Some relativity questions for coleagues

Main Question or Discussion Point

A discussion with nakurusil left open some questions to which answers are highly appreciated.
1. We measure in physics periods and reckon frequencies or vice-versa.
2. Is it possible to teach the Doppler Effect without involving the concept of wave crest?
3. When an Author says that the observer involved in a Doppler Effect receives two successive wave crests remmaining located at the same point, does he make the assumption that the period is "very small"?
I think that the correct answers are usefull for all of us.
In my oppinion in the case 1 the first variant is correct as long as the observer uses his wrist watch, in the second case the concept of wave crest is usefull whereas in the third case the very small (locality) assumption is made.
 

Answers and Replies

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2. Is it possible to teach the Doppler Effect without involving the concept of wave crest?
The displacement 4 vector [itex] x^a = [ct, x, y, z] [/itex] has a known transformation rule, namely that

[tex]\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = x^{a'} = \Lambda^{a'}\mbox{}_{a} x^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix}\ .[/tex]

Given that we have defined

[tex]\Delta \tau = \sqrt{\frac{x^a x^b \eta_{ab}}{c^2}},\, x^a x^b \eta_{ab} > 0[/tex]

we go on to define the energy momentum four vector as the four-velocity multiplied by the particle's mass, the components turn out to be

[tex]p^a = m_0 U^a = m_0 \frac{dx^a}{d\tau} = \left[\frac{E}{c}, p_x, p_y, p_z\right].[/tex]

Either we knew how to transform energy and momentum from one frame to another already, or we have just discovered it since the energy momentum four vector transforms in the same way as the displace 4 vector.

Now, we know for photons in their rest frame, that [itex]E = \hbar \omega[/itex] and that [itex]\vec{p} = \hbar \vec{k}[/itex] and so we go on to write

[tex]k^a = \frac{1}{\hbar} p^a = \left[\frac{\omega}{c}, k_x, k_y, k_z\right].[/tex]

Finally, we know how to transform this vector, since it is ultimately proportional to the four-momentum:

[tex]\begin{bmatrix} \frac{\omega'}{c} \\ k'_x \\ k'_y \\ k'_z \end{bmatrix} = k^{a'} = \Lambda^{a'}\mbox{}_{a} k^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} \frac{\omega}{c} \\ k_x \\ k_y \\ k_z \end{bmatrix}\ .[/tex]

Therefore

[tex]
\begin{array}{rcl}
\frac{\omega'}{c} &=& \gamma \frac{\omega}{c} - \beta\gamma k_x \\
\omega' &=& \gamma (\omega - v k_x)
\end{array}
[/tex]
 
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The displacement 4 vector [itex] x^a = [ct, x, y, z] [/itex] has a known transformation rule, namely that

[tex]\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = x^{a'} = \Lambda^{a'}_{a} x^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix}\ .[/tex]

Given that we have defined

[tex]\Delta \tau = \sqrt{\frac{x^a x^b \eta_{ab}}{c^2}},\, x^a x^b \eta_{ab} > 0[/tex]

we go on to define the energy momentum four vector as the four-velocity multiplied by the particle's mass, the components turn out to be

[tex]p^a = m_0 U^a = m_0 \frac{dx^a}{d\tau} = \left[\frac{E}{c}, p_x, p_y, p_z\right].[/tex]

Either we knew how to transform energy and momentum from one frame to another already, or we have just discovered it since the energy momentum four vector transforms in the same way as the displace 4 vector.

Now, we know for photons in their rest frame, that [itex]E = \hbar \omega[/itex] and that [itex]\vec{p} = \hbar \vec{k}[/itex] and so we go on to write

[tex]k^a = \frac{1}{\hbar} p^a = \left[\frac{\omega}{c}, k_x, k_y, k_z\right].[/tex]

Finally, we know how to transform this vector, since it is ultimately proportional to the four-momentum:

[tex]\begin{bmatrix} \frac{\omega'}{c} \\ k'_x \\ k'_y \\ k'_z \end{bmatrix} = k^{a'} = \Lambda^{a'}_{a} k^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} \frac{\omega}{c} \\ k_x \\ k_y \\ k_z \end{bmatrix}\ .[/tex]


