A scenario to help understand the twin paradox in SR

In summary: The traveling twin begins to see the electromagnetic pulse speed up to faster than every second, due to the Doppler effect, but tempered by relativistic time dilation).The stationary twin sees the same thing, but at presumably less of a rate, so that at some point he sees that his stationary twin is older than him.Step 2, the return trip: So now, on the return trip, the traveling twin begins to see the electromagnetic pulse speed up to faster than every second, due to the Doppler effect, but tempered by relativistic time dilation).The stationary twin sees the same thing, but at presumably less of a rate, so that at some point he sees that his stationary twin is
  • #1
Sophrosyne
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TL;DR Summary
Elaborating on the the twin paradox to help understand it
I am looking at some of the threads on the twin paradox, and getting even more confused. I have been trying to run through the details of what each twin is seeing, and was wondering if I could get some help. I am just trying to imagine how each twin is “seeing” the other twin at each step as the scenario unfolds. Please tell me if I am going wrong at any step in this scenario.

So in my scenario, we have the two twins, each with a powerful telescope through which they can see each other, and also an electromagnetic beacon which emits a pulse every second which the other can receiver. They bid each other farewell and the traveling twin takes off and in short order achieves near-light-speed velocity.

Step 1, the outward journey: the traveling twin travels to a planet 20 light years away. He notes time dilation from the stationary twin’s electromagnetic pulses, so they slow to, say, every 2 seconds instead of one. As he looks back at his twin through his telescope, he sees that his twin has only aged 10 years instead of his 20.

The stationary twin, in the meantime, sees the same thing: he has aged 20 years while the traveling twin has only aged 10, and his electromagnetic beacon frequency also halved for most of the trip. (I am just making up these numbers qualitatively- the exact numbers don’t matter too much- I am looking for a qualitative, intuitive understanding).

Step 2, the return trip: So now, on the return trip, the traveling twin begins to see the electromagnetic pulse speed up to faster than every second, due to the Doppler effect, but tempered by relativistic time dilation). The stationary twin sees the same thing, but at presumably less of a rate, so that at some point he sees that his stationary twin is older than him.

So now, some questions:

1) At what point do the twins realize the stationary twin has aged more? It has to be somewhere as the traveling twin anpproaches Earth again, right? Somewhere mid-flight at some point. Would they both agree that it happened mid-flight as they see each other through their telescopes and keep track of the number of EM pulses, even if they don’t agree on exactly when? IOW, they can both agree that the time for the stationary twin has moved faster on the return trip than for the traveling twin, correct?

2) The traveling twin picks up an alien on that planet on which he he had stopped- one who is used to seeing time dilation when traveling at near light speeds. Wouldn’t this alien be a little surprised to see that the time as measured by the EM pulse at their destination is actually speeding up instead of slowing down? As two objects approach each other at relativistic speeds, the every-second EM pulse from the stationary twin should be slowing down, not speeding up. I am not sure why this should be observed differently by the returning twin as opposed to the alien he picked up who never made the first leg of the journey, but is on the same ship.
———————
Sorry if I am sounding confused here- but that’s because I am!
 
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  • #2
If you have not already done so, read the "Doppler Shift Analysis" section of https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Sophrosyne said:
At what point do the twins realize the stationary twin has aged more?
When they meet at the end of the journey.
You are asking a trick question, probably without intending to. Comparing their ages is only meaningful when the two are in the same place, so "aged more" is a comparison of how much they've each aged between the separation and reunion events. It makes no sense to talk about which one is "aged more" at any other stage of the journey.
IOW, they can both agree that the time for the stationary twin has moved faster on the return trip than for the traveling twin, correct?
No. During the outbound leg they both find (by counting flashes and watching the development of facial wrinkles and greying hair through their telescopes) that their sibling, moving relative to themselves, is aging more slowly. On the return leg they both find that sibling is aging more quickly.

And just to be sure..... you probably do understand this, but you haven't said so in so many words.... The change in age observed through the telescope will always exactly match the number of flashes received. That is, if one twin were use their telescope to take a snapshot of the, then take a second snapshot after receiving 365*24*60*60 more flashes, a comparison of the two snapshots will show the effects of one year of normal aging.
 
