# The twins struggle with Minkowski

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Here below a graphic representing the twin paradox in Minkowski space-time.

The diagram is supposed to be valid for the Frame Of Reference of the non moving observer, spending time doing nothing from point O to point A, while the spaceship travels from O to B, then makes a U-turn and comes back to Earth at point A.

Here is the problem I want to discuss:

When the spaceship arrives at point B, at half the travel, and makes the U-turn, the observer on earth at point H cannot see it simultaneously. He will observe the U-turn after a while, from point M.

At point H, 50 days have passed, and the spaceship is not observed at point B.

At point M, 75 days have passed, the spaceship is observed at point B making the U-turn manoeuvre.

At point A, 100 days have passed, the spaceship is back to Earth.

What is surprising is that the paceship has been observed to travel the distance d in 75 days when going, and 25 days when coming back. The spaceship did not have the same apparent velocity on both parts of the trip.

That's my problem, because I always made the interpretation of the diagram stating that the angle on the graph represents speed. Since the angle of the first blue segment is exactly symmetric with the angle of the second segment, it should mean that both speeds were the same.

How is it possible then that the spaceship has been observed making the return trip in only 25 days?

I hope it is clear till now. There is an inconsistency.

Now, another graph here below:

Here I was deeply thinking the opening statement:

"The diagram is supposed to be valid for the Frame Of Reference of the non moving observer"

We don't care about what is the observation of the traveler, we care only to represent what is observed from Earth's FOR.

What we observe is this:

The spaceship goes for a trip. After 50 days (as observed from Earth's FOR) he makes the U-turn. He comes back after another 50 days.

We observe that on the going trip, the spaceship traveled the measured distance d at speed d/50.

During the return trip, its speed was d/50: the same speed.

What is important is that triangle O-H-B has the same basis and same height with triangle H-A-B. They have the same surface. The angle of the blue segments with the vertical do not represent speed. The ratio between height & basis of the triangle does.

And there is no inconsistency.

Michel.

Edited by michel123456

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Here below a graphic representing the twin paradox in Minkowski space-time.

The diagram is supposed to be valid for the Frame Of Reference of the non moving observer, spending time doing nothing from point O to point A, while the spaceship travels from O to B, then makes a U-turn and comes back to Earth at point A.

Here is the problem I want to discuss:

When the spaceship arrives at point B, at half the travel, and makes the U-turn, the observer on earth at point H cannot see it simultaneously. He will observe the U-turn after a while, from point M.

At point H, 50 days have passed, and the spaceship is not observed at point B.

At point M, 75 days have passed, the spaceship is observed at point B making the U-turn manoeuvre.

At point A, 100 days have passed, the spaceship is back to Earth.

What is surprising is that the paceship has been observed to travel the distance d in 75 days when going, and 25 days when coming back. The spaceship did not have the same apparent velocity on both parts of the trip.

That's my problem, because I always made the interpretation of the diagram stating that the angle on the graph represents speed. Since the angle of the first blue segment is exactly symmetric with the angle of the second segment, it should mean that both speeds were the same.

How is it possible then that the spaceship has been observed making the return trip in only 25 days?

I hope it is clear till now. There is an inconsistency.

The diagram does not represent what is "observed" from the stationary observer's frame of reference, but what "happens" according to this frame of reference. What he "observes" would be determined by light propagation delay and Relativistic Doppler shift, both of which play no role in this diagram (Just because the observer "sees" the ship make the return trip in 25 days doesn't mean that the ship actually made the trip in 25 days according to the stationary observer). IOW, this diagram represents how events transpire according to anyone who is at rest with respect to the frame of reference, regardless of where they are spatially located with respect to the "stationary" observer, not what one observer "sees"..

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The diagram does not represent what is "observed" from the stationary observer's frame of reference, but what "happens" according to this frame of reference. What he "observes" would be determined by light propagation delay and Relativistic Doppler shift, both of which play no role in this diagram (Just because the observer "sees" the ship make the return trip in 25 days doesn't mean that the ship actually made the trip in 25 days according to the stationary observer). IOW, this diagram represents how events transpire according to anyone who is at rest with respect to the frame of reference, regardless of where they are spatially located with respect to the "stationary" observer, not what one observer "sees"..

Think of it twice.

--------------------------

The spaceship goes away. You observe it constantly in your telescope. After 50 days you observe it making a U-turn.

How many days will you have to wait for arrival when you know its speed will be exactly the same?

---------------------------

It is speculated that the "twins paradox" happens at point C of the following diagram:

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What is surprising is that the paceship has been observed to travel the distance d in 75 days when going, and 25 days when coming back. The spaceship did not have the same apparent velocity on both parts of the trip.

