# Who needs photons?

Zz:
…while you may have understood Lorentz transformation, your application of it to the photon frame is faulty. That, in itself, is the crux of your original argument in your first post. You need to first and foremost, understand why this is wrong…
This is valuable advice to me (and future, error prone individuals).

Espen 180:
You have to accept that a frame such that you are [describing] does not exist. It violates the second postulate of SR. You don't even need relativity to see it. An oscillating [EM wave] at rest (zero propagation velocity) doesn't satisfy Maxwell's equations.
This is representative of the issue referred to by Zz, others above and in other forums. Thank you all for your efforts to clarify this. It is incredibly easy for outsiders, such as myself, to make this mistake! I had been applying the concept of Newtonian inertial reference frames which according to Wikipedia (http://en.wikipedia.org/wiki/Inertial_frame_of_reference#Newton.27s_inertial_frame_of_reference) are non-accelerating (i.e. no change in speed or direction) compared to other such frames. Once I started using concepts like absolute speed c, length contraction, time dilation or simultaneity, I had an a priori requirement to adopt relativistic inertial reference frames (RIRFs). Since the rest of you clearly know this, I quote for my own benefit (and future visitors making the same mistake):
Einstein's theory of special relativity, like Newtonian mechanics, assumes the equivalence of all inertial reference frames, but makes an additional assumption, foreign to Newtonian mechanics, namely, that in free space light always is propagated with the speed of light c0, a defined value independent of its direction of propagation and its frequency, and also independent of the state of motion of the emitting body. This second assumption has been verified experimentally…
So, my eyes are open. I now understand why there was so much pushback to my concept of pinholes (photo-induced wormholes) as a replacement for photons. That was based on a mechanism of length contraction in each photon’s own frame c. But if those frames are disqualified as RIRFs, I’m not justified in invoking path contraction to zero.

But help me out please. Suppose two photons A & B pass at right angles. I pick Frame A on photon A (so photon A is “at rest” in Frame A). What is the behavior of photon B in Frame A?

I’m guessing photon B does what it always does in an RIRF. Doesn’t it just zip away at speed c? If that’s the case, are we really saying that Frame A is disqualified as an RIRF just because one photon (photon A) is not traveling at speed c? The whole point of my inquiry is that instead of disqualifying Frame A as an RIRF, we should instead be disqualifying photon A as a photon!

I suggest that in Frame A, photon A is not a photon, it's a collision. With the emitting and absorbing electrons in contact, there is no room for a photon. The frequency of that light goes to zero while the wavelength becomes infinite, as with a static charge. So, as was pointed out, there’s no EM wave, no propagation, just transfer of energy and momentum. But with all the other photons behaving as if Frame A is an RIRF I think the weight of evidence shows it is.

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JesseM
But help me out please. Suppose two photons A & B pass at right angles. I pick Frame A on photon A (so photon A is “at rest” in Frame A). What is the behavior of photon B in Frame A?
There is no relativistic inertial reference frame where this is true. You just quoted something that said "Einstein's theory of special relativity, like Newtonian mechanics, assumes the equivalence of all inertial reference frames, but makes an additional assumption, foreign to Newtonian mechanics, namely, that in free space light always is propagated with the speed of light c0"--this assumption would be violated if we allowed a coordinate system where photon A is at rest to qualify as an "inertial" frame in relativity. Such a coordinate system can be defined, but it is a non-inertial frame, by definition.

...in free space light always is propagated with the speed of light c0"--this assumption would be violated if we allowed a coordinate system where photon A is at rest to qualify as an "inertial" frame in relativity. Such a coordinate system can be defined, but it is a non-inertial frame, by definition.
I hear you. The recurrence of this important message is starting to get through even my thick skull. I’d like to consider photon A a bit further though. Just thinking out loud.

Photon A travels from emitting to absorbing electron with a trajectory. That trajectory can be thought to have a slope, covering a certain span of distance in a certain time. In all relativistic inertial reference frames (RIRFs), the slope MUST be c. (You see, I did hear you, and the others!)

