Understanding the Doppler Effect: Formula and Examples for Light Waves

[Stephanus]

Dear PF Forum,
I wonder about this Doppler formula.
https://en.wikipedia.org/wiki/Doppler_effect#General
##f=(\frac{c+V_r}{c+V_s})f_0##
The speed of sound is 343m/s
Supposed S (Source) moves to the east, toward R(Receiver) at 70 m/s and R moves to the east at 40 m/s. So the formula is:
##f=(\frac{343-40}{343-70})f_0##
But R will say. No, I'm at rest, it's S who moves toward me at 70-40 = 30 m/s. So the doppler factor would be:
##f=(\frac{343}{343-30})f_0##
No, we can't do that. ##\frac{343-40}{343-70} \neq \frac{343}{343-30}##. Okay...
What about blue/red shifted?
Every where we go, the light will always travel at 300 thousands km/s according to us, right.
What is the formula of doppler ratio, for the frequency that is received by the receiver based on the frequency sent by source?
R is at West
S is at East.
If S moves to west at 0.7c and R moves to west at 0.4c?
A: ##\frac{c-0.4}{c-0.7}## or B ##\frac{c}{c-0.3}## or even C ##\frac{c+0.3}{c}##

If S moves to west at 0.7c and R moves to east at 0.4c?
D: ##\frac{c-0.4}{c-0.7}## or E: ##\frac{c+1.1}{c}## or F ##\frac{c}{c-1.1}##
I suspect F is impossible, but then I have no idea at all.
Unless..., there is aether

 

[Mentz114]

Dear PF Forum,
I wonder about this Doppler formula.
https://en.wikipedia.org/wiki/Doppler_effect#General
##f=(\frac{c+V_r}{c+V_s})f_0##
The speed of sound is 343m/s
Supposed S (Source) moves to the east, toward R(Receiver) at 70 m/s and R moves to the east at 40 m/s. So the formula is:
##f=(\frac{343-40}{343-70})f_0##
But R will say. No, I'm at rest, it's S who moves toward me at 70-40 = 30 m/s. So the doppler factor would be:
##f=(\frac{343}{343-30})f_0##
No, we can't do that. ##\frac{343-40}{343-70} \neq \frac{343}{343-30}##. Okay...
What about blue/red shifted?
Every where we go, the light will always travel at 300 thousands km/s according to us, right.
What is the formula of doppler ratio, for the frequency that is received by the receiver based on the frequency sent by source?

The relativistic Doppler effect is reciprocal and depends on the relative velocity ##\beta=v/c##. If the receiver approaches the emitter then the freqency shift is ##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. If they are separating then reverse the sign of ##\beta##.

 

[Stephanus]

The relativistic Doppler effect is reciprocal and depends on the relative velocity ##\beta=v/c##. If the receiver approaches the emitter then the freqency shift is ##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. If they are separating then reverse the sign of ##\beta##.

Ahh, again my hastily question. Of course no matter where we moves, it's the signal from the source that is counted, not from us. Why the hell that we bother with our own signal

 

[jtbell]

But R will say. No, I'm at rest, it's S who moves toward me at 70-40 = 30 m/s. So the doppler factor would be:
f=(343343−30)f0f=(\frac{343}{343-30})f_0
No, we can't do that

Because, from the Wikipedia article that you referred to:

is the velocity of waves in the medium;

is the velocity of the receiver relative to the medium; [...]

is the velocity of the source relative to the medium; [...]

This formula is for classical waves such as sound waves in air (in which case the speeds are relative to the air), ripples on a water surface (in which case the speeds are relative to the water), etc. It does not apply to light, for which you must use the relativistic Doppler effect which makes no reference to a medium, and uses only the relative speed of source and observer, and the the (invariant) speed of light:

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

(Mentz beat me to it while I was struggling with the formatting)

 

[Stephanus]

##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. [..]