...and the above brings us to the well-known 1905 derivation of the relativistic Doppler effect . Very nice, no "crests" :smile:
 
The displacement 4 vector [itex] x^a = [ct, x, y, z] [/itex] has a known transformation rule, namely that

[tex]\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = x^{a'} = \Lambda^{a'}\mbox{}_{a} x^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix}\ .[/tex]

Given that we have defined

[tex]\Delta \tau = \sqrt{\frac{x^a x^b \eta_{ab}}{c^2}},\, x^a x^b \eta_{ab} > 0[/tex]

we go on to define the energy momentum four vector as the four-velocity multiplied by the particle's mass, the components turn out to be

[tex]p^a = m_0 U^a = m_0 \frac{dx^a}{d\tau} = \left[\frac{E}{c}, p_x, p_y, p_z\right].[/tex]

Either we knew how to transform energy and momentum from one frame to another already, or we have just discovered it since the energy momentum four vector transforms in the same way as the displace 4 vector.

Now, we know for photons in their rest frame, that [itex]E = \hbar \omega[/itex] and that [itex]\vec{p} = \hbar \vec{k}[/itex] and so we go on to write

[tex]k^a = \frac{1}{\hbar} p^a = \left[\frac{\omega}{c}, k_x, k_y, k_z\right].[/tex]

Finally, we know how to transform this vector, since it is ultimately proportional to the four-momentum:

[tex]\begin{bmatrix} \frac{\omega'}{c} \\ k'_x \\ k'_y \\ k'_z \end{bmatrix} = k^{a'} = \Lambda^{a'}\mbox{}_{a} k^{a} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} \frac{\omega}{c} \\ k_x \\ k_y \\ k_z \end{bmatrix}\ .[/tex]

Therefore

[tex]
\begin{array}{rc}
\frac{\omega'}{c} &=& \gamma \frac{\omega}{c} - \beta\gamma k_x \\
\omega' &=& \gamma (\omega - v k_x)
\end{array}
[/tex]
Thanks. Very nice. But teaching that way, a learner could ask what is the physical meaning of omega and how do you measure it? I would highly appreciate an explanation.
 
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In this case, omega is proportional to the energy of a photon,, specifically

[tex]E=\hbar \omega[/tex]
 
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Thanks. Very nice. But teaching that way, a learner could ask what is the physical meaning of omega and how do you measure it? I would highly appreciate an explanation.
Pulsation. Closely tied to frequency, the two differ by a [tex]2\pi[/tex] factor:

[tex]\omega=2\pi f[/tex]
 
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A discussion with nakurusil left open some questions to which answers are highly appreciated.
1. We measure in physics periods and reckon frequencies or vice-versa.
Both. Whichever is more convenient.

2. Is it possible to teach the Doppler Effect without involving the concept of wave crest?
Yes, of course. Einstein did it 100 years ago.


3. When an Author says that the observer involved in a Doppler Effect receives two successive wave crests remmaining located at the same point, does he make the assumption that the period is "very small"?
Why would the "Author" claim such a thing?The Doppler effect is valid at all frequencies and at all periods, not just the "very small" ones.
 
omega?

In this case, omega is proportional to the energy of a photon,, specifically

[tex]E=\hbar \omega[/tex]
Thanks again. I think both of us are people involved in teaching in a transparent way. As far as I know the Doppler Effect is associated with the wave character of a wave and not with its corpuscular character. All textbooks present the way in which the Doppler Effect changes the aspect of the successive wave fronts. As far as I know the most transparent definition of the Doppler Effect states: What we have to compare in a Doppler Effect experiment are the time interval between the emission of two successive wave crests by the source measured in its rest frame and the time interval between theirs reception by a moving observer measured in his rest frame. Do you think that when you start teaching the Doppler Effect it is compulsory that the audience know what a photon means? I expect an answer in the limits of netiquette!
 
Pulsation. Closely tied to frequency, the two differ by a [tex]2\pi[/tex] factor:

[tex]\omega=2\pi f[/tex]
Please do not tell me things I know and everibody knows.Why do you not answer the essential part of my question: how do you measure it?

