# Doppler Effect

1. Jun 10, 2013

### Physics_start

I have done a little analysis of Doppler Effect and have tried to compile my thought in a document. My background is not physics thus, I am seeking validation/criticism from this community. I would very much appreciate it if you went through my work and pointed out where I might be wrong.

The file can be found here:
http://commons.wikimedia.org/wiki/File:Visual_Doppler_Effect.pdf [Broken]

Thanks a lot,
Varun

Last edited by a moderator: May 6, 2017
2. Jun 11, 2013

### ghwellsjr

I took a look at your paper. The first diagram I understood very well having made many such similar spacetime drawings. Your second figure is a nightmare. I don't like spacetime diagrams that combine two coordinate systems and I think there is too much learning curve for most people. Why don't you just make separate diagrams like your first one that covers a single coordinate system?

I didn't read past that second figure but I did scan over the remainder of the document to see if I could figure out where you were going with it. I couldn't. But if your point is that a Doppler analysis is frame independent and shows what each observer actually sees, observes and measures, then it is already a well-known concept and easily demonstrated by showing light or radar signals traveling along diagonals and how they start and end at the same events in all Inertial Reference Frames. Is that your main point?

Who is your target audience--people who already know Special Relativity or people who are first learning?

3. Jun 11, 2013

### bobc2

I particularly liked your two space-time sketches--especially the second one, because you did a nice job showing the simultaneous spaces for observers at their specific times along their worldlines. Relativity of Simultaneity is an important concept in special relativity and perhaps your space-time diagram will alert someone to this and encourage them to dig deeper into space-time diagrams rather than just be satisfied with comparing worldlines.

I will continue to read your paper. Thanks for checking in with us.

Last edited by a moderator: May 6, 2017
4. Jun 11, 2013

### Mentz114

A quicker way of answering this is to boost the propagation vector of light in the x-direction, with a velocity $\beta$ in the same direction.

Boosting $u=\omega\ \partial_t \pm k\ \partial_x$ which gives

$\Lambda(\beta)\ u = (\gamma \omega\pm \beta\gamma\ k) \partial_t + (\gamma k\pm \beta\gamma\ \omega) \partial_x$

u is a null vector so ω=k ( with c=1) and so we get

$\omega' = \omega \gamma( 1\pm \beta) = \omega\sqrt{\frac{ (1\pm \beta)^2}{1-\beta^2}}= \omega\sqrt{\frac{(1\pm\beta)(1\pm\beta)}{(1-\beta)(1+\beta)}}$

which is the relativistic Doppler equation.

This is a quote from your posts in the other forum. Unfortunately it does not make sense to me and sounds like dodgy private theory. What do you mean ?

Last edited: Jun 11, 2013
5. Jun 11, 2013

### Physics_start

Thanks for the answer

Hi Mentz114,

Thanks for the answer. But the main point of my paper is not to derive Doppler Effect for frequency and wavelength.

It was to show that the time we experience a wave in (when we are moving) is different from the time the wave was emitted in.

So a bulb that was on for 10 sec and was observed to be on for 10 sec by a an observer at rest relative to the bulb, will seem to be on by a moving observer for the following duration:
10 γ (1+β). For a velocity of c/2, γ = 1.15, β = 1/2. Thus the bulb will seem to be on for:
5 x 1.15 sec = 5.75 sec

My point it that when we speak about SEEING an event as opposed to recording the time it happened. Then the total time we experience is CONTRACTED (when a light source is moving towards an observer). Hence we experience a time contraction EFFECT.

We would experience the opposite of this when the source is moving away from the observer. We would see a time dilation EFFECT.

If we then use these visual equations:
t = t' γ (1 ±β)

Take the reciprocal on both sides, we end up with Doppler effect equations.

Thus, my point is that the Doppler Effect is just a special case of these time contraction and time dilation EFFECTS.

---------

My second point is that for any wave (imagine a hypothetical wave which can travel in vacuum) and has a measured velocity of S according to a stationary observer in space. Then that wave will display frame invariant velocity even without time dilation.

To show this, imagine the wave is emitted as 10 pulses with gaps in between. Let the period of a pulse by 5 sec and for the gaps also be 5 sec. Then we see that the whole signal will be emitted over a time of 10 x 5 s (the pulses) + 9 x 5 s (the gaps). Thus the total time = 95 s. At the send of this our wave would be spread out over a distance of S m/s x 95s = 95 x S m. Let's call 95 x S m d.

Then this entire wave form spread over distance d is moving with velocity S. Thus for a stationary observer, this wave would move past the observer in d/S sec.

If the observer is moving towards the wave, the the observer will cover this distance d in d/S(1+v/S) where v is the velocity of the observer towards the source.

Let t = d/S(1+v/S) and t' = d/S. Where t is the "time" experienced by the moving observer that the train of wave pulses lasted and t' the the time during which the train of pulses was emitted.

