Why do photons allow Doppler shift

[Buckethead]

If we (a detector) are moving toward a star that emits a single photon (due to its distance) and that photon hits our detector, it will be blue shifted. My question is why. If the color of a photon is a reflection of its energy level and since the speed of the photon is always coming at us at c irrespective of the speed at which we are traveling toward the star, then why does the color of the photon change if we increase our speed?

 

[Paul Colby]

In current theory light is composed of electromagnetic waves that have quantized amplitudes. The wave (of which the photon is the energy (and momentum) absorbed by the detector thus changing the waves state by one amplitude step) is Doppler shifted just like a classical wave would be.

 

[Buckethead]

But why, even if we view the photon as a wave, would it be Doppler shifted if its velocity relative to the detector does not change? It seems to me this is equivalent to being in a plane and talking to someone while the plane is in the air. The velocity of the sound wave does not change relative to you (just like light in my example) and as a result its pitch does not change.

 

[Nugatory]

But why, even if we view the photon as a wave, would it be Doppler shifted if its velocity relative to the detector does not change? It seems to me this is equivalent to being in a plane and talking to someone while the plane is in the air. The velocity of the sound wave does not change relative to you (just like light in my example) and as a result its pitch does not change.

If you consider the receiver to be at rest, then the source is moving towards the receiver. This does not affect the speed with which the waves approach the receiver, but every successive wave crest has to cover slightly distance than the one immediately before it. Thus, the time between the reception of two consecutive crests is not the time between their emission; it is somewhat less because the second crest travels a shorter distance so arrives just a bit sooner than it otherwise would. This will probably be clearer if you try drawing a spacetime diagram showing the paths of successive wave crests through spacetime.

That explains the classical Doppler effect. There is an additional relativistic correction from time dilation because the emitter is moving relative to the receiver.

 

[Mister T]

If we (a detector) are moving toward a star that emits a single photon (due to its distance) and that photon hits our detector, it will be blue shifted. My question is why. If the color of a photon is a reflection of its energy level and since the speed of the photon is always coming at us at c irrespective of the speed at which we are traveling toward the star, then why does the color of the photon change if we increase our speed?

The speed is the same but the energy is different. It's the energy, not the speed, that determines the frequency (color).

For massive particles the increase in speed with respect to energy approaches zero as the speed approaches ##c##. In other words, as you approach speed ##c## huge increases in energy produce negligible increases in speed. My point is that the relationship between speed and energy is strange compared to the non-relativistic relationship and the photon is a purely relativistic particle.

 

[Buckethead]

If you consider the receiver to be at rest, then the source is moving towards the receiver.

I'm specifying a single photon which left the star mellenia ago, when the sensor was at rest and then later accelerated to some modest speed toward the star, so the relative speed between the star and the observer couldn't be a factor

The speed is the same but the energy is different. It's the energy, not the speed, that determines the frequency (color).

But why is the energy different? If the speed of the photon never changes relative to the observer, then why would its energy change?

 

[Nugatory]

I'm specifying a single photon which left the star mellenia ago, when the sensor was at rest and then later accelerated to some modest speed toward the star, so the relative speed between the star and the observer couldn't be a factor

There's no quantum mechanics involved in this problem (aside from statistical effects at the receiver, which are a distraction here) so no photons involved - you have a flash of light traveling from the emitter to the receiver. This flash of light is an electromagnetic wave.

If the receiver is approaching the source, then the crests of the wave will be closer to one another using the frame in which the receiver is at rest than using the frame in which the emitter is at rest. This will become clear if you draw a diagram.

But why is the energy different? If the speed of the photon never changes relative to the observer, then why would its energy change?

The light is moving at ##c## in all frames, but the wavelength is shorter in the frame in which the receiver is at rest. Therefore the frequency and the energy are greater. This is just another example of the general fact that kinetic energy is always frame-dependent; the only surprising thing is that the energy carried by a flash of light is dependent on the frequency and wavelength, not the speed.

 

[Buckethead]

There's no quantum mechanics involved in this problem (aside from statistical effects at the receiver, which are a distraction here) so no photons involved - you have a flash of light traveling from the emitter to the receiver. This flash of light is an electromagnetic wave.