As we sample different RIRFs, though time dilates and length contracts, the ratio for the trajectory of photon A is always c. Well, in calculus we used to calculate slopes of curves quite successfully as the span under consideration shrinks to zero. In the case of photon A, I assert that its speed remains c, as path length shrinks, right down to a single point, of “contact”. That photon A becomes undetectable to us in frame A is of no consequence because its energy and momentum are never lost from our view. Nothing is ever lost to us, including the status of Frame A as an RIRF. Photon A never stops traveling at speed c, its path simply gets too short to matter, but the ratio stands. Zero path length occurs in a relativistic inertial reference frame.

That was just me. Here is a similar line of thought I found wandering around PF.
…you can consider what happens if we give progressively lighter particles a fixed amount of kinetic energy before they are to transit from point A to B (which are at rest relative to you). As the the particles become lighter (i.e. having less rest mass) they will have to move faster relative to you (but still less than c) to have the same fixed amount of kinetic energy. From the perspective of each of these particles (ie. from a reference frame at rest relative to each particle) the distance A B has been shortened due to length contraction. In the limit where the mass go to zero, the distance each particle travel also goes to zero, meaning they would travel any finite distance however big in zero time. Now change the particles with ones that are born at A and absorbed at B and you have, more or less, "the life of a photon".
Nonetheless, I’m not really expecting to win any converts here. Your opinions are qualified, mine are not. But aren’t you just a little more interested in the pending spectral analysis of ultracold antihydrogen now than you were before? Why else would Gerald Gabrielse and others spend decades working toward this experiment unless there was some possibility that antiatoms will show us something new? Surprises happen!

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I hear you. The recurrence of this important message is starting to get through even my thick skull. I’d like to consider photon A a bit further though. Just thinking out loud.

Photon A travels from emitting to absorbing electron with a trajectory. That trajectory can be thought to have a slope, covering a certain span of distance in a certain time. In all relativistic inertial reference frames (RIRFs), the slope MUST be c. (You see, I did hear you, and the others!)

As we sample different RIRFs, though time dilates and length contracts, the ratio for the trajectory of photon A is always c. Well, in calculus we used to calculate slopes of curves quite successfully as the span under consideration shrinks to zero. In the case of photon A, I assert that its speed remains c, as path length shrinks, right down to a single point, of “contact”. That photon A becomes undetectable to us in frame A is of no consequence because its energy and momentum are never lost from our view. Nothing is ever lost to us, including the status of Frame A as an RIRF. Photon A never stops traveling at speed c, its path simply gets too short to matter, but the ratio stands. Zero path length occurs in a relativistic inertial reference frame.

That was just me. Here is a similar line of thought I found wandering around PF.

Nonetheless, I’m not really expecting to win any converts here. Your opinions are qualified, mine are not. But aren’t you just a little more interested in the pending spectral analysis of ultracold antihydrogen now than you were before? Why else would Gerald Gabrielse and others spend decades working toward this experiment unless there was some possibility that antiatoms will show us something new? Surprises happen!
Yes, but this doesn't change some of your fundamental confusion here, and using the personal motivation of some researchers doesn't change the possible reality. I also do not believe that the researchers are making the same error in their assumptions that you have been regarding SR.

Yes, but this doesn't change some of your fundamental confusion here, ...error in...assumptions that you have...regarding SR.
I agree there is confusion, understandably so, but is it mine? The model for light, like model for the atom, has been undergoing extensive revision since mankind first had one. Do you imagine we’re finished? Pinholes are not awfully different than photons. They offer collisions. What could be more particle-like than that?

As far as reference frames go, I was quick to recognize and concede to the need for relativistic inertial reference frames (RIRFs), as pinholes rely on path contractions. But as I believe I have shown, the error lies with current theory. A frame in which a photon “would be” at rest is not at all the same as a photon “being” at rest! The instantaneous velocity of a photon, determined as path length shrinks to zero, remains c. Thus, the frame is an RIRF. In such a frame, the photon energy and momentum still exist, no longer between emitting and absorbing electrons but in them (same as billiard balls).

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DrGreg
Gold Member
In such a frame, the photon energy and momentum still exist, no longer between emitting and absorbing electrons but in them (same as billiard balls).
No, if such a "frame" were valid, both the "energy" and "momentum" would be red-shifted to zero.