Then (assuming v is pretty small compared to C, we can take gamma to be 1):
Δt = Δt' (1 + v/S)

Therefore, the frequency will vary as:
fo = fs (1 + v/S)
Where fo is the frequncy observed and fs is the proper frequency of the source.

Now let's consider γ.
Assuming we could start a stop watch when the wave first hit our eyes/ears/sensory instruments and stop the stop watch after n wave lengths had gone past, and recorded the time thus elapsed as ts. Then (equating and multiplying both sides by S):
S Δts = S Δto (1 + v/S)
Where Δts is the time we recorded on our stop watch. Δto is the time n wavelengths were originally emitted by the source.

But S Δts = nλs where λs is the wavelength of the source., and S Δto = nλo, where λo is the wavelength experienced.

Hence:
λo = λs/(1 + v/S) --- Note that this is an EFFECT and not real length contraction. The actual wavelength is preserved
and from before:
fo = fs (1 + v/S)

We get:
fo λo = fs λs

Thus for any wave that has a fixed velocity as observed by a stationary observer should display frame invariant velocity. If the frame velocity has no component of velocity in the direction of the source, then the velocity of the wave as experienced by the observer in the frame will remain as fo λo

If the frame has a component of velocity in the direction of the source, then the velocity measured will be fs λs (where v is taken to be only component of the frame's velocity in the direction of the source).

However, even when the velocity is experienced as fs λs, it is still frame invariant because we have already shown:
fo λo = fs λs

These are the two points I wish to make in my paper:
http://commons.wikimedia.org/wiki/File:Visual_Doppler_Effect.pdf [Broken]

Last edited by a moderator: May 6, 2017
6. Jun 11, 2013

### Physics_start

Since light is continuous, instead of a single 45 degree line to represent a photon, you could draw bands of red, amber and green as the light switches RED (1 sec) AMBER (1 sec) GREEN (1 sec) on the paper. You can see from my diagram that the period during which the photons are emitted contracts to a smaller period in the frame of the observer (observer source distance decreasing at constant rate). Hence there is a time contraction effect

7. Jun 11, 2013

### Mentz114

No I can't see it. Everything you describe is predicted by the Doppler equation. You're free to interpret it as you wish but I can't see anything new here.

8. Jun 11, 2013

### Physics_start

Is the shortening of the duration of the pulses predicted by Doppler equations? If it is then I'm totally fine with that. I thought Doppler only predicted frequency shifts?

9. Jun 11, 2013

### Physics_start

Another thing is photon flux. Since frequency is increasing the number of photons that will be encountered by source and observer moving towards each other will increase.

10. Jun 11, 2013

### Naty1

I am also having this problem:

The first is not a properly stated sentence....The SUBJECT is missing. If I understand what you are trying to say it is incorrect: The speed of light locally is always 'c'. The relative motion of source and observer affects the wavelength observed, not the duration of the overall light signal. A light signal of duration three seconds locally at the source will be observed by a distant observer locally there as also three seconds; but the two separated clocks may not tick at the same rate, and source and observer may record different wavelengths.

There is 'no stationary observer in space'. Do you mean the observer is stationary with respect to the source of the 'signal'? Light, electromagnetic radiation, the CMBR, do have an invariant speed. In order to have a frame invariant velocity, say 'c', space and time must jointly conspire in such a way as to maintain that constant velocity. You can't hold either distance or time fixed: that violates Lorentz invariance....

So if I understand your points, I disagree.

Your opening is somewhat confusing to me :

Do you know 'astronomy' deals with curved spacetime, not flat? Or are you assuming short enough 'astronomical distances' that distance is not expanding? Cartesian, Minkowski coordinates are flat space and flat spacetime, respectively....since that is the math you use, I'd not mention 'astronomy' upfront.

11. Jun 11, 2013

### Naty1

If I understand what you are saying, a different number of photons will arrive than were sent??
THAT would be a miracle. Doppler shift does NOT materialize nor dematerialize the existence of photons.

12. Jun 11, 2013

### Mentz114

Yes. That is how speeding radar works ( well, that's how it worked when they caught me ).

Naty1 has also made some good points. Frequency is not a photon count. Each 'photon' has a frequency.

13. Jun 11, 2013

### Physics_start

14. Jun 11, 2013

### Mentz114

It does. So does the gap between pulses. There is no need to use CAPITALS - we can hear you.

The animations are OK, but they show the pulses all the same width which is not strictly correct.

You might be interested in the Bondi K-calculus, which is a treatment of SR that starts with the observables ( Doppler effects ) and deduces the transformation between frames. A web search gets a lot of hits including this

http://www.math.ku.edu/~lerner/m291/SR_Lecture2.pdf

Last edited: Jun 11, 2013
15. Jun 11, 2013

### Naty1

16. Jun 11, 2013

### Physics_start

@ Naty

No that's not what I am saying. In the rest frame of a observer, the distance between an event divided by the time taken for light from that event to get to the observer is always is always c (both measured in the rest frame of the observer. now when observer is stationary, the source is emitting say x photons per second. Thus x photons reach the observer per second but when x photons reach the observer x photons more are 1 light second behind the position of observer, and another x photons are behind 2 light seconds from the observer.

