The twins paradox and explanation

[Stellar1]

Hello,
I'm reading through my textbook and it claims the explanation for the twins paradox is that the space-traveler must experience acceleration during his journey... but could it not be said that, relative to him, the rest of the universe is accelerating in the opposite direction?

 

[Mentz114]

...the rest of the universe is accelerating in the opposite direction?

I think that's pushing symmetry too far. The two situations require different amounts of energy for a start.

See the other thread on this page about the twins non-paradox.

 

[JesseM]

Hello,
I'm reading through my textbook and it claims the explanation for the twins paradox is that the space-traveler must experience acceleration during his journey... but could it not be said that, relative to him, the rest of the universe is accelerating in the opposite direction?

The twin will experience G-forces when he accelerates, while other parts of the universe that are moving inertially do not, so there is an objective way to decide who has accelerated in SR.

 

[Stellar1]

One more question...

So from the perspective of the traveller, it looks as if time is going by slower for everything around him, correct? (Because in his frame of reference, it is everything else that is moving...)?

 

[JesseM]

One more question...

So from the perspective of the traveller, it looks as if time is going by slower for everything around him, correct? (Because in his frame of reference, it is everything else that is moving...)?

The traveller doesn't have a single inertial rest frame...his inertial rest frame before turning around is different from his inertial rest frame after turning around. In the frame where the traveller is at rest before turning around, the Earth's clock is indeed ticking slower than his before the turnaround, but after the turnaround his speed is even greater than the Earth's in this frame, and so his clock is ticking even slower. Likewise, in the frame where the traveller is at rest after turning around, the traveller is moving even faster than the Earth before the turnaround, so his clock is ticking slower until he turns around. The result is that each of these inertial frames predicts that the traveller's total elapsed time between leaving Earth and returning is less than the time elapsed on the Earth's clocks.

 

[Stellar1]

Ok, forgetting about the paradox as it is written... let's simply make it into this: I have a twin brother. I decide to depart on an epic voyage in a straight line at 0.9c. So, in reference with me, time would seem to me to be going by slower for my brother, is that not so?

 

[JesseM]

Ok, forgetting about the paradox as it is written... let's simply make it into this: I have a twin brother. I decide to depart on an epic voyage in a straight line at 0.9c. So, in reference with me, time would seem to me to be going by slower for my brother, is that not so?

Yes, in your rest frame your brother would be aging more slowly, while in your brother's rest frame you'd be aging more slowly.

 

[Janus]

Ok, forgetting about the paradox as it is written... let's simply make it into this: I have a twin brother. I decide to depart on an epic voyage in a straight line at 0.9c. So, in reference with me, time would seem to me to be going by slower for my brother, is that not so?

Yes, as long as the relative velocity between you and you brother is .9c, then your brother would age slower by your measurement.

 

[Stellar1]

Ok, so if I decided to fly circles around him at 0.9c, given that the radius of my circle was small enough, I could observe him aging more slowely, correct?

 

[JesseM]

Ok, so if I decided to fly circles around him at 0.9c, given that the radius of my circle was small enough, I could observe him aging more slowely, correct?

No, in this case you wouldn't be moving inertially either (inertial motion = constant speed and direction), you'd be experiencing constant G-forces as you moved in a circle. The usual equations of SR like the time dilation equation only work in inertial frames. If you fly in circles around an inertial observer, then no matter what inertial frame one analyzes the problem in, you will age less than the observer at the center with each orbit (and if they are sending out light and you are watching their image, you will see them aging faster than you as well).

 

[Janus]

Ok, so if I decided to fly circles around him at 0.9c, given that the radius of my circle was small enough, I could observe him aging more slowely, correct?

No, because in order to fly in a circle you have to accelerate. (acceleration is a change in velocity and velocity is measured in both speed and direction. Therefore changing direction while maintaining a constant speed is still accelerating. ) The rules dealing with acceleration are different than the rules dealing with inertial motion. You would in fact see your brother aging faster in this scenerio.

 

[Janus]

I've got to learn to type faster.

