In physics class today we were learning about the doppler effect for sound, and the teacher said that blue-shift and red-shift are the exact same thing but for light. This does not make sense to me. The wavelengths of the photons being emitted from a source do not change as the source moves, and so the observer should see the exact same color of light. The property of light compressing in the direction that a photon emitter is moving on spreading out on the other side should just change the rate at which the observer receives photons, meaning that the light will get brighter/darker.

A similar question to this one was asked in this Physics SE post, but their question was more "how could it be possible that the energy of photons is dependent on observers, which is what red-shift would imply?", but mine is more asking why the energy of the particle changes, since seemingly to me it has no reason to.

I'm probably just missing something really simple about special relativity, so please don't go too hard on me in the answers.

  • $\begingroup$ Photons travel at the speed of light always, so the rate you receive them is the same in blue or red shift $\endgroup$
    – Triatticus
    Commented Jan 17, 2021 at 0:45
  • $\begingroup$ @Triatticus I disagree. As the emitter of light moves away, the time it takes for the light to travel between the emitter and the observer increases (since the distance increases), and so if we model the emitter as sending out a finite amount of photons then it is clear to see that the rate of receiving photons really is changing for both the observer and the emitter. Right? $\endgroup$
    – Milo Moses
    Commented Jan 17, 2021 at 0:51
  • $\begingroup$ Say the receiver emits 10 photons, notice that those 10 photons just cross a different distance, but the rate is defined as the number crossing a certain cross sectional area at some point in time. If the emitter is moving away, those 10 photons still cross the same area as the ones for an incoming source. Hence the rate is identical $\endgroup$
    – Triatticus
    Commented Jan 17, 2021 at 0:59
  • 4
    $\begingroup$ Yes, that is true, but the amount of time it takes between when that first photon crosses the cross-section and when the last photon crosses the cross-section will increase with an emitter than is moving away from that cross-section. This must mean that the rate of crossing has decreased. $\endgroup$
    – Milo Moses
    Commented Jan 17, 2021 at 1:01
  • 1
    $\begingroup$ @ChiralAnomaly thank you for your comment. I am thinking of photons as waves traveling in a given direction with a given wavelength. I'm not thinking of the space of all photons being emitted one big wave, however. The redshift/blueshift phenomena seem to me to be acting on that space of all emitted photons, which is what I'm not quite getting $\endgroup$
    – Milo Moses
    Commented Jan 17, 2021 at 2:06

2 Answers 2


The property of light compressing in the direction that a photon emitter is moving on spreading out on the other side should just change the rate at which the observer receives photons...

The relevant aspect of light here - especially in the classical context Relativity is formulated in - is not the quantization that talking about "photons" makes prominent, but that it is spatially extended waves in the EM field. This means that light emitted from a radially moving source changes the rate at which wavefronts arrive at the observer, and the number of wavefront peaks that pass the observer in a given time are what frequency is, and hence the observed frequency of the light changes.

As mentioned in comments, the intensity also changes because the rate of photon impacts changes but this is a separate and independent effect, and is hard to use in those terms because in practice, the number of photons in an emission of a macroscopic amount of energy is hard to determine.

The wavelengths of the photons being emitted from a source do not change as the source moves, and so the observer should see the exact same color of light.

Because Doppler shift of light is a Relativistic effect, you immediately face the question of, as the source moves compared to what? Everyone thinks they are at rest with respect to themselves, and they also think they are "at rest" with respect to the "aether", the proposed analoguous "medium" that carries light in the way that the atmosphere carries sound.

What your teacher said isn't strictly correct, because there's a fundamental asymmetry that exists between sound and light - an emission of sound will change frequency if the source is moving compared to the air around it, as the wavefronts "bunch up" in front of the emitter. (You can imagine this more easily if the source is moving a significant fraction of the speed of sound.) The source can measure the speed of the air whizzing past it and so infer the "true" frequency of the emitted sound as seen in a frame that is at rest w.r.t. the air. This "true" frequency will also change depending on emission direction - emitting in front of you will have the fronts bunch up, emitting behind you will spread them out. A receiver's observation of this "true" frequency is then dependent on its velocity compared to the air, not to the original emitter.