Your "frame" has some peculiar properties
1. photons are at rest
2. photons travel at speed c
3. all distances are zero
4. all times are zero
5. photons have zero energy and momentum
None of this makes sense, which is what everyone has been telling you.

No, if such a "frame" were valid, both the "energy" and "momentum" would be red-shifted to zero.

Your "frame" has some peculiar properties
1. photons are at rest
2. photons travel at speed c
3. all distances are zero
4. all times are zero
5. photons have zero energy and momentum
None of this makes sense, which is what everyone has been telling you.
Thank you, I was pondering how to respond to that one.

TCS
I agree there is confusion, understandably so, but is it mine? The model for light, like model for the atom, has been undergoing extensive revision since mankind first had one. Do you imagine we’re finished? Pinholes are not awfully different than photons. They offer collisions. What could be more particle-like than that?

As far as reference frames go, I was quick to recognize and concede to the need for relativistic inertial reference frames (RIRFs), as pinholes rely on path contractions. But as I believe I have shown, the error lies with current theory. A frame in which a photon “would be” at rest is not at all the same as a photon “being” at rest! The instantaneous velocity of a photon, determined as path length shrinks to zero, remains c. Thus, the frame is an RIRF. In such a frame, the photon energy and momentum still exist, no longer between emitting and absorbing electrons but in them (same as billiard balls).

I don't believe that electrons are like pinholes. They are more like diffraction patterns created by pinholes. As long as the electron is at a constant velocity, it exists as a perfect diffraction pattern/standing wave. If you distort that distribution, you accelerate the particle.

Accelerating an electron, is like moving the central portion of a spherically diffracted wave. That motion is then translated through the diffraction pattern at the speed of light. A photon is the embodiment of that dispersion of the interaction through the electron's diffraction pattern, since that dispersion will alter the way that the elecron interacts with all the other waves.

Accordingly, you can't have a photon reference frame because photons are just the redistribution of energy through a particle's wave pattern.

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No,...
DrGreg, thank you for taking the time to respond. I love your accent! When you say, “No” it has a convincing tone which I couldn’t hope to achieve. I’ll give it a try though. Ahem, “Touch the stars! Pinholes are ours!”

Aw, shucks. Even rhyming and color don’t seem to help.
…if such a "frame" were valid,…
For the benefit of latecomers, the “frame” referred to is a reference frame moving at speed c and parallel to any particular “photon”. Let’s refer to this as "Frame c" in this post and for simplicity, let’s assume the emitting and absorbing particles are both electrons occupying a different frame we’ll name "Frame<c", a relativistic inertial reference frame (RIRF). I put the term “photon” in quotes because I believe they are actually pinholes (photo-induced wormholes) but I will refer to photons normally now.

I’m afraid I see the question of validity settled. It was one of your fellow Brits, who showed us the way to instantaneous velocity, and I do say, it’s worked out quite well! The instantaneous velocity of a photon, determined as its path length shrinks to zero (as frame velocity goes to c), remains c. Frame c is an RIRF.
… both the "energy" and "momentum" would be red-shifted to zero.
If the momentum and energy were locked in a photon perhaps, but not when they are in objects. The emitting and absorbing electrons are massive and therefore can’t be at rest in Frame c. So they must be moving and not just capable of, but obligated to, have energy and momentum in that frame. The difference in their energy is one quantum, as verified in the disposition of their orbitals before and after the collision. With an instantaneous velocity of c giving a path length reduced to a point of contact, it is conceptually much easier to envision a hole than a particle. The pinhole projects as the trajectory of a light ray in Frame<c.
Your "frame" has some peculiar properties
Quantum tunneling, virtual particles, point particles, dark energy, renormalization… Nah! Ain’t nothin peculiar ‘bout physics!

