Photon Matter

[7DSUSYstrings]

2 photons can’t, in a strict sense, collide, or otherwise interact with one another in a way that affects either photon. On both a theoretical and practical level, this is an important concept.

 

Take those same two lasers, HeNe's, and aim them into one another so they intersect at 180deg, i.e. by locking them into the web of a narrow, 90deg aluminum angle, and watch how they cancel each other.

 

Yes, photons can interact and, if you were to switch one of the lasers with a laser of different frequency, you would find deflection detectable with a scintiloscope, or perhaps better if the experiment were performed in a vacuum scintillation tube specially designed to detect an array of deflections.

 

This is because of the chiral nature of the photons I mentioned earlier. One must intrinsically take into account that time, gravity and light travel together...

[CraigD]

Take those same two lasers, HeNe's, and aim them into one another so they intersect at 180deg, i.e. by locking them into the web of a narrow, 90deg aluminum angle, and watch how they cancel each other.

I believe the 180° you mention doesn’t describe the geometric angle of intersection of 2 laser beams of the same frequency, but their phase. Both angles and wave phases are traditionally measured with the same units, such as radians or degrees.

 

Overlapping waves of many kinds, not just light, can produce dramatic and paradoxical-seeming destructive interference – what one might call “canceling each other”.

 

A simple, if impractical to actually construct, example, is two identical lasers pointed directly at one another, with photodetectors small enough to detect but not completely block the beams placed in their path, each connected to a counter that measures the duration that the detector detects a beam. The lasers are positioned so that the distance between them is precisely [imath]\left(n +\frac12 \right) \lambda[/imath] (where [imath]n[/imath] is an integer and [imath]\lambda[/imath] is the wavelength of the laser light) – one of the reasons the pictured setup is impractical for the casual hobbyist to construct is that [imath]\lambda[/imath] for a typical visible light laser is around [imath]10^{-6}[/imath] m, requiring very precise optical equipment to make such small distance adjustments and resist uncontrolled changes in distance due to vibration and other external causes.

If only one of the lasers emits a beam, all 3 detector/counters read the same duration. If the lasers start and stop emitting their beams at nearly the same instant, however, counter A and C read the same, but B reads much shorter, or even zero. Because of optical interference, there is a “canceled” segment of the beam where no light is detected. This zone initially appears at the midpoint between the lasers, widening at the speed of light to eventually include the entire beam, if the lasers are allowed to emit for a long enough duration.

Yes, photons can interact and, if you were to switch one of the lasers with a laser of different frequency, you would find deflection detectable with a scintiloscope, or perhaps better if the experiment were performed in a vacuum scintillation tube specially designed to detect an array of deflections.

It’s important to understand that regardless of interference effects such as the above, the photons of the beams don’t interact, and aren’t affected. Although they’re undetectable within the zone of destructive interference, outside of that they’re detectable exactly as if the destructive interference was never observed. The energy they deliver to an eventual target, their direction of travel, their phase – none of these are at all changed.

 

In physics terms, scintillation refers to a specific kind of interaction between photon bosons and electron fermions, in which the electron gains enough energy to be disassociated with its atom’s nucleus – that is, an ionization event – then becomes re-associated with a nucleus, releasing one or more photons. The ionization event can be caused by a photon, or some other charged particle, such as a fast-moving proton or atomic nucleus. Whatever causes the ionization event is termed ionizing radiation.

 

So, while a scintillation detector is a perfectly good way to detect photons interacting with electrons – as is a more everyday available photoelectric detector – it won’t detect the result of photons interacting with each other such as deflecting photons in a laser beam, because photons simply don’t interact with one another in such a way.

[Qfwfq]

2 photons can’t, in a strict sense, collide, or otherwise interact with one another in a way that affects either photon.

