1
$\begingroup$

If a light source emits light, and due to the inverse square law “energy flux” (number of photons passing through a given area) decreases rapidly. Then at some distance away from the source there would be certain solid angles with 0 photons passing through, right?

I am imagining a shrapnel explosion where the shrapnels are photons.

many images depicting the inverse square law uses this format: enter image description here

I can’t help but get the impression that after a certain distance there will be some squares with 0 arrows passing through.

$\endgroup$
4
  • $\begingroup$ I guess it becomes a question of probability. You have less and less chance of detecting a photon in a certain solid angle, but never zero. $\endgroup$
    – dan-ros
    Commented Dec 28, 2022 at 9:33
  • $\begingroup$ You are right. Anything more specific you are interested in? @dan-ros: Yes, the probability (or the photon detection rate) gets lower with distance and when you actually do a measurement over a certain time interval there will be squares with zero photons measured. $\endgroup$
    – kricheli
    Commented Dec 28, 2022 at 9:36
  • 1
    $\begingroup$ The question Photons from stars--how do they fill in such large angular distances? is closely related if not exactly the same. Also see Some doubts about photons. $\endgroup$ Commented Dec 28, 2022 at 17:49
  • $\begingroup$ @JohnRennie provides very good answers in the cited answers. He says one thing that needs emphasizing: light is neither a wave nor a particle. It's something else entirely, something that has no metaphor or analog in the macroscopic world. You might read "sometimes a wave; sometimes a particle". That gives the wrong idea. It's too simple, and needs explaining to be correct. Light is neither. $\endgroup$
    – garyp
    Commented Dec 28, 2022 at 19:34

1 Answer 1

1
$\begingroup$

There are not many ways to generate EM radiation. Mostly we observe the emission of photons from excited electrons. And very often this takes the form of thermal radiation. For astronomy, there is also radiation produced by the powerful magnetic fields of stars and the electrons deflected in them (Lorentz force or better EM induction).

The fact is that it is always photons - of the most varied energy content - that are emitted and that we also receive as such. While this is not so obvious for close sources, for very distant sources this is the only way to investigate the source. With extremely long exposure times and the registration of single photons, filtered out from a noise of photons until an image of the star emerges.

Then at some distance away from the source there would be certain solid angles with 0 photons passing through, right?

The further away the source and the less radiative it is, the less often a photon will strike a finite surface in a certain period of time. What helps are long-term observations and the enlargement of the detection area.

The whole thing is quite intuitive and I almost suspect that your question was aimed at something deeper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.