I'm good with this since the property of light (wave or particle) depends on the experiment you are performing, so wave it is.

If the receiver is approaching the source, then the crests of the wave will be closer to one another using the frame in which the receiver is at rest than using the frame in which the emitter is at rest. This will become clear if you draw a diagram.

I'm not sure how to draw a diagram to reflect this, and I get what you are saying with regard to viewing this as the receiver at rest, but let's look at this another way. Suppose the emitter and receiver are at rest and the star emits a short pulse. Sometime later before the light reaches the receiver, the emitter accelerates toward the receiver then coasts to a constant speed. I expect the receiver will detect no color change. Now reverse the experiment and instead accelerate the receiver then coast to a constant speed just before the pulse reaches the receiver. I expect the receiver will see a blue shift. First, is this correct? If so I find this curious since when the pulse was emitted, the relative speed between the two was 0. And at the time the receiver finished accelerating and was in coast mode, its speed relative to the emitter is no longer relevant since the pulse is in space between the two. In other words, the emitter could explode and vanish so that all that's left is the receiver and the pulse of light somewhere in space. The receiver is moving at c relative to the pulse of light and yet when it arrives it will be blue.

The only difference between the emitter accelerating and the receiver accelerating is the acceleration itself. But if the accelerations occur only during the time the pulse is between the emitter and receiver, and if both are moving at a constant velocity relative to each other when the pulse is still in between, then in both cases the situation between the pulse and the receiver is identical. The relative speed between the pulse and receiver remains at c in both cases and in both cases the speed of the emitter should be irrelevant since what the emitter is doing in both cases is isolated from the pulse already on its way.

 

[PAllen]

Another way to look at this is purely kinematically. 4 momentum is a vector that transforms per the Lorentz transform. If something has a 4 momentum in an emission frame, it will have the Lorentz transform of that 4 momentum in another frame, e.g. the receiver. If you Lorentz transform (E,p), then specialize to that case of E=p for a massless particle (or light), you get the relativistic Doppler formula. So given that light must have energy and momentum, it must undergo relativistic Doppler between frames due to Lorentz invariance. I should say, you get the Doppler formula applied to E. Then the energy of a photon determines its frequency.

 

[Nugatory]

Suppose the emitter and receiver are at rest and the star emits a short pulse. Sometime later before the light reaches the receiver, the emitter accelerates toward the receiver then coasts to a constant speed. I expect the receiver will detect no color change. Now reverse the experiment and instead accelerate the receiver then coast to a constant speed just before the pulse reaches the receiver. I expect the receiver will see a blue shift. First, is this correct?

Yes.

If so I find this curious since when the pulse was emitted, the relative speed between the two was 0. And at the time the receiver finished accelerating and was in coast mode, its speed relative to the emitter is no longer relevant since the pulse is in space between the two. In other words, the emitter could explode and vanish so that all that's left is the receiver and the pulse of light somewhere in space. The receiver is moving at c relative to the pulse of light and yet when it arrives it will be blue.

If the light is moving to the right in some given frame, and the frequency of the light is ##\nu## as measured by an observer at rest in that frame:
- If the receiver is moving to the left in that frame at the moment of reception then the frequency in the frame in which the receiver is at rest will be ##\nu+a##; the receiver will measure a blueshift.
- If the receiver is moving to the right in that frame at the moment of reception then the frequency in the frame in which the receiver is at rest will be ##\nu-b##; the receiver will measure a redshift.
What happens to the emitter after the light is emitted is irrelevant; all that matter is that the frequency was ##\nu## in the frame in which the emitter was at rest at the moment of emission. Likewise, what happens to the receiver before the light is received is irrelevant; all that matters is the speed of the receiver at the moment of reception.

The only difference between the emitter accelerating and the receiver accelerating is the acceleration itself.

We get different results in the emitter-accelerates and the receiver-accelerates cases because after the acceleration:
- in the first one the receiver is at rest in the frame in which the frequency is ##\nu##.
- in the second one the receiver is moving to the left in the frame in which the frequency is ##\nu##.

