Thread: A special relativity problem

This website is pretty fun. Here's a tricky special relativity problem, whoever can give a solution and justification gets a proverbial cookie.

If a spaceship travels away from earth at relativistic speed (close to the speed of light), turns around, and flies back to earth at relativistic speed, 'time dilation' will occur. To an observer on earth, his watch would be ticking more quickly than a clock on the spaceship. However, remember that from the frame of reference of an observer on the spaceship, the earth could be percieved as 'moving' with respect to the spaceship, so an observer on the ship would think that time on earth is being dilated.

So the question is, if two observers are the same age, and one is on earth and the other is on the spaceship, will either one of the observers be older than the other after the spaceship returns? If so, which observer, and why?

Originally Posted by Arvand

This website is pretty fun. Here's a tricky special relativity problem, whoever can give a solution and justification gets a proverbial cookie.

If a spaceship travels away from earth at relativistic speed (close to the speed of light), turns around, and flies back to earth at relativistic speed, 'time dilation' will occur. To an observer on earth, his watch would be ticking more quickly than a clock on the spaceship. However, remember that from the frame of reference of an observer on the spaceship, the earth could be percieved as 'moving' with respect to the spaceship, so an observer on the ship would think that time on earth is being dilated.

So the question is, if two observers are the same age, and one is on earth and the other is on the spaceship, will either one of the observers be older than the other after the spaceship returns? If so, which observer, and why?

The Earth twin will be older. The trick is in the fact that the spaceship has to turn around in order to return to Earth and must change reference frames to do so. As a result the Relativity of Simultaneity comes into play such that when he turns around, the time on Earth jumps forward from his perspective.

According to him, the Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back, with the end result being the Earth twin having aged more when they meet up again (he ages more than enough during turn around to make up for the slow aging during the rest of the trip.

Nice, its funny because that's the solution I'd suggest, but rarely the one I hear from other people. Others use some sort of reasoning involving the Doppler effect, which is a lot more confusing in my opinion.

If a spaceship travels away from earth at relativistic speed (close to the speed of light), turns around, and flies back to earth at relativistic speed, 'time dilation' will occur. To an observer on earth, his watch would be ticking more quickly than a clock on the spaceship. However, remember that from the frame of reference of an observer on the spaceship, the earth could be percieved as 'moving' with respect to the spaceship, so an observer on the ship would think that time on earth is being dilated.

So the question is, if two observers are the same age, and one is on earth and the other is on the spaceship, will either one of the observers be older than the other after the spaceship returns? If so, which observer, and why?

Isn't this the Twin Paradox?

Originally Posted by Janus

According to him, the Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back, with the end result being the Earth twin having aged more when they meet up again (he ages more than enough during turn around to make up for the slow aging during the rest of the trip.

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

For clarity, I'll post part of Janus' original statement to help explain why.

The Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back,

The twin is undergoing acceleration (or, in this case, deceleration) during turn around which obviously causes the time dilation to decrease, thereby meaning that the twin will start to age more rapidly.

By slowly, I believe Janus is referring to the fact that the twin is moving at a constant velocity, so the time dilation effect will also be constant, meaning that the twin will indeed age slowly.

Admittedly, I am a bit confused by Janus. Why would the Earth twin age slowly and appear to age more rapidly, when the effect of time dilation is actually felt by the twin in the ship?

However, I've answered your question, Arcane, without referring to the twin on Earth or space, so that the answer can then apply to what Janus replies.

Originally Posted by Arcane_Mathematician

Originally Posted by Janus

According to him, the Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back, with the end result being the Earth twin having aged more when they meet up again (he ages more than enough during turn around to make up for the slow aging during the rest of the trip.

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

The planet doesn't age during the slow down, It would age(according to the traveler) when he accelerates up to speed while leaving the Earth. Then the planet ages slowly on the trip out, ending up "older" upon the traveler's arrival.

Originally Posted by Liongold

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

For clarity, I'll post part of Janus' original statement to help explain why.

The Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back,

The twin is undergoing acceleration (or, in this case, deceleration) during turn around which obviously causes the time dilation to decrease, thereby meaning that the twin will start to age more rapidly.

By slowly, I believe Janus is referring to the fact that the twin is moving at a constant velocity, so the time dilation effect will also be constant, meaning that the twin will indeed age slowly.

Admittedly, I am a bit confused by Janus. Why would the Earth twin age slowly and appear to age more rapidly, when the effect of time dilation is actually felt by the twin in the ship?

However, I've answered your question, Arcane, without referring to the twin on Earth or space, so that the answer can then apply to what Janus replies.

You're confused, because you are trying to think in terms of absolute velocity. Time dilation is not something someone "feels", it is something you measure as occurring to the frame that has a relative motion to your frame. Since the has a Relative motion to the traveler's frame, as measured by the traveler, the traveler measures a time dilation of the Earth frame.

