# B Help understanding this derivation of relativistic Doppler

1. Apr 15, 2016

### SiennaTheGr8

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?

2. Apr 15, 2016

### PeroK

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. Apr 15, 2016

### SiennaTheGr8

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. Apr 15, 2016

### PeroK

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.

5. Apr 15, 2016

### SiennaTheGr8

6. Apr 15, 2016

### Staff: Mentor

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. Apr 16, 2016

### PeroK

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. Apr 16, 2016

### vela

Staff Emeritus
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.

9. Apr 16, 2016

### PeroK

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. Apr 16, 2016

### PeroK

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: Apr 16, 2016
11. Apr 16, 2016

### PeroK

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: Apr 16, 2016
12. Apr 16, 2016

### PeroK

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. Apr 16, 2016

### SiennaTheGr8

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. Apr 16, 2016

### vanhees71

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. Apr 16, 2016

### vela

Staff Emeritus
The post #3, @SiennaTheGr8 mentioned avoiding a derivation using four vectors and wave-phase, for whatever reasons.

16. Apr 16, 2016

### vela

Staff Emeritus
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. Apr 17, 2016

### vanhees71

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. Apr 17, 2016

### PeroK

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: Apr 17, 2016
19. Apr 17, 2016

### vanhees71

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. Apr 17, 2016

### Staff: Mentor

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.