I quote from a previous exchange of messages between us.
I stated
I do not speak in terms of frequency but in terms of periods. The periods should be small enough in order to ensure that the velocity in the Doppler formula is an instantaneous one. That is the case in the classic derivation as well, where the frequency in the phase is an instantaneous one and so not measurable.
You answered

]Why makes you persist in this nonsense? It is clearly non-physical.

Consider a scenario presented by
A.P. French, "Special relativity" The MIT Introductory Physics series Nelson 1968) pp.140-144 (am I in a good company in your oppinion?)
who considers a scenario that involves a Sputnik passing at a height h above an obsrvation point O. We shall regard the path as being an approximation to a horizontal straight line, so that the satellite's position can be desribed by
x=Vt ; y=h
We suppose the satelite to have a transmitter that sends out pulses f in its own rest system. Consider two successive pulses that are emitted from from the positions x(1) and x(2) at times t(1) and t(2). The time interval between the pulses is 1/f in the inertial reference frame of the satellite but is greater than this by the time dilation factor g in the observer's frame. The Author continues the calculation you can find looking for the book on your shelter in order to see how he calculates the time interval between the emission of two pulses as detected by the receiver on Earth. And now comes the essential point of the derivation:
Now if the distance x(2)-x(1) is very much less than r(1) (i.e.,if the satellite travels a very small distance during one cycle of its transmitter signals), we can with good accuracy put

and you can find the continuation in your own book.
It is my turn to ask, in your style, why do you persist in the nonsense you support. How would you teach that scenario to your students? Of course all that has contingence with Einstein's derivation if we know how to apply it.
Can you say that French persists in this nonsense'
Teaching has its own deontology. I know that nobody is perfect especially in special relativity. If you answer my thread please analyuse point by point the approach presented above. The situation becomes more complicated if the Sputnik would perform an accelerated motion. I have studied with my humble knowledge that case presented horibile dictu on arXiv.
 
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Please do not tell me things I know and everibody knows.Why do you not answer the essential part of my question: how do you measure it?
Easy on, I don't think it was so obvious from your question (or your wording) that you knew what the omega symbol represented and didn't know how to measure frequency. For low frequency waves (say.. radio) you might normally measure frequency very directly by counting cycles per second (although for extremely low frequencies, you'd instead just time a period). For higher frequencies (eg. light) it may be more convenient to measure wavelength (by diffraction) and velocity, to calculate frequency. For very high frequencies (eg. gamma rays) you would probably find the frequency directly from the energy. Also, once you know the frequency of one wave, you can beat it with an other signal to measure that signal's frequency.
 
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Please do not tell me things I know and everibody knows.Why do you not answer the essential part of my question: how do you measure it?

I quote from a previous exchange of messages between us.
I stated
I do not speak in terms of frequency but in terms of periods. The periods should be small enough in order to ensure that the velocity in the Doppler formula is an instantaneous one. That is the case in the classic derivation as well, where the frequency in the phase is an instantaneous one and so not measurable.
You answered

]Why makes you persist in this nonsense? It is clearly non-physical.

Consider a scenario presented by
A.P. French, "Special relativity" The MIT Introductory Physics series Nelson 1968) pp.140-144 (am I in a good company in your oppinion?)
who considers a scenario that involves a Sputnik passing at a height h above an obsrvation point O. We shall regard the path as being an approximation to a horizontal straight line, so that the satellite's position can be desribed by
x=Vt ; y=h
We suppose the satelite to have a transmitter that sends out pulses f in its own rest system. Consider two successive pulses that are emitted from from the positions x(1) and x(2) at times t(1) and t(2). The time interval between the pulses is 1/f in the inertial reference frame of the satellite but is greater than this by the time dilation factor g in the observer's frame. The Author continues the calculation you can find looking for the book on your shelter in order to see how he calculates the time interval between the emission of two pulses as detected by the receiver on Earth. And now comes the essential point of the derivation:
Now if the distance x(2)-x(1) is very much less than r(1) (i.e.,if the satellite travels a very small distance during one cycle of its transmitter signals), we can with good accuracy put

and you can find the continuation in your own book.
It is my turn to ask, in your style, why do you persist in the nonsense you support. How would you teach that scenario to your students? Of course all that has contingence with Einstein's derivation if we know how to apply it.
Can you say that French persists in this nonsense'


French,Zhang any many others use a lttle infinitesimal calculus in their derivation. This doesn't mean that the Doppler effect is valid only for very large frequencies or infinitely small periods. Whatever put this idea in your brain is wrong.
 
