# B Help with understanding the Doppler effect

#### frostysh

The wind allows you to detect motion with respect to a medium. It doesn't make the rest frame of the medium absolute - it's just the velocity with respect to the medium.
No, the wind is the inertial frame of reference, we can 'jump' between inertial frames, and somekind of a smarty thing called symmetry of Galelean transformation will say that measuring velocity means the same in the all intertial frames.
Doppler's effect is a very different case due to the invariancy of the wavelength and the fact that simple changling of the velocity of the intertial frame of the source of waves will change this invariant — the change of the invariant will be invariant in the all inertial frames of references. At least me understanding it on this level.

#### A.T.

Science Advisor
if medium exist then Doppler's effect can define motion in this medium
That's called 'relative to the medium'.

and this is me calling 'an absolute'.
You calling relative movement 'absolute' doesn't mean it contradicts relativity.

#### Ibix

Science Advisor
Doppler's effect is a very different case
It really isn't. You just need to keep track of the velocities of the source, medium, and receiver in whatever frame you want to work in.

#### Dale

Mentor
I am not understanding of what you saying, but we can measure the wavelength in case of rest observer, and if we know it (and know it because we know the very feature of the medium $\large c$), and the achieving of the speed of the source of the wave is becoming trivial, there not only mathematics, there is physics. Sec I will post the prove which is I not understand but I googled, that Doppler's effect is an invariant.
So, if you look at the link I sent you will see that the general Doppler effect is given by $$\frac{f_a}{f_e}=\frac{1-v_a/c_s}{1+v_e/c_s}\sqrt{\frac{1-(v_e/c)^2}{1-(v_a/c)^2}}$$ where I am using the notation in that paper except for the substitution $\nu=f$ which I use because $\nu$ is visually hard to distinguish from $v$ for me in most fonts, particularly the one in that paper.

If we make the simplifying assumptions $v_a<c_s<<c$ and $v_e<c_s<<c$ then we can solve the simplified formula for $v_e$, giving $$v_e=\frac{f_e}{f_a}(c_s-v_a)-c_s$$ so assuming that $f_e$ and $c_s$ are known a priori then measuring $f_a$ still does not give enough information to determine $v_e$, we also need to know $v_a$ independently somehow.

If you don't make the simplifying assumptions then the same reasoning applies, but I think that there is not a closed form solution and you have to numerically solve it.

#### frostysh

That's called 'relative to the medium'.[/QUTE]Medium in abstraction (as me have pointed) means 'the any medium', no matter which. The only point is the existing of the medium in classical kinematics (of course all is inertial and lineary, no any acceleration, weird stuff etc).
You calling relative movement 'absolute' doesn't mean it contradicts relativity.
As the physical concept, of course not! Doppler's effect exist and it is reality, relativity exist, and this is reality too, but they cannot coexist in the same time in the same model, at least in term of inertial frame of reference, because can pravail some concrete frame corresponding to which with help of Doppler's effect we will be measuring motion.

#### Dale

Mentor
The abstraction there is next: if medium exist then Doppler's effect can define motion in this medium regardles of the inertial frames, and this is me calling 'an absolute'.
I mentioned above the negative connotation of the term absolute above. What you are describing as "absolute" is not the meaning generally ascribed to that term by the rest of the community. By using that term where it does not apply you are being deliberately offensive: going out of your way to use a provocative term that is neither appropriate nor necessary.

#### frostysh

It really isn't. You just need to keep track of the velocities of the source, medium, and receiver in whatever frame you want to work in.
No I don't, I simply measure the difference in wavelengths in the rest case of the source and in my case, and the end, I know what is moving what is not — this difference is an invariant. This invariant will be the same in the any other inertial systems.
So, if you look at the link I sent you will see that the general Doppler effect is given by $$\frac{f_a}{f_e}=\frac{1-v_a/c_s}{1+v_e/c_s}\sqrt{\frac{1-(v_e/c)^2}{1-(v_a/c)^2}}$$ where I am using the notation in that paper except for the substitution $\nu=f$ which I use because $\nu$ is visually hard to distinguish from $v$ for me in most fonts, particularly the one in that paper.