  • #3
Sophrosyne said:
Step 1, the outward journey: the traveling twin travels to a planet 20 light years away. He notes time dilation from the stationary twin’s electromagnetic pulses, so they slow to, say, every 2 seconds instead of one. As he looks back at his twin through his telescope, he sees that his twin has only aged 10 years instead of his 20.
These numbers aren't consistent, which doesn't help anything. If the planet is 20ly away and it takes 20 years of ship time to get there, ##c^2\Delta \tau^2=c^2\Delta t^2-\Delta x^2## tells you that ##\sqrt{2}\Delta x=\Delta t## so the velocity is ##c/\sqrt 2##. That means that the time dilation factor is ##1/\sqrt{2}## and the Doppler factor is ##\sqrt 2-1##. So the traveller will receive ##20(\sqrt 2-1)\approx 8.3## years of light pulses during the outbound trip.
Sophrosyne said:
The stationary twin, in the meantime, sees the same thing: he has aged 20 years while the traveling twin has only aged 10, and his electromagnetic beacon frequency also halved for most of the trip.
This isn't correct. Since the traveller is doing ##c/\sqrt 2##, in the rest frame of the stay-at-home it takes the traveller ##20\sqrt{2}\approx 28.3## years to reach the planet, during which time he will receive ##28.3(\sqrt{2}-1)\approx 11.7## years worth of light pulses. And then it'll be another 20 years (due to light speed delay) before he stops receiving pulses from the outbound trip.
Sophrosyne said:
So now, on the return trip, the traveling twin begins to see the electromagnetic pulse speed up to faster than every second, due to the Doppler effect, but tempered by relativistic time dilation).
The relativistic Doppler effect includes time dilation. So he will receive pulses at a rate of ##1/(\sqrt 2-1)## faster than he emits them, so he'll receive 48.2 years of pulses, for a total of 56.6 years of pulses received during what me measures to be a 40 year journey.
Again, no. For the first 20 years he continues to see the low "travelling away" rate. Only for the last 8.3 years does he see the higher rate, so he ends up seeing 40 years of light pulses in total.
Sophrosyne said:
At what point do the twins realize the stationary twin has aged more?
Depends what you mean. If they know they are doing a twin paradox experiment, each one can determine who will be the older when they work out who will be doing the turn around. In this case, then, they know from the outset.

If they don't know they are doing a twin paradox then they can't know until they meet up, because they can't guarantee that the stay-at-home won't be kidnapped and bundled into a spaceship at the last minute to take his own trip.

What I suspect you mean is, when can one twin look through a telescope and see the other looking the same age he is now? The stay-at-home never does. The traveller can, though, and it's a fairly straightforward intercept calculation to determine when. It is somewhere on the return journey, yes.
Sophrosyne said:
Would they both agree that it happened mid-flight as they see each other through their telescopes and keep track of the number of EM pulses, even if they don’t agree on exactly when? IOW, they can both agree that the time for the stationary twin has moved faster on the return trip than for the traveling twin, correct?
No, and no, because they do not see the same things through their telescopes.
Sophrosyne said:
Wouldn’t this alien be a little surprised to see that the time as measured by the EM pulse at their destination is actually speeding up instead of slowing down?
I would say this alien seems a little confused.
 
  • #4
As for 1)
According to SR you would agree that the twin on Earth observes younger astronaut brother when they meet again.
The spaceship twin transfers inertial frame of reference(IFR) when turning. For this reason simple SR is not applicable for his frame of reference that is synthesized IFRs.

As for radio transmission you proposed, say the travel starts on theri birthday and it takes 60 yrs for the Earth twin and 30 yrs for the astronaut twin. The astronaut watches the Earth twin's annual birthday party video 61 times during the flight including at start and goal. In going-out less frequent pace and in return more frequent pace.