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doppler and frequency

Easy on, I don't think it was so obvious from your question (or your wording) that you knew what the omega symbol represented and didn't know how to measure frequency. For low frequency waves (say.. radio) you might normally measure frequency very directly by counting cycles per second (although for extremely low frequencies, you'd instead just time a period). For higher frequencies (eg. light) it may be more convenient to measure wavelength (by diffraction) and velocity, to calculate frequency. For very high frequencies (eg. gamma rays) you would probably find the frequency directly from the energy. Also, once you know the frequency of one wave, you can beat it with an other signal to measure that signal's frequency.
It is nakurusil who suposed that I do not know what omega means. Your answer is illuminating and I hope that nakurusil will read it in order to find out finally that the measurement of period (frequency) as a fundamental concept, is associated with the concept of two successively received wave crests.
Your answer is an example for people on the Forum for the way in which a question should be answered. My oppinion is that omega in the phase of a wave gains meaning only at high frequencies. Thank you.
 
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My oppinion is that omega in the phase of a wave gains meaning only at high frequencies. Thank you.
Jeez, few ideas but fixed.
Is this because of this paper that has been collecting dust in arxiv? What would it take for you to understand that the Doppler effect applies equally and evenly at all frequencies, that the derivation (unless it is some hockey stuff) is vallid at all frequencies?
You profess respect for Einstein, where in his derivation do you find any reference to "omega in the phase of a wave gains meaning only at high frequencies"?
 
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French,Zhang any many others use a lttle infinitesimal calculus in their derivation. This doesn't mean that the Doppler effect is valid only for very large frequencies or infinitely small periods. Whatever put this idea in your brain is wrong.
I could start in your style, stating that whatever put in your brain the idea you advocate, is wrong. I do not like such a style and I avoid it.
I expected that you will stay away from that thread wayting for the answers of other coleagues without to influence them. That will avoid, from my part the use of your style. Of course the Doppler Effect per se holds at all frequencies, but the formula, in large use, holds only in the case of high frequencies. I quoted French (I am convinced, taking into account the speed at which you answer all threads, that you did not open his book from which I have learned so much) who derives the Doppler shift formula in large use from a scenario in the case of which it is obvious that it is the result of simplificatory assumptions (a given distance is very small, we can with good accuracy put...). I do not know what you mean by little infinitezimal calculus. . I think it is essential in the derivation of the formula in large use a fact obscured by the phase invariance and the use of LT.
Let us end with what I have in my brain:The Doppler Effect by definition exists independent of the magnitude of the involved periods (time intervals between the reception of two successive wave crests.)The formula in large use, considered to account for the Doppler Effect, holds only at very small periods when the observer is supposed to receive two successive wave crests from the same point in space or in the case when the moving source emits two successive wave crests from the same point in space. In a previous message I mentioned two references concerning the locality in the period measurement by accelerating observers. As you can see the concept of nonlocality should be taken into account in the case of the oblique incidence as well.
 
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I could start in your style, stating that whatever put in your brain the idea you advocate, is wrong. I do not like such a style and I avoid it.
I expected that you will stay away from that thread wayting for the answers of other coleagues without to influence them. That will avoid, from my part the use of your style. Of course the Doppler Effect per se holds at all frequencies, but the formula, in large use, holds only in the case of high frequencies.
Of course this is not true, it is just your misinterpretation. If you don't understand the derivation, at least accept Einstein's derivation, or masudr's. Can't you see that there is no such thing as "holding only at hight frequencies"?





I think it is essential in the derivation of the formula in large use a fact obscured by the phase invariance and the use of LT.
Ahh, I see. You call the formulas that are inconvenient to you "obscuring the facts". Couldn't it be that you don't understand any of the derivations, neither Zhang, nor French, nor Einstein, nor masudr?




