I need help with Doppler's effect understanding

[frostysh]

First to say, how ridiculous that on scientific forum I am no see any things that can help me to type formulas... But instead of it a lot smiles.  It is a very fun, but what about Latex or something? My second is connected on the first, I will try to formulate my questions in laconic way and without formulas: is wavelength changing when source of waves moving related to medium?

The point is my 'plank-core' one-dimensional model (the ballistic one) is seems to say "yes it is", the wavelength changing, I have attached files that contain the picture that illustrating the rest and moving case of source, related to the medium. But the very Galilean transformation of the wave-function is saying "not it is not", the wavelength is no changing. Where is the truth? When answering considering that me is no good mathematical level, I was in past student and for now trying to refresh the knowledge and find a jog.

Thanx for the future answers! Of course if it will appear...

Edited July 31, 2019 by frostysh

[Strange]

3 minutes ago, frostysh said:

First to say, how ridiculous that on scientific forum I am no see any things that can help me to type formulas... But instead of it a lot smiles.  It is a very fun, but what about Latex or something?

https://www.scienceforums.net/topic/108127-typesetting-equations-with-latex-updated/

 

[frostysh]

8 minutes ago, Strange said:

https://www.scienceforums.net/topic/108127-typesetting-equations-with-latex-updated/

 

Thanx, I will read and try to post more formulas. There is a way to include the pictures inside the text but not just to attach them on the bottom of post?

[Strange]

29 minutes ago, frostysh said:

My second is connected on the first, I will try to formulate my questions in laconic way and without formulas: is wavelength changing when source of waves moving related to medium?

Note that there are three effects, one of which is the speed of the source relative to the medium. Summarised nicely here:

Quote

For waves that propagate in a medium, such as sound waves, the velocity of the observer and of the source are relative to the medium in which the waves are transmitted.^[1] The total Doppler effect may therefore result from motion of the source, motion of the observer, or motion of the medium. Each of these effects is analyzed separately. For waves which do not require a medium, such as light or gravity in general relativity, only the relative difference in velocity between the observer and the source needs to be considered.

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

[swansont]

1 hour ago, frostysh said:

First to say, how ridiculous that on scientific forum I am no see any things that can help me to type formulas... But instead of it a lot smiles.  It is a very fun, but what about Latex or something? My second is connected on the first, I will try to formulate my questions in laconic way and without formulas: is wavelength changing when source of waves moving related to medium?

The point is my 'plank-core' one-dimensional model (the ballistic one) is seems to say "yes it is", the wavelength changing, I have attached files that contain the picture that illustrating the rest and moving case of source, related to the medium. But the very Galilean transformation of the wave-function is saying "not it is not", the wavelength is no changing. Where is the truth? When answering considering that me is no good mathematical level, I was in past student and for now trying to refresh the knowledge and find a jog.

Thanx for the future answers! Of course if it will appear...

I don't understand why, in your first drawing, the wave is skewed. i.e. the magnitude of the slope at points 3 and 5, for example, are not equal. But in the second drawing they are.

 

[frostysh]

2 hours ago, Strange said:

Note that there are three effects, one of which is the speed of the source relative to the medium. Summarised nicely here:

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

Yes! Thank you. But what about the wavelength in the first case, the motion of source? I think movement of medium related to observer and source in the same frame will be equal of movement of source in one-dimensional case at least. If wavelength (which is invariant in any inertial system) is depends on relative speed of source related to medium, than we have somekind of event that can distinguish this, very particular inertial frame from the others, so it's looks like Doppler effect is an absolute of somekind.

Yes, I have read Wikipedia somehow me poor around a week is trying to understand this effect and don't read Wiki about...  So the question is little bit reorganized: if the wavelength is depends on speed, is inertial frame that connected to moving source can be assumed as privileged due to it's connection to invariant wavelength?

1 hour ago, swansont said:

I don't understand why, in your first drawing, the wave is skewed. i.e. the magnitude of the slope at points 3 and 5, for example, are not equal. But in the second drawing they are.