Yes, right. I should have read other answer in my previous post.
And ##\sqrt{\frac{1+\beta}{1-\beta}}## is different from ##\frac{1+v}{1-v}##. No square root in sound doppler

 

[Mentz114]

Ahh, again my hastily question. Of course no matter where we moves, it's the signal from the source that is counted, not from us. Why the hell that we bother with our own signal

Plugging in some numbers, if ##\beta=1/2## then the frequency ratio is ##\sqrt{\frac{1+1/2}{1-1/2}}=\sqrt{3}## ( better check this !) or ##1/\sqrt{3}## if ##\beta## is -1/2

[thanks to Stephanus for pointing out my arithmetic error]

 

[Stephanus]

Plugging in some numbers, if ##\beta=1/2## then the frequency ratio is ##\sqrt{\frac{1+1/2}{1-1/2}}=\sqrt{2}## ( better check this !) or ##1/\sqrt{2}## if ##\beta## is -1/2

I think if ##\beta = \frac{1}{2}## it's not ##\sqrt{2}##, but ##\sqrt{\frac{1.5}{0.5}} = \sqrt{3}##
And if ##\beta = -\frac{1}{2}##, it is ##\sqrt{\frac{0.5}{1.5}} = \sqrt{\frac{1}{3}}##. Are you sure?

 

[Mentz114]

I think if ##\beta = \frac{1}{2}## it's not ##\sqrt{2}##, but ##\sqrt{\frac{1.5}{0.5}} = \sqrt{3}##
And if ##\beta = -\frac{1}{2}##, it is ##\sqrt{\frac{0.5}{1.5}} = \sqrt{\frac{1}{3}}##. Are you sure?

You're right. I have arithmetic dyslexia.

 

[Stephanus]

The relativistic Doppler effect is reciprocal and depends on the relative velocity ##\beta=v/c##. If the receiver approaches the emitter then the freqency shift is ##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. If they are separating then reverse the sign of ##\beta##.

One more thing.
A moves to west at 0.4c according to rest observer
B moves to west at 0.7c according to rest observer
##s = \frac{u+v}{1+uv}; 0.7 = \frac{0.4+v}{1+0.4v}; v = \frac{5}{12}##
What is ##\beta##?
##0.3 = \frac{18}{60}## or ##\frac{25}{60}## There's slightly different.
In short how to calculate ##\beta##?
Relative to A or relative to "rest" observer that watches A moves 0.4c from him and B moves 0.7c from him.

 

[Mentz114]

One more thing.
A moves to west at 0.4c according to rest observer
B moves to west at 0.7c according to rest observer
##s = \frac{u+v}{1+uv}; 0.7 = \frac{0.4+v}{1+0.4v}; v = \frac{5}{12}##
What is ##\beta##?
##0.3 = \frac{18}{60}## or ##\frac{25}{60}## There's slightly different.
In short how to calculate ##\beta##?
Relative to A or relative to "rest" observer that watches A moves 0.4c from him and B moves 0.7c from him.

The answer depends on whether A starts east or west of B.

But all you need to do is find the relative velocity of A wrt B as seen in A and B coords.

 

[Stephanus]

The answer depends on whether A starts east or west of B.

But all you need to do is find the relative velocity of A wrt B as seen in A and B coords.

Thanks. It's clear for me now

 

[Mentz114]

Thanks. It's clear for me now

I should have said that the relative velocity between two things is invariant. It will be the same in any coords.

 

[Stephanus]

I should have said that the relative velocity between two things is invariant. It will be the same in any coords.

What about this:
B in east of A
A moves to east at 0.4c wrt Rest Observer (R)
B moves to east at 0.3c wrt A
R will see that A moves at (of course) 0.4c and
But, R will see that B moves at ##\frac{0.4+0.3}{1+0.12} = 0.625c##
Even tough we still calculate ##\beta## from "B moves to east wrt A NOT wrt R" right?

 

[vela]

One more thing.
A moves to west at 0.4c according to rest observer
B moves to west at 0.7c according to rest observer
##s = \frac{u+v}{1+uv}; 0.7 = \frac{0.4+v}{1+0.4v}; v = \frac{5}{12}##
What is ##\beta##?
##0.3 = \frac{18}{60}## or ##\frac{25}{60}## There's slightly different.
In short how to calculate ##\beta##?
Relative to A or relative to "rest" observer that watches A moves 0.4c from him and B moves 0.7c from him.

In the relativistic Doppler shift formula
$$f_o = \sqrt{\frac{1-\beta}{1+\beta}}f_s,$$ how is ##\beta## defined? Look that up, and you'll have the answer to your question.

 

[Mentz114]

What about this:
B in east of A
A moves to east at 0.4c wrt Rest Observer (R)
B moves to east at 0.3c wrt A
R will see that A moves at (of course) 0.4c and
But, R will see that B moves at ##\frac{0.4+0.3}{1+0.12} = 0.625c##
Even tough we still calculate ##\beta## from "B moves to east wrt A NOT wrt R" right?