The formula in large use, considered to account for the Doppler Effect, holds only at very small periods when the observer is supposed to receive two successive wave crests from the same point in space or in the case when the moving source emits two successive wave crests from the same point in space.[/B]
Do you realise how ridiculous this is?
 
doppler doppler what is that

Jeez, few ideas but fixed.
Is this because of this paper that has been collecting dust in arxiv? What would it take for you to understand that the Doppler effect applies equally and evenly at all frequencies, that the derivation (unless it is some hockey stuff) is vallid at all frequencies?
You profess respect for Einstein, where in his derivation do you find any reference to "omega in the phase of a wave gains meaning only at high frequencies"?
few ideas but fixed?
On the basis of the reciprocity principle I can consider the same thing about you and we know where such people belong. Do you work in a university medium where such a style is used? We here avoid it!
As far as I know (J.D. Jackson, Classical Electrodynamics John Wiley 1962) Ch.4 (bed company?) starts with: Consider a plane wave and Einstein does the same. You know (benefit of doubt you never give me) that the plane wave assumption is associated with a very large distance between source and observer or infinite large source. That fact makes the formula we obtain under such conditions questionable holding only in that case! Why not to consider a spherical wave as French does and to show that the formula wich accounts for the Doppler shift in the case of a plane wave can be recovered now only if we make some assumptions concerning the magnitude of distances and of the involved periods i.e a combination of very large distance and very high frequency assumptions. Is that so hard to understand? In what concerns my dusty paper on arxiv thank you for having it made known. It was consulted and even quoted and contains an ideea I defend with my humble powers. Is that a criminal offense? Please let me know where did you publih your clean papers, which entitle you to play, under anonimity, the part of an omniscient scientist. Where are they quoted?
I invite the coleagues to participate at the discussion without being intimidated by your style, against which I am imune.
I know so far one participant on the Forum who repudiates your style. Probably there are others as well.
 
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A discussion with nakurusil left open some questions to which answers are highly appreciated.
1. We measure in physics periods and reckon frequencies or vice-versa.
2. Is it possible to teach the Doppler Effect without involving the concept of wave crest?
3. When an Author says that the observer involved in a Doppler Effect receives two successive wave crests remmaining located at the same point, does he make the assumption that the period is "very small"?
I think that the correct answers are usefull for all of us.
In my oppinion in the case 1 the first variant is correct as long as the observer uses his wrist watch, in the second case the concept of wave crest is usefull whereas in the third case the very small (locality) assumption is made.
Because the phase [tex]k_\mu x^\mu[/tex] of a plane wave is a scalar and
[tex]x^\mu[/tex] is a 4-vector, [tex]k^\mu=[\omega;{\bf k}][/tex] is a 4-vector. Then [tex]\omega'=\gamma(\omega+{\bf v\cdot k})[/tex], which is the Doppler effect. Who needs crests or small periods?
 
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Because the phase [tex]k_\mu x^\mu[/tex] of a plane wave is a scalar and
[tex]x^\mu[/tex] is a 4-vector, [tex]k^\mu=[\omega;{\bf k}][/tex] is a 4-vector. Then [tex]\omega'=\gamma(\omega+{\bf v\cdot k})[/tex], which is the Doppler effect. Who needs crests or small periods?
He'll ignore your post because it doesn't conform with his "discovery". He did the same with masudr's identical post.
He thinks that he has discovered a flaw in the relativistic Doppler paradox or in its derivation. No argument will change his mind.Not even the fact that it contradicts the original Einstein derivation.
 