That is why when the observer decides to move towards the source, the observer will encounter a path full of photons and hence flux > x photons per second.

I agree the above line is misleading and serves no real purpose. Will remove it.

My point about the above is if you measure velocity exclusively by measuring frequency and wavelength, then the product of these is constant across frames as I have shown.

Mentz114 and Naty you guys seem to disagree on this one....

17. Jun 11, 2013

### Mentz114

This is built into the Lorentz transformation. Light always travels at c. You assumed this ( the 45o lines ), so it is no surprise it reappears tautologously.

18. Jun 11, 2013

### Physics_start

No I don't assume this. See my analysis of a hypothetical wave which travels at speed S in vacuum according to an observer at rest relative to the wave. The question is will this hypothetical wave display frame invariant velocity? And my answer is yes because?

If v is not small. The gamma factor can be added in as well. But it would still result in fo λo = fs λs

Last edited: Jun 11, 2013
19. Jun 11, 2013

### Naty1

ok....the energy of light is observer dependent.....sure....for example, momentum depends on relative velocity of the observer

20. Jun 11, 2013

### Naty1

physics posts
What did Mentz say about this? All I could find was:

and I agree.

21. Jun 11, 2013

### Physics_start

Could you guys provide me any pointers to links or sites where it is clearly shown how the entire time we measure visually by seeing events occur includes the (1 +- beta) factor (in addition to the gamme factor). I think that would conclude this discussion :D and help me to move on....

Thanks for the help so far!

22. Jun 11, 2013

### Mentz114

Going back to my first post, we can see how a non-null propagation vector transforms, i.e. $\omega\ne k$.

$\Lambda(\beta)\ u = (\gamma \omega\pm \beta\gamma\ k) \partial_t + (\gamma k\pm \beta\gamma\ \omega) \partial_x$.

Now $v=\lambda\omega = \omega / k$ so

$v'=\frac{\gamma \omega\pm \beta\gamma\ k}{\gamma k\pm \beta\gamma\ \omega}=\frac{ \omega\pm \beta\ k}{ k\pm \beta\ \omega} = \frac{v\pm\beta}{1\pm v\beta}$
I'm not sure this is right*. But it is true that any non-null vector will not have the same speed in the moving frame.

*I must be getting tired, this is the relativistic velocity addition rule. I rest my case.

I also removed some surplus factors of 2pi

Last edited: Jun 11, 2013
23. Jun 11, 2013

### Physics_start

Mentz, this is greek to me. I am not as advanced in the theory as you probably. Could you give that in layman's speak please, with perhaps a resource link where I could read up more on the equations you just described

24. Jun 11, 2013

### Mentz114

But it is elegant, concise and true Greek.

I'll be happy to explain it to you. You need to know what a 4-vector is. I'm working in Minkowski coordinates (t,x,y,z) and the $u$ thing is a velocity 4-vector pointing in the x-direction. So it describes something moving or propagating in the x-direction. Is that OK so far ?

25. Jun 11, 2013

### pervect

Staff Emeritus
This is all true, and something we've been discussing on another thread (or attempting to, at least). It's called relativistic beaming, or sometimes "the headlight effect" in the literature of relativistic visualziation. But it's basically just the doppler effect in disguise.

For some more on the headlight effect see http://arxiv.org/pdf/physics/0701200v1.pdf

The angular size change is real, but not of interest in this thread. What we're interested in is the factor of D in intensity due to the doppler effect.

Here's my simple explanation of why the effect has to occur.

Consider a classical pulsed wave. Suppose we modulate it, we turn it off and on. If we use engineering terms, we call the frequency of the base wave the "carrier" frequency, and the frequency with which we turn it off and on, the modulating frequency.

When we look at the modulated wave from an object in relative motion, both the carrier frequency AND the modulating frequency get red or blue shifted simultaneously by the same amount.

Let's take an example:

Suppose we have a 1 mhz carrier, and the modulation is such that it's 1000 cycles "on" and 1000 cyles off.

This makes the modulating frequency 500 hz, as the on period is 1000 us, and the off period is 1000us, making the toal period 2000 us.

If we observe this modulated wave from a relatively moving observer, the number of carrier pulses in a wavepacket does NOT change.

Example. If we have a 2:1 doppler shift, our 1 mhz carrier gets blue-shifted to 2 mhz.

WIth 1000 cycles in the "on" part of the wave, the on period is now 500us, and the off period is 500us, so our total period is 1000 us. Thus our modulating frequency gets blue-shifted by an identical amount as the carrier frequency. This must happen to keep observables (such as the number of carrier pulses) constant.