With light, we have a problem because it turns out the aether doesn't actually exist. Everyone, even people travelling at different velocities, thinks they are "at rest" with what ought to be an aether - specifically, they observe no change in the round-trip time in any direction around them, as discovered by the Michelson-Morley experiment. If the aether did exist, then light should take a different amount of time to travel in line with our aether velocity compared to perpendicular to it, and this does not happen.

And if there's no medium, then there's no "true frequency", and so we can't talk about the emitted wavefronts "bunching up" like we did with sound. Unlike with sound, where both source and receiver agree on their speeds relative to the medium, the "true frequency" and thus how they observe the latter, the story of EM Doppler shift splits in two depending which frame you look at:

  1. In the source frame, you are at rest, and so no shift occurs in the emission. However, if you watch a receiver moving at relativistic velocity through your emitted wave, their observation will be shifted because their own velocity carries them through the wavefronts faster (or slower) than they would be naturally delivered by the light's own velocity if they were "at rest." (...in your frame)
  2. In the receiver frame, you are at rest, but the source is not. As a result, the emission still travels at the speed of light, and the source still "shoots" them at the correct rate(*), but they end up squashed together (or stretched apart) within the wave that propagates through space because the source has physically moved a significant distance between each wavefront it transmits. It can never overtake the previous wavefront, because that'd involve travelling faster than light, but it can get very close and subsequently emit a wave that has very small spatial gaps between wavefronts, and thus has a high frequency. This is a similar story to the one about Doppler shifts in air, but the key difference is that different observers will disagree on how much "bunching" happened depending on their velocity relative to the source.

(*) If the source is moving very quickly, it might also be undergoing time dilation, but I'm neglecting that for now.

It just so happens that the magnitude of these effects exactly correspond to each other, and both source and receiver can arrive at correct predictions of what frequency the other observed the emission to have. The upshot is that the source not observing its emission to change frequency as it moves, while true, is not the whole story - how the receiver interacts with the wavefronts "already in space" has to be considered in the source's frame.

  • $\begingroup$ Thank you so much for your answer! I'm going to have to read it over a few more times to really get it, but I can tell that this answers my question. Thank you so much for your time. $\endgroup$
    – Milo Moses
    Commented Jan 18, 2021 at 2:55

Another effect - or better phenomenon - is the measurement of a wavelength shift in a changing gravitational potential.

... it was clear
that the frequency of light should not change from place to place,
since waves from a source with a fixed frequency keep the same frequency everywhere.
One way around this conclusion would be if time itself were altered—if clocks at different points had different rates.

This was precisely Einstein's conclusion in 1911. He considered an accelerating box, and noted that according to the special theory of relativity, the clock rate at the "bottom" of the box (the side away from the direction of acceleration) was slower than the clock rate at the "top" (the side toward the direction of acceleration).

Einstein thus proposed a formula that backed up what Laplace had already formulated:

The gravitational weakening of light from high-gravity stars was predicted by John Michell in 1783 and Pierre-Simon Laplace in 1796, using Isaac Newton's concept of light corpuscles and who predicted that some stars would have a gravity so strong that light would not be able to escape.

If we take the speed of light as a reference for the space-gravity metric, the time at the event horizon of a black hole is zero. And for a star with a gravitational metric slightly smaller than a black hole, the speed of light - as seen from a distant third observer! - is very low.

However, for reasons that are not clear to me, the Pound-Rebka experiment is presented differently.
It is not the change in the temporal deceleration or acceleration of the source and receiver with respect to their position in the earth's gravitational potential that is given as the reason for the experimental results, but the change in the frequency of the light. As a result, the energy content changes.

From a methological point of view, this is problematic. It implies that both time dilation and energy change occur when the gravitational potential changes. I would prefer to maintain the energy content as constant and focus on the time dilation. That is exactly what you noticed.

  • $\begingroup$ I don't understand how this answers my question. If you were trying to tell me that time dilation exists, then I was already aware of that. $\endgroup$
    – Milo Moses
    Commented Jan 17, 2021 at 17:17
  • $\begingroup$ @ÁrpádSzendrei c is invariant for any not accelerated observer at some location in space but not for observers at different locations. To claim both time dilation and energy change is possible but you get two changing from gravitational potential metrics (time and energy). Explaining the Pound-Rebka experiment with the different positions of the emitter and the receiver only and not of any energy change should be the better way. $\endgroup$ Commented Jan 18, 2021 at 5:23

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