When the traveler reaches the end of his trip and reverses course, he changes reference frames.(from one with a relative motion away from the Earth to one with a relative motion towards.) It is this change of reference frames that, from his perspective, causes the Earth to jump forward in age.

IOW, while the two twins will agree as to which is older when they meet up, they will not agree as to how this came about, and there is no way to say that one Twin's version of events is more correct than the other.

Originally Posted by Arcane_Mathematician

Originally Posted by Janus

According to him, the Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back, with the end result being the Earth twin having aged more when they meet up again (he ages more than enough during turn around to make up for the slow aging during the rest of the trip.

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

The relativity of simultaneity is the key to everything. In order to see this you have to imagine each objects connected to a string of clocks going out ahead of them and behind them in the direction of motion, and moving along with that object. Now according to those on the earth the string of clocks attached to the earth will always read the same time as the clocks on the earth, and according to those on the ship the clocks attached to the ship will always read the same time as the clocks on the ship (as long as the velocity remains constant). This is not what they actually see because they need to wait for the light to reach them and that distorts everything, so from now on when I use the words "look" and "see" I really mean when they include a calculation to take into account the time it took the light to reach them.

So as the ship approaches the speed of light, when those on the ship look at the clocks attached to the earth they all read different times and when those on the earth look at the clocks attached to the ship they also read different times. Lets say the ship is traveling at 86.6% of the speed of light with respect to the earth. Now if those on the ship look at one particular clock attached to the earth then the time on that clock will be passing more slowly recording only half an hour for every hour that passes on the ship. Likewise when those on the earth look at one particular clock attached to the ship then the time on that clock will be passing more slowly recording only half an hour for every hour that passes on the earth.

HOWEVER when the ship looks at the clocks attached to the earth they don't all read the same but have different times on them. So if instead of looking at only ONE clock the ship looks at whatever clock attached to the earth which is next to the ship at that time, then the time observed in that way will NOT be slower but faster with two hours passing for every hour that passes on the ship. The same is true of those on earth looking at whichever clock attached to the ship happens to be closest to the earth at the time (as long as those clocks keep moving).

What you have to be careful about is in using the phrase "at the same time". We can talk about what those on the ship see and what those on there earth see but what you cannot do is compare what they are seeing "at the same time" because that makes no sense. This will also cause trouble if we talk about a whole string of clocks suddenly moving or stopping, because all of them doing so "at the same time" will have an ambiguous meaning.

Anyway if the ship's clock reads 10 hours after starting, and the velocity has been 86.6% of the speed of light then the clock attached to the earth but nearest the ship will read 20 hours after the starting time, but according to those on the ship the clocks actually on the earth will read 5 hours after starting.

If the ship comes to a stop when the ship clocks read 10 hours, then here is where everything changes for those on the ship. Those on the ship see the clocks attached to the earth all change until they agree with the time on the clock that is closest to the ship, 20 hours. This suggests that we should ask what those on the earth see when their clocks read 20 hours after starting. What they see is all the clocks attached to them read the same time and it is the clocks attached to the ship which all have different times. For the earth people at this time, the clocks on the ship reads 10 hours and if you calculate what the clock attached to the ship which is nearest them would read, you get 40 hours.

But that does not make sense because it would mean that this clock would have to have been traveling 30 hours past the time when we have said that the ship stopped. And thus if we really want to talk about all the clocks attached to the ship stopping at the same time then what those on the earth would see is that 5 hours after the ship left, a clock attached to the ship that reads 10 hours will stop next to the earth, even though they have to wait another 15 hours for the ship itself to stop, and all the clocks attached to the ship between them and the ship will have stopped moving when they read 10 hours but that means that these clocks will have been waiting with the earth various amounts of time for the ship to stop and will not read the same time when the ship stops. The clock nearest the earth would then read 25 hours. So as a result, those on the ship would see all these clocks all suddenly change to read different times just as they see all the clocks attached to the earth change to read the same time of 20 hours.

Well that would be one way of interpreting the meaning of all these clocks "stopping at the same time". Another way would be to have those on the earth see all the clocks attached to the ship stopping at the same time as when they see the ship stop. In that case those on the ship would still see the times on these clocks all suddenly changing not only to different times but to different locations relative to the earth (all closer together) and the times would be different from the situation described above. For example the clock nearest the earth would read 40 hours rather than 25 hours.

In any case, perhaps the thing to notice is that those on the earth do NOT see any clocks either those attached to the earth or those attached to the ship suddenly changing anything. It is only for the ship that suddenly decelerates which changes how it must measure space and time in a rather abrupt manner. One of the most significant changes is the distance it sees between itself and the earth, which stretches to twice what the distance seemed to be while it was moving.