Doppler Effect

Because the phase [tex]k_\mu x^\mu[/tex] of a plane wave is a scalar and
[tex]x^\mu[/tex] is a 4-vector, [tex]k^\mu=[\omega;{\bf k}][/tex] is a 4-vector. Then [tex]\omega'=\gamma(\omega+{\bf v\cdot k})[/tex], which is the Doppler effect. Who needs crests or small periods?
Thank you. Would you start teaching the Doppler Effect presenting the equations above? At which level? Considering that the audience knows the Doppler Effect from a person who teaches special relativity with human face?
IMHO it is compulsory to say what we compare in a Doppler Effect experiment (proper periods measured in the rest frame of the source and of the observer respectively and that the measurement of periods involves crests or to explain how we measure omega. A participant on the forum gave an illuminating answer), that the derivation starts with the phase of a plane wave which is no more then a mathematical construction which holds only in the case of the very large source-observer).
I have on my desk a Russian version of Jackson's electrodynamics from which I quote:
The derivation of the Doppler shift formula involves the invariance of the phase of a plane wave. Does that introduce a limitation of the area for which the equation accounts?. The observer located at a given point in space counts the number of wave crests which arrive at his location probably in order to measure the period he detects.
If you read further you will see that the Author imposes the condition that
the moving observer receives all the wave crests from the same point in space what some relativists call "very small period assumption" because that is possible only in that case! We find a simillar situation in the case when source and observer perform accelerated motions. Some authors use in this case the Doppler shift formula in large use and replace in it V with the instantaneous velocity of the accelerating observer (very small period assumption) others consider that during the reception of two successive wave crests the velocity changes, the observer receives two successive wavecrests from two different points in space
The situation becomes more evident in the case of a spherical wave which is in my oppinion more physical. A derivation presented by French shows that in the case of a well defined scenario (not plane wave) we can derive the Doppler Effect shift formula you propose (not longitudinal, oblique incidence??) only if we make the very large distance and very small period assumptions.
I think that such a presentation does not put under question Einstein's special relativity because it does not conflict with his two postulates.
That is my humble point of view. Deontology of teaching requires the clear presentation of the assumptions made during the derivation ,and in order to respect Einstein's relativity, without violate the two postulates.
For all that nakurusil considers that my point of view is a poor and fix idea.
Consider please that the humble point of view above is presented to you by a student you teach and explain him (me) where is he (I) wrong with patience, point by point and without to humiliate him (me). Please have the patience to do so. I think many participants (I) on the Forum will benefit from it.
Special relativity with human face
 
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few ideas but fixed?
On the basis of the reciprocity principle I can consider the same thing about you and we know where such people belong. Do you work in a university medium where such a style is used? We here avoid it!
As far as I know (J.D. Jackson, Classical Electrodynamics John Wiley 1962) Ch.4 (bed company?) starts with: Consider a plane wave and Einstein does the same. You know (benefit of doubt you never give me) that the plane wave assumption is associated with a very large distance between source and observer or infinite large source. That fact makes the formula we obtain under such conditions questionable holding only in that case! Why not to consider a spherical wave as French does and to show that the formula wich accounts for the Doppler shift in the case of a plane wave can be recovered now only if we make some assumptions concerning the magnitude of distances and of the involved periods i.e a combination of very large distance and very high frequency assumptions. Is that so hard to understand? In what concerns my dusty paper on arxiv thank you for having it made known. It was consulted and even quoted and contains an ideea I defend with my humble powers. Is that a criminal offense? Please let me know where did you publih your clean papers, which entitle you to play, under anonimity, the part of an omniscient scientist. Where are they quoted?
I invite the coleagues to participate at the discussion without being intimidated by your style, against which I am imune.
I know so far one participant on the Forum who repudiates your style. Probably there are others as well.


OK,

Let's try to convince you in a different way. Einstein says in his paper:

"In the system K, very far from the origin of co-ordinates, let there be a source of electrodynamic waves, which in a part of space containing the origin of co-ordinates may be represented to a sufficient degree of approximation by the equations"

So, he uses a certain thought experiment in order to derive the well known formula for the Doppler effect.
1. Does this mean that the formula applies only for the conditions in his thought experiment? Yes or No?

2. Does it mean that if French, Zhang and a bunch of other author use different thought experiments (and arrive to the same formula), the Doppler formula applies only for the specific conditions of their thought experiments? Yes or No?

3. Einstein used a thought experiment in order to derive [tex]\Delta E= c^2 \Delta m[/tex] . Does his formula apply only for the situation of a particle emitting two photons at opposition? Yes or No?

4. Do the specific conditions of a thought experiment used for the derivation of a formula restrict the applicability of the formula to the conditions spelled out by the thought experiment? Yes or No?
 