Time space diagram plot in Earth IFR

1688993899738.png


Speed of rocket is ##\pm \frac{\sqrt{3}}{2}c##, go-return. The last orange and green lines show its diagrams. Target and turning point is about 26 light-year distance from the Earth. Around turning the astronout get 4 year birthday video of his brother twin on Earth. Then he is 15 yrs old. Other 56 videos he sees during return journey. The paces for the astronauts to get the videos are
Go: 4/15=0.27 video/year
Return: 56/15= 3.73 video/year

I do not understand 2) good enough to comment.
 
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  • #5
Thank you, that graph is helpful.

As far as question 2: as one object approaches another at relativistic speeds, each sees the other as time dilated, ie, their click as running slower.

But when the traveling twin is returning towards earth, the alien will be seeing time speeding up on the object on which he is approaching at relativistic speeds. Wouldn’t he find that strange?

I chose the alien because part of the twin paradox is that the returning twin necessarily has a history of accelerating away from the original destination, and then accelerating back towards it, and this frame shift is what creates the paradox. But the alien that he boards on the other planet has no awareness of that twin’s travel history. He just got on the ship. And he will be seeing a very bizarre thing happening: time speeding up towards an object at which she is approaching with relativistic speed.

How would he explain it?
 
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Sophrosyne said:
But when the traveling twin is returning towards earth, the alien will be seeing time speeding up towards the object on which he is approaching at relativistic speeds. Wouldn’t he find that strange?
No. He'll see the pulses arriving at short intervals, but if he corrects for the changing travel time of each pulse he'll find that they are emitted at longer intervals. The naive Newtonian Doppler effect overwhelms the time dilation for a net blueshift.
Sophrosyne said:
And he will be seeing a very bizarre thing happening: time speeding up towards an object at which she is approaching with relativistic speed.
It's important to distinguish between what you see and what you calculate. The alien (and the travelling twin) will see pulses coming in close together in time, but will calculate that they were emitted far apart in time.
 
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  • #7
Sophrosyne said:
TL;DR Summary: Elaborating on the the twin paradox to help understand it

I am just trying to imagine how each twin is “seeing” the other twin at each step as the scenario unfolds.
Can you clarify your intention:

1) sometimes in relativity when people use the word “seeing” they literally mean the images they would visually see (magnified with a telescope if needed).

2) sometimes in relativity when people use they word “seeing” they mean the positions and times as calculated in the observer’s inertial frame.

Which do you intend? I think you intend 1) but I am not sure
 
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  • #8
Here are four Minkowski diagrams - similar to the diagram @anuttarasammyak posted above, but with a scale fixed so that one light year on the x axis is the same size as one year on the t axis. Light thus propagates on lines of 45° slope.

The first shows just the path of stay-at-home (red) and traveller (green) in the rest frame of the stay-at-home. As per my calculations above you can see that turnaround happens about 28.3 years after departure and return happens at 56.6 years.
1689009768302.png

Now let's get the stay-at-home to send a signal every two years. These are drawn as fine red lines
1689009958030.png

You can see that the slope of the signals is 1, so they are moving at light speed. You can also see the red and blue shift, because the distance along the green line between the arrival of red pulses is much larger on the outbound trip than on the inbound one, so only 8 of the 27 pulses sent arrive on the outbound leg. You can also see that the arrival rate changes instantly at turnover.

Now let's get the traveller to send a signal every two years - fine green lines this time.
1689010320323.png

If you count carefully you'll see that there are 19 signals sent (the 20th is sent as the traveller returns home, so isn't visible). And what the stay-at-home sees is very different to what the traveller saw - the frequency change doesn't happen until long after the half-way point of the experiment. That difference is the fundamental reason why "what the stay-at-home sees" isn't inconsistent with "what the traveller sees". Although both see the same difference in frequency, consistent with the principle of relativity, they see the "fast" and "slow" frequencies for different lengths of time because they do different things.

Finally, here's a Minkowski diagram showing the same set of red signals and only showing the emission events of the green signals. The red lines are the light showing every even-numbered birthday of the stay at home twin and the green crosses are the even numbered birthdays of the traveller. You can see that on the outbound leg the traveller sees the stay-at-home's birthdays lagging behind his; when he turns around they begin to catch up and eventually pass.
1689011183060.png

If I'm counting right, the light from the stay-at-home's 14th pulse (i.e. 28th birthday) arrives after the traveller's 14th pulse (28th birthday), but the stay-at-home's 15th pulse arrives before the traveller sends his fifteenth - so the traveller sees the stay-at-home looking older than him for something like the second half of the return journey.

