Help understanding this derivation of relativistic Doppler

  • #1
459
161
I'm looking at George Smoot's derivation on pp. 2-3 here: http://aether.lbl.gov/www/classes/p139/homework/five.pdf

It's elegant and succinct, but I'm having trouble understanding the very last step. Using the Lorentz transformation, he gets this relationship:

##dt = dt^\prime \gamma (1 + \frac{v}{c} \cos \theta^\prime)##,

and I'm with him so far.

But then he ends up with:

##\nu = \nu ^\prime \gamma (1+ \frac{v}{c} \cos \theta^\prime)##,

where ##\nu## is the symbol for frequency.

This confuses me. Aren't frequency and period (i.e., time) inversely proportional, meaning that ##\nu## and ##\nu^\prime## should switch places? Or is there some subtlety I'm missing about what the ##t##'s represent here, and how they relate to frequency?
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
I don't follow what he's doing at all. He seems to be saying that a light ray is at ##(t', x')## and this corresponds to ##(t, x)##, and calculates ##t## in terms of ##t'##. But, he doesn't calculate ##x##, so that could be anywhere. And this is for a single pulse.

To work out the Doppler effect, I would expect a pair of pulses ##t'## apart arriving at a common point ##x## a time ##t## apart. That would give you the relationship between the frequencies.

And, yes, with the formulas he has, he ought to invert them to get the formula for frequency.

If I were you I'd work it out myself and see what you get. I'm not convinced he's correct.

I would do it by energy-momentum transformation as that seems cleaner, given that its the frequency (hence energy) of a photon we are dealing with.
 
  • #3
459
161
Thanks for the response, PeroK.

For anyone reading: I'm trying to derive the relativistic Doppler shift (for arbitrary angle) directly from the Lorentz transformation, and as simply as possible. No four-vectors, no energy-momentum relation, and preferably no wave-phase. (I have my reasons.) That Smoot derivation is exactly the kind of thing I'm looking for, except apparently there's something wrong with it(?). Any help in that direction would be appreciated.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
I don't see how you could do it without the energy momentum. If you take a conventional, relativistic approach you actually get the relative frequency of incident particles. For photons that would give an increased or decreased intensity in addition to the energy/frequency shift of each photon.

I certainly don't see that the argument in the pdf proves anything about the energy/frequency of photons.
 
  • #6
Dale
Mentor
Insights Author
2020 Award
30,855
7,455
Thanks for the response, PeroK.

For anyone reading: I'm trying to derive the relativistic Doppler shift (for arbitrary angle) directly from the Lorentz transformation, and as simply as possible. No four-vectors, no energy-momentum relation, and preferably no wave-phase. (I have my reasons.) That Smoot derivation is exactly the kind of thing I'm looking for, except apparently there's something wrong with it(?). Any help in that direction would be appreciated.
I would use the Bondi k-calculus approach. @robphy may be able to direct you to an appropriate resource.

However, the most direct way to do it is to write down the equation of a plane wave and then Lorentz transform it and simplify.
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
Thanks for the response, PeroK.

For anyone reading: I'm trying to derive the relativistic Doppler shift (for arbitrary angle) directly from the Lorentz transformation, and as simply as possible. No four-vectors, no energy-momentum relation, and preferably no wave-phase. (I have my reasons.) That Smoot derivation is exactly the kind of thing I'm looking for, except apparently there's something wrong with it(?). Any help in that direction would be appreciated.
Here's an argument. For a massive particle you have:

##E = mc\frac{dt}{d\tau}##

So:

##E/E' = \frac{dt}{dt'}##

If this also holds for a photon, then ##E/E' = \nu/\nu' = \frac{dt}{dt'}##.

Again, I think you have to appeal to the frequency being related to energy.

Perhaps there is an intuitive argument that argues directly that ##\nu/\nu' = \frac{dt}{dt'}##, but I don't see it.
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,881
1,455
This confuses me. Aren't frequency and period (i.e., time) inversely proportional, meaning that ##\nu## and ##\nu^\prime## should switch places? Or is there some subtlety I'm missing about what the ##t##'s represent here, and how they relate to frequency?
I think you're misinterpreting what ##t## and ##t'## represent. They're the time that has elapsed since the pulse was emitted as measured by observers in S and S' respectively.

For simplicity, let's assume ##\theta'=0##. Then you have
$$\frac{dt}{dt'} = \gamma(1+\beta) = \sqrt{\frac{1+\beta}{1-\beta}} > 1$$ where ##\beta = v/c##, which says clocks in S run faster than clocks in S'. As he wrote, "The frequency can be considered the beats of the clock," so you must have ##\nu > \nu'##.

