Doppler Analysis of Twin Experiment: Valid Argument?

[Kairos]

in a discussion on the twin experiment, I read that an author (Darwin) proposed that the non-inertial traveler twin sees the transition between the redshifts and blueshifts at the very moment when he turns around, while the inertial twin sees this transition with a delay, which means that for him the perception of redshifts exceeds that of blueshifts, and therefore he receives less flashes than his brother. Does this argument seem valid to you?

 

[Ibix]

It's one way of doing the analysis, yes. It's reasoning from direct observables (the received flashes) to predict another direct observable (clock readings at meet-up), which is a very experimentally oriented approach.

 

[Kairos]

Thank you. This delay was not obvious to me.

 

[Ibix]

The traveller sees a change in frequency immediately when they turn around because the frequency change is caused by them accelerating. The stay-at-home doesn't see a change in frequency until the first pulse emitted by the traveller after turn around reaches home, which is after half the journey time. So the stay-at-home has more time receiving redshifted pulses than the traveller.

 

[Kairos]

Okay. And if we suppose that instead of turning back, the traveler stops. He returns to the same inertial reference frame as his brother, but at a distance. Based on the number of redshifted pulses received, he is also younger in this case?

 

[Ibix]

Ambiguous, because the traveller and stay-at-home are not in the same place so cannot compare clocks without first agreeing a simultaneity convention. Using the obvious Einstein synchronisation in their mutual rest frame the traveller would be younger, yes. You can use Doppler analysis to establish this, but you still need the simultaneity convention.

By the way "returns to the same inertial reference frame" is meaningless. Everything is always in all reference frames. You could say "returns to rest in his brother's rest frame" or "he has the same inertial rest frame as his brother".

 

[Kairos]

Thank you for these precisions. The calculations based on the number of pulses received are very convincing but what bothers me is that the intervals between these pulses are not taken into account for calculating the durations. If the signal transmitted between the twins is an electromagnetic wave, there should be a compensation between the frequencies and the periods which are just inverse; no?

 

[Ibix]

but what bothers me is that the intervals between these pulses are not taken into account for calculating the durations

How are you calculating duration if you aren't counting the time between the pulses?

 

[Nugatory]

The calculations based on the number of pulses received are very convincing but what bothers me is that the intervals between these pulses are not taken into account for calculating the durations.

The interval between when two consecutive pulses are emitted, or when they are received? The first is measured by a clock with the traveler and is unchanging; the second is measured by a clock with stay-at-home so tells us nothing about the passage of time as measured by the traveler.

 

[Kairos]

the stay-at-home perceives fewer periods than his brother, but he perceives these periods on average larger than his brother. That's why I was wondering if it would make any difference if formulating the duration as Delta t= nT (where n is smaller but T is larger)

 

[Dale]

but he perceives these periods on average larger than his brother.

This is incorrect. The frequencies/periods for the redshift are symmetrical. This can be seen easily with the relativistic Doppler shift formula.

Delta t= nT (where n is smaller but T is larger)

n is smaller but T is the same.

 

[Kairos]

I agree that the redshifted periods are the same and the blueshifted periods are the same, but on average the received <T> is larger for the stay at home, given that he receives more redshifted and less blueshifted ones than the traveller

 

[Dale]

Ok, but the point of the Doppler formula is to explain it using only direct observables. We therefore use the relativistic Doppler formula to get the actual period/frequency. At no time does either twin ever receive any actual signal at the average frequency/period.

You can only calculate these averages you are interested in after you have already determined the total time for each observer. So it makes no sense to then try to predict the total time from them.

 

[Kairos]

Certainly, but can we not add the periods: delta t= T1+T2+T3+...+Tp received by the traveler and Delta t'= T'1+T'2+...+T'q received by the stay-at-home (q<p), and then give each its actual value: either redshifted (Tr) or blueshifted (Tb)

 

[Dale]

You can do it but it doesn’t tell you anything that you don’t already know from the direct observations. So what is the point?

Again, the purpose of the Doppler shift explanation is to explain the asymmetry in terms of the actual direct observations. The average period discussion thus misses the whole point.

