is in scintillating form as he describes his interactions with some paragons of Dunning-Kruger pathology.
indio007
1 / 5 (3) Mar 22, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here:


I might as well post the rebuttal.

Gerardus 't Hooft, Nobel Laureate,
On Black Hole Perturbations

Stephen J. Crothers

ABSTRACT
Professor Gerardus 't Hooft, Nobel Laureate in Physics, Editor-in-Chief of the
journal Foundations of Physics, has again brought attention to my work on
black hole theory, big bang cosmology, and General Relativity, by means of
his personal website, providing me thereby with the opportunity to address his
most recent comments, particularly on black holes. Black hole universes are
either asymptotically flat or asymptotically curved, by definition, and so there
can be no universe containing multiple black holes. All alleged big bang
universes are not asymptotically anything.

http://vixra.org/...41v2.pdf
Stevepidge
1 / 5 (3) Mar 22, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here: http://www.staff....ons.html


@Protoplasmix That link is simply spectacular! Everyone who has had interactions with those passionate deniers of science who afflict the forums at PO and elsewhere should read it, even if they don't know much about (or care much about) relativity. Old Gerard, who is a good writer to begin with, is in scintillating form as he describes his interactions with some paragons of Dunning-Kruger pathology.


Gerardy poo is a hack pushing determinism. In other words, what the hell is the point in living life. Boring as shit.
DarkLordKelvin
4.2 / 5 (5) Mar 22, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here
I might as well post the rebuttal.

Gerardus 't Hooft, Nobel Laureate,
On Black Hole Perturbations

Stephen J. Crothers

ABSTRACT
Professor Gerardus 't Hooft, Nobel Laureate in Physics, Editor-in-Chief of the
journal Foundations of Physics, has again brought attention to my work on
black hole theory, big bang cosmology, and General Relativity, by means of
his personal website, providing me thereby with the opportunity to address his
most recent comments, particularly on black holes. Black hole universes are
either asymptotically flat or asymptotically curved, by definition, and so there
can be no universe containing multiple black holes. All alleged big bang
universes are not asymptotically anything.
That is addressed on GvH's site as well.
DarkLordKelvin
4.3 / 5 (6) Mar 22, 2015
debunked my ass. You can't debunk ric=0
um, Crothers' assertions, such as 'black holes do not exist', have been shown to be in error, perhaps most notably by 't Hooft here: http://www.staff....ons.html


@Protoplasmix That link is simply spectacular! Everyone who has had interactions with those passionate deniers of science who afflict the forums at PO and elsewhere should read it, even if they don't know much about (or care much about) relativity. Old Gerard, who is a good writer to begin with, is in scintillating form as he describes his interactions with some paragons of Dunning-Kruger pathology.


Gerardy poo is a hack pushing determinism. In other words, what the hell is the point in living life. Boring as shit.


Yeah .. Nobel Prizes in Physics .. you can't even drive through Stockholm these days with your windows down, else one of those will come flying through into your lap.
Dethe
1 / 5 (4) Mar 22, 2015
perhaps most notably by 't Hooft here: http://www.staff....ons.html
for example this
"Einstein's equations for gravity are incorrect,
they have no dynamical solutions, and do not imply gravitational waves as described in numerous text books."
Is fully relevant to my posts above. Until the only time dimension is this one, which represents the space-time metric, then this space-time metric must be time invariant (i.e. static). It means no gravitational waves, also no motion and changes of massive objects, etc. Just frozen universe....

It also means, if something is moving, propagating and/or changing within space-time metric, it also violates the relativity..
Dethe
1 / 5 (4) Mar 22, 2015
Not many physical systems have cylindrical symmetry, but that's not the point. The point is now that these solutions contradict L's claim.
It's also explained above already (just search for word "cylindrical" in discussion). The cylindrical metric is hyperdimensional, despite proff. 't Hooft is willing/able to realize it or not.

I'm not saying that the general relativity or whatever else theory is incorrect. All theories are better or worse approximations of reality, that's all.. Each of them is also based on particular observational perspective, for example general relativity is based on intrinsic perspective of space-time blobs. It's a good wide scale approximation, but every other perspective (extrinsic in particular) would violate it.
chardo137
not rated yet Mar 22, 2015
The most interesting aspect of this seems to have been missed by the writer. What technique was used to confirm that the photons of different wavelengths were, in fact, emitted by the source at the same time? I have no problem with the meanings of the headline and the content of the article. I also wonder whether any of the data collected in this study could be used to shed light on the possible correction to the overall speed of light involving the gravitational polarization of the quantum vacuum (which could explain the anomalous timing in the detection of neutrinos from supernova 1987 A).
baudrunner
1 / 5 (3) Mar 22, 2015
Sigh*. People still got it wrong about photons and gravity.