 

[robphy]

Along the lines of @PAllen 's comment, here are some energy-momentum diagrams of photons in different frames of reference.
(These were taken from my post to a different question on another site https://physics.stackexchange.com/questions/362125/momentum-conservation-with-photons/363478#363478 )

In the rest frame of the source, light signals of the same frequency are emitted in the forward and backward direction.
(The original question I responded to asked about the velocity of the source after emission.)
The diagram visualizes the conserveration of 4-momentum problem:
$$
\tilde P_{fin}+\tilde {k_1}+\tilde {k_2}=\tilde P_{init}
$$
in the rest frame of the source,
and in the lab frame [which observes the source moving].

Note, in the lab frame,
the forward light-signal's 4-momentum is increased
(compared to that signal's 4 momentum in the rest frame)
and the backward light-signal's 4-momentum decreased (similarly).
Thus, in the lab frame, the light-signals have different frequencies.

 

[phyzguy]

Now reverse the experiment and instead accelerate the receiver then coast to a constant speed just before the pulse reaches the receiver. I expect the receiver will see a blue shift. First, is this correct? If so I find this curious ...

Yes, this is correct. I'm surprised you find this curious. Suppose I throw a baseball at you at 50 miles per hour. While the baseball is in flight, you accelerate towards it to 200 miles per hour. Is it surprising that the baseball will hit harder than if you stayed stationary? I think not. The analogy is not perfect, since light moves at a constant speed, but it shows that the receiver's speed can matter. As Nugatory has explained, the receiver's speed at the point of absorbing the pulse is what matters, because it causes the wave crests to arrive closer together.

 

[PAllen]

I'm good with this since the property of light (wave or particle) depends on the experiment you are performing, so wave it is.

I'm not sure how to draw a diagram to reflect this, and I get what you are saying with regard to viewing this as the receiver at rest, but let's look at this another way. Suppose the emitter and receiver are at rest and the star emits a short pulse. Sometime later before the light reaches the receiver, the emitter accelerates toward the receiver then coasts to a constant speed. I expect the receiver will detect no color change. Now reverse the experiment and instead accelerate the receiver then coast to a constant speed just before the pulse reaches the receiver. I expect the receiver will see a blue shift. First, is this correct? If so I find this curious since when the pulse was emitted, the relative speed between the two was 0. And at the time the receiver finished accelerating and was in coast mode, its speed relative to the emitter is no longer relevant since the pulse is in space between the two. In other words, the emitter could explode and vanish so that all that's left is the receiver and the pulse of light somewhere in space. The receiver is moving at c relative to the pulse of light and yet when it arrives it will be blue.

The only difference between the emitter accelerating and the receiver accelerating is the acceleration itself. But if the accelerations occur only during the time the pulse is between the emitter and receiver, and if both are moving at a constant velocity relative to each other when the pulse is still in between, then in both cases the situation between the pulse and the receiver is identical. The relative speed between the pulse and receiver remains at c in both cases and in both cases the speed of the emitter should be irrelevant since what the emitter is doing in both cases is isolated from the pulse already on its way.

The situation is not symmetric at all. What the emitter does after emission is obviously wholly irrelevant, as is what the receiver does after reception. If the emetter changes speed before emission, that will have an effect. If the receiver changes speed before reception, that will have an effect. Both of these before/after statements are invariant because they events along one world line (timelike).

 

[jartsa]

But why, even if we view the photon as a wave, would it be Doppler shifted if its velocity relative to the detector does not change? It seems to me this is equivalent to being in a plane and talking to someone while the plane is in the air. The velocity of the sound wave does not change relative to you (just like light in my example) and as a result its pitch does not change.

When you do a small acceleration, everything in the universe changes shape (for you). The change is very small for slow moving objects, not very small for fast moving objects. Lorentz-contraction formula can be used to calculate the change of shapes of all things, except those 'things' that move at the speed of light. If we use 0.99999c as the speed that the light pulse moves, we get almost correct results using the Lorentz-contraction formula.

And the derivation of a general contraction formula is trivial.Oh yes this was about energy. During your acceleration there is a (gravitational) potential difference between every point of the universe (for you), and when you do a small acceleration, everything in the universe changes energy (for you). The amount of change is the original energy of the object multiplied by the potential between the starting position and end position of the object.