You can go through the derivation of the regular Doppler effect to figure out the difference in wave periods for an observer at rest and a moving observer as measured in S. To get the relativistic Doppler effect, you then just have to Lorentz transform to figure out what the observer in S' would experience.
 
  • Like
Likes PeroK
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
"The frequency can be considered the beats of the clock," so you must have ##\nu > \nu'##.
I don't follow this "beats of the clock" argument. If the photon has oscillated ##n## times in ##t'##, then the frequency would be ##n/t'## and you would get the inverse relation.

How do you think about the "frequency" of a photon in this context?
 
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
Idea: the pdf is wrong. In order to observe a photon in the S frame where the origins coincide and the source frame is moving to the right, then the photon must have velocity ##-c## (or the relative velocity be ##-v##).

That changes the Lorentz formula to:

##\frac{dt}{dt'} = \gamma (1 - \beta)##

The argument in the pdf relates to an observer at the origin, in which case the source is moving away.

PS If you want to see what the photon is doing for an observer at a point ##x##, then the origins for the observer and source no longer coincide and you need to add a "leading clocks" term to the calculations.
 
Last edited:
  • #11
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
Here's the calculation for a source moving away. The source emits a photon when the origins coincide. The photon oscillates ##n## times when it reaches:

##(t', x')## in the S' (source) frame or ##(\gamma(1 + \frac{v}{c})t', \gamma(1 + \frac{v}{c})x')## in the S (observer) frame.

Therefore, the relative wavelength ##x## is longer than ##x'##, but the frequency is lower in the S frame.

For a source moving towards an observer, you either trust that the maths will work out and put ##-v## or ##-c## into the equation, or the source must start from a distance from the common origin and the calculations are a bit more complicated ...

PS I think my last two posts are not right. This goes the wrong way compared to the energy-momentum approach. I don't know what the frequency of a photon can be, other than a measure of its energy!
 
Last edited:
  • #12
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
Okay, I think I sorted out the nonsense in my previous posts. Apologies!

1) Using Energy-Momentum, you get the increased energy/frequency when both ##c## and ##v## are positive (source approaching from left). And you get the decreased energy/frequency when ##c## is positive and ##v## is negative (source receding to the left).

2) Using the wave equation, you get the same result.

3) Trying to interpret the frequency of a photon as some sort of intrinsic change of state of the photon, leads to the mess I got myself into. For this reason, I still don't see a direct argument that equates any "frequency" with the ratio ##dt/dt'## (relating the relative times the photon is at a given position).

I've gone full circle back to my original thought. Without considering energy, I don't see how he concludes that ##\nu/\nu' = dt/dt'##.
 
  • #13
459
161
I appreciate all the responses. A lot to think about.

Incidentally, I thought the second PDF I linked to has a rather clever derivation.
 
  • #14
vanhees71
Science Advisor
Insights Author
Gold Member
17,053
8,162
That's all way too complicated. You only need to consider plane waves, and it's even sufficient to consider scalar plane waves
$$\phi(t,\vec{x})=A \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}).$$
Now let's write this in manifestly covariant form (using ##c=1## for simplicity)
$$\phi(x)=A \exp(-\mathrm{i} k_{\mu} x^{\mu}).$$
Here ##(k^{\mu})=(\omega,\vec{k})##.

This is a scalar field and thus under a boost ##{\Lambda^{\mu}}_{\nu}## you have
$$\phi'(x')=\phi(x)=\phi(\hat{\Lambda}^{-1} x').$$
Then you have
$$\phi'(x')=A \exp(-\mathrm{i} k \cdot \hat{\Lambda}^{-1} x'),$$
but you can use that for any Lorentz transformation and any two vectors ##a## and ##b##
$$\hat{\Lambda} a \cdot (\hat{\Lambda} b)=a \cdot b.$$
Now set
$$a=k, \quad b=\hat{\Lambda}^{-1} x'$$
Then you get
$$\phi'(x')=A \exp(-\mathrm{i} (\Lambda k) \cdot x')=A \exp(-\mathrm{i} k' \cdot x').$$
Thus in the new frame you have the four-wave-vector
$$k'=\hat{\Lambda} k.$$
This is clear also a priori, but I think the above derivation is quite illuminating.

Now take the boost in ##x^1## direction. Thus we have
$$\omega'=\gamma(\omega-\beta k^1), \quad k^{\prime 1}=\gamma(k^1-\beta \omega).$$
Now for a massless field (as for photons) ##|\vec{k}|=\omega## and thus with ##\cos \theta=\vec{\beta} \cdot \vec{k}/(|\vec{\beta}||\vec{k}|)## you get
$$\omega'=\gamma \omega (1-\beta \cos \theta).$$
The inverse transformation always follows by making ##\beta## to ##-\beta##, i.e., you have
$$\omega=\gamma \omega '(1+\beta \cos \theta)$$
as claimed.
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,881
1,455
That's all way too complicated. You only need to consider plane waves, and it's even sufficient to consider scalar plane waves…
The post #3, @SiennaTheGr8 mentioned avoiding a derivation using four vectors and wave-phase, for whatever reasons.
 