Not directly, in the sense that there is no photon-photon interaction term in the lagrangian, but they interact indirectly via the fermion-antifermion pairs in their Fock space. The electromagnetic field is almost but not quite perfectly linear. The cross section is exceedingly low and of course an energy of at least [imath]2m_{\rm e}[/imath] is necessary, but two photons can pair produce by way of the fermion of one coinciding with the antifermion of the other. In QED this actually is the mechanism, in a formal sense, but it is ordinarily accomplished in circumstances involving the field of a heavy nucleus which give a reasonable cross-section.

 

Take those same two lasers, HeNe's, and aim them into one another so they intersect at 180deg, i.e. by locking them into the web of a narrow, 90deg aluminum angle, and watch how they cancel each other.

Apart from the fact that in this case they would give an alternating sequence of constructive and destructive interference, called a standing wave, this is not what we mean when talking about interaction and it isn't the mechanism of pair production.

 

This is because of the chiral nature of the photons I mentioned earlier. One must intrinsically take into account that time, gravity and light travel together...

This is a Strange Claim.

[7DSUSYstrings]

This is a Strange Claim.

 

So was Fock space at one time. I suppose you are going to tell me that if we split a beam then reflect them back into themselves, one will magically reverse its direction of rotation?... or are you going to tell me there is no photon rotation?... or are you going to tell me that Fock space is two dimensional?... one dimensional? I suppose the photons simply freeze in mid air (or vacuum) and that the photoelectric effect works equally well at atmospheric pressure?

 

Imagine this as sound. The standing wave will progressively regenerate and increase in amplitude, but it doesn't simply cancel. If you think it does, I recommend attending a Van Halen concert...:phones:

 

I'm certain this discussion will become a comparison of how the "top down" approach to quantum physics views time as a single dimension, where the "bottom up" approach views it as nine.

[Qfwfq]

Read the rules about unsupported claims and try to make sense when you post.

 

If you think it does

I did not say the standing wave simply cancels, I recommend you to understand what folks say before you reply. Besides, if it did cancel to zero, there'd be no standing wave after all.

[Janus]

QM's standard model says the proton is made of quarks. It has no explanation for the mass of the electron. A collision between two beams of photons will produce an electron anti-electron pair.

 

Since you mention electrons, let's consider them. We know the mass of the electron, and we also know the maximum size an electron can have. We also know how much energy the mass of the electron equates to. If you consider a photon of the same energy you will find that its wavelength (effective size) is about 10 times that of an electron.

[Little Bang]

The wavelength of a photon can be infinitely short.

 

E = fh and E = MC^2 so f = MC^2/h. You can calculate a frequency for the mass of the electron which will give a wavelength of a photon with the same frequency.

[Qfwfq]

Wavelength is not the "effective size" of the associated corpuscle. The energy is related to frequency, it's momentum that's related to wavelength.

 

Since you mention electrons, let's consider them. We know the mass of the electron, and we also know the maximum size an electron can have. We also know how much energy the mass of the electron equates to. If you consider a photon of the same energy you will find that its wavelength (effective size) is about 10 times that of an electron.

I'm not sure what you are taking as the maximum size an electron can have, but the highest energy scattering experiments have shown no spatial extent; it is pointlike as far as we can tell. There is no upper limit on its wavelength. Only composite bodies are known to have a geometric extension, as expected by motion of their constituent parts. If it isn't enough that tiny atoms can emit and absorb in the visible range, in NMR the tiny tiny proton absorbs photons in the microwave range.

[Janus]

The wavelength of a photon can be infinitely short.

Not for any energy of a photon.

 

E = fh and E = MC^2 so f = MC^2/h. You can calculate a frequency for the mass of the electron which will give a wavelength of a photon with the same frequency.

 

The wavelength of a photon is inversely proportional to its energy and has a fixed value for any photon of a given energy. For an photon of equal energy equivalnce to an electron, that wavelength is much larger than an electron. Dividing the electron into more photons would require photons of even longer wavelengths.

 

Electrons are not "made up" of photons.

[Qfwfq]

Dividing the electron into more photons would require photons of even longer wavelengths.

As I said somewhat simultaneously to your post, this does not support the matter. Erasmus00 supplied an excellent point concerning the spin value of composite bodies. It can be expanded on, if anybody would like.