Typically fast moving objects, like light pulses, have a large distance between their starting position and end position.

 

[Mister T]

But why is the energy different? If the speed of the photon never changes relative to the observer, then why would its energy change?

Ask yourself why you expect that the energy should depend on the speed? In non-relativistic physics we expect the energy to be proportional to the square of the speed, but that is just the non-relativistic approximation. It's flawed and the flaws become more and more apparent the closer you get to speed ##c##. At speed ##c## there is no dependence at all. Photons of various energies all have the same speed.

 

[Wes Tausend]

Skinnier, but taller photons. :)

 

[Buckethead]

Thanks everyone for contributing to the answers. I've read everyone's post carefully, (understanding some more than others) and enjoy that it is indeed clearing my head a little. I understand the conclusions and have no reservations that red/blue shift occur under the stated circumstances, but am still struggling with this from a strictly logical point of view. Here is my most concise way of putting it:

Emitter and receiver have zero relative velocity. Emitter strikes out, then both emitter and receiver accelerate in the same direction and cease acceleration some time later, all while the pulse of light is somewhere in between the two. At all times the emitter and receiver have zero relative velocity. The pulse reaches the receiver and is blue shifted. During the entire time, the pulse should also have had a velocity of c relative to both the emitter and receiver. Therefore, since the only players in question are the emitter, the pulse, and the receiver and since their relative velocities never changed there should not have been a blue shift.

Do I see a flaw in my argument? Yes. From a third frame watching this whole thing from the side the acceleration caused the relative velocities between the pulse and the receiver to change hence we have blue shift. So there is a way out, but I still don't understand why, if the velocity between the receiver and the pulse never changed from the perspective of the receiver, why it would see blue shift. Not sure I'll ever really understand this.

 

[Ibix]

At all times the emitter and receiver have zero relative velocity.

Not true in relativity from the receiver's perspective due to the relativity of simultaneity - see Bell's spaceship paradox.

 

[PAllen]

I am not sure what the confusion is, but I'll try another clarification. The only thing that matters for Doppler in SR is the relative velocity between the emitter at the emission event and the receiver at the corresponding reception event. What the emitter or receiver do before or after these events is wholly irrelevant (for that reception event). Simultaneity is irrelevant.The Doppler then results from the Lorentz boost for this relative velocity. The analysis can be done in any frame because the relative velocity described is frame invariant.

 

[jartsa]

If we use 0.99999c as the speed that the light pulse moves, we get almost correct results using the Lorentz-contraction formula.

I mean we pick some speed close to c as the the speed that the light pulse moves according to the observer before the observer has accelerated. When the observer accelerates some small amount the speed that the light pulse moves according to the observer changes a tiny amount.

 

[jartsa]

I am not sure what the confusion is, but I'll try another clarification. The only thing that matters for Doppler in SR is the relative velocity between the emitter at the emission event and the receiver at the corresponding reception event. What the emitter or receiver do before or after these events is wholly irrelevant (for that reception event). Simultaneity is irrelevant.The Doppler then results from the Lorentz boost for this relative velocity. The analysis can be done in any frame because the relative velocity described is frame invariant.

Why does the light source seem to become time dilated according to an observer that accelerates relative to the light source?

If light was emitted by a non-time dilated star, how does the light become indistinguishable from a light emitted by a time dilated star, just by acceleration of the observer?

I think the only thing that matters for Doppler in SR is the relative velocity between the emitter and the observer at the observation event. And observer decides what the relative velocity is, light source's opinion is not needed.

Oh yes, a wise observer would say: "I do not know the current velocity of the distant star". So it is the observer's current opinion about the star's past velocity that determines the redshift that the observer observes currently.

 

[Nugatory]

I think the only thing that matters for Doppler in SR is the relative velocity between the emitter and the observer at the observation event.

As with the classical Doppler effect, the relative velocity of the emitter and the observer at the time of observation is irrelevant. (And note that because the emitter and receiver are not colocated, the phrase "relative velocity between the emitter and the observer at the observation event" makes no sense - they aren't both be at the observation event).

