# B Why do photons allow Doppler shift

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1. Oct 28, 2017

### Staff: Mentor

There are many frame-dependent properties. Frequency is one of them. Thus, "frame dependent frequency" and "frame dependent property" are different things in the same way that elephants and animals are different things: An elephant is an animal, but there are animals that are not elephants.
That might be the root of your misunderstanding right there. The frequency in any given frame is defined as $\nu=1/\Delta{t}=c/\lambda$ where $\Delta{t}$ is the time between the arrival of successive wave crests and $\lambda$ is the distance between successive wave crests. This quantity is well defined and can be calculated in any frame as long as you know what it is in some frame.

2. Oct 28, 2017

### Staff: Mentor

3. Oct 29, 2017

### PAllen

I already answered the 'why' accepting that you wanted to consider a particle point of view rather than a wave point of view. Any particle carries an energy and momentum. The combination of these form a Lorentz covariant vector. This means that the energy and momentum are frame dependent, transforming via the Lorentz transform. The norm of this vector is the mass, which is zero in the case of a photon. The behavior of this vector is the same for a photon, a proton, or a baseball. If emitted with some energy and momentum per some emission frame, no matter where or when the particle is measured, its energy or momentum will be given by the Lorentz transform corresponding to the velocity of the detector in the original emission frame, applied to the energy/momentum as specified in the emitter frame.

Thus what any particle encodes (to use your terminology) is an energy and momentum for any possible measurement frame. It is best to think of this as one vector property, with different expression in different frames. The only thing specific to a photon is that its energy is proportional to its frequency = color (and its mass is zero). Note that no information about the motion of the emitter is really encoded. What is encoded is this vector property specifying how the the particle would be measured in any frame. One that happens to be the same as the emission event rest frame, will measure the same energy as the original emitter would.

Maybe it will be clearer to forget the emission frame and just talk about a photon that per some arbitrary frame would have energy E and momentum p. Then because it has no mass, the magnitude of the momentum is E (remember E2 - p2 = m, which is zero). Then, the momentum 4 vector for the light can be written E(1,k) ,where k is a unit 3 vector in the spatial direction of the light's (photon, if you will) motion. Now imagine an arbitrary detector with 4-velocity γ(1,v), where v is the 3-vector velocity of the detector. Then the energy measured by the detector will be the time component of the momentum vector expressed in the detector's rest frame, which is just given by the vector product of the momentum 4-vector and detector 4-velocity, generally: γE(1-k⋅v) . In the co-linear case, with v and k in the same direction, this simply becomes γE(1-v), which with trivial algebra is E√((1-v)/(1+v)), which is the standard relativistic Doppler factor applied to energy. Then, in the case of a photon, this factor applies to frequency as well.

4. Nov 14, 2017

Sorry for the delayed response. I've been trying to digest all this to see where the flaws in my logic are. It occurred to me that the heart of my question does not require relativistic speeds and in fact that confuses the issue. After fooling around on a spread sheet it turns out that if the speed between the emitter and receiver is only 1000MPH, the Newtonian part of the Doppler shift formula is much much larger than the relativistic part. (about 136,000:1) I originally thought that the problem could be solved, as many here have mentioned, if you just consider that energy of the photon or the frequency of the pulse are changed between the transmitter and receiver by the Lorentz factor. However, that is only a very small part of it. Let's look at this at much slower speeds.

A ship (A) is stationed at a place where a pulse of light of duration 1 second (duration as measured by ship A) is coming from the left, and a ship (B) coming from the right at v=1000MPH collide. Ship B measures the speed of the incoming light and of course it is c. It also measures the duration. Ship A however calculates that ship B and the light beam have collided at c+v. It also therefore calculates that the duration of the pulse as seen by ship B must be 1-v/c. According to Doppler theory ship B will also measure the duration to be 1-v/c. The question is, how can this be? For one thing, all measuring instruments on A and B match up to within the Lorentz factor. In other words, there is insignificant length contraction and time dilation. It's easy to see why ship A's measurement is accurate. But ship B cannot measure 1-v/c because the incoming pulse is colliding from B's frame at c, not c+v. It might seem like ship B is measuring a shorter pulse (shorter by 1-v/c), but again, A has measured the pulse and it is exactly 1 second.

