# Do observers always agree on the number of waves in a beam of light?

• B
• Herbascious J
Herbascious J
TL;DR Summary
If a beam of light with a specific wavelength travels in a space that is some multiple of that wavelength, do all observers measure the same number of waves/crests in that space?
Imagine there is an experiment setup on a train. A laser, with a specific wavelength of light, is aimed at a target. The target is at a distance from the laser of some multiple of the wavelength. Let's say 10cm for the target distance, and the light's wavelength is 1cm, so when a pulse of light is emitted from the laser, exactly 10 crests or waves are created before hitting the target. Importantly, the laser is pointed forward in the direction of travel of the train. The question is; according to Special Relativity, an observer on the train will always count 10 waves/crests as the laser pulse travels between the laser and the target. Is this always still true to any observer outside the train regardless of relative velocity to the train, that they will also always count 10 waves, regardless of how noticeable relativistic effects are, like length contraction of the device and lengthening of the lights wavelength, etc?

I'm pretty sure the relativistic Doppler shift formula was stated in your last thread, and you know the length contraction formula, and you should be able to do an intercept calculation to work out the beam flight time. That's all the maths you need. What do you think?

The correct statement is that the phase is a relativistic invariant at each event. So at any event all observers will agree on the phase.

Herbascious J said:
TL;DR Summary: If a beam of light with a specific wavelength travels in a space that is some multiple of that wavelength, do all observers measure the same number of waves/crests in that space?

The question is; according to Special Relativity, an observer on the train will always count 10 waves/crests as the laser pulse travels between the laser and the target
Because of the relativity of simultaneity, different frames will have different sets of events that they call the wave at a given moment in time. As @Ibix mentioned, with Doppler shift and length contraction you can calculate the answer. Knowing that phase is a relativistic invariant at each event doesn’t give you a shortcut for this question

Herbascious J said:
TL;DR Summary: If a beam of light with a specific wavelength travels in a space that is some multiple of that wavelength, do all observers measure the same number of waves/crests in that space?

Imagine there is an experiment setup on a train. A laser, with a specific wavelength of light, is aimed at a target. The target is at a distance from the laser of some multiple of the wavelength. Let's say 10cm for the target distance, and the light's wavelength is 1cm, so when a pulse of light is emitted from the laser, exactly 10 crests or waves are created before hitting the target. Importantly, the laser is pointed forward in the direction of travel of the train. The question is; according to Special Relativity, an observer on the train will always count 10 waves/crests as the laser pulse travels between the laser and the target. Is this always still true to any observer outside the train regardless of relative velocity to the train, that they will also always count 10 waves, regardless of how noticeable relativistic effects are, like length contraction of the device and lengthening of the lights wavelength, etc?

It is not surprising that you use non-standard terminology, but its imprecision can cause some confusion.

I'm going to use the language and terminology of space-time diagrams in an attempt to clear things up, but you might have to draw a few such diagrams to make sense of it.

A beam of light on a space-time diagram is just a straight line representing how the light travels. Usually, the horizontal and vertical scale is chosen such that the beam of light travels at a 45 degree angle. This would be called the "world-line" of the light beam in standard terminology.

The places (and times) where the light beam "starts" and "stops" is marked by two events - represented by points on the space-time diagram. An event has both a place that it occurs, and a time that it occurs. I will call these two special start and stop events the "endpoint events'. A space-time diagram is usually drawn to scale in the frame of a specific observer. The choice of frame is not something physical, the choice of frame depends on who or what is making the observations.

So, there are three concepts that need to be understood here. "Events", which occur at a time and place, "worldlines", a set of events that can represent a beam of light (and other things, too), and a "frame of reference" or "coordinate systems" or "observer". I'll use these terms interchangeably, one might argue about subtle differences.

The key point here is that different observers / different frames represent the exact same physical situation with different space-time diagrams. This is similar to the way that one might have different maps of the same territory. The map is not the territory, the map represents the territory. The space-time diagram is similar to a map, but it includes representations of time as well as space. And the space-time diagram is a representation of what , for lack of a better term, one might call "reality".

We can now discuss what features of the space-time diagram are observer independent, and which depends on a choice of observer / frame of refererence.

Peaks of the electric field mark off regular intervals along the "beam of light", the straight 45 degree line on the space-time diagram. The regularity of the intervals is agreed upon by all observers - i.e. in all frames of reference. Any observer who draws such a space-time diagram will mark regular intervals long it to represent the peaks.

The "distance" between the two endpoint events, and the "time" between two events, in special relativity, does depend on the choice of observer, however.

This is a bit abstract, but I hope it answers the question and helps you to understand.