 

[Kairos]

it tells me is that Delta t and Delta t' are identical despite the asymmetry and although p and q are different.

What direct observations are you talking about?

 

[Dale]

it tells me is that Delta t and Delta t' are identical despite the asymmetry and although p and q are different.

That seems incorrect anyway. Please show your derivation.

What direct observations are you talking about?

I am talking about the actual signals received, their frequency and their number. Those are the direct observations.

 

[Kairos]

That seems incorrect anyway. Please show your derivation.

Take a velocity of v=4*c/5 and set the atomic watches on a period of T0= 1 hour, which gives a redshifted period of Tr=3 and a blueshifted period of Tb=1/3. Take a distance of 100 light-hours between the start and the U-turn. The apparent distances covered per period being v*Tr for the outward journey and v*Tb for the return and taking into account the shift of perception of the U-turn discussed above, the traveller counts for the stay-at-home 25 periods during the outward journey and 225 periods during the return journey and the stay-at-home counts for the traveller 45 periods on the outward journey and 45 periods on the return journey. This gives during the entire trip a total duration of 150 hours for the traveler and 150 hours for the stay-at-home. The same.

 

[PeterDonis]

I read

Where? Please give a specific reference.

proposed that the non-inertial traveler twin sees the transition between the redshifts and blueshifts at the very moment when he turns around, while the inertial twin sees this transition with a delay

This didn't need to be "proposed"; it's obvious as soon as you look at the worldlines of the light pulses being emitted by each twin and how they intersect the worldlines of the twins themselves. For a good if brief discussion, see the Usenet Physics FAQ article on the twin paradox; the two most relevant pages are the Doppler Shift Analysis [1] and Figure 2 of the Too Many Analyses page [2].

[1] https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

[2] https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html

 

[Kairos]

Where? Please give a specific reference.

The wikisource cites this reference: Darwin, C.G. THE CLOCK PARADOX IN RELATIVITY. Nature 180:976-977, Nov.9,1957.

 

[Kairos]

This didn't need to be "proposed"; it's obvious

indeed this seems to be universally accepted, but I preferred to have your opinion in this forum. This article and the Figure 2 of the "Too Many Analyses" page describe the asymmetry of the number of pulses but do not calculate the resulting duration. See the previous posts.

 

[robphy]

Certainly, but can we not add the periods: delta t= T1+T2+T3+...+Tp received by the traveler and Delta t'= T'1+T'2+...+T'q received by the stay-at-home (q<p), and then give each its actual value: either redshifted (Tr) or blueshifted (Tb)

Take a velocity of v=4*c/5 and set the atomic watches on a period of T0= 1 hour, which gives a redshifted period of Tr=3 and a blueshifted period of Tb=1/3. Take a distance of 100 light-hours between the start and the U-turn. The apparent distances covered per period being v*Tr for the outward journey and v*Tb for the return and taking into account the shift of perception of the U-turn discussed above, the traveller counts for the stay-at-home 25 periods during the outward journey and 225 periods during the return journey and the stay-at-home counts for the traveller 45 periods on the outward journey and 45 periods on the return journey. This gives during the entire trip a total duration of 150 hours for the traveler and 150 hours for the stay-at-home. The same.

So, it seems the point of the last sentence
is that, via the Doppler effect, both twins agree that 150 hours elapsed on the non-inertial twin's wristwatch (that is, the non-inertial astronaut sent 150 signals and the inertial astronaut received those 150 signals),
which confirms what is known by time-dilation, applied piecewise.
A different calculation would confirm that the inertial astronaut sent 250 signals and the non-inertial astronaut received those 250 signals,

Note: the point of the clock effect is that elapsed wristwatch-time between two events depends on the spacetime-time path taken of that wristwatch... in this case, the inertial astronaut wristwatch had 250 hours elapse and the non-inertial wristwatch had 150 hours elapse.

(The twin paradox is making the mistaken-assertation
that "being able to be considered at rest" means "being able to be considered inertial",
in the hope of denying that there is a clock effect.)