 

[Stellar1]

Ok, I guess I'm just not yet at the level to deal with acceleration.

 

[Jorrie]

Ok, I guess I'm just not yet at the level to deal with acceleration.

You don't have to. I found that the easiest way to wrap my head around some of the issues of the 'twin paradox' is to consider time intervals as per the Lorentz transformations. It tells us that an inertial observer that is present at two events will always record less time between the events (i.e., proper time) than what any inertial observer not present at the same two events will record.

In the case of the twins, there are four events that define the inertial phases of the away-twin's voyage: departure, start turnaround, end turnaround and arrival back home. Only the away-twin is present at the two turnaround events and will hence record the lesser time interval between them and the departure and arrival events respectively. The accelerating period between the start and stop turnaround events can be insignificant if the trip is very long. It is however important to note that for the twins to reunite, at least one of them must accelerate at some point in time, but it can usually be ignored in the calculations.

 

[Xeinstein]

No, because in order to fly in a circle you have to accelerate. (acceleration is a change in velocity and velocity is measured in both speed and direction. Therefore changing direction while maintaining a constant speed is still accelerating. ) The rules dealing with acceleration are different than the rules dealing with inertial motion. You would in fact see your brother aging faster in this scenario.

Could you explain why I see my brother aging faster in this scenario?

 

[tim_lou]

Could you explain why I see my brother aging faster in this scenario?

imagine a rocket accelerating constantly upward, in my reference frame. One person at the top of the rocket sends a light signal to the bottom. Since the rocket is accelerating up, each consecutive light pulse needs to travel progressively less distance. Suppose that the signal at the top looks like tick... tick... tick, then the signal received at the bottom might look like tick.tick.tick.tick.

Similarly, when you are going in circle, you accelerations inward, and suppose your brother sends you his heart beats as signals... i.e. tick... tick... tick, you would see something like tick.tick.tick.tick. i.e. your brother is aging faster according to you.

 

[Xeinstein]

imagine a rocket accelerating constantly upward, in my reference frame. One person at the top of the rocket sends a light signal to the bottom. Since the rocket is accelerating up, each consecutive light pulse needs to travel progressively less distance. Suppose that the signal at the top looks like tick... tick... tick, then the signal received at the bottom might look like tick.tick.tick.tick.

Similarly, when you are going in circle, you accelerations inward, and suppose your brother sends you his heart beats as signals... i.e. tick... tick... tick, you would see something like tick.tick.tick.tick. i.e. your brother is aging faster according to you.

In your scenario, the direction of acceleration and that of velocity are the same.
But in this scenario, the distance between me and my brother does not change with time
Can we still use Doppler effect to explain it?

 

[Xeinstein]

No, because in order to fly in a circle you have to accelerate. (acceleration is a change in velocity and velocity is measured in both speed and direction. Therefore changing direction while maintaining a constant speed is still accelerating. ) The rules dealing with acceleration are different than the rules dealing with inertial motion. You would in fact see your brother aging faster in this scenerio.

In the CERN muon storage ring experiment, where the muons had a time of decay depending only on their velocity (in agreement with the time dilation formula) despite the fact that their acceleration was of 10^18g.

 

[JesseM]

In the CERN muon storage ring experiment, where the muons had a time of decay depending only on their velocity (in agreement with the time dilation formula) despite the fact that their acceleration was of 10^18g.

No one is saying that the rate of time dilation at any given moment depends on acceleration; in any inertial frame, a clock moving at velocity v at a given moment will be slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$ at that moment. It is nevertheless true that if two clocks move apart and then reunite, the one that accelerated will always show less total time; the total time can be found using the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} \, dt$$, where v(t) is the velocity as a function of time in the inertial coordinate system you're using, and this integral has the property that if you take the v(t) for two objects which start and stop and the same positions and times, the one with v(t) = constant will always yield a greater value than the one with a varying v(t). This can be thought of as a "geometrical" property of spacetime, similar to the fact that when you draw two paths between a pair of points on a piece of paper, the one with constant slope in whatever coordinate system you use (i.e. the path that's a straight line) will always have a shorter total length than the one whose slope is not constant (the path that is not totally straight)--"a straight line is the shortest distance between two points". Similarly, a straight worldline has the longest proper time between two points in the uncurved 4D spacetime of special relativity.