 

This is because I didn't know how make formulas on this forum when I posted it, there is no matter. I draw it in GIMP. It's not a wave but an abstraction of one-dimensional wave which the mathematical description like that \[ \large A \left(x, t \right) = A_{max} \sin{2 \pi \left(\dfrac{x}{\lambda} - \nu t \right)} \] where \( \large x \) — is a displacement, can be larger or smaller than zero, \( \large \lambda \) — the wavelength and \( \large \nu \) — is a frequency. There is no wave pictured, the projection of that 'cores' on absciss axis (it is a distance line between source and observer) is a traveling in space values of property such as density.

 

P. S. No working scaling formulas such as \large, \Large, etc... 

Edited July 31, 2019 by frostysh
I have read Wikipedia!

[swansont]

31 minutes ago, frostysh said:

 

This is because I didn't know how make formulas on this forum when I posted it, there is no matter. I draw it in GIMP. It's not a wave but an abstraction of one-dimensional wave which the mathematical description like that

It looks like a wave ( a sine wave), and it's incorrect.

 

Quote

A(x,t)=Amaxsin2π(xλ−νt)

where x — is a displacement, can be larger or smaller than zero, λ — the wavelength and ν — is a frequency. There is no wave pictured, the projection of that 'cores' on absciss axis (it is a distance line between source and observer) is a traveling in space values of property such as density.

Sorry, not understanding what you mean here. If it has a wavelength and a frequency it's a wave.

If there is relative motion between source and observer, the peaks arrive faster (if moving toward each other) so the frequency goes up, but the distance between the peaks — the wavelength — is the same. Frequency goes down if they are moving apart.

[frostysh]

13 minutes ago, swansont said:

It looks like a wave ( a sine wave), and it's incorrect.

 

Sorry, not understanding what you mean here. If it has a wavelength and a frequency it's a wave.

If there is relative motion between source and observer, the peaks arrive faster (if moving toward each other) so the frequency goes up, but the distance between the peaks — the wavelength — is the same. Frequency goes down if they are moving apart.

An abstraction means we are taking into account only very general properties or somekind — there is no matter the shape of the curve on the picture. It is one-dimensional wave, the picture is two-dimensional abstract model in which we are imagine the values of density of medium for an example as the shooting cores, when are too many core it will be a line. The curve is skew because me little bit fast trying to draw it, but as I said it's no matter. And why in your quote of my formula the displacement and wavelength is multiplied? The argument of sine-function must be dimensionless, it's so called transcendential funcion (OMG! I am not forget the whole university study which was long ago...), your formula that you quote have no physical meaning at all.

You are particularly wrong about Doppler's effect, wavelength is changing when source is moving related to medium. And I think this process like an event, invariant, so it's breaking Galilean relativity.

Edited July 31, 2019 by frostysh

[Bufofrog]

1 hour ago, frostysh said:

wavelength is changing when source is moving related to medium. And I think this process like an event, invariant, so it's breaking Galilean relativity

Why do you think this breaks Galilean relativity?

[swansont]

1 hour ago, frostysh said:

An abstraction means we are taking into account only very general properties or somekind — there is no matter the shape of the curve on the picture. It is one-dimensional wave, the picture is two-dimensional abstract model in which we are imagine the values of density of medium for an example as the shooting cores, when are too many core it will be a line.

I don't know what you mean by core.

1 hour ago, frostysh said:

 

The curve is skew because me little bit fast trying to draw it, but as I said it's no matter. And why in your quote of my formula the displacement and wavelength is multiplied?

Probably a rendering error of the code. I didn't make any changes and the equation was not important in my response.

1 hour ago, frostysh said:

 You are particularly wrong about Doppler's effect, wavelength is changing when source is moving related to medium.

Yes. I was thinking in terms of light (no medium), not in general. 

1 hour ago, frostysh said:

And I think this process like an event, invariant, so it's breaking Galilean relativity.

Not sure why you would think that.

[dimreepr]

18 minutes ago, Bufofrog said:

Why do you think this breaks Galilean relativity?

I think that might be the wrong answer, and, strangely the wrong question...

[Mordred]

Well in one of the posts above it was mentioned that wavelength was invarient. It isn't how one measures wavelength is observer dependent and will vary between observers.