What is your point ?

If two inertially moving objects have relative velocity ##\beta## then they have this in all frames. If you think you've found differently then you made a mistake.

I can't see what the problem is.

 

[vela]

I should have said that the relative velocity between two things is invariant. It will be the same in any coords.

This isn't correct. Stephanus's example clearly shows that the B's velocity relative to A as seen by A is (5/12)c whereas the observer in the rest frame would say B's velocity relative to A is (3/10)c.

 

[Mentz114]

This isn't correct. Stephanus's example clearly shows that the B's velocity relative to A as seen by A is (5/12)c whereas the observer in the rest frame would say B's velocity relative to A is (3/10)c.

OK, thank you.

What I should have said was 'The relative velocity between A and B as measured in A and B coords is invariant'.

(I did say that but then I crossed it out for some reason.)

 

[vela]

OK, thank you.

To clarify, I know what you were trying to say, and you're correct that all observers would get the same value for ##\beta## to plug into the formula. It depends on what exactly one means by "relative velocity". If you take "B's velocity relative to A" to mean B's velocity in A's rest frame, you're indeed correct. On the other hand, the phrase could also mean "how much faster B is moving compared to A" as seen in a given reference frame." Now that I think about it, perhaps your interpretation is more common.

 

[Mentz114]

To clarify, I know what you were trying to say, and you're correct that all observers would get the same value for ##\beta## to plug into the formula. It depends on what exactly one means by "relative velocity". If you take "B's velocity relative to A" to mean B's velocity in A's rest frame, you're indeed correct. On the other hand, the phrase could also mean "how much faster B is moving compared to A" as seen in a given reference frame." Now that I think about it, perhaps your interpretation is more common.

I take the definition given by Malament which is valid when the worldlines coincide and involves spatial and temporal projections (this is the first definition you give). But, as you say different interpretations are possible.

 

[PAllen]

This isn't correct. Stephanus's example clearly shows that the B's velocity relative to A as seen by A is (5/12)c whereas the observer in the rest frame would say B's velocity relative to A is (3/10)c.

No. You are confusing separation speed versus relative speed. Separation speed is coordinate dependent, relative speed is invariant. It is defined from the dot product of two 4 velocities, which is obviously invariant. This dot product gives gamma for the relative speed. This is related to all the idiocy spouted about superluminal relative speeds. These are invariably separation speeds not relative speeds. Separation speeds have no upper limit, even in flat spacetime. See the Milne model for an example.

 

[Stephanus]

What is your point ?

If two inertially moving objects have relative velocity ##\beta## then they have this in all frames. If you think you've found differently then you made a mistake.

I can't see what the problem is.

No, not a problem now. What matters are A speed and B speed. It's between A and B. What the other observers (either rest or moving) don't matter. It's A who receives B signal not the other observer. It's B speed relative to A that matters. I should have pondered five more minutes before asked.

 

[Stephanus]

This isn't correct. Stephanus's example clearly shows that the B's velocity relative to A as seen by A is (5/12)c whereas the observer in the rest frame would say B's velocity relative to A is (3/10)c.

But I realize now that ##\beta = \frac{5}{12}## not ##0.3##

 

[harrylin]

Dear PF Forum,
I wonder about this Doppler formula.
https://en.wikipedia.org/wiki/Doppler_effect#General
##f=(\frac{c+V_r}{c+V_s})f_0##
The speed of sound is 343m/s
Supposed S (Source) moves to the east, toward R(Receiver) at 70 m/s and R moves to the east at 40 m/s. So the formula is:
##f=(\frac{343-40}{343-70})f_0##
But R will say. No, I'm at rest, it's S who moves toward me at 70-40 = 30 m/s. So the doppler factor would be:
##f=(\frac{343}{343-30})f_0##
No, we can't do that. ##\frac{343-40}{343-70} \neq \frac{343}{343-30}##. [..]

Yes you can do that BUT you must account for time dilation. This was discussed recently by me and Janus in a thread that you may have missed:
#6 and #7 as well as #13

 

[Stephanus]

Yes you can do that BUT you must account for time dilation. This was discussed recently by me and Janus in a thread that you may have missed:
#6 and #7 as well as #13

Miss?? It's likely "Don't fully understand". That's why, now I'm concentrating with Doppler, instead of Measuring distance.
And once I saw Mentz114 answer.