Nugatory and Herbascious J
Ibix said:
I'm pretty sure the relativistic Doppler shift formula was stated in your last thread, and you know the length contraction formula, and you should be able to do an intercept calculation to work out the beam flight time. That's all the maths you need. What do you think?
Ok, I did try it, and it worked. Thank you for the guidance. Admittedly, I think my approach was unconventional but the results fit perfectly. Both observers see the same number of waves from emission to target.

pervect said:
Peaks of the electric field mark off regular intervals along the "beam of light", the straight 45 degree line on the space-time diagram. The regularity of the intervals is agreed upon by all observers - i.e. in all frames of reference. Any observer who draws such a space-time diagram will mark regular intervals long it to represent the peaks.
That is actually very clear, thank you. My intuition thought it must be so, but I was wavering. This is much more obvious when looked at that way.

Herbascious J said:
Both observers see the same number of waves from emission to target.
Intuitively that sounds wrong to me, although I haven't worked it through formally. Post your maths and we'll have a look.

Here is the math I attempted. This is a first attempt at calculating SR so it's a bit daunting and I'm not sure if I'm handling the variables correctly...

A train is traveling past an observer on a platform at velocity (v, equal to 0.1c). On the train a pulse of light is fired from a laser, in the direction of travel, at a mirrored target. The pulse leaves the laser and meets the target after (t) amount of time as seen by observers on the train. They also see that the light pulse has a wavelength (λ, equal to 1 meter) that is precisely equal to the distance between the laser and the target (L). The observer on the platform sees the distance between the laser and the target as (L') and the time from the emission of the light pulse to the time it meets the target is (t'). The question to be answered is if the observer on the platform also sees one and only one wavelength of light between the point of emission and the target. The wavelength as measured by the observer on the platform will be λ'.

As the train moves at velocity (v=0.1c), the observer on the platform sees the train travel a distance of (D) in the time (t') in takes the light pulse to go from the laser to the target. So, v=D/t' which gives D=vt'. The distance the pulse of light must travel (d) relative to the observer on the platform is the distance to the target L' plus the distance the train travels, because the target is moving away from the on coming pulse of light as seen by the observer on the platform during that time. So, d=L'+D. The speed of light is c=d/t' for the pulse of light as seen by the observer on the platform. This gives t'=d/c. I would like to point out that this t' is not the same t' which the time dilation equation of SR gives, instead it's just the time the observer measures. I'll address this later.

The distance the pulse travels as seen by the observer on the platform d=L'+D can be combined with D=vt' and t'=d/c to yield d=L'/(1-v/c). Using the length dilation equation of SR...

L'=0.99498743710662 meters
and then
d=1.105541596785133 meters

In this calculation we are checking to see if d is in fact identical to λ'. The Doppler equation for SR in the direction of travel is...

Which yields λ'= 1.105541596785133. So λ'=d, and we can see that both observers, on the platform and on the train, agree that exactly one cycle of the lights frequency was observed before meeting the target.

As mentioned earlier, t' is not t' as used in the time dilation equation of SR. I believe this is because there is a simultaneity shift that occurs at the laser relative to the target as see by the observer on the platform. I think that means that time is flowing differently in the direction of travel compared to the opposite direction. To test this, I decided to also calculate the time it would take for a pulse to return to the laser, reflected off of the mirrored target. By reversing the sign of v to -v, this yields...

d'=0.904534033733291

This should be the distance the light must travel relative to the observer on the platform on the return path.

And using t'=d/c from above, with a value for c=3x10^8m/s...
t''=0.301511344577764 x10^-8 seconds (the return time from target back to laser)
t'=0.368513865595044 x10^-8 seconds (the time originally observed from laser to target)

The total time round trip is t'+t''=0.670025210172808 x10^-8 seconds

Observers on the train should see the light pulse travel 2 meters total equaling...

0.666666666666667 x10^-8 seconds

The time dilation equation of SR...

Total time round trip as measured by the observer on the platform is

0.670025210172808 x10^-8 seconds

This is in agreement with the above calculations so I think it's correct. I'm not sure about my use of the equations and if there isn't a much simpler way of doing this. Thank you for looking at my post and pushing me to try it, any critical response is very appreciated.

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Dale
Your intercept calculation appears to be correct, but you have the wrong expression for the Doppler shift. Imagine a source moving in the +x direction with velocity ##v## and emitting light in both the ##+x## and ##-x## direction. At time zero it starts emitting a wave in both directions and at time ##\delta t## it finishes. The fronts of the waves are at ##\pm c\delta t## and the source is at ##v\delta t##. Which pulse has the shorter wavelength, the forward going or the backward going pulse? So which one has wavelength ##\lambda\sqrt{\frac{c+v}{c-v}}##?