P.S. All this was worked out mathematically using the lorentz transformations. Perhaps you can use these to work out what would happen if instead of the ship stopping, it is the earth that suddenly accelerates the to same velocity as the ship.

Originally Posted by Janus

Originally Posted by Arcane_Mathematician

Originally Posted by Janus

According to him, the Earth twin ages slowly on the trip out, ages a lot very rapidly during turn around, and then ages slowly again on the trip back, with the end result being the Earth twin having aged more when they meet up again (he ages more than enough during turn around to make up for the slow aging during the rest of the trip.

Okay, I have an issue here. Why is the twin aging slowly on the return trip? It doesn't make sense to me, and I'll explain my position:

Take a distant planet, say, 500 light years away. If an observer were to travel to that planet (no return trip), how could the planet age slowly on the trip there and yet become so aged in the slow down? I don't really get it.

The planet doesn't age during the slow down, It would age(according to the traveler) when he accelerates up to speed while leaving the Earth. Then the planet ages slowly on the trip out, ending up "older" upon the traveler's arrival.

So, the quick-aging occurs during the acceleration towards the object. Why is this?

Originally Posted by Arcane_Mathematician

Originally Posted by Janus

The planet doesn't age during the slow down, It would age(according to the traveler) when he accelerates up to speed while leaving the Earth. Then the planet ages slowly on the trip out, ending up "older" upon the traveler's arrival.

So, the quick-aging occurs during the acceleration towards the object. Why is this?

That is correct. It all happens when the ship is turning around in order to head back to the earth. However it is only "quick aging" from the perspective of calculations made by those on the ship because they change how they measure space and time. They don't actually see such things happen because you cannot really see things that are happening that far away and changing your velocity only changes the rate at which you receive light from what you are looking at. There are some rather big distortion in what you see due to traveling at such velocities but that is a whole other topic called the aberration of light (for which there are a few simulators including mine).

While Janus makes his/her attempt to explain all this with moving pictures I will continue my own explanation with a few diagrams:

This is the senario of the ship leaving earth at 86.6% of the speed of light where both have a string of clocks attached to them front and back. The ship clocks (c1,c2,c3,c4) will be in red and the earth clocks in blue:

From earth point of view

...c1.........................c2.........................ship........................c3...........................c4.
30h<----8.66lh---->15h<----8.66lh---->_0h<----8.66lh--->-15h<----8.66lh--->-30h
..c1...........................c2........................earth........................c3...........................c4
_0h<----8.66lh---->_0h<----8.66lh---->_0h<----8.66lh---->_0h<----8.66lh---->_0h

then after 10hours earth time

..............................c1...........................c2.........................ship...........................c3.
___<----8.66lh---->35h<----8.66lh---->20h<----8.66lh---->_5h<----8.66lh--->-10h
..c1...........................c2.......................earth.......................c3.........................c4
10h<----8.66lh---->10h<----8.66lh---->10h<----8.66lh--->10h<----8.66lh--->10h

then after 20hours earth time

.................................................. ...........c1...........................c2.........................ship
___<----8.66lh---->___<----8.66lh---->40h<----8.66lh---->25h<----8.66lh---->10h
.c1...........................c2........................earth.......................c3..........................c4
20h<----8.66lh---->20h<----8.66lh---->20h<----8.66lh--->20h<----8.66lh--->20h

h is for hours and lh is for light hours


Then from the ships point of view

...c1.........................c2.........................ship........................c3...........................c4.
_0h<----4.33lh---->_0h<----4.33lh---->_0h<----4.33lh---->_0h<----4.33lh---->_0h
..c1...........................c2.......................earth........................c3..........................c4
-15h<---4.33lh--->-7.5h<---4.33lh---->_0h<----4.33lh--->7.5h<----4.33lh---->15h

then after 5hours ship time

...c1.........................c2.........................ship.........................c3...........................c4.
_5h<----4.33lh---->_5h<----4.33lh---->_5h<----4.33lh---->_5h<----4.33lh---->_5h
c2..........................earth........................c3..........................c4
-5h<----4.33lh---->2.5h<----4.33lh---->10h<----4.33lh--->17.5h

then after 10hours ship time

.c1...........................c2.........................ship.........................c3...........................c4.
10h<----4.33lh---->10h<----4.33lh---->10h<----4.33lh---->10h<----4.33lh---->10h
earth........................c3.........................c4
_5h<----4.33lh--->12.5h<---4.33lh---->20h


After the ship stops then ship and earth agree again

.................................................. .................................................. ....................ship
<------------------17.32lh----------------->___<----------------17.32lh--------------->10h
..c1...........................c2........................earth........................c3...........................c4
20h<----8.66lh---->20h<----8.66lh---->20h<----8.66lh---->20h<----8.66lh---->20h

I left out the other clocks attached to the ship because of the ambiguity in how they would stop.