[Evan haines II]

Evan haines II  asked the Naked Scientists:
   Me and one of my friends (at school) have been contemplating the problem for quite a few months now, we've each come up with various solutions but none could bare close examination, the problem is this:

Although light dosn't act like particles or waves, could Newton's 3rd law be applied to photons because they bounce?
What do you think?

[Vern]

Newton's Third Law I think is:
For every action, there is an equal and opposite reaction.
To me it seems that this would apply to light. Poincare was calculating the effect of light bouncing off a target back around the turn of the century.

[yor_on]

momentum?

[A Davis]

Photons mathematically have a cylindrical solution its like a tube the wave is also rotating so the final solution actually looks like a coiled spring. One could imagine this coil being compressed at a surface and the energy in the compression being released when it bounces away. Food for thought.

[yor_on]

Interesting A D.
Here is an artistic interpretation of how a elementary particle might look like.
http://physicsworld.com/cws/article/news/23369

We can only guess at how a photon looks like, as I understands it.
We can look at electromagnetic waves (consisting of photons) and draw conclusions from there.
A electromagnetic wave has two properties. One electrical and one magnetic, perpendicular to each other.
That means that they are placed at a 90 degree angle to each other
(somewhat like two 'sine waves', one vertical, the other horizontal but bound together) .
http://rh5.clemson.edu/ropermtn/EMbasics.php

Maybe one could see it as a propagating electro-magnetic field, constantly creating itself as a pulsating 3D-bubble, as its 'two sine waves' interact with each other?

There is also the possibility that what we see as being photons just are expressions of 'something else', not existing as free 'entities' at all.
That as we only can prove them as existing when they 'interact' with something, like a eye:)

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But that doesn't really explain their 'particle like' properties as I see it.
On the other hand I can't see them as being the same as matter.
So?

[Vern]

A electromagnetic wave has two properties. One electrical and one magnetic, perpendicular to each other.
That means that they are placed at a 90 degree angle to each other
(somewhat like two 'sine waves', one vertical, the other horizontal but bound together) .
http://rh5.clemson.edu/ropermtn/EMbasics.php

I made a little computer program that makes that kind of photon. It looks kinda like this:


Red represents the positive electric field; blue the negative electric field; and the other two colors are the magnetic field north and south polarities.

[Vern]

But that doesn't really explain their 'particle like' properties as I see it.
On the other hand I can't see them as being the same as matter.
So?

To me it makes sense to think of the maxima of the wave where change is greatest as looking like a particle where the planes of the electric and magnetic cross.

[yor_on]

looking like, or being, a 'particle'?

[Vern]

momentum?

I think it was momentum Poincare was studying, but I couldn't find my reference.

[Vern]

I suspect that the EM wave is saturated at its peak amplitudes. That is what gives quantum phenomena to everything. If we determine the mass of a particle as m = hv / cc we're saying a photon is potential mass. The potential is realized any time we confine the photon. Such as bouncing it around inside a mirrored box or between atoms in any piece of mass. Here's Isaic Newton's take on it.

"Are not gross Bodies and Light convertible into one another, and may not Bodies receive much of their Activity from the particles of Light which enter their Composition?
The changing of Bodies into Light, and Light into Bodies, is very conformable to the Course of Nature, which seems delighted with Transmutations. [...] And among such various and strange transmutations, why may not Nature change Bodies into Light, and Light into Bodies?“

IsaacNewton - Optics 1704, Book Three, Part 1 Qu.30

[lyner]

Newton's Third Law gives you the Momentum Conservation law directly.
Photons exhibit momentum.
So yes it should apply.

Vern:
Have you any 'size' for the picture you have drawn for a photon? How does it relate to the wavelength of the em, concerned?

[Vern]

Vern:
Have you any 'size' for the picture you have drawn for a photon? How does it relate to the wavelength of the em, concerned?

Hi; yes the size in wavelength is from the forward tip to the rearward tip. This is the classical photon that was taught in school back some 40 years ago when I went to school.