  • #16
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,881
1,455
I don't follow this "beats of the clock" argument. If the photon has oscillated ##n## times in ##t'##, then the frequency would be ##n/t'## and you would get the inverse relation.

How do you think about the "frequency" of a photon in this context?
You want to think of each oscillation at a point in space as a beat of a clock. As seen in S, in the time n beats occurs at x'=0, ##n/(1-\beta)## beats occur at x=0. (Again, I'm assuming ##\theta=0## for simplicity.) The numbers are different because the light wave has to catch up with x'=0 which is moving with speed ##\beta## in S. This is just the regular non-relativistic Doppler effect. In the relativistic version, the factor of ##\gamma## appears due to time dilation.
 
  • #17
vanhees71
Science Advisor
Insights Author
Gold Member
17,053
8,162
The post #3, @SiennaTheGr8 mentioned avoiding a derivation using four vectors and wave-phase, for whatever reasons.
Why should one avoid the most appropriate and clear method to derive a result? There may be didactical reasons for making a hard subject even harder to understand by avoiding the appropriate mathematics, which I don't understand ;-).
 
  • #18
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
You want to think of each oscillation at a point in space as a beat of a clock. As seen in S, in the time n beats occurs at x'=0, ##n/(1-\beta)## beats occur at x=0. (Again, I'm assuming ##\theta=0## for simplicity.) The numbers are different because the light wave has to catch up with x'=0 which is moving with speed ##\beta## in S. This is just the regular non-relativistic Doppler effect. In the relativistic version, the factor of ##\gamma## appears due to time dilation.
If we have a source of photons moving towards an observer, then the observer will receive more photons per second (according to the Doppler formula). But, also, the energy/frequency of each individual photon will be greater.

With ##\theta = 0## the two formulas are the same. I understand the beats of the clock argument with respect to the first of these but not with respect to the energy/frequency of an individual photon. Unless you treat it as a wave packet, where the frequency of the wave packet itself is also increased.

That said, I'm not sure the original pdf did either of these calculations.
 
Last edited:
  • #19
vanhees71
Science Advisor
Insights Author
Gold Member
17,053
8,162
That's a very delicate issue. If you have a well-defined photon-number Fock state (note that these are non-interacting photons) then the Lorentz boost of such a state is again a Fock state of the same photon number. There are not more or less photons for different observers in that case. What changes are the wave four-vectors according to the transformation rule of a four-vector as demonstrated in my previous posting. It's however better to first understand the classical electrodynamical theory (Maxwell equations), which is the paradigmatic example of a relativistic quantum field theory (including being an Abelian gauge theory).
 
  • #20
Dale
Mentor
Insights Author
2020 Award
30,855
7,455
Why should one avoid the most appropriate and clear method to derive a result?
Because the OP specifically requested it.

A method of answering is only "clear" if the questioner understands the method. If I gave you an elegant answer that required fluency in ancient Sanskrit then it would not be clear to you regardless of how appropriate it seems to me.

Answering at the right level is hard to do and something I am still working on improving too.
 
  • #21
vanhees71
Science Advisor
Insights Author
Gold Member
17,053
8,162
But the quoted pdf in #1 was very confusing (at least to me). Isn't it better to show the simple way first? Then you might be able to analyze the other more confusing derivation from this perspective much easier!
 
  • #22
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,190
8,222
But the quoted pdf in #1 was very confusing (at least to me). Isn't it better to show the simple way first? Then you might be able to analyze the other more confusing derivation from this perspective much easier!
I think that's what the whole post was about: whether the logic in the pdf is sound. Personally, I'm not convinced by it. But, if someone gets a formula that corresponds to the formula they are looking for and they do some hand waving and say QED, then it's difficult to argue perhaps!
 
  • #23
459
161
But the quoted pdf in #1 was very confusing (at least to me). Isn't it better to show the simple way first? Then you might be able to analyze the other more confusing derivation from this perspective much easier!
I appreciated your contribution. Thanks!
 

Related Threads on Help understanding this derivation of relativistic Doppler

Replies
25
Views
13K
Replies
4
Views
2K
Replies
3
Views
669
Replies
4
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
7
Views
791
  • Last Post
Replies
12
Views
3K
Replies
17
Views
4K
Replies
3
Views
4K
Top