 Normal everyday Doppler that doesn't include time dilation effects will follow the Galilean transformation rules, however once relativity effects become significant then one must use the relativistic Doppler formulas to account for this.

The transformation rules for the latter case being the Lorentz transformations of SR.

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

For example the speed of sound is non relativistic in Earth's atmosphere so the normal Doppler works fine.

Edited July 31, 2019 by Mordred

[frostysh]

14 hours ago, Bufofrog said:

Why do you think this breaks Galilean relativity?

It's a good question, of course the Galilean transformation (so-called Galilean group in Group Theory) is far more general, and the 'break' is not the word. I mean some inertial systems which connected to the source will be privileged if medium exist, because we can connect the very relative speed related to medium with a very non relative and the very invariant wavelength.

14 hours ago, swansont said:

I don't know what you mean by core.

Probably a rendering error of the code. I didn't make any changes and the equation was not important in my response.

Yes. I was thinking in terms of light (no medium), not in general. 

Not sure why you would think that.

Cores is a modeling object in ballistic model of one-dimensional wave, it's mean a particular value of wave process which is travelling trough space (trough a line in one dimensional case), I don't know how to explain it to you more good, just google 'one-dimensional wave' or take any book about waves. I don't know what kind of error and code you talking about either, I just drawn a picture in GIMP, first one and the second one, no any 'errors' or you mean your formulas in post? I don't know why it is so, actually the formulas typing on this forum is a pretty awful to me at least...  I have said clearly "we have a medium", in case of light there is entirely different physical meaning because we cannot connect inertial frame with a vacuum, it's obviously.

14 hours ago, dimreepr said:

I think that might be the wrong answer, and, strangely the wrong question...

Well, I am trying to understand it actually.

10 hours ago, Mordred said:

Well in one of the posts above it was mentioned that wavelength was invarient. It isn't how one measures wavelength is observer dependent and will vary between observers.

 Normal everyday Doppler that doesn't include time dilation effects will follow the Galilean transformation rules, however once relativity effects become significant then one must use the relativistic Doppler formulas to account for this.

The transformation rules for the latter case being the Lorentz transformations of SR.

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

For example the speed of sound is non relativistic in Earth's atmosphere so the normal Doppler works fine.

Wavelength is NOT observer dependent, wavelength is an invariant, it's only source-medium dependent (that why I think Galilean relativity 'break'). You can realize it by yourself if you will do Galilean transformation of one-dimensional wave, it's the case like a source of waves disappeared and left only the medium and the very wave itself. I have done this transformation, and I will post it there, but after me will be familiar how to type formulas on this forum to avoid this errors and stuff. 

Edited August 1, 2019 by frostysh

[Mordred]

You have a different meaning to invariant quantities then to standard definition.

An invariant quantity means all observers agree on the value. The constant c is one example under relativity. All observers in every reference frame will measure the same value.

This isn't the case with wavelength different observers will measure different wavelength values hence the terms redshift and blueshift.

 So how can you claim wavelength is invariant under that definition ?

Take the following relations

[latex]\frac{\Delta_f}{f} = \frac{\lambda}{\lambda_o} = \frac{v}{c}=\frac{E_o}{E}=\frac{hc}{\lambda_o} \frac{\lambda}{hc}[/latex]

[latex]f=\frac{c+v_r}{c+v_s}f_o[/latex]

This is the Doppler shift formula

c=velocity of waves in a medium
Vr is the velocity measured by the source using the source’s own proper-time clock(positive if moving toward the source
vs is the velocity measured by the receiver using the source’s own proper-time clock(positive if moving away from the receiver)

The above are for velocities where the source is directly away or towards the observer and for low velocities less than relativistic velocities. A relativistic Doppler formula is required when velocity is comparable to the speed of light.

The relativistic Doppler is given by

[latex]v_{observed }=v_{source}\sqrt{\frac{1+\frac {v}{c}}{1-\frac{v}{c}}}[/latex]

This describes a variant wavelength and hence a variant frequency which also means the corresponding energy is also variant between  different observers.

In the last case c is the constant  c

By the way  the definition of invariant and variant  quantities do not change between relativity or Galilean transformations.