You can pick any frame at all, and in that frame the wave will have a particular frequency. In other frames, it will have a different frequency.

Thus, if the wave has a given frequency in some frame, and the receiver is in motion in that frame, then in the frame in which the receiver is at rest at the moment of the reception event there will a Doppler shift in one direction or the other.

The only reason that the speed of the emitter ever comes into the picture is that it is particularly convenient to calculate the frequency of the wave in the frame in which the emitter is at rest at the moment of emission. Once we've established what that frame is, it is irrelevant whether the emitter remains at rest in that frame or not; we know what the frequency of the wave is in that frame. And from there we can ask the next question: is the receiver at rest in that frame at the moment of reception? If not, there's non-zero Doppler.

 

[Mister T]

Emitter and receiver have zero relative velocity. Emitter strikes out, then both emitter and receiver accelerate in the same direction and cease acceleration some time later, all while the pulse of light is somewhere in between the two. At all times the emitter and receiver have zero relative velocity. The pulse reaches the receiver and is blue shifted. During the entire time, the pulse should also have had a velocity of c relative to both the emitter and receiver. Therefore, since the only players in question are the emitter, the pulse, and the receiver and since their relative velocities never changed there should not have been a blue shift.

I'll try to re-write the scenario, eliminating the acceleration portion of the thought experiment.

Emitter and receiver have zero relative velocity, and are at rest relative to an Observer A, but are in motion along a line joining them according to Observer B. The pulse reaches the receiver and is blue shifted is B's frame^*. During the entire time, the pulse should also have had a velocity of c relative to both the emitter and receiver. Therefore, since the only players in question are the emitter, the pulse, and the receiver and since their relative velocities never changed there should not have been a blue shift in A's frame.

^*Edit: Note that the receiver will not show this blue shift, and B will reckon it's because the receiver is in motion relative to him.

 

[PAllen]

Why does the light source seem to become time dilated according to an observer that accelerates relative to the light source?

Read what wrote carefully. The acceleration of the receiver, per se, is irrelevant. A consequence of the acceleration is that the receiver has a relative velocity at reception event compared to emitter at emission event. This is the sole source of redshift/blueshift between the front and back of an accelerating rocket.

If light was emitted by a non-time dilated star, how does the light become indistinguishable from a light emitted by a time dilated star, just by acceleration of the observer?

This is not meaningful. What is a time dilated star?? There is no such category in SR.

I think the only thing that matters for Doppler in SR is the relative velocity between the emitter and the observer at the observation event. And observer decides what the relative velocity is, light source's opinion is not needed.

Completely wrong. The motion of the emitter 'at reception time' is irrelevant and ambiguous - whose simultaneity? Doppler is an invariant phenomenon that does not involve simultaneity at all. There are two events (one emission event on the world line of the emitter, one reception event on the world line of the receiver). These world lines have velocities at these events. The two velocity values are frame dependent. The relative velocity is frame independent in SR.

Oh yes, a wise observer would say: "I do not know the current velocity of the distant star". So it is the observer's current opinion about the star's past velocity that determines the redshift that the observer observes currently.

This thread is wholly about SR, not GR, so let's not muddy it with cosmology GR consideration. Relative velocity is unambiguous in SR.

 

[PAllen]

As with the classical Doppler effect, the relative velocity of the emitter and the observer at the time of observation is irrelevant. (And note that because the emitter and receiver are not colocated, the phrase "relative velocity between the emitter and the observer at the observation event" makes no sense - they aren't both be at the observation event).

In SR, relative velocity at a distance is unambiguous. I agree completely with everything else you wrote.

 

[Dale]

I still don't understand why, if the velocity between the receiver and the pulse never changed from the perspective of the receiver, why it would see blue shift.

I think you have to just let go of this concept. The quantity you are focusing on is the instantaneous relative velocity of two spatially separated objects. This quantity never has any physical significance in any scenario whatsoever. Thinking that it does will lead to confusion such as this. The measured Doppler shift cannot depend on it, because no physical quantity can. It is a frame-dependent quantity, and no reference frame is privileged.

Does that make sense? You must empty your mind of this incorrect idea before you can fill it with the correct idea.

