B Why do photons allow Doppler shift

[Buckethead]

By that you mean there's a portion of the derivation of the Doppler shift formula that's the same as the derivation that makes use of the Newtonian approximation?

Yes.

Relativistic physics is not something separate from Newtonian physics. Newtonian physics is just an approximation that works well enough under certain conditions, it can't tell you anything that relativistic physics can't also tell you

Yes, I understand that. Because my examples use slow spaceship speeds, Its fine to just use the Newtonian portion of the Doppler formula. But either way the end result is very nearly the same.

 

[PeterDonis]

You are still talking as if the relative velocity between the receiver and the light pulse changes.

No, I'm not. The only relative velocity I've mentioned is the relative velocity between the source and the receiver. The relative velocity of the receiver and the thingie that is sent from the source to the receiver doesn't matter as long as it's the same for two successive thingies. That does not require that the speed of the thingie be frame-independent; the relative speed of receiver and thingie can be different for different receivers. It just has to be the same for the same receiver with two successive thingies from the same source.

Please explain to me how the energy of the light pulse, or bullets, or pulses of sound can change if its velocity relative to you does not change.

We are not talking about a change in energy. We are only talking about a change in frequency, i.e., how often each thingie arrives. (It is true that, for light, energy is proportional to frequency, but we haven't gotten there yet; that's a separate question from how frequency itself behaves.)

My understanding is that there is the relativistic (Lorentz transform) amount and then there is the Newtonian Doppler shift.

If we are talking about SR, you have this backwards. As far as SR is concerned, there is one Doppler shift. It can be useful to split that one shift up, conceptually, into a "Newtonian" piece and a "relativistic" piece (the latter involving things like time dilation that don't appear in Newtonian physics), but that's a feature of human models, not of the actual physics.

However, before you even try to understand the SR version with light, you need to understand the simpler Newtonian version, where we just have thingies going from source to receiver and we assume any SR effects like time dilation are negligible.

 

[Ibix]

You are still talking as if the relative velocity between the receiver and the light pulse changes.

Are you perhaps not taking into account relativistic velocity addition? In a frame where the source is moving the separation rate between source and light pulse is not c, but in the frame where the source is at rest it is c. You can't ignore this by keeping the source velocity low because it's the speed of light that is being transformed. That's non-Newtonian by definition.

 

[PeroK]

True, a medium is not necessary, as I'm OK with baseballs in space. However it's the same thing. You are still talking as if the relative velocity between the receiver and the light pulse changes. Please explain to me how the energy of the light pulse, or bullets, or pulses of sound can change if its velocity relative to you does not change.

In fact, all you need is a diagram for the receiver's frame of reference. Try with the source moving away, as that might be easier to see.

Note that, as @Ibix has pointed out, the separation velocity between the source and the light wave (in the receiver's frame) lies between $0$ and $2c$. You may be confusing "relative" velocity with "separation" velocity here.

Take a source, receiver and an light wave and 1D motion.

The velocity of the light wave "relative" to the source/receiver means the velocity in the reference frame of the source/receiver. This is invariant (the same in both frames).

However, the separation velocities are:

Source frame: $c-v$ (speed at which the light wave is catching up the receiver)

Receiver frame: $c+v$ (speed at which the light wave is moving away from the source).

Where $v$ is the relative velocity of the source and the receiver.

 

[Dale]

This is totally a Newtonian kinematic viewpoint. It implies that the wave is traveling in a medium and the ship is moving through that medium at some velocity causing the crests of the waves to be closer together. This works fine for sound, but I don't see how it can for light because c is frame independent.

Not one word of @Nugatory implied a medium in any way. That is simply something that you mentally added on your own. All we can do is recommend that you not do that.

@Nugatory simply picked a frame that was convenient and analyzed the problem completely in that frame. That is a general approach that you should use. Pick a frame. It can be any frame that you think will simplify the math. Common choices are the rest frame of the emitter, the rest frame of the receiver, or the frame where they are moving at equal speeds in opposite directions. Just pick one and work it through in that frame.