Photons don't actually exist. They were invented by physicists to explain the exchange of information between two particles in a propagating medium. They created a whole new class of subatomic "particle" they called a charge carrier, but totally ignored Ocham's razor. You see, particles of like charges repel, and particles in the propagating medium have electron shells which will repel the electron shells of adjacent particles in a chain reaction that eventually reach the retinas in the observer's eyes, exchanging their characteristic information throughout this chain. Philip Bucksbaum of Michigan University proved that an infinite amount of information can be stored in an electron cloud. The particles in a propagating medium carry the modulations of the characteristics of the source of photonic excitations as well as the characteristics of the reflecting surfaces of the ambient waves.
baudrunner
1 / 5 (3) Mar 22, 2015
Physicists like to get too deep. It gets them respect.

And at the same time, they are trying to determine just how a Volkswagen works by blowing it up into a zillion pieces, which is essentially what is happening in particle colliders. They would have more luck understanding the truth if they instead create those theoretical particles and then put them together. The physics world is rife with untestable theories and untenable ideas, not to mention misleading and just plain bad science, but we have to take their words for it because their walking papers qualifies them. I don't know why they think themselves more qualified to theorize incredulities than any other person on the street in light of all the pollution and misinformation that they have been exposed to in the colleges and universities. Reality is plenty practical, and makes plenty of sense to those who know the truth about things, but like I said, simplicity is not on the agenda.
baudrunner
1 / 5 (3) Mar 22, 2015
Incidentally, this "spacetime foam" idea is on the right track. I call those discreet units of spacetime foam "iotons", and refer to "iotonic space". In the absence of any matter at all in the Universe, light cannot propagate, because it is matter that propagates light. Also, in the absence of any matter at all in the Universe, iotonic space has a tendency toward uniformity with all iotons lined up the same way, because they have orientation. Matter displaces space, the way a swimmer displaces water in a pool. Remembering that iotons want to re-occupy the space displaced by an object, there is an ordered orientation of iotons with respect to a center of mass. This orientation has a 180° phase relationship with the iotons around a nearby body, and the result is a cancellation of iotons between them. The net result is that the bodies come closer together. That is gravity.
baudrunner
1 / 5 (3) Mar 22, 2015
So, in effect, we have the mutable density of iotonic space and the mutable density of the particle medium of propagating photonic waves to contend with when we are analyzing spacetime and gravity. That should keep the relativity technicians occupied for awhile.

I have concluded that space, time, and matter are continually created at the periphery of the expanding Universe. The "Big Bang" is an ongoing, perpetual event. When iotonic space is cancelled, iotons spontaneously fill the increased distance between bodies "pulled" apart by their attraction to other nearer bodies. Introduce time, and latency is introduced in this process. In the case of a black hole, the geometry and dynamics of the local iotonic space have created a well of sorts, in which iotons are constantly cancelling each other out, causing a great attraction. Possibly a great mass travelling at extreme velocities through an already iotonically disturbed spacetime is the seed that begets black holes.
Protoplasmix
5 / 5 (4) Mar 22, 2015
What technique was used to confirm that the photons of different wavelengths were, in fact, emitted by the source at the same time?
Paraphrasing from the paper, GRB090510 was one of the brightest GRBs ever detected. It had a short duration of ~1second. The photons had very high energy (up to ~13 GeV) and the light curve had a fine temporal structure with ~10 millisecond spikes. GRBs, as such, are ideally suited for testing the idea of stochastic variations in the speed of light. They're high energy bursts of photons of short duration from very far away.

As for the neutrinos from SN 1987A, good question, they were emitted before the photons, see http://clarkplane...a-1987a/
Protoplasmix
5 / 5 (2) Mar 22, 2015
I might as well post the rebuttal.
You might as well, since you lack the necessary skills to produce the correct evaluation, which is obvious from the fact that those skills are exactly what you need to recognize the correct evaluation.

@DLK, spot on re Dunning-Kruger effect, worth noting that they propose the obvious solution -- exposure to the training required for a particular skill. Education, quite the concept.