5. Nov 14, 2017

### PeroK

@Buckethead it seems to me your problem is trying to think of frequency, energy and momentum of the photon as an absolute property of the photon, independent of reference frame. The energy of a photon is not related to its speed (which is frame independent) but to its frequency (which is frame dependent).

If you're moving towards a light source, you cannot measure the same photon energy as the source. However that energy is produced, it cannot be the same in the two frames. Energy must transform between frames.

Any interaction of a photon with a massive object (including the star that produced it) must have a balanced energy-momentum equation in every frame. The energy of the photon, therefore, must be frame dependent and cannot be an absolute instrinsic property of the photon.

In terms of the mechanism, it seems to me this was perfectly explained by @Nugatory in post #4.

6. Nov 14, 2017

### Staff: Mentor

I have no idea what you are trying to calculate here.

If you are willing to do the calculation, then just do the full fledged Doppler calculation. Say the beginning of one cycle is emitted at t=0 so the beginning of the next cycle is emitted at t=T where T is the period (1/f). Then simply calculate when the beginning of each cycle is recieved and subtract to get the recieved period.

You can simplify it even further if you use Bondi’s “k calculus” approach.

7. Dec 4, 2017

I don't think I am. I understand that in Newtonian kinematics the momentum of an object is determined by (for example) the relative velocity between the object and another object when they collide. And it appears we are applying a similar idea to light.
OK good. Since the speed is always c this makes sense that the momentum must then be defined by the lights frequency.
Yes, this appears to be the case.
Yes, again this appears to be the case.
Actually, to me, no, it doesn't. Here is my problem.
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. 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. Regardless of how fast you run toward the ball, if the ball always hits you at 50MPH, how can its energy change?

The first thought (for me) is that it's a relativistic problem. But if the ship in question is traveling at a slow speed (1000MPH for example), the frequency shift due to Newtonian kinematics (non-relativistic portion of the Doppler shift formula) is far greater than that determined by the Lorentz formula. So something else is going on.

I do understand that the Doppler shift formula does actually work, so I'm not questioning that its use is proper, but the non-relativistic portion of the Doppler shift formula is Newtonian and I don't know how it can be applied to light if c is frame independent. The issue becomes more clear if you take into account that the Doppler shift felt by the receiver is dependent on the relative speed of the emitter when there is no link whatsoever between the emitter and receiver as would be in the case of sound for example (where the medium for the sound wave determines the relative velocity of the emitter and receiver). If there is no medium to relate the speed of the two, which there isn't, then the only way around this is....well actually, I can't see any way around this.

8. Dec 4, 2017

### Staff: Mentor

No, it doesn't. There's no need to postulate a medium. All you need is that the source emits wave crests at some fixed frequency by its clock, and that the source moves closer to the receiver by some distance in between emitting any two successive wave crests. They don't even have to be wave crests; they could be bullets, or very quick pulses of sound. The reasoning is the same: the time between two successive thingies being received by the receiver is different than the time between two successive thingies being emitted by the source, because of the relative motion.

The only difference the frame independence of $c$ makes, if we're talking about purely non-transverse motion (i.e., the relative motion of the source and receiver is all along the line between them) is in the quantitative details of how much Doppler shift there is for a given relative speed. (If you add transverse motion to the mix, things get more complicated, because SR predicts a Doppler shift for tranverse motion while Newtonian mechanics does not. But I don't think we need to get into those complications here.)

9. Dec 4, 2017

### Mister T

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?

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.

10. Dec 4, 2017

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.
My understanding is that there is the relativistic (Lorentz transform) amount and then there is the Newtonian Doppler shift. For slow speeds we can easily dismiss the relativistic amount. It's also my understanding that the Newtonian Doppler shift for frame independent and frame dependent things are the same. So what is the quantitative difference that you are referring to?

11. Dec 4, 2017

Yes.
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.

12. Dec 4, 2017

### Staff: Mentor

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.

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.)

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.

13. Dec 5, 2017

### Ibix

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.

14. Dec 5, 2017

### PeroK

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.

15. Dec 5, 2017

### Staff: Mentor

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

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.

16. Dec 5, 2017

### Mister T

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

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.

17. Dec 5, 2017

### Mister T

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 $c$.

Last edited: Dec 5, 2017
18. Dec 7, 2017

### 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.)