I elaborated on this analogy to plane geometry a bit in post #8 of this thread:

Here's an analogy--on a 2D sheet of paper, draw two points, a "starting point" A and a "finishing point" B, and then draw two paths between them, one a straight line and the other a bent line. Now draw x and y coordinate axes, with the y-axis parallel to the the straight line. To get some specific numbers, let's say the starting point A is at x=0, y=0 and the finishing point B is at x=0, y=8, and the bent path consists of two straight line segments at different angles, the first of which of which goes from A to a point C at x=3, y=4, while the second line segment goes from C to B. Note the y-coordinates of the two points A and B, in this case y=0 and y=8, and then for any y-coordinate in between these two values, like y=4, there will be a unique point on each path with this y-coordinate. So you can ask about the distance along each path that you'd need to travel to get to the point on the path that has that y-coordinate; let's invent a term for that distance, like "partial path length". For example, at coordinate y=4, the "partial path length" along the straight path would have to be 4, while the "partial path length" on the bent path would larger, in this case 5 (the distance from point A to point C). If you look at the y-coordinate of the finishing point B, y=8, then the "partial path length" at y=8 would just be equal to the total length of the path from the starting point to the finishing point. In this case the "partial path length" for the straight path at y=8 would be 8, while the "partial path length" for the bent path would be 10.

Now, keep the same two paths between the same two points, but redraw your x and y axes so the y-axis is no longer parallel to the straight path--for example, we might draw the y-axis so it's parallel to the line segment joining A and C. Now the coordinates of the starting point A and the finishing point B for each path won't be the same--if we place the origin so that A still has coordinatex x=0, y=0, then the finishing point B will now have coordinates x=0, y=5.12. It's still true that "partial path length" for each path at the y-coordinate of the finishing point, y=5.12, must just be the total length of each path, which won't have changed just because we picked a different coordinate system, so it'll still be 8 for the straight path and 10 for the bent path. But at some earlier y-coordinate, since the lines of constant y are now at different angles, they'll intersect the two paths at different points so the "partial path length" at this y-coordinate will be different--for example, at y-coordinate y=2.56 in this coordinate system, the "partial path length" on the straight path would be 4 (just like the partial path length at y=4 in the previous coordinate system), while the "partial path length" on the bent path would be 2.56. Notice that while in the previous coordinate system the "partial path length" of the straight path was always smaller than the bent path at a given y-coordinate, in this coordinate system the "partial path length" of the straight path can actually be larger for certain values of y, although both coordinate systems agree that the total path length between A and B is shorter for the straight path.

All of this is pretty closely analogous to the situation in relativity, with different coordinate systems on the paper being analogous to different inertial reference frames in relativity, the y-coordinate being analogous to the coordinate time t in a given frame, and the "partial path length" at a given y-coordinate being analogous to the proper time T accumulated by a particular clock at a given coordinate time t. Just as both coordinate systems agreed on the value of the "partial path length" at the y-coordinate of point B where the two paths reunite, so different frames in relativity will always agree on the value of the proper time read by each twin's clock at the t-coordinate where they reunite at a single point in space. But hopefully you would agree that there is no single true answer to the question of which path is accumulating "partial path length" faster before they reach point B--this is entirely coordinate-dependent, you can get different answers depending on how you orient your y-axis and none is more "objectively true" than any other. In the same way, I'd say there's no single true answer to the question of which twin is accumulating proper time faster (or 'aging faster') before they reunite at a single point in space.