OK,

Let's try to convince you in a different way. Einstein says in his paper:

"In the system K, very far from the origin of co-ordinates, let there be a source of electrodynamic waves, which in a part of space containing the origin of co-ordinates may be represented to a sufficient degree of approximation by the equations"

So, he uses a certain thought experiment in order to derive the well known formula for the Doppler effect.
1. Does this mean that the formula applies only for the conditions in his thought experiment? Yes or No?
A yes or no answer does not suffice. Honest as he always was Einstein tells us that he makes an assumption that others call plane wave approximation. Using a scenario in which the distance between source and observer is not very high say an aeroplane flying at a law altitude it is my rigtht to check if the formula holds. I learned from him "never stop thinking". Does that mean that I have objections concerning the validity of his theory? And if I conclude that we should derive for each scenario an adequate Doppler formula do I offend Albert of course allways correctly using his two postulates.? Does his formula apply in the case of accelerated motion? Should we replace in that case the velocity of the observer with his instantaneous magnitude. with its instantaneous magnitude?
I can be right or I can be wrong. (Nobody is perfect in special relativity:rofl:) Experiment should decide!


2. Does it mean that if French, Zhang and a bunch of other author use different thought experiments (and arrive to the same formula), the Doppler formula applies only for the specific conditions of their thought experiments? Yes or No?
Again yes or no answer does not close our point of view. IMHO if you read with attention French (Zhang) you will see that in order to recover Einstein's formula we should make two assumptions: first Einstein's assumption (large distance) and thevery small period assumptions as well. Please think with patience about that statement. A comment on it is important for me.

3. Einstein used a thought experiment in order to derive [tex]\Delta E= c^2 \Delta m[/tex] . Does his formula apply only for the situation of a particle emitting two photons at opposition? Yes or No?
That question is out of our discussion. I will think about it. So many derivations of it convinced me that not. My convinction is that the addition law of relativistic velocities and the concept of proper mass are sufficient to do the job.

4. Do the specific conditions of a thought experiment used for the derivation of a formula restrict the applicability of the formula to the conditions spelled out by the thought experiment? Yes or No?
Definitely YES
I propose you an armistice, the single imposed condition is to use an academic style!
 
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A discussion with nakurusil left open some questions to which answers are highly appreciated.
1. We measure in physics periods and reckon frequencies or vice-versa.
I understand and agree so far Bernhard.
2. Is it possible to teach the Doppler Effect without involving the concept of wave crest?
Maybe, depending on what you mean by "wave crest". But off the cuff I think the answer is no. The dictionary defines "Doppler Effect" as follows - http://www.m-w.com/cgi-bin/dictionary?va=Doppler%20effect [Broken]
Doppler Effect: a change in the frequency with which waves (as of sound or light) from a given source reach an observer when the source and the observer are in motion with respect to each other so that the frequency increases or decreases according to the speed at which the distance is decreasing or increasing
For there to be a frequency there must be some sort of repetitive disturbance on the wave. E.g. you might be thinking of a puire sinusodial wave. Howeve you could just as well send out a series of pulses. As you move away from the source the time beteen the pulses will become longer and longer with increasing speed of seperation. As you move toward the source then the time between pulses constantly becomes shorter and shorter. This isn't exactly a sine wave but it does have the characteristic of what you were refering to I believe.
3. When an Author says that the observer involved in a Doppler Effect receives two successive wave crests remmaining located at the same point, does he make the assumption that the period is "very small"?
That would change with the observer but the observer might have some expectations according to what he's observing. There is nothing inherent in the wave to tell him what the source is doing just from the wave's frequency.
I think that the correct answers are usefull for all of us.
In my oppinion in the case 1 the first variant is correct as long as the observer uses his wrist watch, in the second case the concept of wave crest is usefull whereas in the third case the very small (locality) assumption is made.
Bernhard - Sorry I didn't get to this before. I wasn't posting at that time and I'm currently in the midst of reviewing a new EM text for errors. Nice opportunity. I get a free autographed text and I get to refresh my memory all for nothing. Probably a book I would have bought anyway.

So if you see me slow to respond please keep in mind that I'm busy with this task. Okay my friend?