 

[Kairos]

I am afraid that there is a confusion between the signals and the hours.. According to my simple above calculation which I believe complies with the laws of special relativity, the inertial twin has received 90 signals from its non-inertial twin and that the non-inertial twin has received 250 signals from the inertial twin. This result is obviously not symmetrical; but introducing the intervals beween these signals gives a symmetrical result: the inertial twin sees that 150 hours have elapsed on the non-inertial twin's watch and the non-inertial twin sees that 150 hours have elapsed on the inertial twin's watch.

 

[PeterDonis]

Take a velocity of v=4*c/5 and set the atomic watches on a period of T0= 1 hour, which gives a redshifted period of Tr=3 and a blueshifted period of Tb=1/3. Take a distance of 100 light-hours between the start and the U-turn. The apparent distances covered per period being v*Tr for the outward journey and v*Tb for the return and taking into account the shift of perception of the U-turn discussed above, the traveller counts for the stay-at-home 25 periods during the outward journey and 225 periods during the return journey and the stay-at-home counts for the traveller 45 periods on the outward journey and 45 periods on the return journey. This gives during the entire trip a total duration of 150 hours for the traveler and 150 hours for the stay-at-home. The same.

Your logic here is incorrect. The "periods" do not change duration; there is no such thing as a "redshifted period" and a "blueshifted period": each "period" between two successive signals is the same amount of proper time (1 hour) for the observer sending the signals.

In the scenario you describe here, the traveling twin emits 150 signals (not 90--there are not 45 traveler periods during each of the outbound and return legs, there are 75--see further comments below) total during his trip, and the stay at home twin emits 250 signals. That means the traveling twin experiences 150 hours of proper time and the stay at home twin experiences 250 hours of proper time. Geometrically, this corresponds to the length of the traveling twin's path through spacetime being 150 hours, and the length of the stay at home twin's path through spacetime being 250 hours. (These "lengths" are given in time units because they are timelike worldlines.)

Let me expand your description of the scenario somewhat to illustrate the points I have just made. Here is a more detailed description of the traveling twin's experience:

The traveling twin, on his outbound leg, receives 25 signals from the stay at home twin. He receives these signals over a period of 75 hours by his own clock. (75 comes from the Minkowski metric formula for the length of the traveling twin's outbound leg: $\sqrt{125^2 - 100^2} = 75$. That means he receives one signal every 3 hours, for a Doppler redshift factor of 3. Note, however, that this says nothing about how much time elapses on the stay at home twin's clock between signals. That time is still 1 hour, not 3 hours. The Doppler redshift factor of 3 gives the ratio of the interval between signals being received, by the receiver's clock, to the interval between signals being sent, by the sender's clock.

On the traveling twin's return leg, he receives 225 signals from the stay at home twin, over a period of 75 hours by his own clock. That means he receives one signal every 1/3 hours, for a Doppler blueshift factor of 1/3. Again, this does not mean the sender's clock marks only 1/3 of an hour between signals; it only means the receiver's clock marks 1/3 of an hour between signals.

So the traveling twin receives a total of 250 signals from the stay at home twin, over a period of 150 hours by his own clock. There's the time difference.

Now let's look at the stay at home twin's experience. For the first 225 hours by his clock, he receives 75 signals from the traveling twin. This gives a Doppler redshift factor of 3. For the last 25 hours by his clock, he receives another 75 signals from the traveling twin. This gives a Doppler blueshift factor of 1/3. So he receives a total of 150 signals from the traveling twin, over a period of 250 hours by his own clock. Again, there's the time difference, and both twins agree on it.

 

[PeterDonis]

According to my simple above calculation which I believe complies with the laws of special relativity

It doesn't. See my post #24 just now.

 

[Mister T]

in a discussion on the twin experiment, I read that an author (Darwin) proposed that the non-inertial traveler twin sees the transition between the redshifts and blueshifts at the very moment when he turns around, while the inertial twin sees this transition with a delay, which means that for him the perception of redshifts exceeds that of blueshifts, and therefore he receives less flashes than his brother. Does this argument seem valid to you?

Yes. Do a search on youtube for Hewitt Twin Trip. It's an old cartoon that makes use of the Doppler effect to explain the twin paradox. It will give you a good concrete example of what's been discussed in this thread.