@Fleetfoot, always glad when you have the chance to provide some of that "exposure" -- DLK's question was very good, I failed to think outside the 'Dyson sphere' and consider the radiation-dominated era, duh...
Dethe
1 / 5 (4) Mar 22, 2015
Photons don't actually exist. They were invented by physicists to explain the exchange of information between two particles in a propagating medium
On the contrary, they're a consequence of the foamy nature of space-time (the vibrations of this foam makes it more dense in similar way, like the shaking of soap foam). It all works together.
neutrinos from SN 1987A they were emitted before the photons
It gives support to theories of neutrons emanated with black holes (like this one at the central area of Milky Way).
DarkLordKelvin
3.7 / 5 (6) Mar 22, 2015
The most interesting aspect of this seems to have been missed by the writer. What technique was used to confirm that the photons of different wavelengths were, in fact, emitted by the source at the same time?
That is indeed an interesting question, and the way the PO summary article is written seems to indicate it was overlooked in the study, when of course it was not. If you read the full article (sadly paywalled), you see that the authors modeled the data using a stochastic distribution of photon speeds around c, with a std. dev. proportional to photon energy. The hypothesis was that the "quantum foam" should have a disproportionate effect on higher energy photons, producing a measurable broadening of their energy distribution compared to lower energy photons. Their analysis showed that there is no such broadening to within a very small tolerance .. they claim any such effects must be smaller than about 60% of the Planck scale with 99% certainty.
swordsman
2.3 / 5 (3) Mar 23, 2015
Very amusing. The article says nothing, but reaches conclusions based on nothing. Laced with "quantum" notations to make it seem believable. Comments above correlate to the material.
indio007
1 / 5 (1) Mar 23, 2015
GRBs, as such, are ideally suited for testing the idea of stochastic variations in the speed of light.

Variations in the "official" speed of light are well documented. However , I'm sorry to report that no variation can ever be found in the future. The meter is now defined as a derivative of the distance light travels over a certain time.
Mathematically, there can be no variation.
Protoplasmix
5 / 5 (4) Mar 23, 2015
If you read the full article (sadly paywalled) ...
The webpage "Spacetime foam and the speed of light" has a link to the full article at the bottom of the page: http://www.relati...ht.shtml
For convenience (and the sake of education) here's the link to the pdf file:
http://www.relati...ight.pdf

And a correction, GRB090510 photon energy was up to ~31 GeV, not 13, oops.
indio007
1 / 5 (2) Mar 23, 2015
Maybe you guys don't understand how big of a fail ric=0 is. Ric = 0 for "empty space" that is "outside a body".
The field equation is highly non-linear and describe a single object present in the model.
You can't simply add another mass (even an identical one) and compute a solution.
There is no TRUE way to figure out the value of what the ricci tensor is for the two masses.
You can't recover the value by reversing because again Ric = 0. A complete expansion of the field equations show that in the single body case 0 is a divisor (let's pretend you can divide by 0). So reversal multiplys by zero killing the whole endevour. Ric can not be recovered to calculate the curvature.
indio007
1 / 5 (3) Mar 23, 2015
Crothers has several proof here
https://www.youtu...HHXaPrWA
't Hooft fails to address all or maybe of them useing strawman arguments and ad hominems .
Protoplasmix
5 / 5 (3) Mar 23, 2015
Very amusing. The article says nothing, but reaches conclusions based on nothing. Laced with "quantum" notations to make it seem believable. Comments above correlate to the material.
You will likely appreciate the recent post over at BackReAction --
http://backreacti...ime.html

In the comments there, Zeph is given some sage advice about his question on the mass of massless photons. I may have to take the rest of the day off if I can't stop chuckling...
Comphuman
not rated yet Mar 23, 2015
Noting that photons coming here from a great distance are not inhibited by quantum foam does NOT "confirm" Einstein's General Theory of Relativity.

It could also "confirm" another thesis where ethers and foams do not exist, and where the observed data match the theory. I recommend this YouTube slide show: youtu.be/6YmHdTMdmHc
baudrunner
1 / 5 (2) Mar 23, 2015
On the contrary, they're [photons] a consequence of the foamy nature of space-time (the vibrations of this foam makes it more dense in similar way, like the shaking of soap foam). It all works together.
No, you don't understand. Photonic waves have no relationship to iotonic space. Iotons are not particles, they occupy what in effect represents the true Planck volume, the smallest unit of space that can be occupied by a volume. In order for anything to exist, it must have a differential attribute. The only justification for an ioton's existence is that they have the attribute of phase.

My choice of words describing iotonic space in the absence of any matter- "lined up" - was a bad one. I should have said that all iotons in such a space would be "in phase".