The relativistic Doppler effect is reciprocal and depends on the relative velocity ##\beta=v/c##. If the receiver approaches the emitter then the freqency shift is ##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. If they are separating then reverse the sign of ##\beta##.

I remember the ##\sqrt{\frac{1+\beta}{1-\beta}}##
[Edit:] Wow, that's not my thread.
https://www.physicsforums.com/threads/measuring-distance-speed-and-clock.821115/

 

[Mentz114]

No, not a problem now. What matters are A speed and B speed. It's between A and B. What the other observers (either rest or moving) don't matter. It's A who receives B signal not the other observer. It's B speed relative to A that matters. I should have pondered five more minutes before asked.

Yes, I was going to point that out. Obviously only the relative velocity as measured by the receiver and emitter are relevant.

In fact, measuring the Doppler shift is measuring the relative velocity indirectly.

 

[Stephanus]

Thanks for all the answers.
But I need to ask something here regarding
A: ##f = \sqrt{\frac{1+v}{1-v}}f0## and
B: ##f = \frac{1+v}{1-v}f0##. No square root here.
I copy this from my previous thread.

V = 0.6c. Above is the ST diagram as we see on our computer screen.
This is what B see

Events  Time (A)    Diff     Freq    Read (B)

E1           700     ..          1      -1100
E2           900     200    1 -> 4       -900
E3           950      50         4       -700
E4          1125      …                     0
E5          1175      50         4        200

Time(A) is the time when B receives the bouncing signal.
Read (B) is the data that is read by B when B sent the signal.
According to this, the frequency different conforms equation B. ##f = \frac{1+V}{1-V}##
Supposed B sends a red light to G at 400THz, what will G see? Violet at ##\sqrt{4} * 400THz###?
What about sound?
Supposed B moves at 0.6mach, what will G hear if B sends 440hz (Middle A), 880hz (A1)?
So I need confirmation, please
A: ##f = \sqrt{\frac{1+v}{1-v}}f0## for signal frequency?
B: ##f = \frac{1+v}{1-v}f0## for RECEIVING frequency?
C: What equation for sound signal frequency, A or B?
Thanks

 

[Mentz114]

Changed my mind. No reply.

I can't find a delete option for posts

 

[PAllen]

For sound (or light if there were a transmission medium with associated rest frame, analogous to air; that is one of the pre-relativistic models), you have two velocities to worry about: relative velocity of emitter and receiver, and velocity of medium relative to either emitter or receiver (you only need to specify one of these, obviously, for a complete problem description). I will use c for signal speed through medium, whether sound or light per old theories. I will ignore any relativistic effects for this first discussion. I will ignore the general case, and describe two important special cases:

1) Emitter moving through medium directly away from receiver, receiver stationary in medium. Then, if the emitter emits at frequency f, the receiver will receive at f' given
by: f' / f = c / (c+v)

2) Receiver, moving through medium directly away from emitter, emitter stationary in medium. If emitter emits at f, receiver will receive at f' given by:
f' / f = (c-v)/c. Note that in this case, v=c is a limiting case because at v=c, f'=0, because the signal can never catch the receiver. Obviously, v > c has the same issue.
There is no such limit for a moving emitter.

3) A 'mirror': this is just a moving receiver, then moving emitter. The effects compound, giving:

f'/f = (c-v)/(c+v)

Looked at from this starting point, relativity is different in just the following two ways:

a) The special case of emitter or receiver stationary with respect to medium becomes the general case - all frames behave the same say.
b) There is a time dilation correction in addition to the finite signal speed of the classical formulas.

1) Emitter's frequency must be divided by gamma to make it correspond to receiver's clocks, before the classical effects apply. This leads to :

√ ((c-v)/(c+v) after algebra.

2) We expect from relativity that this scenario will be symmetric. Indeed, adjusting this case to account for receiver clocks slow by gamma (f' as computed per emitter needs to be replace by f' as actually measured by receiver divided by gamma). This leads to the same formula, as expected, after algebra.

3) For the mirror, it remains true that you simply have compounding of (1) and (2) [the mirror can be modeled as a receiver that then immediately emits], leading to the same formula as classically. This is not a coincidence, because you can argue, correctly, that we don't need to worry about time dilation for the mirror because we don't care about clocks moving with the mirror. We only care about the complete description from the rest frame of emitter/receiver, with mirror moving. Then the pair of classical special cases accurately describes the general case.