But seriously, consider the pairing behavior of electrons. There are orbital filling, covalent bonds, ionic bonds even the Cooper pairs in metallic superconductors to illustrate the tendency that electrons pair in space for various lengths of time. To the extent that Einstein attributed dimensionality to time, there may also be a tendancy that electrons pair in time over various distances in space. While I explain light as pinholes with slope c. The “quantum non-local connection” of the EPR paradox seems to offer us a spinhole (spatial interconnecting wormhole), with slope zero to explain entanglement. So, it’s not just outsiders who envision the utility of a "tiny wormholes" model. As I recall, John Wheeler’s quantum foam is also filled with them.
1. photons are at rest
Photons always travel at speed c in RIRFs, even Frame c. But there is no obligatory distance which must be traveled.
2. photons travel at speed c
Yes they do! They all do. That’s why Frame c is an RIRF.
3. all distances are zero
By "all distances", I assume you refer to those components of distance parallel to Frame c. The perpendicular components are unaltered. Now for the parallel components. While the Lorentz transformations do not explicitly contain this restriction, it is intuitive and consistent with observation that its effects, while independent of path length, do not extend beyond the emitting and absorbing particles. A solution might be to adjust the definition of “opacity” to restrict the transforms externally. In addition, pinholes, unlike photons are very narrow. The transform need only be considered within it. It is the unrestricted nature of current electric and magnetic field descriptions which have lead us to think otherwise.
4. all times are zero
Nothing ages in Frame c. And there is nothing to say transfer of energy between to colliding electrons takes any time at all. This is equivalent to the question, “Does a photon accelerate from an emitting electron or is it “born” at speed c? My impression is distinctly the latter for the photon model.
5. photons have zero energy and momentum
Photons have E= hf, as always but only as far as they travel (see reply to 1.)
None of this makes sense,…
You are obviously in very good company. I hope it makes more sense now.
…what everyone has been telling you.
“What do you care what other people think?”, R. Feynman, in a book title no less! Do you think he meant it?

Please excuse my sometimes, “playful” wording on a subject you obviously take seriously. I take it from your PF name, that you have an advanced degree in physics. I respect your accomplishment and value your opinion. I don’t think you should change that opinion based upon anything I’ve written. Demand proof! I believe Dr. Gabrielse and colleagues will be as pivotal to pinhole theory as Sir Arthur Eddington was to Relativity (description above, diagram attached).

Can you imagine the blunder though, if without testing, we continue in the assumption that light (from normal matter) is exactly the same as antilight (from antimatter). There could be entire distant galaxies of antimatter out there and their dim rays would be entirely invisible to us unless we adapt telescopes to detect the characteristic secondary gamma emissions resulting from remote annihilations. Here's to not keeping ourselves in the dark. Good day.

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A question was posed to me from outside PF. I will volunteer and respond to it here since, in my view, it represents the last, logically independent argument from Relativity against pinholes (photo-induced wormholes) as an alternative to photons. That is, I believe all arguments would challenge based upon validity of reference frame (as relativistic), length contraction (path = 0), time dilation (to infinity) or simultaneity. The following question is based upon the last of these and has not yet been raised here.
Suppose you’re right that “Frame c” (moving at light speed) actually is relativistic. It seems that you have still painted yourself into a corner regarding simultaneity. Many times you stressed that the electrons make “contact”, “collision” or have “zero path length” through a pinhole. But when there is an event such as contact, occurring in one place at one time, it must be that all observers agree on the event. But when a picture is taken of the night sky, no one agrees that any star is touching the camera!
This is a good point. I could give a wimpy answer like, “There aren’t really enough star electrons touching the camera for anyone to witness.” But that wouldn’t get to the heart of this issue. To be sure, Wikipedia in the first paragraph in, “Relativity of Simultaneity” says:
Where an event occurs in a single place-for example, a car crash-all observers will agree that both cars arrived at the point of impact at the same time.
I assert that a speed of light reference frame (Frame c) is a relativistic inertial reference frame because its motion is uniform and all photons have speed c (even those traveling zero distance). However, it remains true that no massive particle can achieve speed c. Thus, as has been stated here (#10) and in other forums, no observer can enter Frame c. Frame c is, in fact, the only place that we can be sure all observers will agree they did not see the collision! If you roll it about in your mind a bit, I think you’ll find that this is really quite consistent with what the questioner and Wikipedia have said.

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I assert that a speed of light reference frame (Frame c) is a relativistic inertial reference frame because its motion is uniform and all photons have speed .

Ok,

Can you derive the transformations between your "Frame c" and the other inertial frames?
All the other inertial frames are connected by the Lorentz transforms, derive the transform that connects your "Frame c" with the rest.