Yes, it does. To echo Janus, it does not matter when the event is observed - it matters when the observed event occurs. At day 75, the Earth observer observes that the spaceship has changed direction at day 50. He knows that it takes 25 days for the signal to reach him, from that distance of 25 light-days. (Obviously, all times and distances are in the Earth's rest frame. They will be different in the traveling twin's outgoing and incoming rest frames.)

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Yes, it does. To echo Janus, it does not matter when the event is observed - it matters when the observed event occurs. At day 75, the Earth observer observes that the spaceship has changed direction at day 50. He knows that it takes 25 days for the signal to reach him, from that distance of 25 light-days.

Forget the 100 days. Say you know nothing. You don't know the speed of the spaceship, you don't know the distance, you know nothing about its time dilation or length contraction. You are only observing in your telescope.

After 50 days you observe it in your telescope making a U-turn.

How many days will you have to wait for arrival when you know its speed will be exactly the same?

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Forget the 100 days. Say you know nothing. You don't know the speed of the spaceship, you don't know the distance, you know nothing about its time dilation or length contraction. You are only observing in your telescope.

After 50 days you observe it in your telescope making a U-turn.

How many days will you have to wait for arrival when you know its speed will be exactly the same?

If you don't know the speed of the spaceship, then all you can know is that it will be some amount of time less than 50 days.

None of this so far has anything to do with time/space dilation or the twin paradox. You could set up the same thing with a traveler sending letters through the mail.

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You seem to accept that the velocity of the spaceship (as observed from earth) is different when coming back.

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You seem to accept that the velocity of the spaceship (as observed from earth) is different when coming back.

No, the Earth observer will calculate the same velocity for both legs of the journey. That he receives the signals from the return journey in a shorter amount of time is just a consequence of the doppler effect. It's not really moving faster.

Again, this is true of any situation where a moving body is sending signals faster than they are moving. You can do the same thing with the sound of a speeding car driving away and back again - you'll be hearing the driving away part of the journey for longer than the returning part, even though it is moving the same speed in both directions. (And in fact, if the car was moving faster than sound, you wouldn't hear the return journey until after it had already arrived!)

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Here is how to describe the situation:

At day 75, we are looking "through" the spaceship, observing its position far away although it is closer to us. The spaceship is supposed to run behind its own image. We have an object 25 days from us, that we cannot see, but we observe its image behind it.

Is that what you are assuming?

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Sure, I suppose. When Earth receives the signal from when the spaceship was farthest away at day 50, it will be day 75.

I'm not really comfortable with focusing on the "observed position," though. The observer doesn't really think the spaceship is there - he knows his information is 25 days out of date.

I'm also not really comfortable with using spaceships and light signals, etc. Not because it's inaccurate, but because it might give the impression that that is what the twin paradox is all about. So far in this topic, relativity has been irrelevant. You could do the same thing with the sound of an airplane, or postcards from a world traveling friend. You receive the postcard days after it was sent. There is no confusion that your friend is still in the post office mailing it as you are reading it. You know the signal has taken time, and your friend has had more adventures in the meantime, that you will have to wait for the next postcard to hear about.

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Sure, I suppose. When Earth receives the signal from when the spaceship was farthest away at day 50, it will be day 75.

I'm not really comfortable with focusing on the "observed position," though. The observer doesn't really think the spaceship is there - he knows his information is 25 days out of date.

I'm also not really comfortable with using spaceships and light signals, etc. (...)

Your comfort is not part of the theory.

-------------------

That was sarcasm.

-------------------

Sure, I suppose. When Earth receives the signal from when the spaceship was farthest away at day 50, it will be day 75.

Without sarcasm now.

Your words reflect what you know and what you have learned. Not what you are thinking. I hope.

Edited by michel123456
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When I say I'm not comfortable, I don't mean that it's wrong. I mean that I'm worried it will lead to misunderstandings. For example, the title of this thread references the twin paradox and Minkowski diagrams, neither of which have much bearing on the original post, which could just as easily be about listening to an airplane fly overhead.

Your words reflect what you know and what you have learned. Not what you are thinking. I hope.

What do you hope I'm thinking?

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When I say I'm not comfortable, I don't mean that it's wrong. I mean that I'm worried it will lead to misunderstandings. For example, the title of this thread references the twin paradox and Minkowski diagrams, neither of which have much bearing on the original post, which could just as easily be about listening to an airplane fly overhead.

What do you hope I'm thinking?

That it is not physically possible to observe behind an object, except made of glass, of course.

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That it is not physically possible to observe behind an object, except made of glass, of course.

What does that have to do with anything?

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How is it possible then that the spaceship has been observed making the return trip in only 25 days?

It is because you are using the term "observed" in terms of what would be recorded by a camera held by A and not in terms of what is actually occuring relative to the reference frame of A.

What you "see", optically in special relativity is not what one would measure. This is due to the finite speed of light and relativity of simultaneity.