 

[Kairos]

Your logic here is incorrect. The "periods" do not change duration; there is no such thing as a "redshifted period" and a "blueshifted period": each "period" between two successive signals is the same amount of proper time (1 hour) for the observer sending the signals.

In the scenario you describe here, the traveling twin emits 150 signals (not 90--there are not 45 traveler periods during each of the outbound and return legs, there are 75--see further comments below) total during his trip, and the stay at home twin emits 250 signals. That means the traveling twin experiences 150 hours of proper time and the stay at home twin experiences 250 hours of proper time. Geometrically, this corresponds to the length of the traveling twin's path through spacetime being 150 hours, and the length of the stay at home twin's path through spacetime being 250 hours. (These "lengths" are given in time units because they are timelike worldlines.)

Let me expand your description of the scenario somewhat to illustrate the points I have just made. Here is a more detailed description of the traveling twin's experience:

The traveling twin, on his outbound leg, receives 25 signals from the stay at home twin. He receives these signals over a period of 75 hours by his own clock. (75 comes from the Minkowski metric formula for the length of the traveling twin's outbound leg: $\sqrt{125^2 - 100^2} = 75$. That means he receives one signal every 3 hours, for a Doppler redshift factor of 3. Note, however, that this says nothing about how much time elapses on the stay at home twin's clock between signals. That time is still 1 hour, not 3 hours. The Doppler redshift factor of 3 gives the ratio of the interval between signals being received, by the receiver's clock, to the interval between signals being sent, by the sender's clock.

On the traveling twin's return leg, he receives 225 signals from the stay at home twin, over a period of 75 hours by his own clock. That means he receives one signal every 1/3 hours, for a Doppler blueshift factor of 1/3. Again, this does not mean the sender's clock marks only 1/3 of an hour between signals; it only means the receiver's clock marks 1/3 of an hour between signals.

So the traveling twin receives a total of 250 signals from the stay at home twin, over a period of 150 hours by his own clock. There's the time difference.

Now let's look at the stay at home twin's experience. For the first 225 hours by his clock, he receives 75 signals from the traveling twin. This gives a Doppler redshift factor of 3. For the last 25 hours by his clock, he receives another 75 signals from the traveling twin. This gives a Doppler blueshift factor of 1/3. So he receives a total of 150 signals from the traveling twin, over a period of 250 hours by his own clock. Again, there's the time difference, and both twins agree on it.

Thank you for your long explanation which I don't perceive for the moment but which I will study with interest. I am very surprised by your numbers; for me the ratio of the number of signals received by the non-inertial vs. the inertial is 250/90 = 1-(v/c)^2 (without square root)...

 

[Nugatory]

I am very surprised by your numbers; for me the ratio of the number of signals received by the non-inertial vs. the inertial is 250/90 = 1-(v/c)^2 (without square root)...

That ratio is just the ratio of the proper times along the two paths between the separation event and reunion event (because every emitted signal is received, and the number of signals emitted is determined by the proper time along the emitter’s worldline). That is most easily calculated using the inertial coordinates in which stay-at-home is at rest…. And at a casual glance it looks as if you’ve calculated the ratio of the squares of the proper times not the ratio of the proper times, so you owe yourself a square root.

 

[PeterDonis]

for me the ratio of the number of signals received by the non-inertial vs. the inertial is 250/90 = 1-(v/c)^2

Then you are doing the math wrong.

without square root

Which does not make sense. The square root has to be there; that's how spacetime geometry works. It's the same as the Pythagorean theorem, except that there is a minus sign in the metric instead of a plus sign. If you have a right triangle with sides 3 and 4, you don't say the length of the hypotenuse is 25. You have to take the square root.

That ratio is just the ratio of the proper times along the two paths between the separation event and reunion event (because every emitted signal is received, and the number of signals emitted is determined by the proper time along the emitter’s worldline).

No, it isn't. The stay at home twin's total path length is 250 hours, and the traveling twin's total path length is 150 hours: 75 hours on the outbound leg and 75 hours on the return leg. I computed the 75 hours explicitly in post #24. So the ratio of the proper times is 250 to 150, not 250 to 90.