Can you derive the transformations between your "Frame c" and the other inertial frames?
I use the limiting case where velocity of the frame approaches c. Time stops (goes to infinity), path length goes to zero and contact between emitting and absorbing particle occurs. I believe the particle aspects of light are not an accident or mysterious but a simple collision.

Controversy arises when people mistakenly presume a photon is “at rest” in Frame c. That is resolved two ways. First, a “photon’s” instantaneous velocity remains c as path length shrinks to zero. Zero path length is not itself problematic as there is no obligatory distance required for any photon in relativistic inertial reference frames (RIRFs). But with path length zero a photon is obviously superfluous. So second, photons were never more than a very useful accounting tool in any frame (in my view). I replace photons with pinholes (photo-induced wormholes).

There is room for debate about the meaning of “contact” as the position and velocity of electrons is uncertain and their diameters (let alone surfaces) are indeterminate for what are often referred to as “point particles”. But a functional definition is still available to us. For this discussion, “contact” is: sufficient proximity for transfer of kinetic energy (and momentum). In the test case illustrated in #34 asserting a remote collision between electron and positron (via light transmission) contact is: sufficient proximity to result in annihilation. I offer this as both necessary and sufficient to the confirmation of pinholes. Except for this discussion, such an occurrence is, to my knowledge, completely unanticipated. To the extent that contact may "work" at distances greater than zero (Planck length?), the discussion only shifts toward more established RIRFs and pinholes.

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I use the limiting case where velocity of the frame approaches c.
So, where are your formulas for the transformation I asked you? Can you show them?

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l=ct, l'=c't', c=c' {...goose bumps}

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l=ct, l'=c't', c=c' {...goose bumps}
There are not transforms, these are nothing.

reilly
Faradave -- There's a good reason why the concept of photon has survived for over a century -- the concept is immensely useful, spanning many branches of physics, and it passed many tests with flying colours. Simply stated, there is no reason to challenge the idea of a photon -- there's no evidence at all that the concept is wrong.

To gain any legitimacy, you must show 1. that your ideas agree with experiment, and 2. the specifics of where current theory is not correct and your theory is.
You have yet to do this.

Lot's of luck.
Regards,
Reilly Atkinson

Light from the sun is slowed when entering water. What is its speed after emerging from the water?

Light from the sun is slowed when entering water. What is its speed after emerging from the water?
c.

BTW, light is not slowed down in water, its path is lengthened. This is a complex phaenomenon.

We know that light is slowed when passing through water. Is this slowing cumulative? What is the speed of the same light upon emerging from the water?

Btw: I disagree. But let me put it another way. When I see objects that are under water, like a tasty tuna, I know that this light has passed through a given distance, let's say 100 feet. 100 feet is the same distance whether in air or water. Is the light from the tuna faster after leaving the water?

We know that light is slowed when passing through water. Is this slowing cumulative?
No.

What is the speed of the same light upon emerging from the water?

Btw: I disagree.
That's your problem. Taking a physics class might help.

But let me put it another way. When I see objects that are under water, like a tasty tuna, I know that this light has passed through a given distance, let's say 100 feet.
Actually light in the water has taken a complex, zig-zag path, not a straigh one like in vacuum.

100 feet is the same distance whether in air or water.
No, see above.

Is the light from the tuna faster after leaving the water?
No. It is still c.

We know that light is slowed when passing through water. Is this slowing cumulative? What is the speed of the same light upon emerging from the water?
This shows that you have literally no idea what Relativity in its special or general formulation is. None at all, which is alright, but not if you're trying to debunk photons, instead of learning basic physics.

...these are nothing.
They were quite something to Dr. Einstein, especially the last one, c=c'.

DrGreg
Gold Member
l=ct, l'=c't', c=c' {...goose bumps}
There are not transforms, these are nothing.
Faradave, what is being asked for here are equations that express l' and t' in terms of l and t (although I'd prefer to use the symbol x instead of l), not just for the motion of photons but for any arbitrary events.

Hint: the Lorentz transform between two inertial frames is conventionally given by the equations

$$t' = \gamma(t -vx/c^2)$$
$$x' = \gamma(x - vt)$$​

where $\gamma = 1/\sqrt{(1-v^2/c^2)}$.