 

[Nugatory]

In SR, relative velocity at a distance is unambiguous. I agree completely with everything else you wrote.

Ah - yes, of course you're right.

 

[sweet springs]

since the speed of the photon is always coming at us at c irrespective of the speed at which we are traveling toward the star, then why does the color of the photon change if we increase our speed?

Say a particle of non zero mass m has four velocity ##(u_0,u_1,u_2,u_3)## of norm 1. Multiplied by m ##m(u_0,u_1,u_2,u_3)## is four-momentum ##(p_0,p_1,p_2,p_3)## of norm m. Momentum is given by velocity.
Photon with zero mass does not have four velocity but four-momentum ##(p_0,p_1,p_2,p_3)## of norm zero. You cannot expect proportional relation of four-velocity and four-momentum for zero mass particle as you do for non zero mass particle. Momentum varies even if speed is the same c.

Wave picture is shared among zero and non zero particles. Four wave number ##(k_0,k_1,k_2,k_3)## = ##\frac{1}{\hbar}(p_0,p_1,p_2,p_3)## for both de-Broglie wave and light wave.

 

[robphy]

Hopefully this diagram will help clarify what many of us have been trying to say.

These diagrams are doing double duty as a spacetime diagram and as energy-momentum diagrams.
We draw these diagrams on rotated graph paper
to help visualize the light-signals and to visualize the unit tickmarks on observer worldlines.

The key idea is that:
once the light-signal with 4-momentum p_emitted is emitted by the instantaneous emitter at E,
it travels along to event R on E's future light-cone along the direction of p_emitted.
The emitter had influence on this 4-momentum only at E---what the emitted did before or after E doesn't matter.
The 4-momentum of the light-signal arrives at event R (call it p_received)
and can be measured by a instantaneous receiver at event R.
What the receiver did before R or does after R doesn't affect that measurement at R.

[Of course, since R and E are lightlike related,
there is no frame of reference where R and E are simultaneous... and
there is no frame of reference that can visit both R and E.]

For completeness...

In the emission diagram,
the source is moving with velocity [itex]v_{emitter,lab}=-5/13[/itex] and emits a forward light-signal [with velocity +c] with energy=6.
The lab frame measures that energy to be 4 units (thus redshifted).
Indeed [itex]k_{emitter,lab}=\sqrt{\frac{1+v_{emitter,lab}}{1-v_{emitter,lab}}}=\sqrt{\frac{1+(-5/13)}{1-(-5/13)}}=\frac{2}{3}[/itex].

In the reception diagram,
the lab frame would still measure that energy to be 4 units (no Doppler shift).
Since the receiver is moving with velocity [itex]v_{receiver,lab}=3/5[/itex], the receiver would measure energy=2.
Indeed [itex]k_{receiver,lab}=\sqrt{\frac{1+v_{receiver,lab}}{1-v_{receiver,lab}}}=\sqrt{\frac{1+(3/5)}{1-(3/5)}}=2[/itex].

From the emitter frame,
[itex]v_{receiver,emitter}=4/5[/itex] (check: [itex]v_{re,em}=\frac{v_{re,lab}-v_{em,lab}}{1-v_{re,lab}v_{em,lab}}=\frac{(3/5)-(-5/13)}{1-(3/5)(-5/13)}=4/5[/itex])
so, the relative-doppler factor is
[itex]k_{re,em}=\sqrt{\frac{1+v_{re,em}}{1-v_{re,em}}}=\sqrt{\frac{1+(4/5)}{1-(4/5)}}=3[/itex]. (check: [itex]k_{re,em}=\frac{k_{re,lab}}{k_{em,lab}}=\frac{(2)}{(2/3)}=3[/itex]).

From the receiver frame,
[itex]v_{em,re}=(-v_{re,em})=-4/5[/itex]
and [itex]k_{em,re}=(1/k_{re,em})=1/3[/itex].
This explains the received redshifted light-signal of energy 2, compared to the emitted energy 6.

