# Doppler effect and the twin paradox

Hi.
I am learning special Relativity and I am seeing that the answer to the twin paradox involves the Doppler effect, and I am struggling to understand what is going on.

First let me explain what I know about the Doppler effect.
If an object produces waves (might it be light waves or sound waves),then if the object is moving then it will "squeeze" the waves in front of it thus resulting in a higher frequency, while the waves in the back of it will get "stretched" and this will result in a lower frequency.
This is what I know about the Doppler effect.

Yet I don't understand how it applies to the twin paradox, since the doppler effect deals with "frequency" not "speed" which is constant for any frequency even for radio waves or gamma rays.
So what effect does the Doppler effect have on the time it takes for the light to arrive to the twin?

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The twin 'paradox' has no relation to the doppler effect, which is calculated entirely from an instantaneous relative velocity. The differential ageing phenomenon, which creates the twin 'paradox' in the minds of people who do not understand relativity is easily understood if the concept of proper time, defined as an integral along a worldline is mastered.

Why don't you go away and study a bit more. Most of your questions on this forum have displayed a profound misunderstanding of SR.

The twin 'paradox' has no relation to the doppler effect, which is calculated entirely from an instantaneous relative velocity. The differential ageing phenomenon, which creates the twin 'paradox' in the minds of people who do not understand relativity is easily understood if the concept of proper time, defined as an integral along a worldline is mastered.

Why don't you go away and study a bit more. Most of your questions on this forum have displayed a profound misunderstanding of SR.
So can you give me a good source that covers all of these?

ghwellsjr
Gold Member
There are two different kinds of doppler effect. The one you are describing is the normal one which applies to waves that travel at a fixed speed in a medium independent of the speed of the sources of the waves and the destinations of the waves. The other one is called relativistic doppler which is a combination of the first one, plus the reciprocal effect of time dilation between two observers, which results in the doppler frequency being independent of any medium.

When we are talking about the twin paradox, we are assuming that each one has an identical clock that ticks at the same rate when they are at rest with respect to each other and that they can see the ticking of each other's clock; think of a clock that flashes a light once per second. When we put them in relative motion, it takes time for the once-per-second light flashes to travel between them (they are both sending out flashes and observing the other one's flashes).

Now if there were an absolute medium to carry the flashes between them and if there were no time dilation, then we could think of one of them at rest in this medium and the other one traveling away through this medium.

Let's say that the traveling twin is going at 90% of the speed of light. What will have happened after ten seconds? Well, the stationary twin will have emitted eleven flashes (counting the first and the last ones) but the traveling twin will be 9 light-seconds away and will just now be seeing the flash that was emitted 9 seconds earlier, which is the second flash (the first one he saw just as he was leaving). The other nine flashes are still trying to catch up to him.

In the mean time, the traveling twin has also emitted eleven flashes but a lot of them have already been seen by the stationary twin because they were emitted when the traveling twin was much closer to him.

So the doppler frequencies for the two twins, assuming a stationary medium to carry the light and no time dilation, are different. By doppler frequency, we mean the measured rate at which the flashes are detected by each twin of the other twin's clock flashes compared to their own clock tick rates (which are the same in this pretend example).

But, we can't ignore time dilation and what happens in reality, is that the traveling twin's clock slows down, it ticks more slowly than the stationary twin's clock. This has two effects: first it means that the traveling twin will percieve the stationary twin's clock flashes to be coming in faster because he is comparing them to his slow clock and, second, it means that he will be emitting his flashes at a slower rate, and so the stationary twin will see the traveling twin's clock flashes coming in at a slower rate. It turns out, wouldn't you know, that these two effects result in each twin seeing the other's clock as ticking at exactly the same slow rate compared to their own.

And it turns out, wouldn't you know, that no matter how fast the traveling twin is going compared to the stationary twin, they will both observe each other's clocks as going slow compared to their own. As a matter of fact, we could have both twins traveling in opposite directions at the same or different speeds and still, they would each observe the other's clock going at the same slow rate compared to their own.

What this means is that the relativistic doppler is independent of the relative speeds of the observers with respect to any medium. Relativistic doppler depends on only one speed, the relative speed between the two observers and is identical for the two observers, whereas the normal kind of doppler is dependent on two speeds, the speed of each observer with respect to the stationary medium and produces two different doppler frequencies, one for each observer.