 

[PeterDonis]

your long explanation which I don't perceive for the moment

Please note that my explanation is straightforward, standard Special Relativity. If it does not seem obvious to you, that means you probably need to spend more time learning straightforward, standard Special Relativity.

 

[Nugatory]

No, it isn't. The stay at home twin's total path length is 250 hours, and the traveling twin's total path length is 150 hours: 75 hours on the outbound leg and 75 hours on the return leg. I computed the 75 hours explicitly in post #24. So the ratio of the proper times is 250 to 150, not 250 to 90.

I was talking about the $1-(v/c)^2$ part…. I have no idea where the 250 and the 90 came from.

 

[PeterDonis]

I was talking about the $1-(v/c)^2$ part….

Which is wrong. The correct ratio of proper times is $\sqrt{1 - (v/c)^2}$. In this case, $v/c = 4/5$, so the ratio of proper times is $\sqrt{1 - (4/5)^2} = 3/5$, or 250 to 150. This is basically the same calculation that I did in post #24, except that I used the corresponding $t$ and $x$ coordinate values for the outbound leg in the stay at home twin's rest frame, which are 125 and 100. That gives a proper time for the traveling twin on the outbound leg of 75. Then the 125 and 75 get doubled to get the values for the entire trip.

I have no idea where the 250 and the 90 came from.

They came from the OP's incorrect belief that the ratio of proper times is $1 - (v/c)^2$, without the square root.

 

[Nugatory]

The correct ratio of proper times is $\sqrt{1 - (v/c)^2}$.

Which is exactly what I said: when he wrote $1-(v/c)^2$ he owed himself a square root.

 

[PeterDonis]

Which is exactly what I said: when he wrote $1-(v/c)^2$ he owed himself a square root.

Ah, sorry, I missed the last part of your post #28.

 

[Dale]

Take a velocity of v=4*c/5 and set the atomic watches on a period of T0= 1 hour, which gives a redshifted period of Tr=3 and a blueshifted period of Tb=1/3. Take a distance of 100 light-hours between the start and the U-turn. The apparent distances covered per period being v*Tr for the outward journey and v*Tb for the return and taking into account the shift of perception of the U-turn discussed above, the traveller counts for the stay-at-home 25 periods during the outward journey and 225 periods during the return journey and the stay-at-home counts for the traveller 45 periods on the outward journey and 45 periods on the return journey. This gives during the entire trip a total duration of 150 hours for the traveler and 150 hours for the stay-at-home. The same.

It is hard to see where you went wrong since you didn't post the derivation, just your final numbers, but this is not entirely correct. Using units where $c=1$ and $v=4/5$ we do have a Doppler redshift factor of 3 and a Doppler blueshift factor of 1/3. Everything below is done in the stay at home twin's frame which is inertial and the coordinates are given as $(t,x)$.

The worldline of the home twin is $$r_{home}=(t,0)$$ while the worldline of the traveling twin is $$r_{twin}=
\begin{cases}
(t,t \ 4/5) & t \le125 \\
(t,200-t \ 4/5) & 125\le t
\end{cases}$$

The worldline of the n-th outgoing (from home to traveling twin) light ray is $$r_{out}=(t,t-n)$$ while the worldline of the m-th ingoing (from traveling to home twin) light ray is $$r_{in}=\begin{cases}
(t,3 \ m-t) & m\le 75 \\
(t,200+m/3-t) & 75 \le m
\end{cases}$$

With $\Delta s^2=-\Delta t^2+\Delta x^2$ and the above worldlines we can confirm that the proper time between each emission and the next is 1, and we can confirm that the proper time between each reception and the next is either 3 or 1/3. So, the Doppler shift formula is correct.

There are a total of 250 emissions from the home twin and a total of 150 emissions from the traveling twin corresponding to a time dilation factor of 5/3, as expected for $v=4/5$. The traveling twin receives 25 signals on the outbound leg (25*3=75) and a total of 225 signals on the inbound leg (225/3=75) so they received redshifted signals for half the trip and blueshifted signals for half the trip. The home twin receives a total of 75 redshifted signals (75*3=225) and a total of 75 blueshifted signals (75/3=25) so they received redshifted signals for most of the trip and blueshifted signals for only the last little bit of the trip.

The times are not equal.