If you consider a circular hoop, circular in the rest frame of the hoop, passing an observer at relativistic speed then length contraction in the direction of motion would present that hoop to the observer as an ellipse. That is correct. But when light transit times are included in the calculation, the observer, and his camera, would "see" a circle. Roger Penrose was the first to recognize this.

The "paradox" that you have noted is the result of failing to discriminate between "what you see" and "what you get". WYSIWYG does not hold in relativity.

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The "paradox" that you have noted is the result of failing to discriminate between "what you see" and "what you get". WYSIWYG does not hold in relativity.

It's not even necessary to invoke relativity here. Let the diagrams in the OP represent not spaceships sending light signals, but airplanes sending sound signals through the air. It looks the same, because it's not a relativity problem. It's a completely mundane observation about signals from a moving source, namely that by the time they arrive, the source is no longer in the same location as when they were sent. (Duh.)

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It's not even necessary to invoke relativity here. Let the diagrams in the OP represent not spaceships sending light signals, but airplanes sending sound signals through the air. It looks the same, because it's not a relativity problem. It's a completely mundane observation about signals from a moving source, namely that by the time they arrive, the source is no longer in the same location as when they were sent. (Duh.)

In mundane observations, airplane or letters examples, you are making addition of speeds, something not allowed here.

For example: the letters.

Lets say your brother goes away, and every 2 days write a letter that makes half the time to come to you. That is because the time for the signal (the letter) is less than the time for travel.

The first day, you get no letter.

the 2nd day, your brother writes the letter.

the 3rd day, you get the first letter. (2days of travel for your brother + one day for the letter).

The next letter you will get 6 days from day zero (4 days brother+2days letter)

The following at day 9 (6 days brother+3 days letter)

The formula is, with $D$ number of days to wait, $N$ number of days to travel of brother, $C$ number of days for letter.

$D=2N+C$

In fact, C is the speed of the signal: in this example it is the speed of the letter, in not "mundane" example, it is the speed of light.

IOW you have speed summation, not allowed in Relativity.

The example of letters must be dismissed, as any other "mundane'" example.

In the OP, there are spacetime diagrams, these are not mundane examples.

I may be wrong though.

The "paradox" that you have noted is the result of failing to discriminate between "what you see" and "what you get". WYSIWYG does not hold in relativity.

I guess so.

There is something that bothers me very much in standard diagram:

It is stated from the beginning that speed to go is the same as speed to come back. That is not negotiable.

So we are looking in our telescopes, and after 50 days, we observe the spaceship making the U-turn. Till that moment, there is no problem. we can calculate the distance that the spaceship traveled, it is $X$

The speed, as calculated from Earth, is $X/50$. Of course, we don't know what the spaceship measured, neither what another observer on another planet measured, we have only our measurement from our FOR. Maybe even the value of $X$ is wrong, and will be corrected afterwards, but what matters is that distance $X$ to go is the same with distance $X$ to come back, because it is the U-turn point.

Then it is assumed that the spaceship will come back in only 25 days!

It means that its apparent speed, as measured from our FOR, is $X/25$, which is different from the measured speed at departure.

But that was not negotiable.

Where am I wrong?

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So we are looking in our telescopes, and after 50 days, we observe the spaceship making the U-turn.

No, we don't. We can reconstruct from the data that this happened, but on day 50, looking through the telescope will not show the spaceship making a U-turn. c is finite. We will not have the information about what happened on day 50 until day 75.

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In mundane observations, airplane or letters examples, you are making addition of speeds, something not allowed here.

But that hasn't made any difference. There are no Lorentz transformations in the OP, and a diagram of sending letters would look the same. "On day 75, you receive your friend's letter that he sent on day 50, telling you he was returning. All the letters he sends on the return journey you receive in the next 25 days."

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In mundane observations, airplane or letters examples, you are making addition of speeds, something not allowed here.

For example: the letters.

Lets say your brother goes away, and every 2 days write a letter that makes half the time to come to you. That is because the time for the signal (the letter) is less than the time for travel.

The first day, you get no letter.

the 2nd day, your brother writes the letter.

the 3rd day, you get the first letter. (2days of travel for your brother + one day for the letter).

The next letter you will get 6 days from day zero (4 days brother+2days letter)

The following at day 9 (6 days brother+3 days letter)

The formula is, with $D$ number of days to wait, $N$ number of days to travel of brother, $C$ number of days for letter.

$D=2N+C$

In fact, C is the speed of the signal: in this example it is the speed of the letter, in not "mundane" example, it is the speed of light.

No, it isn't, it is a time. All the values in this equation are times.

IOW you have speed summation, not allowed in Relativity.

There was never ever summation of speed in these equations, only summations of time.

The example of letters must be dismissed, as any other "mundane'" example.

There is nothing wrong with the example.

In the OP, there are spacetime diagrams, these are not mundane examples.