The precise calculation is that, if the stay at home emits a pulse at time ##T## it is received when its x coordinate equals the x coordinate of the returning twin - that is, when ##c(t-T) = D-v(t-D/v)##, where ##D## is the distance to the planet and ##v## is the speed of the ship. The traveller's age at that point is ##\mathrm{age} = t/\gamma##. If we want the age of the traveller at reception to equal the age of the stay-at-home at emission, we just set ##\mathrm{age}=T## and eliminate ##t## from those two equations. That gets us that $$T=\frac{2D}{(c+v)\gamma-c}$$Subbing in the values for this case, we get (consistent with the graphical method above) that 28.3 is the age at which the traveller sees the stay-at-home overtake him in age.

The stay-at-home always sees the traveller looking younger, of course.
 
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  • #9
Ibix said:
It's important to distinguish between what you see and what you calculate. The alien (and the travelling twin) will see pulses coming in close together in time, but will calculate that they were emitted far apart in time.

On the return leg of the journey, the alien is going to see the stationary twin as younger than the one sitting next to him. Then he is going to see that stationary twin visibly aging faster in his telescope than the traveling twin he is sitting next to- as he supposedly sees the stationary twin’s time frame slowed down relative to himself.

This is what I’m trying to wrap my head around. It’s not intuitively making sense right now. I am not sure how to picture how this would look like to the alien. He would be seeing the stationary twin’s time time slowed down but the twin aging faster? That doesn’t make any sense.
 
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  • #10
Sophrosyne said:
On the return leg of the journey, the alien is going to see the stationary twin visibly aging faster in his telescope than the traveling twin he is sitting next to- as he supposedly sees the stationary twin’s time frame slowed down relative to himself.
This is the problem - you are using "see" in two different senses.

Looking through the telescope he literally sees the stay-at-home aging faster. You can see this in my last Minkowski diagram, where the red lines are arriving more frequently than the green crosses.

However, the alien should know that the distance between him and the stay-at-home is falling and the speed of light is finite. So he knows that the stay-at-home twin is older now than he was when the light the alien sees left Earth. He could compute the travel time of light and calculate the age of the stay-at-home now. If he does that, he'll find that the stay-at-home's calculated age is increasing more slowly than the twin he's sat next to.

Many sources use "see" to mean both what I've called see and what I've called calculate. This appears to have confused you. All you need to do to your sentence I've quoted above is take into account this confusion and then correct one of the words: "On the return leg of the journey, the alien is going to see the stationary twin visibly aging faster in his telescope than the traveling twin he is sitting next to- as he supposedly calculates the stationary twin’s time frame slowed down relative to himself." That is the correct statement. It's only confusing if you use "see" to mean two different things.
 
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  • #11
OK, I think I am starting to see what you are saying.

So right now, using the word “see” as in what you can see with your eyes, the alien would see events speeding up on earth as he approaches, Even though the calculated time on earth will be slower.

Am I understanding this correctly?

But then how is this different than two spaceships approaching each other at relativistic speeds? I thought they would each see each other’s time slowed down.

If you could tune in to the viewing screen of the spaceship approaching earth vs the spaceship approaching another ship, would you be able to tell that one had undergone a frameshift, and the other had not, depending on how they’re seeing with their eyes the movement of time on the object to which they were approaching?
 
  • #12
Sophrosyne said:
But then how is this different than two spaceships approaching each other at relativistic speeds? I thought they would each see each other’s time slowed down.
It isn't different. They both see each other moving fast and calculate that they are moving slow. And many sources will confuse the two and use see where they should use calculate.
 
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  • #13
Sophrosyne said:
So right now, using the word “see” as in what you can see with your eyes, the alien would see events speeding up on earth as he approaches, Even though the calculated time on earth will be slower
Yes. This is “see” meaning what you visibly see directly with your eyes or with a telescope. My 1) above

Sophrosyne said:
I thought they would each see each other’s time slowed down
This is “see” meaning the positions and times as calculated in the observer’s inertial frame. This is my 2) above.