It is the same definition

Start with the Galilean transformations 

[latex]x=\acute {x}-vt [/latex]

[latex]y=\acute {y}[/latex]

[latex]z=\acute {z}[/latex]

[latex]t=\acute {t}[/latex]

Then apply the equations on the following

https://www.google.com/url?sa=t&source=web&rct=j&url=http://users.physics.harvard.edu/~schwartz/15cFiles/Lecture21-Doppler.pdf&ved=2ahUKEwjG8NSM--DjAhURsp4KHbGmCtkQFjAQegQICBAB&usg=AOvVaw2-wPdgX3Qsq1K7kWG6zuG0

This applies [latex]c_s [/latex] for the speed of sound. (Non relativistic Doppler it describes how wavelength varies  (variant) between observers 

 

 

 

Edited August 1, 2019 by Mordred

[Strange]

2 hours ago, frostysh said:

It's a good question, of course the Galilean transformation (so-called Galilean group in Group Theory) is far more general, and the 'break' is not the word. I mean some inertial systems which connected to the source will be privileged if medium exist, because we can connect the very relative speed related to medium with a very non relative and the very invariant wavelength.

If there is a medium you can tell if you or the source are moving relative to it. That doesn't make it a privileged frame f reference. You can measure the speed of your relative to the road, that doesn't make the road "privileged".

The medium (or the road) could be moving relative to something else (and that would not be directly detectable) so it isn't a privileged frame of reference.

[Mordred]

Also don't confuse Galilean invariance with invariant quantity.

Galilean invariance states the laws of physics are the same in all inertial frames. Though Newton believed in an absolute frame. This does not mean measured quantities are also invariant.

Forgot to add that above

Now let's apply that.

Take two observers one observer is moving. One is not. Remove all other references except the wavelength itself.

Neither observer will be able to tell who is moving or who is not. They will only be able to tell they are moving towards or away from each other. (Assume they are transmitting the same signal ) after all you must know the original frequency/wavelength to know if there is a redshift.

So mathematically 

[latex] ( x=\acute {x}-vt)=(\acute {x}=x-vt) [/latex] either x or primed x can be the moving observer. There is no distinction or change that makes one frame more priveliged than the other as the relations are symmetric.

Edited August 1, 2019 by Mordred

[Bufofrog]

8 hours ago, frostysh said:

It's a good question, of course the Galilean transformation (so-called Galilean group in Group Theory) is far more general, and the 'break' is not the word. I mean some inertial systems which connected to the source will be privileged if medium exist, because we can connect the very relative speed related to medium with a very non relative and the very invariant wavelength

The wavelength is not invariant.  There are any number of ways to determine your relative speed to another inertial frame, sound is not somehow special.  

In the title of the thread you asked for help to understand the doppler effect.  I suggest you listen to the help that has been offered.  I fear you won't however.

Edited August 1, 2019 by Bufofrog

[frostysh]

12 hours ago, Mordred said:

You have a different meaning to invariant quantities then to standard definition.