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George Jones
Staff Emeritus
Gold Member
So can you give me a good source that covers all of these?
Try "A Traveler's Guide to Spacetime: An Introduction to the Special Theory of Relativity" by Thpmas A. Moore.

The Doppler effect can be used to calculated the different ages of the twins. I have done this in

but the stuff there might make sense only after reading something like the reference that I gave above.

The Doppler effect can be used to calculated the different ages of the twins. I have done this in

but the stuff there might make sense only after reading something like the reference that I gave above.

Two twins, Alfred and Betty, are together on the planet Omicron 7. After synchronizing their watches, Betty sets off on a return trip from Omicron 7 to Earth and back to Omicron 7, and Alfred remains the whole time on Omicron 7. The distance between Omicron 7 and Earth is 3.75 lightyears in the (approximately) inertial reference frame of Omicron 7 and the Earth. Betty takes the most direct route and moves at a constant speed of 3/5 lightspeed during both the outgoing and incoming segments of the trip. Both the outgoing an incoming legs of Betty's trip take 5 years according to her watch.
(my emphasis).

Betty may use this method, because she can see her clock. But how did you know what would be on her clock at the turning point ? By calculating the proper interval sqrt( 6.252-3.752) = 5. Using the same formula we can also calculate the elapsed time on Alfred's clock. If you hadn't used the proper time formula, we could only say that Betty's clock showed N years at the turn, and so Alfred's clock would show a total of 2N+N/2 = 5N/2 years, which is not so impressive is it ?

It is not necessary nor sufficient to use the doppler formula to calculate elapsed times, so I stand by the sentiments in my first post. Without actually working out a proper time, it is not possible to calculate elapsed times.

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hprog said:
Yet I don't understand how it applies to the twin paradox, since the doppler effect deals with "frequency" not "speed" which is constant for any frequency even for radio waves or gamma rays.
So what effect does the Doppler effect have on the time it takes for the light to arrive to the twin?
As I've said, it doesn't apply to the twin paradox. The differential ageing is because the proper times of worldlines may be different, thus the clocks travelling those worldlines may have different elapsed times.

In order to calculate the relativistic doppler formula, it is necessary to use the proper time, not the other way round. The space-time diagram I've attached shows how it's done geometrically, without any assumptions except the proper time formula. 'cT' and 'ct' are proper lengths.

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It is not necessary nor sufficient to use the doppler formula to calculate elapsed times, so I stand by the sentiments in my first post. Without actually working out a proper time, it is not possible to calculate elapsed times.
Not sure if you do but if you are saying that the home staying twin cannot calculate the differential aging using the elapsed time on his clock and the average received frequency of light from a beacon sending a fixed frequency carried by the traveling twin I would have to disagree.

pervect
Staff Emeritus
It's possible to derive the entire Lorentz transform from the assumption that there is a doppler shift and that it depends only on relative velocity, plus the usual assumption that the speed of light is constant. This is done by using radar methods - the doppler shift information ultimately gives you everything you need to know to do a radar type "distance vs time" analysis.

If this is of interest, one place this is done is Bondi's old book, "Relativity and Common Sense".

ghwellsjr
Gold Member
It's possible to derive the entire Lorentz transform from the assumption that there is a doppler shift and that it depends only on relative velocity, plus the usual assumption that the speed of light is constant. This is done by using radar methods - the doppler shift information ultimately gives you everything you need to know to do a radar type "distance vs time" analysis.

If this is of interest, one place this is done is Bondi's old book, "Relativity and Common Sense".
It's possible to derive the entire Lorentz transform and the relativistic doppler from just the usual assumption that the speed of light is constant. I think you need to elaborate. Why don't you tell us what is in Bondi's book?

pervect
Staff Emeritus
It's been a while, but the short version goes something like this:

Let's assume you have to bodies moving away from each other, so the doppler shift factor is greater than 1. Call the doppler shift factor k. Then if we set t=0 as the time when both observers are at the same point, then a signal emitted from traveller A at time T by her clock will be recieved at time k*T by traveller B. And by symmetry, a signal emitted from traveller B at time T by his clock will be received at time k*T by traveller A.