To spell out the analogy:

1. a given set of xy axes on the paper = a given inertial coordinate system in SR

2. y-coordinate on xy axes = time-coordinate in inertial coordinate system

3. two paths on paper = two worldlines in SR

4. "partial path length" of a given path as a function of y = elapsed time on clock moving along a given worldline as a function of time-coordinate in inertial coordinate system

5. the fact that the rate at which the partial path length is growing at a given y-coordinate depends only on the slope of the line at that y-coordinate = the fact that the rate a clock's elapsed time is growing (i.e. its instantaneous rate of ticking) as a function of the coordinate's system time-coordinate depends only on its speed in that coordinate system

6. The fact that the length of a path between two points that has a non-constant slope will always end up being greater than the length of a path between the same two points with a constant slope = the fact that the elapsed time on a clock that goes between two points in space time with a non-constant speed will always end up being less than the the elapsed time on a clock that goes between the same two points with a constant speed

7. The fact that the statement about geometry (6) can be restated without reference to a particular set of xy axes, and without reference any notion of "slope" or "instantaneous rate that partial path length is growing" in that coordinate system, just by saying "a straight line is always the shortest path between two points" = the fact that the statement about SR in (6) can be restated without reference to a particular inertial frame, and without reference to any notion of "speed" or "instantaneous rate that a clock is ticking" in that frame, just by saying "an inertial worldline always gives the greatest elapsed time between two points".

 

[Xeinstein]

No one is saying that the rate of time dilation at any given moment depends on acceleration; :

I think Janus did say that

No, because in order to fly in a circle you have to accelerate. (acceleration is a change in velocity and velocity is measured in both speed and direction. Therefore changing direction while maintaining a constant speed is still accelerating. ) The rules dealing with acceleration are different than the rules dealing with inertial motion. You would in fact see your brother aging faster in this scenerio.

 

[JesseM]

I think Janus does

No, I am sure you're misinterpreting him--what specific quote of his suggests to you he's saying the rate of time dilation at a given moment depends on acceleration, as opposed to just saying that the total time elapsed on an accelerating clock over some section of its path will be less than the total time elapsed on an inertial clock whose path has the same endpoints?

edit: the statement "The rules dealing with acceleration are different than the rules dealing with inertial motion. You would in fact see your brother aging faster in this scenerio" doesn't imply that, if that's what you're suggesting. The quote is just saying that you can't apply the usual rules of time dilation to an accelerating observer's own non-inertial reference frame, but you can analyze the problem from the perspective of an inertial reference frame like the frame of the observer at the center of the orbit, and if you calculate when light signals from the center observer's clock reach the eyes of the orbiting observer in this frame, you still find that the orbiting observer sees the ticks of the central observer happening more quickly than ticks of his own clock.

 

[thestien]

ok i might get told off for this but i thought posting here is better than starting a new topic.
my main problem starts with the fact that time isn't constant?
so in my brain i can see that both twins in the paradox will always age the same due to the length of time that has past.

i imagine a 4d world with 3 co-ord's call them x,y,z and a time reference called t.

now if I am at location 1x,1y,1z and the time is set to t=0.
after then clicks of time t=10 i would of aged by 10 units of time.
now my twin starts at location 1x,1y,1z and moves along the x-axis for 10000 units but it only took him 1 unit of time his new position is 10000x,1y,1z,1t.
my location is still 1x,1y,1z,1t.
although he may of moved to fast for me to of seen him it has only been 1 unit of time that has past.
then he slows down turns around and moves back to me which takes a little longer so 1.5time units have past on the return journey so now for both me and my twin t=2.5
and we both at 1x,1y,1z,2.5t.

now if I am right in thinking about SR then infact i would be at sayt=2.5 and my twin could be at t=<2.5

and that is where i fail to understand what happens because logic forces me to believe that time can only be a constant and that time isn't relative to the observer but constantly advancing sorry if this is the wrong place to post this?

 

[Fredrik]

You would both be at the same coordinates, obviously (since we're talking about one particular event where you meet again after being separated for a while). But during his trip, his aging rate is slower than yours, from your point of view. It is also a fact that during the trip, your aging rate is slower than his, from his point of view. Those two facts seem contradictory, but they really are not, because of how simultaneity works in special relativity.

I suggest that you try to learn a little about space-time diagrams, and in particular how to determine what sets of events are simultaneous to a moving observer. Then try to understand the space-time diagram for the twin "paradox" that I have posted in several places, most recently in this thread. (Post #3).