Despite I very don't like a typing formulas on this forum, I will try. We have one-dimensional wave which have wavelength, frequency, and the amplitude which can be density for an example, the mathematics says that in simplest of situation we can describe this wave in model by the next equation: \[ G \left(x, t \right) = G_{max} \sin{2 \pi \left(\dfrac{x}{\lambda} - \nu t \right)}, \] where \( G_{max} \) — the maximum value of density in a waver, \( x \) — displacement from some particular point, in can be greater or less than zero. Let's say this density will be when we are looking on wave from an observer which is stationary related to the medium. Let's say we also have another observer which is moving with constant speed \( v \) related to the medium, and this observer see the density like that \[ g \left(X, t \right) = g_{max} \sin{2 \pi \left(\dfrac{X}{\Lambda} - \Gamma t \right)}, \] where \( X, \Lambda, \Gamma  \) — displacement from the new point, new wavelength and the new frequency correspondingly. So what density will see the first observer in the point \( x \) after some time \( t \) and the second one in the point \( X = x - vt \)? Due to laws of conversation and time interval invariancy of classical mechanics this density must be the same of course, because the point on like is the same, and if we choose (we can choose point deliberately) with maximum, this is mean \( G_{max} = g_{max} \). Or will no be the any physical meaning at all. Okay, now we can make the equality, equation in this, very particular point \[ G \left( x, t \right) = g \left(x - vt, t \right)  \] \[ G_{max} \sin{2 \pi \left(\dfrac{x}{\lambda} - \nu t \right)} =  g_{max} \sin{2 \pi \left(\dfrac{\left(x - vt \right)}{\Lambda} - \Gamma t \right)}, \] of course the arguments of sinus-functions is not equal when the value of this functions equal! Bu we can chose the interval where it will be true, the point is the sinus function is so-called periodical one. And we can rewrite the very equation in the next way \[ \left(\dfrac{x}{\lambda} - \nu t \right) = \left(\dfrac{\left(x - vt \right)}{\Lambda} - \Gamma t \right) \] \[ \left(\dfrac{x}{\lambda} - \nu t \right) = \left(\dfrac{x}{\Lambda} - \left(\Gamma + \dfrac{v}{\Lambda} \right) t \right), \] let's make some reassembling \[ \left( \Gamma + \dfrac{v}{\Lambda} \right) = \Gamma', \] at least dimensional analysis saying that indeed it's some kind of frequency \[ \dfrac{1}{T} + \dfrac{\dfrac{L}{T}}{L} = \dfrac{1}{T}, \] and it's will be no other frequency than \( \nu \), because we have only two systems which making transformation and \( \Gamma' \neq \Gamma \) which is obviously, so let's rewrite our fine equation again \[ \left(\dfrac{x}{\lambda} - \nu t \right) = \left(\dfrac{x}{\Lambda} - \nu t \right), \] and, as easy to see the wavelengths is equal!

Again, the wavelength is an invariant due to Galilean transformation, it's intuitively can be understand because the wavelength have dimension of \( [L] \) and length is an invariant the same as time interval.

Edited August 1, 2019 by frostysh

[Mordred]

No the laws of physics are the same however you still measure different wavelengths between observers

That is NOT an invariant quantity. If all observers measured the same quantity you would never have redshift to begin with...

[frostysh]

10 hours ago, Strange said:

If there is a medium you can tell if you or the source are moving relative to it. That doesn't make it a privileged frame f reference. You can measure the speed of your relative to the road, that doesn't make the road "privileged".

The medium (or the road) could be moving relative to something else (and that would not be directly detectable) so it isn't a privileged frame of reference.

A good mention and try but we can imagine it from the less abstract way — assume that our model is a whole Universe, filled with this very medium (of course only theoretically, okay even hypothetically), and we somehow know the wavelength of source which is in the rest related to this medium — and the rest i connected to invariant, so it's privileged, no any other invariant with same value can appear except this will be equal invariants.

[Mordred]

There is no Eather or medium that was tested to null results via Michelson and Morley experiment. Spacetime isn't a medium

Edited August 1, 2019 by Mordred

[frostysh]

1 minute ago, Mordred said:

No the laws of physics are the same however you still measure different wavelengths between observers

That is NOT an invariant quantity. If all observers measured the same quantity you would never have redshift to begin with...

Do you see formulas? Do you can disapprove it? You can try. Redshif is term which corresponds to the light, light have no require medium in which it propagating, we cannot use the classical theory there with a good physical meaning of this very theory because we cannot connect vacuum to the inertial frame... *facepalm* So your mention about red shift is meaningless too...

[Mordred]

Doppler shift applied to any wavelength including sound waves. It is not restricted to light. I already provided a paper showing this detail.

Edited August 1, 2019 by Mordred

[frostysh]

11 minutes ago, Mordred said:

There is no Eather or medium that was tested to null results via Michelson and Morley experiment

I mean hypotetically, it's just analogy that is less abstract to the one where we will imagine this very system of source-medium-observer, and the of infinite number of the inertial systems in the Universe. Of course I can type a nonsense, so beware, because I can create a new 'physics' and even believe in it, not saw my own mistake, it's sometimes happens... 

4 hours ago, Bufofrog said:

The wavelength is not invariant.  There are any number of ways to determine your relative speed to another inertial frame, sound is not somehow special.  