If we make the simplifying assumptions $v_a<c_s<<c$ and $v_e<c_s<<c$ then we can solve the simplified formula for $v_e$, giving $$v_e=\frac{f_e}{f_a}(c_s-v_a)-c_s$$ so assuming that $f_e$ and $c_s$ are known a priori then measuring $f_a$ still does not give enough information to determine $v_e$, we also need to know $v_a$ independently somehow.

If you don't make the simplifying assumptions then the same reasoning applies, but I think that there is not a closed form solution and you have to numerically solve it.
Cool formulas, but the first one I have very poorly barely understand, the second one no giving any physical explanations, if looking just on mathematics of this formula we will not understand that corresponding to the interiatl frames of observers the wavelength is an invariant. Of course in terms of mathematics and unknown terms this equation no different from the other in case of solving it, if too many unknows we cannot resolve it.

And by the way, usually peoples describing what symbols have what meanings in formulas, because can be missunderstandings... What is means of $c_{s}, f_{e}, v_{a}, f_{a}$???
I mentioned above the negative connotation of the term absolute above. What you are describing as "absolute" is not the meaning generally ascribed to that term by the rest of the community. By using that term where it does not apply you are being deliberately offensive: going out of your way to use a provocative term that is neither appropriate nor necessary.
How do you to name the effect that can be used to destroy the equallity of inertial frames of references?

#### Ibix

Science Advisor
This invariant will be the same in the any other inertial systems.
Just like the wind speed your observer would measure. Everyone will agree his measurement, and everyone will agree that it is his speed relative to the medium. That doesn't make the rest frame of the medium special.
How do you to name the effcet that can be used to destroy the equallity of inertial frames of references?
A misunderstanding of what you are doing.

#### frostysh

Just like the wind speed your observer would measure. Everyone will agree his measurement, and everyone will agree that it is his speed relative to the medium. That doesn't make the rest frame of the medium special.
A misunderstanding of what you are doing.
The problem is that wavelength is not relative to the frame of reference like speed of the wind... (Cannot find facepalm smile.) The invariant is something that is not changing due to change of intertial frame. The speed of wind will change, this is not an invariant.
The speed of wave propagation in medium is an invariant to any inertial frame too, because it's a very property of medium itself. What about your wind?

#### A.T.

Science Advisor
Doppler's effect exist and it is reality, relativity exist, and this is reality too, but they cannot coexist in the same time in the same model,
Sure they can. And your word games don't change that.

#### A.T.

Science Advisor
The speed of wave propagation in medium is an invariant to any inertial frame
"In medium" already defines the frame relative to which the speed is measured in, so adding 'to any inertial frame' makes no sense, and is just gibberish.

#### frostysh

Sure they can. And your word games don't change that.
This is more philosophical question but as you saw, the Doppler's effect is invariant and clearly showing which frame moving, which not. So in this terms 'rest' and 'moving' is not relative, if you can make such an abstraction, you can and model think as a whole Universe is filled by some medium — the end of relativity then in this model, but of course it is not true, the vacuum is not a medium (obviously) and no any 'ether' observed during hundred of years so the relativity rulezz.

#### Ibix

Science Advisor
The problem is that wavelength is not relative to the frame of reference like speed of the wind
However, both measured frequency and wave speed (relative to the reference frame, not the medium) are frame dependent. Thus there is no problem with changing inertial frames - as long as you remember to transform all relevant physical quantities.
The speed of wave propagation in medium is an invariant to any inertial frame too, because it's a very property of medium itself.
The speed relative to the medium is invariant, sure. The speed relative to the frame isn't.
What about your wind?
The speed of the wind relative to itself is zero always - an invariant just like the speed of sound relative to the medium. The speed of the wind relative to the frame is frame dependent, of course, just like the speed of sound relative to the frame.