If you get stuck, just show us your work and we can get you unstuck

For example imagine someone throws a baseball at you at 50MPH. There is a certain momentum in that ball relative to you. Now imagine you run at 10MPH toward the pitcher and he throws again, however, because the baseball is light he throws the ball at 40MPH. It still hits you at 50MPH and its pulse width (the width of the ball going past you) is the same as before and therefore its frequency is the same and likewise its momentum is the same.

Doppler isn’t directly about the energy or momentum of waves/photons/baseballs. It is about frequency. How much time is between two successive waves/photons/baseballs? Please ignore any question of energy or momentum and focus exclusively on the time. Calculate the time between two, not the energy or momentum of one. If you don’t have at least two with a given time between emissions then you are not doing a Doppler problem.

 

[Herman Trivilino]

By that you mean there's a portion of the derivation of the Doppler shift formula that's the same as the derivation that makes use of the Newtonian approximation?

Yes.

But that derivation is just a heuristic device. There are other ways to do the derivation.

my examples use slow spaceship speeds, Its fine to just use the Newtonian portion of the Doppler formula. But either way the end result is very nearly the same.

In a practical way it's okay, but the Newtonian viewpoint is entirely wrong. You can't patch it up by adopting it as a valid viewpoint to which relativistic "corrections" are added only when necessary. That approach works only in practice. The relativistic viewpoint is a worldview that works at all speeds.

 

[Herman Trivilino]

True, a medium is not necessary, as I'm OK with baseballs in space. However it's the same thing. You are still talking as if the relative velocity between the receiver and the light pulse changes.

As he points out, he's not doing that. Perhaps what you mean here is the the relative speed between the source and the baseball changes when the baseball is thrown. Note that when the light pulse is created its speed becomes $c$.

 

[SiennaTheGr8]

I think the Doppler effect for light is easiest understood in terms of period and wavelength first.

Say you have a receiver at rest, with a source of light approaching head-on at (normalized) speed $\beta$.

From the source's perspective, a time $c \Delta t$ elapses during the emission of a single cycle (crest to crest). That's the wave's period $cT$ in the source's rest frame: $cT = c \Delta t$. But for light, the period $cT$ is equal to the wavelength $\lambda$, so we actually have $c \Delta t = \lambda$.

By time dilation, that $c \Delta t$ is equivalent to $\gamma c \Delta t$ in the receiver's frame: $c \Delta t^\prime = \gamma c \Delta t = \gamma \lambda$ (where I've designated the receiver's frame the primed frame).

So in the receiver's frame, the source emits a second wavefront a time $c \Delta t^\prime = \gamma \lambda$ after emitting the first. What we want to know is: what's the distance between the source and the first wavefront at this moment? The answer will tell us the wavelength $\lambda^\prime$ in the receiver's frame.

Well, during this time $\gamma \lambda$, the first wavefront moves closer to the receiver a distance $\gamma \lambda$. But the source itself also moves toward the receiver during this time, a distance $\gamma \lambda \beta$. Thus, the first wavefront is "ahead" of the source by a distance $\gamma \lambda - \gamma \lambda \beta$ when the second wavefront is emitted:

$\lambda^\prime = \gamma \lambda ( 1 - \beta )$.

Then since wavelength and frequency $\nu$ are inversely proportional, we have:

$\nu = \gamma \nu^\prime (1 - \beta )$.

For a receding source, the minus sign becomes a plus.

For the general case, the source might not be approaching or receding head-on. Then we use only the component of the source's displacement that's parallel to the light wave's direction of propagation (as opposed to the total distance the source travels). This isn't difficult, but it involves defining an angle, and that can be done in either frame, so one must be mindful of signs and primes. IIRC, Einstein's original derivation uses the angle in the source's frame, but most other treatments I've seen use the angle in the receiver's frame. Can lead to some confusion. (Anyway, the angles are related by the aberration formula.)