Is reflection instantaneous?

Question

I was wondering while reading "On the Electrodynamics of moving bodies" by Albert Einstein (1905) (Translated to English).

In the paper, he describes the time as being:

by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A. Let a ray of light start at the “A time” $t_{\rm A}$from A towards B, let it at the “B time” $t_{\rm B}$ be reflected at B in the direction of A, and arrive again at A at the “A time” $t'_{\rm A}$.

It sounds completely normal however the word "reflected" always intrigued me as normally when I bounce a tennis ball (for analogy of reflection), there is this time duration from with it moves downwards then it moves up towards my hand again, I noticed during this time the ball will be deforming in order to give it time to get the time to "spring" back up towards me.

Similarly, I am wondering even though light travels at $c$ in all time-frames and of course all-reference frames, will light not have any time to change it's direction? If there is $0$ time difference does this not lead to $\infty$'s from being raised like it does to baryonic matter?

I asked this question as I know that any light-ray\photon has a given momentum proportional to its energy like so: $\vec{p} = \frac{E}{c}$ and normally change in momentum direction causes the deformation.

I cannot manage to comprehend how the direction of the photon just changes without any time being passed it seems very nonphysical. I can imagine an light wave being reflected how ever not at a normal incidence (where $\theta = {0}$ degrees as the entire waves would simply collide and annihilate each other).

If there is an time difference it may lead to following situations, the speed for $c$ would change from $t_B - t_A = t'_{\rm A} - t_B$ into this: $t_B - t_A \ne t'_{\rm A} - t'_B$

The change is from $t_B$ to $t'_B$ where $t'_B$ is the new time from when the direction of the light is again heading towards point $A$ after the brief delay between the reflection. This therefore creates a inequality between both expression therefore we cannot calculate $c$ in that way as Einstein proposed as the time difference between $t_A$ and $t_B$ is not invariant and therefore we can say lights velocity is not invariant as Einstein concluded.

Before few get wrong message, I am not criticizing nor suggesting Einsteins work is wrong in fact just trying to get hang of this concept.

1st einstein's observation does not need the light to be reflected, one can very well get the same relation with 2 photons one in one direction and the other in the opposite (no reflection required), as far as reflection is concerned, yes it would take some time depending on material and frequency — Nikos M., Oct 28 '14 at 16:44
Obviously but I'm saying such a scenario would or should at least in my opinion create a problem like I have stated — , Oct 28 '14 at 16:45
let me re-state, the relation that einstein (special rel) got does not depend on photons reflecting. It just needs two photons one in one direction and the other in the opposite direction (because both would have to travel with speed $c$). So reflection is not needed nor necessary and there is no problem. i would say "reflection" in the paper is used metaphoricaly — Nikos M., Oct 28 '14 at 16:48
It may have but it clearly says to be reflected at B in the direction of A, which is unclear for few newbie scientists but nevertheless, I'm asking about the nature of reflection itself rather than Einsteins work. Furthermore, I'm asking whether the "reflection" based derivation would work without having a very very very small amount of inaccuracy due to the delay of reflection? — , Oct 28 '14 at 16:53
yes your comment is correct (imo). BUT Sp. Rel. here uses an idealisation of a process, and this can be made more rigorous as such: Consider all photon paths from A to B and B to A and from this set, use the minimum path (which is unique and happens to be the one in which the "reflection" is instantaneous or in other words two photons used as my prev comments) — Nikos M., Oct 28 '14 at 16:57
Please don't take this the wrong way because I'm simply curious. What kind of answer you are expecting? — Alfred Centauri, Oct 28 '14 at 17:53
@AlfredCentauri I am expecting an answer that first of all answers how does light reflect without changing its velocity and how can the light reflect in NO time at all, if not how much time does it take and surely about the deformation of mirror which leads to extra distance being covered by the light, which again leads to an inconsistency of accuracy. Next, I hopefully will need a response that clarifies how Einsteins derivation of Lorentz transforms using reflection, works. — , Oct 28 '14 at 17:55
Rohan, so an answer along the lines of "the reflection time can be made arbitrarily small relative to the travel time and thus can be idealized to zero without changing the physical result" is not what you're looking for? — Alfred Centauri, Oct 28 '14 at 17:58
@AlfredCentauri Sorry, its not what I am looking for, as even I worked this and said its nearly negligible but yet gives us very small inaccuracy on level of $10^{-20}$ or such level. Furthermore, the answer has not answered: how does light reflect without changing its velocity and how can the light reflect in NO time to the object and moreover it has not answered about deformation or any sort of other time delays that could be caused that may lead to light travelling slightly more thus increasing time. — , Oct 28 '14 at 18:02
in geometrical optics (where light is seen as a particle or line) the reflection is instantaneous in the opposite angle, in wave mechanics (or electromagnetism) where light is seenas a wave (or wave-particle) the reflection depends on the micro-interactions between electrons which then generate a photon in the opposite direction (reflection) and this takes some time. Both of these are irrelevant in the derivation of the relation in einstein's paper (as per previous comments) — Nikos M., Oct 28 '14 at 18:23
@NikosM. That helped in fact first 2 parts helped me very much, however it does because as you said "this takes some time", in that time the light is not really travelling $c$ in distance as its "changing" course by interacting and as you said it takes time, therefore causes that the time is being added into it which changes the average velocity of the light. Furthermore, how can the reflection be instantaneous? There must be some level of deformation or else it leads to $\infty$ being raised and angular momentum being not conserved. This deformation should lead to the light travelling more. — , Oct 28 '14 at 22:57