#### A.T.

Science Advisor
Doppler's effect is invariant and clearly showing which frame moving, ...
... moving relative to the medium.

#### Dale

Mentor
Cool formulas, but the first one I have very poorly barely understand, the second one no giving any physical explanations, if looking just on mathematics of this formula we will not understand that corresponding to the interiatl frames of observers the wavelength is an invariant.
The wavelength is not an invariant, and in any case if $v_a\ne 0$ then I am not sure how the absorber can measure the wavelength. What formula are you using for that?

Of course in terms of mathematics and unknown terms this equation no different from the other in case of solving it, if too many unknows we cannot resolve it.
Yes, that is the point made by @A.T. and myself earlier in the thread.

And by the way, usually peoples describing what symbols have what meanings in formulas, because can be missunderstandings... What is means of $c_{s}, f_{e}, v_{a}, f_{a}$???
The meanings of the symbols is described in the paper that I linked to as I explicitly stated "where I am using the notation in that paper" with one exception which I described above.

How do you to name the effect that can be used to destroy the equallity of inertial frames of references?
It doesn't matter because this is not such an effect. By trying to use that name where it does not apply you are being deliberately and inappropriately provocative.

#### frostysh

However, both measured frequency and wave speed (relative to the reference frame, not the medium) are frame dependent. Thus there is no problem with changing inertial frames - as long as you remember to transform all relevant physical quantities.

The speed relative to the medium is invariant, sure. The speed relative to the frame isn't.

The speed of the wind relative to itself is zero always - an invariant just like the speed of sound relative to the medium. The speed of the wind relativet to the frame is frame dependent, of course, just like the speed of sound relative to the frame.
The speed of wind is not connected to the wavelength of wind, and wavelength is and invariant. Do you understand that wavelength is the same as mass, as volume, I don't know, as the time interval in the classical mechanics? At least me understand this (if I am wrong, correct me). So your wind is useless analogy...
... moving relative to the medium.
Omg... And to which your medium is relative? To observer inertial frame? To source? — There is Galelean transformation, the end of story. Speed of source relative to the medium (not the speed of medium relative to the source) changing the very invariant, what do you not understand?
The wavelength is not an invariant, and in any case if $v_a\ne 0$ then I am not sure how the absorber can measure the wavelength. What formula are you using for that?
Very please, read the page two few sentences — Galilean Transformation of Wave Velocity. Do you anderstand which is means 'invariant'? This is a very fundamental thing (I mean Galelean transformation), you can scribe your formula but it's meneangles because the basis of your formulas is that very transformation, actually in the article case only of the moving observe, but it's no matter.
The meanings of the symbols is described in the paper that I linked to as I explicitly stated "where I am using the notation in that paper" with one exception which I described above.
It doesn't matter because this is not such an effect. By trying to use that name where it does not apply you are being deliberately and inappropriately provocative.
Okay, not problem.

#### Ibix

Science Advisor
The speed of wind is not connected to the wavelength of wind
Wind does not have a wavelength.
So your wind is useless analogy
I think you don't understand my point. Measuring the speed of the wind simply tells you your speed relative to the medium, whether you regard yourself or the medium or both as moving. That measurement is invariant - everyone will agree it. That does not make your speed invariant, merely your measurement of speed relative to the medium.

Sure, the wavelength is frame invariant. However, as I said, neither the frequency nor wavespeed is frame invariant. Thus your measure of frequency simply gives you your speed relative to the medium (assuming you know the speed of sound relative to the medium in order to subtract it out).

Do you understand that wave speed is frame variant? Because that seems to me to be the bit that you keep missing.

#### Dale

Mentor
Very please, read the page two few sentences — Galilean Transformation of Wave Velocity. Do you anderstand which is means 'invariant'?
As I already said, physics is described by the Lorentz transform, not the Galilean transform. Wavelength is not an invariant. This has been well known now for more than 100 years. Equation 1 in that paper has been known to be incomplete for that same period.

