Time Dilation Puts Two Observers at different Times?

[JDude13]

Okay.

A man in a spaceship travels at
$$\beta=\frac{\sqrt{3}}{2}$$
giving him a Lorentz Factor of
$$\gamma=2$$
and a time dilation of
$$t'=2t$$

Another man sits "stationary" in space.

After traveling for one of the stationary observer's seconds the moving observer passes by the stationary observer. The stationary observer measures the moving observer's clock at 0.5s because of the time dilation.

However if the moving observer were to measure his clock at 0.5s he would measure the stationary observer's clock at 0.25s.

Then if the stationary observer were to measure the moving observer's clock at 0.25s he would measure 0.125s. Ad infinitum.

This makes no sense to me. Could someone help me out?

 

[DaveC426913]

They are moving wrt each other. Other than the moment of closest approach, how do you propose they will get all those subsequent measurments? The delay of their observations as they recede from each other will affect their measurments of simultaneity.

 

[darkhorror]

in the ships frame of reference 1 second in the spaceship is .5 to the stationary platform.

in the platform's frame of reference 1 second on the platform is .5 seconds on the ship.

 

[JDude13]

in the ships frame of reference 1 second in the spaceship is .5 to the stationary platform.

in the platform's frame of reference 1 second on the platform is .5 seconds on the ship.

so their clocks are faster AND slower?

 

[harrylin]

so their clocks are faster AND slower?

They explain what they observe differently - based on their choice of reference system.

What an observer of S' interprets as a slower clock of S, is explained by an observer of S as a wrong interpretation due to wrong clock synchronization of S' (and vice-verse). Thus , the first thing to understand is relativity of simultaneity:

http://en.wikipedia.org/wiki/Relativity_of_simultaneity

Historically this was also the first thing that was understood.

Cheers,
Harald

 

[ZikZak]

After traveling for one of the stationary observer's seconds the moving observer passes by the stationary observer. The stationary observer measures the moving observer's clock at 0.5s because of the time dilation.

OK, but implicit in this statement is that there was some kind of synchronization procedure that occurred before the two observers arrived at the same place to compare clocks. How did they do that?

However if the moving observer were to measure his clock at 0.5s he would measure the stationary observer's clock at 0.25s.

No. If the moving observer's clock reads 0.5 seconds when it is at the same location as the stationary clock when it reads 1 second, then that is a spacetime event (something which happens at a single location in spacetime) that all observers everywhere must agree on. To say otherwise would be a denial of external reality. So no. If the spaceship and the stationary clocks are in the same place at the same time, and they will read 0.5 and 1 second as measured by one observer, then they must do so as measured by all observers.

The problem is to ask what happened during the attempt at synchronization. How do you suppose the spaceship and stationary observer synchronized their clocks when they were far apart and moving relative to one another?

 

[JDude13]

OK, but implicit in this statement is that there was some kind of synchronization procedure that occurred before the two observers arrived at the same place to compare clocks. How did they do that?



No. If the moving observer's clock reads 0.5 seconds when it is at the same location as the stationary clock when it reads 1 second, then that is a spacetime event (something which happens at a single location in spacetime) that all observers everywhere must agree on. To say otherwise would be a denial of external reality. So no. If the spaceship and the stationary clocks are in the same place at the same time, and they will read 0.5 and 1 second as measured by one observer, then they must do so as measured by all observers.

The problem is to ask what happened during the attempt at synchronization. How do you suppose the spaceship and stationary observer synchronized their clocks when they were far apart and moving relative to one another?

lets say they met up and syncronised their clocks and then one of them took a year-long journey at 5m/s to get half a light-second away. :P

 

[DaveC426913]

Jdude, the thing that you're missing is that there is no "objective time" and there is no agreement on "simultaneous events"; they are different for each observer.

Each spaceship observes what he observes as if he is at rest in his own reference frame. Once the observations are made, the results will reconcile. It works out because, in order to do the comparison, one orf both of the craft must decelerate and rendenzvous, which will change their passage of time wrt each other.