Again, photonic waves are sourced by matter and propagated by matter particles in the particle medium.
Protoplasmix
5 / 5 (5) Mar 23, 2015
Crothers has several proof here
https://www.youtu...HHXaPrWA
't Hooft fails to address all or maybe of them useing strawman arguments and ad hominems .
k, one last appeal to your common sense, indio -- the Global Positioning System wouldn't work as accurately if we were to use the assertions of Crothers, because relativity. Also, the orbital period of the two neutron stars in PSR B1913+16 would be unchanged using the "Crothers metric" -- do you know which theory is required to calculate the observed rate of inspiral for that system, indio? Hulse & Taylor used the general theory of relativity, quite successfully, you may have heard.
TimLong2001
1 / 5 (1) Mar 25, 2015
@indio0071 / 5 (2) Mar 19, 2015
Time as a dimension has a big problem. It's a point. The past is gone and the future hasn't happened. What's left? The point like present. spacetime does not represent reality.

CORRECT! What Einstein was saying was that time arises through motions in space due to interaction of charged particles. The only thing that exists is THE PRESENT STATE OF THE UNIVERSE (Minkowski's "loaf" notwithstanding. Photons have mass, though vanishingly small. Pair formation of 1.0216 MeV gammarays is the key to understanding the dynamics of the photon of electromagnetic radiation. An electron (negatron, of matter) and a positron (antimatter) are formed when the threshold gammaray decays. These slightly smaller masses in stable photons, throughout the full spectrum of electromagnetic radiation, occur when opposite charges of equal mass bind by charge attraction. A "finding" at the LANL Plasma research facility was that opposite charges interact at right angles.
TimLong2001
1 / 5 (1) Mar 25, 2015
(cont'd): The charge attraction, balanced by the centrifugal force of the right-hand rule rotation, locks the specific energy photon into a fixed wavelength due to the equal masses associated with the opposite charges. BTW, the massive photon is responsible for the background red shift, heretofor attributed to expansion. There are numerous causes, i.e., Poynting Robertson Effect of a rotating particle, absorption and re-radiation, gravitational lensing, interference effects (both constructive and destructive). Hubble was correct in his original proclamation that the factor R relating to red shift may be attributable to other factors than the proposed "theologically consistent" expansion described in the big bang hypothesis.
DarkLordKelvin
4.2 / 5 (5) Mar 25, 2015
Photons have mass, though vanishingly small.
No, they don't, for reasons already covered in this thread.
Pair formation of 1.0216 MeV gammarays is the key to understanding the dynamics of the photon of electromagnetic radiation. An electron (negatron, of matter) and a positron (antimatter) are formed when the threshold gammaray decays.
Except pair production can't happen from a gamma in empty space, it must occur near a massive body (e.g. nucleus), for momentum conservation.
These slightly smaller masses in stable photons
what does that mean?
throughout the full spectrum of electromagnetic radiation, occur when opposite charges of equal mass bind by charge attraction
photons do not "contain" massive particles with opposite charges; if they are of sufficient energy, they can enable massive particles to come into existence, either briefly as virtual particles, or potentially permanently if momentum and energy can be simultaneously conserved.
Dethe
1 / 5 (3) Mar 26, 2015
for reasons already covered in this thread
These "reasons" are based on assumption, the photons have zero mass, so we are facing circular reasoning here. The math formulas cannot oppose the experimental physics.
pair production can't happen from a gamma in empty space
It can happen at presence of CMB, which represent a space-time curvatures here.
photons do not "contain" massive particles with opposite charges; if they are of sufficient energy
They're in dynamic equilibrium with particle-antiparticle pairs, as explained above.
Fleetfoot
5 / 5 (3) Mar 26, 2015
The math formulas cannot oppose the experimental physics.


The experiments confirm the prediction:

http://en.wikiped...ton_mass

All experimental results are consistent with zero mass.
Fleetfoot
5 / 5 (3) Mar 26, 2015
Time as a dimension has a big problem. It's a point. The past is gone and the future hasn't happened.


Only if you make the philosophical assumption that only the present exists. It is well known that presentism is incompatible with relativity, see for example Skow's attempts to reconcile them.

What's left? The point like present. spacetime does not represent reality.


Experiments tell us that relativity is correct so it is presentism as a philosophy that fails to correspond to reality.
Dethe
1 / 5 (2) Mar 29, 2015
All experimental results are consistent with zero mass
Zero REST mass. Can you spot the difference? Apparently not..
Fleetfoot
5 / 5 (2) Mar 29, 2015
All experimental results are consistent with zero mass.
Zero REST mass. Can you spot the difference? Apparently not..


I can see the difference, apparently you can't read. As I said correctly, all experiments are consistent with "zero mass", not "zero rest mass". Presumably you are not aware that mass is invariant.

The archaic concept of "relativistic mass" (if that was what you were thinking about) isn't actually the mass, it is the sum of the mass plus the kinetic energy of a massive particle divided by c^2.

You really should try learning some basic physics before posting, you can't be an effective troll unless your posts have some veneer of credibility.

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