#### frostysh

Wind does not have a wavelength.
And that is the reason!
I think you don't understand my point.
I just distracted by the nonsense which is typing the other users, or at least I think it is nonsense... So maybe.
Measuring the speed of the wind simply tells you your speed relative to the medium, whether you regard yourself or the medium or both as moving. That measurement is invariant - everyone will agree it. That does not make your speed invariant, merely your measurement of speed relative to the medium.
The problem the length is not the speed, the wavelength.
Sure, the wavelength is frame invariant. However, as I said, neither the frequency nor wavespeed is frame invariant.
Okay, let's measure the wavespeed, in my frist post it's $\large \left(c + v \right)$.
Thus your measure of frequency simply gives you your speed relative to the medium (assuming you know the speed of sound relative to the medium in order to subtract it out).
Of course.
Do you understand that wave speed is frame variant? Because that seems to me to be the bit that you keep missing.
Yes, that thing is an invariant (I have speciall y not named it by a single symbol), but if this thing $\large \left(c + v \right)$ related to the movement of the source of the waves corresponding to the medium (not vice versa) this very relative speed which you called 'wave speed' producing a very not relative in the inertial frames wavelength, by measuring this wavelength, and knowing the wavelength of the rest case (the very property of the medium), we can distinguish where is the movement, in observer or in the source. This is logic of mine, simple as that.

Last edited:

#### Ibix

Science Advisor
we can distinguish where is the movement, in observer or in the source
No - you can distinguish where there is movement relative to the medium.

#### frostysh

As I already said, physics is described by the Lorentz transform, not the Galilean transform. Wavelength is not an invariant. This has been well known now for more than 100 years.
You are literaly cannot read the name of topic, which is called 'classical Doppler effect', the classical mechanics is a complited self-defined theory, Galelean transformation is no need any Lorentz's transformation to describe the the classical casess. So in the name of WHAT you continuously trying to connect Special Relativity there, it is an enigma for myself, with the all respect...

#### frostysh

No - you can distinguish where there is movement relative to the medium.
And your medium relative to what? Oh my god... We have closed, defined system: medium, source, observer. On this system we using Galelean transformation, the end. You can imagine it abstractly — expand the medium to the size of the Universe!

#### Dale

Mentor
You are literaly cannot read the name of topic, which is called 'classical Doppler effect', the classical mechanics is a complited self-defined theory, Galelean transformation is no need any Lorentz's transformation to describe the the classical casess. So in the name of WHAT you continuously trying to connect Special Relativity there, it is an enigma for myself, with the all respect...
Well, you say "classical" in the title, but then in the text you use words like "absolute", "event", "invariant", "frames", and "observer" throughout all of your posts, and you never specify $v<c_s<<c$. All of which essentially require answers to be in the context of relativity if they are to be physically correct answers. I don't know how it is possible to address a question in those terms without relativity.

#### Ibix

Science Advisor
And your medium relative to what
You can detect the motion of the source relative to the medium or the observer relative to the medium. Or you can see it as the motion of the medium relative to the source or the medium relative to the observer.

Get a source and a receiver and set them in motion relative to the medium and record whatever you want to record - wavelengths, frequencies, whatever. Now do the same experiment (same velocities with respect to the medium and everything) in a moving train. You will get the same results, even though the experiment is doing 60mph with respect to the other one.

There is nothing absolute about this experiment. It is only measuring speeds relative to a medium, which may be in motion with respect to other parts of the medium.

#### Dale

Mentor
On another note, if you can measure the wavelength and the frequency and if $c_s$ is known then I think you can make two equations in two unknowns and solve for both the emitter and the absorber’s speed.

My analysis above presumed only measuring the frequency, since that is the usual meaning of the Doppler shift.

### Want to reply to this thread?

"Help with understanding the Doppler effect"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving