# Privileged frame

1. Jul 14, 2006

### Born2Perform

i know those type of question are posted every day but it is short:

one twin is on a spaceship and the other on a planet, moving with relative speed.

each twin sees the other's time slow down until the simmetry is not broken, and the moving twin get discovered.
but doesn't it mean that in every simmetrical sistem there is always a privileged frame, also if we cannot know it without acceleration?
isn't this against the relativity principle?

2. Jul 14, 2006

### pervect

Staff Emeritus
You seem to be missing the essential symmetry of the problem.

Either twin can visit the other. If the earthbound twin is Alice and her brother is Baker, Alice can leave the Earth on a spaceship to visit her brother, or her brother can turn around to revisit the Earth.

The delta-v requirement will be the same in either case.

Assuming that Alice and Baker were originally the same age when Baker left Earth,
If Alice is the one who travels, she will be the younger twin at their reuniion, she will be the younger twin. If Baker turns around, the reverse will be true.

There is no essential assymetry in the problem, nor is there a preferred frame.

Last edited: Jul 14, 2006
3. Jul 14, 2006

### Born2Perform

What i'm missing is this: ok it is simmetrical, but in the end, if we know that one twin must stay firm and the other move, for sure the one that is firm doesn't become younger in comparison to the other.

so, if the guy on the spaceship says that his brother is becoming younger, he is not right: because he is the one that is moving, even if he doesn't know it, so his telescope view is the wrong frame, in fact he will realize later it with an acceleration.

in know i'm completly wrong, but for me this seems ok...

4. Jul 14, 2006

### pervect

Staff Emeritus
This is an improvement, but it's better to take the "wait and see" approach.

The "wait and see" approach says that simultaneity of distant events is a relative concept, not an absolute one. Therfore one just waits until he/she gets here to see who is older and who is younger.

At zero distance, there is no uncertainity about simultaneity - at a distance of 'n' light years, the uncertainity is +/- n years.

Another way of saying this: suppose we have to events, A and B. We wish to determine which came first.

Basically, if a light beam can reach from event A to event B, everyone will agree that event B is later than event A. If a light beam cannot reach from A to B different observers will have different opinions on which came first.

5. Jul 14, 2006

### actionintegral

I think born2perform, like me, is asking "help me to understand this" type questions.

I think he is asking "you and me are at rest and have synchronized watches". I start moving and we compare watches while I keep moving.
My watch is running slower than yours. I can see that with my own eyes.
Therefore I ABSOLUTELY MUST be moving. And you ABSOLUTELY MUST be at rest.

6. Jul 14, 2006

### Born2Perform

I could not have expressed to me better.
and 2 distant watches could be sinchronized, in fact if i put a lamp in the center, and 2 rays start at the same time from the lamp, when they reach the clocks the clocks will start, so they can be synchronized.

really i cannot solve this...what i'm thinking of this problem is like: "the truth exists but we cannot know it"
i'm just sayng that one frame is true and the other modified..

Last edited: Jul 14, 2006
7. Jul 14, 2006

### pervect

Staff Emeritus
And I am (and have been) trying to explain that simultaneity is relative. To keep things really simple, I'm saying that you can't "really" compare your watch with a distant watch. You are assuming that you can do this - that is the root of your problem.

When I say "really" I am making a philosphical statement, which I will make more precise - you can't compare your watch with a distant watch in a manner that is indepedent of the observer. So when I say "really" I actually mean "in an observer independent manner".

To use the famous example from Einstein

http://www.bartleby.com/173/9.html

two events (lightning strokes) which are simultaneous in one frame of reference are NOT simultaneous in another moving frame of reference.

8. Jul 14, 2006

### MeJennifer

There really is no problem or paradox.

The problem usually is that two different things are intermixed.

Relativistic measurements

When objects X and Y change their distance at a constant rate they both will observe that the clocks of their counterparts will run slower, not just clocks, the length (in the direction of travel) will be shorter and their (relativistic) masses will be greater. In actuality none of those properties change, their simply appear to be changed due to relativistic effects. So in other words in relativity clocks never run slower, they only appear to run slower!

Paths in space-time

How much time something takes to travel from A to B depends on how we travel in space-time. If we travel on a geodesic which is the shortest path in space-time the elapsed time will be maximized. However other paths will take less time. Note that it is not about time running slower or faster but simply on how much time is spent.

Think about a trip from A to B comparing a short way and a long way with the same speed. The amount of time you will have traveled is obviously greater for the longer path.
Would you, if you compare the odometers, conclude that time slowed down for the shorter path? No you would not right? It is simply that you spent less time.

Now with time in space-time it is exactly the opposite! A longer path will have less time spent than a shorter path. That is counterintuitive for our sublight experiences but if you consider the Minkowski (you can use another form as well where the sign is different but it really does not matter) metric:

$$ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2$$

you see that to calculate the distance between two events is different than in our classical idea of space.

If we simplify this metric (only one spatial dimension (s) and make c=1) for the sake of explanation we would get:

$$ds^2=t^2 - ds^2$$

Now for two observers who separate and meet later we would have a similar distance in space-time.

So it becomes like a playing match between time and space.

If one travels a lot of space (s) it would mean there less amount for time (t) and if one travels no space (s) it would mean that time is maximum.

So, I hope you can see that if you compare clocks of different objects with the same distance in space-time they do not necessarily have to show the same elampsed time, it really depends on how much space they travelled. It will turn out that some travelers simply spent less time than others.

So with regard to the twin problem:
When forces, such as the electro-magnetic force of a rocket engine (the burning of kerosene) cause an object to accelerate it is no longer traveling on a geodesic. And thus the elapsed time will definitely be shorter than if it had traveled on a geodesic.

So when A accelerates away from B it means that A is no longer traveling on a geodesic (if both were already not traveling on a geodesic it would also work) and if they would later meet and compare their clocks, it would turn out that less time has elapsed for the traveler. Not, and this is important, because his clock went slower, it did not, but simply because he spent less time traveling through space-time.

Last edited: Jul 14, 2006
9. Jul 16, 2006

### yogi

This is a falacious - you cannot make this experiment - we can synchronize all the clocks in the earth Frame (which includes the clock at the turn around destination) and you will in fact discover at the time the traveling twin arrives at the destination and stops (the turn around point), the traveling twin's clock reads less than all of the clocks in the earth frame. This does not mean that during transit the traveling twin can look at his own clock and determine that it is running slower than the earth clocks - what occurs is that the traveling clock accumulates less time - because of the interrelationship between space and time. The space interval and the temporal interval for the traveler lead to a different amount of lapsed time than the clocks of the earth frame which experience only a temporal change. From the standpoint of the earth frame, the traveling twin's clock appears to be running slow; from the perspective of the traveling clock, the earth clocks appear to be running slow. So as between two clocks in relative motion, neither can determine which has been put in motion until the trip is complete. In the case of the twins, half the time is lost during the outward journey, the other half is lost during the return voyage if the outbound and inbound velocities are the same.

Einstein's assertion that: "Moving clocks run slow." is somewhat misleading (at least w/o further explanation). What should be said is that: "If one of two synchronized clocks is put in motion, the one put in motion will accumulate less time during the spacetime interval. The spacetime interval is invariant, and it is this principle that resolves the apparent paradox of the twins and other questions re relativity

Last edited: Jul 16, 2006
10. Jul 16, 2006

### MeJennifer

Very true!

The whole "clocks running slow" idea is a constant source of confusion. It implies the idea of absolute time.

11. Jul 16, 2006

### JesseM

What do you mean by "spacetime interval"? Normally the spacetime interval is proportional to the proper time along the worldline, but obviously all clocks tick by the same amount in a given amount of proper time along their own worldline. Also, it doesn't matter which is "put into motion" (accelerated) initially, all that matters is which one had the straight (inertial) path between the time they started moving apart and the time they reunited, and which one had the bent (non-inertial) path--a straight path between two points in spacetime will always have a greater proper time. If one clock is initially accelerated to get it moving away from a second clock, but after that it moves inertially while the second clock drifts away for a while and then accelerates to turn around and catch up with it, it will be the second clock that shows less time, not the first one that was initially "put into motion".

12. Jul 16, 2006

### JesseM

There is no priveliged inertial frame here, you can analyze this situation from any inertial frame you like and you'll get the same answer to which clock will have read less time when they meet. For example, you could pick the inertial frame where the ship was at rest during the outbound leg while the planet was moving away from it, then after the ship turned around it was moving even faster than the planet during the inbound leg. This frame will say that the earth's clock was ticking slower during the outbound leg, but the ship's clock was ticking even slower than the earth's during the inbound leg, with the net result that the ship's clock has elapsed less time when they reunite, by exactly the same amount that you'd predict if you analyzed it from the frame where the earth was at rest. We can go through a simple numerical example of this if you want.

13. Jul 17, 2006

### yogi

All true - but in the simple example I wanted to get across, there is no need to bring the two clocks together at the starting point - one needs only to consider a one way trip to alpha centuri (AC) in which there are no bent paths. When the traveling twin clock B arrives at AC and stops he will find his B clock reads less than every other clock in the frame from which he departed - not only the clock his brother maintains on earth but a clock on AC. The AC clock and the earth clock always read the same - and the B clock will read less than AC clock upon arrival

If you don't believe this, ask a high speed pion.

14. Jul 17, 2006

### Ich

:rofl: :rofl:

15. Jul 17, 2006

### Born2Perform

Just one more issue:

Going back on twins: One twin leaves the Earth accelerating, and the other stays on the planet. so we know who is moving.

Then the twin on the spaceship, that is moving but is not accelerating more, dies, and his son takes control of the spaceship.

SO, we know the spaceship is moving, but the son doesn't: for him the system is symmetrical, and he has not a way to discover the truth, that his ship is moving.

What you say?

Last edited: Jul 17, 2006
16. Jul 17, 2006

### HallsofIvy

Staff Emeritus
The statement "he has not a way to discover the truth, that his ship is moving" is incomplete. Relative to what frame of reference? His ship is moving relative to some frames of reference, not moving, relative to others (in particular, a person is alway "not moving" relative to his own frame of reference). It simply isn't true to say it is "the truth, that his ship is moving". That's a fundamental fact of relativity (indeed, the reason for the name).

If, say, one twin accelerates, leaving the other sitting in a "non-moving" spaceship (non-moving relative to the frame in which both were originally stationary- not accelerating), then, eventually, stops accelerating, each of the twins would see the other moving relative to himself- and, therefore, each would see the other's clock moving slower than his. That's not a paradox. The only way there could be a paradox would be to have the two twins come back together again. That would require one of the twins to accelerate again so they can come back together. That could be done in either of two ways: the twin who first accelerated could "decellerate" until he was going back to the "stationary" twin and then "accelerate" to a stop (relative to his twin) or the originally "stationary" twin could accelerate to catch up with the other and then decelerate to a stop (relative to his twin). In the first case, a detailed calculation, including the accelerations, would show that the "accelerating" twin would be younger than the other. In the second case, the same calculation would determine that their clocks would read the same after all the accelerations and decellerations.

If the situation involves one spaceship that continues accelerating, it would be easy for the son of the twin on the non-accelerating spaceship to determine that the other is accelerating and so his clock would continue slowing relative to the other person's. A person on the accelerating spaceship could take himself to be stationary (in his own frame of reference) but would have to accept a "gravitational" force equal to ma which would still cause his clock to slow.

17. Jul 19, 2006

### yogi

After the Father croaks, the Son has no information re the history - so its correct to say that as far the Son and the Brother of the father, both view the other as moving - if each sets up a two clock experiment in their own frame to measure the rate of a clock in the other frame, each well conclude that the clock in the other frame runs slow. But this is a different problem and its endemic of all twin type paradox(s). The traveling twin is moving to a new point in spacetime - he will experience a temporal distance ct' and a spatial distance (vt) - the twin that stayed at home only experiences a temporal distance (ct) from start to finish. This is true whether the traveler stops somewhere along the way or turns around and returns - all the clocks in the frame of the stay-at-home twin read the same. The watch carried by the traveling twin will actually read less whenever the traveling twin stops and his watch is compared with any clock in the original frame - no matter where it is located. This is the surprising result Einstein derived - what is hard to accept is that the final physical difference in the clock readings (an actual time difference) was arrived at by Einstein using mathematics that followed from the observing of times and distances in other frames. The fact that Einstein arrived at physical time difference in a situation involving two inertial observer, neither having a preferred frame - is amazing - Einstein called it a peculiar result.