I may be wrong though.

I guess so.

There is something that bothers me very much in standard diagram:

It is stated from the beginning that speed to go is the same as speed to come back. That is not negotiable.

So we are looking in our telescopes, and after 50 days, we observe the spaceship making the U-turn. Till that moment, there is no problem. we can calculate the distance that the spaceship traveled, it is $X$

The speed, as calculated from Earth, is $X/50$. Of course, we don't know what the spaceship measured, neither what another observer on another planet measured, we have only our measurement from our FOR. Maybe even the value of $X$ is wrong, and will be corrected afterwards, but what matters is that distance $X$ to go is the same with distance $X$ to come back, because it is the U-turn point.

Then it is assumed that the spaceship will come back in only 25 days!

It means that its apparent speed, as measured from our FOR, is $X/25$, which is different from the measured speed at departure.

But that was not negotiable.

Where am I wrong?

THis has been explained to you over and over again. The ship travels at a fixed speed back to the Observer, light travels at a fixed speed of twice the value. If the ship takes 50 days reach the observer from the turn around point, the light will take 25 days. All this means is that the ship is following the light back to the observer and arrives 25 days after the image does. This in no way means that the observer will conclude that the ship only took 25 days to make the trip.

As an aside, you seem to be confusing "frame of reference" with "point of view". A frame of reference is not fixed to a particular observer or his position. A frame of reference would include any observer at rest with that frame, no matter where he is located. In other words, If that other planet you mentioned were at rest with respect to the Earth, it would be in the same frame of reference, and any spacetime diagram drawn from it would look exactly the same as the one drawn from the Earth.

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IOW, the basic flaw is when I get a signal is not the same as when the signal originated (because c is not infinite)

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The time you need to cross a specific distance is the expression of a speed. One hour to cross 1km, or one second to cross 1km, is speed. It means 1 km/hour and 1km/sec respectively. In the letter example, the speed of the letter is 1 [distancetraveledbybrother]/day, or 1 day.

No, it isn't, it is a time. All the values in this equation are times.There was never ever summation of speed in these equations, only summations of time.There is nothing wrong with the example.

THis has been explained to you over and over again. The ship travels at a fixed speed back to the Observer, light travels at a fixed speed of twice the value. If the ship takes 50 days reach the observer from the turn around point, the light will take 25 days. All this means is that the ship is following the light back to the observer and arrives 25 days after the image does. This in no way means that the observer will conclude that the ship only took 25 days to make the trip. (...)

Bolded where it hurts. What I cannot swallow is that I observed the ship going slowly, and coming back in a blink of the eye, evoluting at the same speed.

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The time you need to cross a specific distance is the expression of a speed. One hour to cross 1km, or one second to cross 1km, is speed. It means 1 km/hour and 1km/sec respectively. In the letter example, the speed of the letter is 1 [distancetraveledbybrother]/day, or 1 day.

Bolded where it hurts. What I cannot swallow is that I observed the ship going slowly, and coming back in a blink of the eye, evoluting at the same speed.

It's because your measurement of the time (or distance, depending on how you look at it) is flawed. It violates the Einstein clock synchronization protocol. http://en.wikipedia.org/wiki/Einstein_synchronisation

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It's because your measurement of the time (or distance, depending on how you look at it) is flawed. It violates the Einstein clock synchronization protocol. http://en.wikipedia.org/wiki/Einstein_synchronisation

I ment all observations done from FOR of Earth.

To show I am not completed stubborn.

If at the U-turn point, a flash light is emitted, I can imagine a spaceship tracking behind the flash, and arriving at Earth a few instant after the flash arrival.

But that is not exactly what happens.

In fact the spaceship flashes constantly.

The spaceship constantly follows constant flashes.

I must have wood in my head.

Edited by michel123456
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Bolded where it hurts. What I cannot swallow is that I observed the ship going slowly, and coming back in a blink of the eye, evoluting at the same speed.

When an ambulance drives by you, you know that the 'dee doo dee doo' siren is letting out 'dee doo's at a constant rate, yet you hear more 'dee doo's in the 100 yards the ambulance travels toward you than in the 100 yards it travels after it has passed you (travelling away from you).

You don't have to use yards either.... you can use time:

When an ambulance drives by you, you know that the 'dee doo dee doo' siren is letting out 'dee doo's at a constant rate. But when the ambulance is travelling toward you, it takes 25 seconds to hear 25 'dee doo's, and when the ambulance is traveling away from you it takes 40 seconds to hear 25 'dee doo's

If you know the 'dee doo's are being emitted at a constant rate, how can this discrepancy exist?

analogous to:

It is stated from the beginning that speed to go is the same as speed to come back. That is not negotiable

It means that its apparent speed, as measured from our FOR, is $X/25$, which is different from the measured speed at departure.

But that was not negotiable.

Where am I wrong?

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