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When emitted from an atom, does a photon propagate through space-time as a sphere, in all directions, as a wave or as a directed point particle? I think it's a wave and not a directional point particle and certainly not at the same time. I think that as the Photons travel through space-time as a wave, in a spherical manner and collapses via some sort of interaction (reflection, refraction, absorption) then and only then can a directional point be determined. therefore I think the particle nature of light is simply a by-product of wave interactions, they only exist as a way to show interactions of wave functions or put in another way, light is simply the interactions of wave functions.

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marked as duplicate by John Rennie, user108787, ACuriousMind, user36790, Wolpertinger Sep 3 '16 at 20:01

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    $\begingroup$ Possible duplicate of Some doubts about photons $\endgroup$ – John Rennie Sep 3 '16 at 6:34
  • $\begingroup$ Related physics.stackexchange.com/q/55109 $\endgroup$ – user108787 Sep 3 '16 at 8:35
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    $\begingroup$ Why do you think that something that is not observable is more realistic than something we observe? The wave behaviour of photons is concluded from the intensity distribution behind edges. Even if you shot single photons on an single edge, you will see after a lot of shots some intensity distribution of individual photon impacts. Perhaps it would be better to conclude that there is an interaction between the photons field and the field of the surface electrons of the obstacle and the quantized behaviour of this common field is what we see as an intensity distribution on the observers screen. $\endgroup$ – HolgerFiedler Sep 3 '16 at 9:30
  • $\begingroup$ Yes, you have the right idea, although expressed in language that we don't normally use. The article that @JohnRennie posted has more precise language. $\endgroup$ – garyp Sep 3 '16 at 15:37
  • $\begingroup$ The short answer is that matter and photons have a wave/particle duality. It depends on the experiment as to whether you observe a particle or a wave. $\endgroup$ – MaxW Sep 3 '16 at 18:19
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Can the Particle nature of light be explained by wave interactions?

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When emitted from an atom, does a photon propagate through space-time as a sphere, in all directions, as a wave or as a directed point particle? s.

There is confusion here in the concepts "photon", and "light".

The photon is a quantum mechanical entity, i.e. its existence in space obeys quantum mechanical rules and a wave equation defines its wavefunction, a solution of a quantized form of Maxwell's equations. The Psi*Psi of the solutions of this equation for given boundarry conditions give the probability density of finding a "photon" hit at an (x,y,z,t).It is the probability density for the photon that has sinusoidal/wave properties.

therefore I think the particle nature of light is simply a by-product of wave interactions, they only exist as a way to show interactions of wave functions or put in another way, light is simply the interactions of wave functions.

Light is an emergent quantity from an enormous confluence of photons with energy h*nu. To see how this happens one needs quantum field theory. It is not surprising that the classical electromagnetic wave will have as a frequency in its amplitude the nu of the photon energy, because both are solutions of maxwell's equations.

The photons in a light beam do not interact, (very small probability) they are superimposed as wavefunctions and when complex squared the electric and magnetic fields which define the wave properties of light appear. It is the quantum nature that is the underlying framework. The same is true for classical waves emerging from the underlying quantum mechanical level of atoms and molecules

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does a photon propagate through space-time

A massless photon is not propagating through spacetime, because its lightlike spacetime interval is zero. That means that there is nothing between the point of emission and the point of absorption. The propagation from one point in spacetime to the same point is meaningless, there is nothing which is propagating.

However, it is the wave of a photon which is being observed as propagating through spacetime.

therefore I think the particle nature of light is simply a by-product of wave interactions,

A massless photon is the transmission of a momentum from one point in spacetime to an adjacent point (in spacetime). That means, particle characteristics are transmitted directly from one atom to an adjacent atom within spacetime. The wave is a by-product of this transmission because the zero spacetime interval can correspond to billions of light years in space. And while the transmission is directly from atom to atom in spacetime, no observer can observe the lightlike interval, but instead all observers observe - as a sort of "by-product" - a wave as an intermediate between both atoms. A wave of a massless photon is the observable witness of the invisible zero interval.

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  • $\begingroup$ Light propagates just fine through spacetime. The invariant interval between two events on the light-like world line is zero, but that does not mean that there is no propagation through spacetime. I don't quite get what you are saying in the second part. $\endgroup$ – garyp Sep 3 '16 at 15:34
  • $\begingroup$ @garyp "the light-like world line is zero, but that does not mean that there is no propagation through spacetime" How does that work propagating over an interval that is zero? I would be interested to know how we might have to imagine that. $\endgroup$ – Moonraker Sep 3 '16 at 15:53
  • $\begingroup$ A pulse of light leaves point A at time $t_1$. It arrives at point B one second later. The distance between point A and point B is $3\times 10^8$ m, and the elapsed time between the two events is 1 s. The pulse has propagated through spacetime. Yet the interval between those two events is zero. $\endgroup$ – garyp Sep 3 '16 at 18:32
  • $\begingroup$ No. I can imagine a propagation through space over a certain distance or possibly a propagation through time over a certain time period. But I cannot imagine to "propagate" over an empty zero interval, (space, time or spacetime interval). This does not make sense. $\endgroup$ – Moonraker Sep 3 '16 at 19:38
  • $\begingroup$ Sensible or not, it's true. The invariant interval (squared) is $(\Delta x)^2 - c^2(\Delta t)^2$. The two events are separated from each other in space and time. But if a light pulse can join them, the interval is zero. $\endgroup$ – garyp Sep 3 '16 at 20:20

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