[A Davis]

To yor_on and Vern you are both neglecting rotational polarisation all photons have rotational polarisation otherwise polaroid glasses wouldn't work. The normal solution is double rotation, which produces a circular tube within a tube so i did cheat a bit with the spring analogy it's for single rotational polarisation for the P1(cos(theta) solution but higher order solution do produce springs within springs even with double rotational polarisation P4(cos(theta) is a good example.

[Vern]

To yor_on and Vern you are both neglecting rotational polarisation all photons have rotational polarisation otherwise polaroid glasses wouldn't work.

The photon depicted in the graphic is vertically polarized; in that the electric field is vertical; it could just as well be horizontal or even spinning; I can't see how this is neglecting it. I didn't mention it, but I know it is so as you say. Your spring analogy is interesting; I'll have to think about it.

[lyner]

Vern:
Have you any 'size' for the picture you have drawn for a photon? How does it relate to the wavelength of the em, concerned?

Hi; yes the size in wavelength is from the forward tip to the rearward tip. This is the classical photon that was taught in school back some 40 years ago when I went to school.

That's' what I suspected.
It's a very poor place to start from if you want to advance understanding. Unless there is an infinite bandwidth, photons can't be regarded as taking up just one cycle. The interaction of a photon with an atom ( or something) must take time because it is a resonance phenomenon - the effect needs time to build up. So the photon, if it's going to exist on the way, must be a wavetrain of finite length. I haven't actually heard a really consistent argument for the existence of photons anywhere else than where they interact at either end of their journey.
The models presented in School are nearly all 'lowest common denomator' and passed on through the curriculum via a teacher who may (a very unfair generalization, I know) have never seriosly thought it through. A teacher from the 1960s may well have never had the benefit of being taught what was the state of understanding at the time.
If you want a good model then go for one from modern publications and then you need to read between the lines and note how non-commital Sciectists are when describing how things 'actually are'. That's often because there isn't and can't be an actual answer.

[lyner]

To yor_on and Vern you are both neglecting rotational polarisation all photons have rotational polarisation otherwise polaroid glasses wouldn't work. The normal solution is double rotation, which produces a circular tube within a tube so i did cheat a bit with the spring analogy it's for single rotational polarisation for the P1(cos(theta) solution but higher order solution do produce springs within springs even with double rotational polarisation P4(cos(theta) is a good example.

That's yet another problem with insisting on the existence of photons 'on the way from a to b'. It involves more and more intricate descriptions compared with the more straightforward wave description. It puts me in mind of the desperate attempts to describe planetary orbits in terms of a complicated series of 'perfect' circles (epicycles) rather than ellipses.
Polarization is much more straightforwardly included using the wave model.

[Vern]

That's yet another problem with insisting on the existence of photons 'on the way from a to b'.

Very interesting; this is the first I have seen of the notion that photons only exist at each end of their journey. I know that they can only be observed when they become mass.

If you allow one photon to be more than one wave length you lose the logical connection between relativity phenomena and the speed of light. I know that this is the current thinking in the scientific community, but it seems a great loss to me.

Edit: If a photon exists as more than one wave length, then how many wave lengths does it take? Infinite?

[lightarrow]

To yor_on and Vern you are both neglecting rotational polarisation all photons have rotational polarisation otherwise polaroid glasses wouldn't work.

There isn't any need of circular polarization to describe a polaroid's functioning:
http://en.wikipedia.org/wiki/Polarizer
http://en.wikipedia.org/wiki/Polarization

[Vern]

The interaction of a photon with an atom ( or something) must take time because it is a resonance phenomenon - the effect needs time to build up.

Yes; I agree; that's why I suspect a photon exists as just one cycle. One cycle takes just exactly the amount of time it takes to absorb a photon.

If the photon does not exist and only its effects exist, there still has to be communication between the transmitting atom and the receiving atom doesn't there. Otherwise how does the receiving atom "know" a packet of energy is available.

[Vern]

Unless there is an infinite bandwidth, photons can't be regarded as taking up just one cycle.

I think this is one of the hazards of dwelling too much on the uncertainty principle. This is true if you insist on statistical analysis of the wave. Statistical analysis might not be as necessary in the real world as it is in the world of Quantumania. []