So, if traveller A emits a radar pulse at time T, it will be received at time KT by traveller B. If traveller B reflects the radar pulse, or transmits a reply at time KT, it will be recieved at time k^2*T by traveller A (because we know a signal sent at time T' will be received at time kT', and T' = kT).

From this traveller A can conclude that at the midpoint of transmission and reception, i.e. (T + k^2T)/2, the distance to traveller B was half the travel time - c(K^2-1)T/2 to be precise.

Since at T=0 the two travellers were coincident, the velocity of traveller B must be
c*(K^2-1)/(K^2+1), for example if K=2, v = 3/5 c

Also, if traveller B transmits a timestamp along with the reply, traveller A can see immediately that traveller's B's clock is not running at the same rate - the time (K^2+1)/2 T by traveller A's reckoning corresponds to time K*T by traveller's B clock.

This tells us a lot already, it gives us the relativistic doppler shift equation if we solve for K as a function of v (rather than what we're given, v as a function of k), and the value for time dilation.

With a little more work, you can work out the complete Lorentz transform with only a bit more algebra, by the same mechanism of tracing the radar signals on the space-time diagram

I've gone over the explanation awfully fast, the book presents it much more slowly and with diagrams - and some other historical material. But it's still pretty short and sweet, and it's very concrete.

Not sure if you do but if you are saying that the home staying twin cannot calculate the differential aging using the elapsed time on his clock and the average received frequency of light from a beacon sending a fixed frequency carried by the traveling twin I would have to disagree.
I'm saying it can't be done from the doppler information only.

Notice that two pieces of independent information are being used - the doppler readings and an elapsed time.

Therefore my statement
It is not necessary nor sufficient to use the doppler formula to calculate elapsed times,
should be taken to mean that additional information is required, in the form of an elapsed time. As you can see from the derivation of the doppler formula above, it is a ratio of elapsed times, and so on it's own can only give back a ratio of elapsed times.

Being able to reconstruct LT's etc from empirical data cannot alter this mathematical fact.

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Try "A Traveler's Guide to Spacetime: An Introduction to the Special Theory of Relativity" by Thpmas A. Moore.

The Doppler effect can be used to calculated the different ages of the twins. I have done this in

but the stuff there might make sense only after reading something like the reference that I gave above.
Thanks for the information.
Do you have anything online as well?

ghwellsjr
Gold Member
The twin 'paradox' has no relation to the doppler effect, which is calculated entirely from an instantaneous relative velocity.
It is not necessary nor sufficient to use the doppler formula to calculate elapsed times, so I stand by the sentiments in my first post. Without actually working out a proper time, it is not possible to calculate elapsed times.
As I've said, it doesn't apply to the twin paradox.
...
In order to calculate the relativistic doppler formula, it is necessary to use the proper time, not the other way round.
I'm saying it can't be done from the doppler information only.

Notice that two pieces of independent information are being used - the doppler readings and an elapsed time.

Therefore my statement should be taken to mean that additional information is required, in the form of an elapsed time. As you can see from the derivation of the doppler formula above, it is a ratio of elapsed times, and so on it's own can only give back a ratio of elapsed times.
Mentz,

I find your adamant statements on this thread to be totally at odds with my understanding of relativistic doppler and the Twin Paradox. We are assuming that each twin is broadcasting an identical signal, such as a repetitive "tick" flash at some constant interval. All each twin has to do is count their own outgoing "tick" flashes and the incoming "tick" flashes from the other twin to keep track of the accumulated times during the round trip. Of course, when they reunite, their respective counts will be the same, that is, each one's outgoing total count matches the other one's incoming total count (although the outgoing and incoming for each observer is different, of course).

The significance of the relativistic doppler is that it allows each observer to observe the time dilation on the other one's clock. The ratio of the incoming tick rate divided by the outgoing tick rate is the doppler factor. [EDIT: Actually, I'm a bit confused on this point and I may have defined the reciprocal of the doppler factor. It may in fact be the ratio of the time interval between the incoming ticks divided by the time interval between the outgoing ticks. I'm sure someone will straighten me out on this point.] There is a one-to-one correspondence between this number and the relative speed between the two observers. From the speed, each observer can calculate the time dilation of the other observer's clock. And all through the trip, except during the momentary turn-around, each observer will see the other observer's clock as running slow by the same factor.

Except by taking advantage of the relative doppler formula, how does each observer know what the time dilation is of the other observer?

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