Pete
 
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Definitely YES
I propose you an armistice, the single imposed condition is to use an academic style!
I wish you formatted your posts correctly. I will do it for you:

nakurusil said:
OK,

Let's try to convince you in a different way. Einstein says in his paper:

"In the system K, very far from the origin of co-ordinates, let there be a source of electrodynamic waves, which in a part of space containing the origin of co-ordinates may be represented to a sufficient degree of approximation by the equations"

So, he uses a certain thought experiment in order to derive the well known formula for the Doppler effect.
1. Does this mean that the formula applies only for the conditions in his thought experiment? Yes or No?
bernhard said:
A yes or no answer does not suffice. Honest as he always was Einstein tells us that he makes an assumption that others call plane wave approximation. Using a scenario in which the distance between source and observer is not very high say an aeroplane flying at a law altitude it is my rigtht to check if the formula holds. I learned from him "never stop thinking". Does that mean that I have objections concerning the validity of his theory? And if I conclude that we should derive for each scenario an adequate Doppler formula do I offend Albert of course allways correctly using his two postulates.? Does his formula apply in the case of accelerated motion? Should we replace in that case the velocity of the observer with his instantaneous magnitude. with its instantaneous magnitude?
I can be right or I can be wrong. (Nobody is perfect in special relativity) Experiment should decide!
Experiment has decided: it is called the Ives-Stilwell experiment and it has proved the formula [tex]f'=f \gamma(1-\beta cos \theta)[/tex] beyond any doubt.

nakurusil said:
2. Does it mean that if French, Zhang and a bunch of other author use different thought experiments (and arrive to the same formula), the Doppler formula applies only for the specific conditions of their thought experiments? Yes or No?
bernhard said:
Again yes or no answer does not close our point of view. IMHO if you read with attention French (Zhang) you will see that in order to recover Einstein's formula we should make two assumptions: first Einstein's assumption (large distance) and thevery small period assumptions as well. Please think with patience about that statement. A comment on it is important for me.
Turns out that Ives-Stilwell experiment and its many reenactments decided the issue.


nakurusil said:
3. Einstein used a thought experiment in order to derive [tex]E=mc^2[/tex] graphic is being generated. Reload this page in a moment. . Does his formula apply only for the situation of a particle emitting two photons at opposition? Yes or No?
bernhard said:
That question is out of our discussion. I will think about it. So many derivations of it convinced me that not. My convinction is that the addition law of relativistic velocities and the concept of proper mass are sufficient to do the job.
Yes, think about it. While you are at it, remember that your answer needs to be consistent with the one you gave to question 1.


nakurusil said:
4. Do the specific conditions of a thought experiment used for the derivation of a formula restrict the applicability of the formula to the conditions spelled out by the thought experiment? Yes or No?
bernhard said:
Definitely YES
Really? Can you expand on your answer?
 
doppler no more

I wish you formatted your posts correctly. I will do it for you:
Experiment has decided: it is called the Ives-Stilwell experiment and it has proved the formula [tex]f'=f \gamma(1-\beta cos \theta)[/tex] beyond any doubt.






Turns out that Ives-Stilwell experiment and its many reenactments
decided the issue.
I think we discussed so far the Doppler Effect. Do you think that we should give up teaching the Doppler Effect presenting instead of it the experiment you quote? Your answer is what the Latins called ignoratio elenki, avoiding the essence





At that point of our discussion I consider that we should stop it. It was for me a sad experiment. I never thought that a scientific discussion can arrive to such a low moral level. I mention that I have learned nothing from you. I wait for the oppinion of other participants and please do not discourage them by telling that I have a few and fix ideas and that I consider that Einstein was wrong. I consider that big Albert is our equal opportunity employer and I do not know whom will he fire first.
 
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I wish you formatted your posts correctly. I will do it for you:
Experiment has decided: it is called the Ives-Stilwell experiment and it has proved the formula [tex]f'=f \gamma(1-\beta cos \theta)[/tex] beyond any doubt.






Turns out that Ives-Stilwell experiment and its many reenactments
decided the issue.
I think we discussed so far the Doppler Effect. Do you think that we should give up teaching the Doppler Effect presenting instead of it the experiment you quote? Your answer is what the Latins called ignoratio elenki, avoiding the essence
You asked for experimental proof, I gave you experimental proof, the Ives-Stilwell experiment has confirmed the Einstein formula for relativistic Doppler effect. I thought that you might get interested in this and admit (finally) that you do not have a valid argument.
If you indeed hope to find flaw with the derivation and the conclusion, you would need to find or run an experiment that confirms your point. Do you have any ?
 
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