If you are going to be posting maths here it's well worth learning how to use the LaTeX feature. There's a guide linked below the reply box. The only gotcha is that if you are the first person on a page to use LaTeX the renderer doesn't load. To wake it up, type some LaTeX, preview the post and refresh the page while in preview.

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Herbascious J
The title says "do observers always agree on the number of waves in a beam of light", but it was actually a pulse of light, according to the text? Right? I did not read carefully though.

To be sure, I consider both cases:

Of course observers always agree on the number of waves in a pulse of light. Like observers always agree on the number of legs of a centipede.

Of course observers disagree on the number of waves of beam of light, that are inside a railway tunnel, when said long beam goes through said short tunnel. Like observers disagree on how many cars of a long train going through a short tunnel, are inside the tunnel.

Last edited:
Dale
jartsa said:
Of course observers always agree on the number of waves in a pulse of light. Like observers always agree on the number of legs of a centipede.
Some care is needed with this statement, because it's strictly only true for an eternally travelling pulse. If there's an emitter and a receiver I can always find a frame that will regard the end of emission as after the start of reception, so they will not agree on the number of waves in flight because the "pulse" is a "beam" to them. How implausibly fast such a frame is moving does depend on how short the rest measure of emission time is compared to the rest length of the path.
jartsa said:
Of course observers disagree on the number of waves of beam of light, that are inside a railway tunnel, when said long beam goes through said short tunnel. Like observers disagree on how many cars of a long train going through a short tunnel are inside the tunnel.
I believe this is the case under consideration.

You are analogising the null-travelling light pulse/beam to timelike objects (a centipede/train). It's not a perfect analogy, since light lies on the boundary between timelike and spacelike. That's why the number of legs on a centipede is frame invariant but the number of waves in a pulse is only invariant for most frames.

I said above that I'd reasoned about this intuitively - and as I said in @Herbascious J's last thread, Minkowski diagrams and learning to interpret them are, IMO, the key to understanding special relativity.

Here's a Minkowski diagram of the train. As always, time is up the page and units are picked so that ##c=1##.

This is drawn in the rest frame of the train. The blue vertical lines are the ends of the train and the orange diagonal lines represent the locations of wave crests as they travel left to right. Along any horizontal line ("an instant of time in this frame") there are always ten waves in flight. This is easiest to count along the ##x## axis where it is labelled, but it is true anywhere - one crest is emitted just as the one ten before is absorbed.

But the Lorentz transform says that ##t'=\gamma(t-vx/c^2)##, and we can rearrange to get ##t=\frac 1\gamma t'+\frac v{c^2}x##. That is, a line of constant primed time ("an instant of time in the primed frame") is a sloped line. We can add the line a frame doing 0.6c calls "time zero" to our diagram - it's the fine grey line.

If you count along it, you'll see that it intersects fewer than ten wavecrests - I count four (which will be the Doppler factor squared). So this frame does not see the same number of waves. If you aren't sure you can visualise this in your head you can consider the case where the velocity is very high. Here's 0.95c:

Here the grey line is very nearly parallel to the orange ones, and by considering even higher speeds we can push the grey line closer and closer to the wavecrest passing through the origin. So if you can sketch this in your head, you can quite easily convince yourself that the number of waves must be frame variant.

@jartsa raised a point about a pulse of light versus a beam. The above case is a beam - the laser was turned on before ##t=-20## and is still on at ##t=12##, so there is no doubt that in this region all frames agree that the laser is on and the target is illuminated. But you can consider the case where we flip the laser on and off in a very short timescale. That's illustrated here, with just two full waves between three crests:

In this case, many frames will agree that there are exactly two waves in flight at once. Here are three grey lines illustrating "an instant of time in a frame" for three frames. You can see that they all pass through exactly three orange lines, so they agree the number of waves:

But, again, we can consider the case where the frame velocity is very high, and the "instant of time" for that frame is nearly parallel (as drawn in this frame) to one of the orange lines. Frames whose definition of "simultaneous with the last wave crest emission" fall within the blue region below see the same number of crests in the pulse; those in the red region see a beam and the number of wave crests depends on speed.

You can see the same experiment in a frame moving very quickly to the right:

Here, the train is moving fast to the left. Consequently it is length contracted and moving nearly at the same speed as light, so the laser emission point is moving rapidly to the left and the reception point is rushing towards each emitted wavecrest, and it is absorbed before the end of the pulse is emitted - so frames like this see something more like the original beam case than the pulse case.

I hope that's helpful. You do need to know the Lorentz transforms to be able to draw Minkowski diagrams, but I found drawing them to be the thing that really made SR "click" for me.

DrGreg and Dale
Ibix said:
I found drawing them to be the thing that really made SR "click" for me
Same here.

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