score 5 · Answer 1 · edited Apr 13 '17 at 12:39

You are quite right: Einstein is using highly idealised concepts in a thought experiment. In general ALL interactions between light and matter take nonzero time. But Einstein is justified in this approach because:

As stated by Nikos M in his comment:

"Einstein's observation does not need the light to be reflected, one can very well get the same relation with 2 photons one in one direction and the other in the opposite (no reflection required)"

But even so, if there is a nonzero time, you simply scale the thought experiments size so that the absorption / re-emission time is arbitrarily small compared to the light's time of flight in the experiment. The great Karl Weierstrass might have said something along the lines of: for every $\epsilon>0$, there exists a scale of experiment $L$ big enough that the deductions hold true within a precision given by $\epsilon$. There is no requirement for us to realise the experiment practically: thought experiments probe principles, so you can make $\epsilon$ as small as you like.

To get back to your main question: reflexion from a metallised or, if you're after higher optical quality suitable for the laboratory, system of dielectric layers actually absorbs a photon and re-emits a new one. Same for any dielectric: we seem to observe light slowing down because actually we don't have light inside a dielectric, we have a quantum superposition of free photons and excited matter states. Or, you could say, photons are repeatedly losslessly absorbed and re-emitted in dielectrics. These processes ALWAYS take nonzero time and I probe these ideas more carefully in my answer to the question "Do Photons Accelerate Upon Creation"

score 2 · Answer 2 · answered Oct 28 '14 at 23:11

2

One way to imagine computing the reflection delay is to assume that the reflection process takes place within the first skin depth of the material. Taking numbers straight from the Wikipedia article, a copper conductor reflecting a 100 MHz (radio) wave would do so within its outermost 6.6 μm. In vacuum, light crosses 6.6 μm in about 20 femtoseconds.

Skin depths shrink with higher frequencies, so for visible light the reflection delay will be even shorter. There is an asymptotic limit to the skin depth for very high frequencies.

In general it's probably safe to assume that the reflection delay time is negligible as long as the reflecting layer of the mirror has negligible thickness compared with the total light travel distance.

answered Oct 28 '14 at 23:11

rob

89,569

Presumably this could then be checked by an experiment using a different number of mirrors to create the same path length over which the speed is measured. Surely someone has tried this? – ProfRob Oct 28 '14 at 23:17
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Femtosecond timing raises more subtle issues than I'm prepared to list here. If you wanted to measure this effect you'd prepare a special material where reflection is "slow" and measure that. – rob Oct 29 '14 at 04:45
I believe that the reason light reflection is instantaneous, is the fact that photons have no mass. – Guill Oct 31 '14 at 17:14

score 0 · Answer 3 · answered Oct 08 '20 at 12:49

The importance of any delay or advance at reflection for the thought experimenst has been discussed adequately in other answers.

Reflection and transmission at interfaces of media that show linear response is described by the Fresnel equations. For perpendicular incidence from vacuum the intensity reflectance is $$R2 = \left|\frac{1-n}{1+n}\right|^2~.$$ An ideal mirror has infinite n and the phase change upon reflection is 180 degrees. Note that n is a complex number so for an ideal metallic mirror it is the imaginary part related to conductivity that is infinite. For a non-ideal mirror the phase change is different from $\pi$ and there is a delay upon reflection.

Is reflection instantaneous?

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