 

[Buckethead]

I am not sure what the confusion is, but I'll try another clarification. The only thing that matters for Doppler in SR is the relative velocity between the emitter at the emission event and the receiver at the corresponding reception event. What the emitter or receiver do before or after these events is wholly irrelevant (for that reception event). Simultaneity is irrelevant.The Doppler then results from the Lorentz boost for this relative velocity. The analysis can be done in any frame because the relative velocity described is frame invariant.

This is a very concise description of what happens and sums things up nicely regarding the question of "what happens". My frustration is the "why". This morning I came up with a better example of my concern:

An inertial emitter sends out a pulse of red light. The pulse itself, traveling away at c has no doppler shift information as that is undefined until a reception is made. This is in contrast to a sound wave for example where the relative velocity between the emitter and the air determines the frequency of the sound wave. Again I want to clarify that the pulse has no information in it that would indicate the velocity of the source. Many years later a ship arrives out of warp from the Delta Quadrant and smacks right into the pulse. The pulse hits the ship at the expected speed of c from the frame of the ship. The ships sensors measure the frequency of the pulse and find it is blue in color. Considering that there is no way for the light pulse to have encoded any speed information from the emitter (since it will always be emitted at c) why is the pulse now blue when it was emitted with a red color.

 

[Buckethead]

I see a great many new posts just came in since PAllens post was posted that I did not see when I wrote my most recent response, so I need to go through all of them. So unless my recent post raises a new concern, no need to respond until I can get through all these new posts with hopefully a new sense of enlightenment.

 

[Nugatory]

The pulse itself, traveling away at c has no doppler shift information as that is undefined until a reception is made.

It does, however, have a frame-dependent frequency; and it has this frame-dependent property whether it eventually encounters a receiver or not.

You are thinking of the frequency in the frame in which the emitter was at rest at the moment of emission as the "real" or unshifted frequency. It's not; that frame is no more special than any other frame.

 

[Mister T]

Considering that there is no way for the light pulse to have encoded any speed information from the emitter (since it will always be emitted at c) why is the pulse now blue when it was emitted with a red color.

When you say it was emitted with a red color what do you mean? That in the rest frame of the emitter it was red?

 

[Buckethead]

When you say it was emitted with a red color what do you mean? That in the rest frame of the emitter it was red?

Yes.

 

[Buckethead]

You can pick any frame at all, and in that frame the wave will have a particular frequency. In other frames, it will have a different frequency.

Key point.

It does, however, have a frame-dependent frequency; and it has this frame-dependent property whether it eventually encounters a receiver or not.

Now we're getting to the meat and potatoes and I appreciate that you said this. I need to offer up an amusing way to look at this. (PLEASE read this for amusement only*)

It occurs to me that frame dependent frequency and frame dependent property might be two different things and I offer up my definition that frame dependent frequency refers to two frames, the emitter during emission and the receiver during reception. Combined, these result in a specific frequency upon reception. I further offer the definition that frame dependent property as used here should be called an emitter frame dependent property and is a property with a value set by the emitter during emission only. A frame dependent frequency cannot be a property of the light pulse because the receiving frame is still undefined while the pulse is in transit but a emitter frame dependent property can be.

it makes sense that the light pulse would contain an (emitter) frame dependent property that would be a constant which is set at the time of emission. It is now possible for the pulse to roam freely with no strings attached to the emitter as long as it holds this constant. The receiver also has an inherent frame dependent property, the value of which is not determined by the emitter or the pulse.

The final frequency of the light pulse is a function of both the emitter frame dependent property value encoded in the pulse and the frame dependent property value of the receiver.

The emitter frame dependent property can be encoded into the light pulse's frequency and would be different than the frequency of the emitter because it would contain information both about the frequency of the emitter hardware and of the emitter frame dependent property value.

The only problem with this idea is that I don't know what might determine the emitter frame property value during emission or the receiver property value during reception. Again, my apologies for going out on a limb here. I know that is frowned upon but I couldn't help myself. This is just a really fun thought.

*To everyone reading please do not construe this as a speculation as to what I think might be happening behind the scenes, but instead please read it only as an interesting analogy to what actually is happening, such as a change in momentum as was offered up earlier or some other mechanism. It just seems to me that what I'm going to mention here is an interesting way to sort through this from a purely logical standpoint in a world where "that's just the way it is" is used too much.