[edit: To be pedantic, I should note that the pre-relativistic formulas given first, when used for sound, theoretically still need a correction for gamma. However, for v similar to the speed of sound, this correction is less than 1 part per trillion, so the formulas as given first may be used for any practical purpose for sound.]

 

[Stephanus]

For sound (or light if there were a transmission medium with associated rest frame, analogous to air; that is one of the pre-relativistic models), you have two velocities to worry about: relative velocity of emitter and receiver, and velocity of medium relative to either emitter or receiver (you only need to specify one of these, obviously, for a complete problem description). I will use c for signal speed through medium, whether sound or light per old theories. I will ignore any relativistic effects for this first discussion. I will ignore the general case, and describe two important special cases:

Yes, very good logic. You explained it to me very clear.

1) Emitter's frequency must be divided by gamma to make it correspond to receiver's clocks, before the classical effects apply. This leads to :

√ ((c-v)/(c+v) after algebra.

[edit: To be pedantic, I should note that the pre-relativistic formulas given first, when used for sound, theoretically still need a correction for gamma. However, for v similar to the speed of sound, this correction is less than 1 part per trillion, so the formulas as given first may be used for any practical purpose for sound.]

This once just popped up. When I type.
1 per triilion, not million? 340/s as opposed to 300 million / sec.

 

[Stephanus]

Oh, ##\frac{V_{sound}^2}{V_{light}^2}##. Okay 1 per trillion.

 

[Mentz114]

But I need to ask something here regarding
A: ##f = \sqrt{\frac{1+v}{1-v}}f0##
''
''

This Doppler stuff is a bit tricky so I've attached some of my old notes (tidied up) where I worked out the Doppler equation from a ST diagram. It turns out to be just algebra and a bit of simple geometry. This calculation has not been checked but it works out so neatly I'm pretty confident it's right.

 

[PAllen]

This Doppler stuff is a bit tricky so I've attached some of my old notes (tidied up) where I worked out the Doppler equation from a ST diagram. It turns out to be just algebra and a bit of simple geometry. This calculation has not been checked but it works out so neatly I'm pretty confident it's right.

That's correct. You use period instead of frequency, but it is all correct.

 

[Stephanus]

That's correct. You use period instead of frequency, but it is all correct.

Ahh, "PERIOD", I didn't know the term. I usually use "the frequency of receiving signal". So it's period or interval. Thanks.
And many thanks to you Mentz114.

 

[Stephanus]

This Doppler stuff is a bit tricky so I've attached some of my old notes (tidied up) where I worked out the Doppler equation from a ST diagram. It turns out to be just algebra and a bit of simple geometry. This calculation has not been checked but it works out so neatly I'm pretty confident it's right.

First, before I study it. I want to ask something.

I try to continue your worldline in the past (blue line) it touches the bottom of x axis, not at coordinate (0,0) as indicated by the red line. I add the line myself.
##{\tau_E}^2 = (t2-t1)^2 - (x2 - x 1)^2 (1)
= (t2 - t1)^2 - (vt2 - vt1)^2 (2)##
So, x2 = vt2 and x1 = vt1.
Shouldn't the wordline originate from (0,0)? It's not that I like to criticize, I'm afraid I miss something here. But if you just draw it not to scale, but to make better understanding, then, it's all right with me.

 

[Mentz114]

First, before I study it. I want to ask something.
View attachment 86485

I try to continue your worldline in the past (blue line) it touches the bottom of x axis, not at coordinate (0,0) as indicated by the red line. I add the line myself.
##{\tau_E}^2 = (t2-t1)^2 - (x2 - x 1)^2 (1)
= (t2 - t1)^2 - (vt2 - vt1)^2 (2)##
So, x2 = vt2 and x1 = vt1.
Shouldn't the wordline originate from (0,0)? It's not that I like to criticize, I'm afraid I miss something here. But if you just draw it not to scale, but to make better understanding, then, it's all right with me.

The worldline (W) can be anywhere. I did not mark it but the bottom left corner is (0,0) as you deduced.

The calculation does not depend on the values of ##x_1## nor ##x_2## as long as they are not equal.

The calculation is correct so you have misunderstood something.