# Some relativity questions for coleagues

1. Feb 12, 2007

### bernhard.rothenstein

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.

2. Feb 12, 2007

### masudr

The displacement 4 vector $x^a = [ct, x, y, z]$ has a known transformation rule, namely that

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

Given that we have defined

$$\Delta \tau = \sqrt{\frac{x^a x^b \eta_{ab}}{c^2}},\, x^a x^b \eta_{ab} > 0$$

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

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

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 $E = \hbar \omega$ and that $\vec{p} = \hbar \vec{k}$ and so we go on to write

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

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

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

Therefore

$$\begin{array}{rcl} \frac{\omega'}{c} &=& \gamma \frac{\omega}{c} - \beta\gamma k_x \\ \omega' &=& \gamma (\omega - v k_x) \end{array}$$

Last edited: Feb 12, 2007
3. Feb 12, 2007

### nakurusil

...and the above brings us to the well-known 1905 derivation of the relativistic Doppler effect . Very nice, no "crests"

4. Feb 12, 2007

### bernhard.rothenstein

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.

5. Feb 12, 2007

### masudr

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

$$E=\hbar \omega$$

6. Feb 12, 2007

### nakurusil

Pulsation. Closely tied to frequency, the two differ by a $$2\pi$$ factor:

$$\omega=2\pi f$$

Last edited: Feb 12, 2007
7. Feb 12, 2007

### nakurusil

Both. Whichever is more convenient.

Yes, of course. Einstein did it 100 years ago.

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.

8. Feb 12, 2007

### bernhard.rothenstein

omega?

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!

9. Feb 12, 2007

### bernhard.rothenstein

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.

]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.

10. Feb 12, 2007

### cesiumfrog

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.

Last edited: Feb 12, 2007
11. Feb 12, 2007

### nakurusil

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.

Last edited: Feb 12, 2007
12. Feb 12, 2007

### bernhard.rothenstein

doppler and 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.

13. Feb 12, 2007

### nakurusil

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"?

Last edited: Feb 12, 2007
14. Feb 12, 2007

### bernhard.rothenstein

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.

15. Feb 12, 2007

### nakurusil

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"?

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?

Do you realise how ridiculous this is?

16. Feb 13, 2007

### bernhard.rothenstein

doppler doppler what is that

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.

17. Feb 13, 2007

### Meir Achuz

Because the phase $$k_\mu x^\mu$$ of a plane wave is a scalar and
$$x^\mu$$ is a 4-vector, $$k^\mu=[\omega;{\bf k}]$$ is a 4-vector. Then $$\omega'=\gamma(\omega+{\bf v\cdot k})$$, which is the Doppler effect. Who needs crests or small periods?

Last edited: Feb 13, 2007
18. Feb 13, 2007

### nakurusil

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.

19. Feb 13, 2007

### bernhard.rothenstein

Doppler Effect

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

20. Feb 14, 2007

### nakurusil

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 $$\Delta E= c^2 \Delta m$$ . 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?