Unfortunately a lot of sources use 2), which I think is really confusing. I don’t think “see” should be used in that sense at all. But it often is.

Time dilation is not something that is visible (see 1). It is something that is calculated (see 2). What is visible (see 1) is Doppler shift. You take the visible Doppler shift and correct for the light travel time to get time dilation.

Doppler shift is an optical effect. Time dilation is not. Using the word “see” for both causes your confusion and also leads some people to mistakenly believe that time dilation is an optical illusion
 
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  • #14
Sophrosyne said:
But then how is this different than respond two spaceships approaching each other at relativistic speeds? I thought they would each see each other’s time slowed down.
When two spaceships or the two twins are approaching one another they both SEE the other twin aging more quickly. This is the Doppler effect.
They both also CALCULATE that the other twin's clock is running slow because the speedup is less than what they expected just from the Doppler effect.

When two spaceships or the two twins are moving away from each other they both SEE the other twin aging more slowly. This is the Doppler effect.
They both also CALCULATE that the other twin's clock is running slow because the slowdown is greater than what they expected just from the Doppler effect.

Now go back and do what we told you to do back at the beginning of this thread:
If you have not already done so, read the "Doppler Shift Analysis" section of https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
In the twin paradox, the earthbound twin SEES a long period of the travelling twin aging more slowly followed by a short period of the travelling twin aging more quickly so is not surprised to find that the traveler ends up younger. This is because the during the first part of the return leg the earthbound twin is still SEEing light that was emitted by the traveller during the outbound leg - the traveler is most of the way home before earthbound SEEs the turnaround.
 
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Got it- I think. I have to chew on this a little more though to make sense of it. Thank you for your responses. They were helpful!
 
  • #16
Dale said:
Yes. This is “see” meaning what you visibly see directly with your eyes or with a telescope. My 1) aboveThis is “see” meaning the positions and times as calculated in the observer’s inertial frame. This is my 2) above.

Unfortunately a lot of sources use 2), which I think is really confusing. I don’t think “see” should be used in that sense at all. But it often is.

Time dilation is not something that is visible (see 1). It is something that is calculated (see 2). What is visible (see 1) is Doppler shift. You take the visible Doppler shift and correct for the light travel time to get time dilation.

Doppler shift is an optical effect. Time dilation is not. Using the word “see” for both causes your confusion and also leads some people to mistakenly believe that time dilation is an optical illusion
I see. This distinction you are making between between “seeing” and “calculating” is hugely helpful. Thank you for the clarification. This is making more sense now.

They don’t really make that distinction in the usual popular explanations of this, For those who actually want to think about it, it can make for a lot of confusion.
 
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  • #17
Sophrosyne said:
Got it- I think. I have to chew on this a little more though to make sense of it. Thank you for your responses. They were helpful!
Special Relativity is not about what you literally see. Note that in classical physics, a clock moving towards you will appear to run fast and a clock moving away from you appear to run slow. That's nothing to do with Special Relativity.

Note also that time dilation is not necessarily related to differential ageing. The twin paradox is about differential ageing.
 
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  • #18
PeroK said:
Special Relativity is not about what you literally see. Note that in classical physics, a clock moving towards you will appear to run fast and a clock moving away from you appear to run slow. That's nothing to do with Special Relativity.

Note also that time dilation is not necessarily related to differential ageing. The twin paradox is about differential ageing.
If you've not seen the phrase "differential aging" often, it refers to the re-united twins showing different signs of aging.

One has undergone more aging than the other and shows a greater elapsed time on his wrist watch because his path through spacetime had a larger cumulative time-like length.
 
  • #19
For example: consider the twins in the same circular orbit about a star or planet, but going in opposite directions. Each twin is moving relative to the other and each is continuously time dilated relative to the order. But, every time they pass each other their clocks show the same elapsed time.

That's time dilation without any differential ageing.
 
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  • #21
Do a YouTube search for Hewitt Twin Trip. It's the same scenario described in your OP with different numbers.
 

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