In the title of the thread you asked for help to understand the doppler effect.  I suggest you listen to the help that has been offered.  I fear you won't however.

You are mistaken. You can imagine the standalone wave in medium without the source, and left the only observer and the medium, the wavelength is the length — no matter of observer speed (inertial frame) we will measure the same length. The dimension of wavelength can also torch up your intuition, it's — \( [L] \).

8 minutes ago, Mordred said:

Doppler shift applied to any wavelength including sound waves. It is not restricted to light. I already provided a paper showing this detail.

Yes of course, but it's no so simple as can be on the first look, and paper not showing it in detail either (at for my opinion), there is a physical meaning in the model must be or will be no any point to speak about physics, it's will just abstract mathematical metamorphoses, when you saying "redshift" you must imagine the measurement without the very medium... And the inertial frames must apply somehow there, and so on, it's easily to mistaken there something, at least on my point of view.

The point is the formula of Doppler effect derivated from the Lorentz's transformation have different physical meaning of inertial frame than the one from classical Galilean transformation.  At least on my current level of understanding, because the both of this very transformations is more fundamental than the inertial frames in the physical models. I can mistaken, so if anybody knows something about this, actually not a trivial case, me very good to read it.

Edited August 1, 2019 by frostysh

[Mordred]

In Galilean relativity the primary differences is that the speed of light had no speed limit. Time itself is absolute. The principle of invariance however remains the same. The definition of an inertial frame is also identical.

We know today that Newtonian/ Galilean physics is wrong on two major points. The speed of light is limited and so is all information exchange by the constant c.

We also know that time is not absolute and that the absolute frame does not exist. An Eather has medium like properties that if it did exist would cause directional drag aka the one way speed of light tests. All tests on c show there is no directional difference in the speed of light between two observers.

In Lorentz transformations you have both length contraction and time dilation however in each reference frame the laws of physics such as Pythagoras theorem remain the same (per the reference frame) this is the same within the reference frame for Galilean relativity. 

 Conjecturing some medium or Eather is very commonly done by far more posters than you realize you'd be amazed how often it crops up.

Now two observers that see different densities do apply in gravitational redshift but why would the two observers see different densities in non relativistic Doppler ?

Here is the details for the Michelson Morley test for an Eather

https://www.google.com/url?sa=t&source=web&rct=j&url=https://faculty.etsu.edu/gardnerr/5310/5310pdf/dg2-2.pdf&ved=2ahUKEwjWksX-tuLjAhUFj54KHVpeBC0QFjAAegQIAhAB&usg=AOvVaw3fVRCzr82fWGmyXI9J2AD3

Ok let's try this we will stick to sound waves for now and show why  moving relative to the wavefronts changes the wavelength.

First you have a wave that emits a wavelength [math]\lambda[/math] this wavelength emits a continuous sequence of wavefronts given at a rate of [latex]\frac{1}{\nu}[/latex] which is it's given frequency so let's say the source is moving slowly towards you the observer at speed [latex]\mu[/latex] now each time the source transmits it's signal it has moved a distance [latex] \mu t[/latex] this changes the distance of the frequency wavefronts from 

[latex]\frac{1}{\nu}[/latex]

Which now becomes 

[math]\lambda=\frac{c}{v}-\mu t=\frac{c}{\nu}-\frac{\mu}{\nu}=\frac{c}{\nu}(1-\beta)[/math]

It's apparent frequency becomes

[math]\acute{\nu}=\frac{c_s}{\nu}=\frac{\nu}{1-\beta}[/math]

There you have it the source is moving so each time it emits a frequency wave pulse its position is different thus changing the frequency. You do not require any medium to get this effect nor does this in any way violate Galilean relativity (however sound waves itself required a medium) the effect itself doesn't require a medium  if the density of the two emitters location differs from the observer location the speed of sound will vary on route and you will need to account for that seperately.

However the generalized Doppler effect is based on the medium density being the same between the observer and emitter and the only cause is from a moving source.

In essence it is simply addition of velocities for classical physics.

 

Edited August 1, 2019 by Mordred