If you walk through your scenario one step at a time and apply the time dilation factors at each time (For example: observer A uses a telescope to check spaceship B's clock - they are 1 light second apart; observer B uses a telescope to check spaceship A's clock - but they are now 2 light seconds apart) the paradox will go away.

 

[ZikZak]

lets say they met up and syncronised their clocks and then one of them took a year-long journey at 5m/s to get half a light-second away. :P

Then in order to get up to speed, he had to accelerate, and in doing so, changed reference frames. Changing reference frames changes simultaneity.

Let's make sure we know the distance he started at. He started not 1/2 light second away from the earth, but sqrt(3)/2 light seconds away from Earth in order to get to Earth in 1 second at the given speed.

Suppose that there is a referee at the starting line, at rest WRT the Earth. At the moment the spaceship launches, then, both the pilot and referee agree that their clocks read zero. However, once at speed, the pilot will disagree that the referee's and Earth's clocks are synchronized. He will observe the referee's clock right next to him reading zero, but at the same time, the Earth's clock will read $\beta L = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4}$.

The distance to Earth is length contracted to $l/\gamma = \frac{\sqrt{3}}{4}$, a distance covered in time $t=d/\beta = 1/2$ second on ships' clocks. During the trip, the Earth clock is indeed slowed and advances by an amount $\Delta t = t/\gamma = (1/2)/2 = 1/4.$ But that clock didn't start at zero! It started at 3/4. So when the ship arrives at Earth, the Earth's clock reads 1 second. No problem, and everyone agrees.

 

[harrylin]

lets say they met up and syncronised their clocks and then one of them took a year-long journey at 5m/s to get half a light-second away. :P

Don't you care to check the references we gave?!

As we all stressed, the first thing to understand is relativity of simultaneity; it's useless to discuss mutual time dilation before that.

For example, the traveler accelerates in no time (for simplicity) and sets up a reference system in which he is in rest. Then, in the beginning, time dilation cannot yet play a role. Do you think that he will assign the same times as the stay-at-home to the Earth and himself?
If so, please study Einstein's train thought experiment that was mentioned in the relativity of simultaneity article. You can find it here: http://www.bartleby.com/173/9.html

 

[grav-universe]

We can determine what the traveling observer measures on the clock of the stationary observer from the stationary frame itself, but that requires more than just time dilation, it includes length contraction and the simultaneity difference involved, the latter being the key point here. Let's say the traveling observer passes the stationary observer when both observers' clocks read T = 0. According to the stationary observer, the traveling observer continues on until 1 second passes upon the stationary observer's clock. If the time dilation of the traveling observer is .5, then the stationary observer now views .5 second on the traveling observer's clock.

Okay, so according to the stationary observer, the traveling observer has now traveled a distance of x = v t = sqrt(3) / 2 light-seconds, so let's say the traveling observer is in a ship that is sqrt(3) / 2 light-seconds long as measured from the stationary frame. Then according to the stationary observer, when the traveling observer's clock at the front of the ship reads sqrt(1 - (v/c)^2) t = .5 second, a clock at the back of the ship directly coincides with the stationary observer and will read sqrt(1 - (v/c)^2) t + d v / c^2, where the stationary observer views the clock at the back of the traveller's ship to read a simultaneity difference of d v / c^2 greater than the clock at the front and d is the ship's length according to the traveling observer. The reason for this simultaneity difference is so that the traveling observer will measure the same speed of light in either direction along the length of the ship, and in all directions, which is another topic in itself, but the traveling observer says that his clocks read the same since he can't tell the difference as there is no absolute way to synchronize clocks at a distance, and the M-M experiment tells us that the two way time for light away and back over the same distance within an arbitrary inertial frame to a single clock is the same in every direction, so we can synchronize the clocks in each inertial frame to measure an isotropic speed of light by setting the clocks at the other end of the distance the light is reflected to and from accordingly.

So getting back on topic, the clock at the back of the ship reads sqrt(1 - (v/c)^2) t + d v / c^2. The stationary observer measures the length of the ship to be v t, the same as the distance travelled, so accounting for length contraction, the traveling observer measures the length of his own ship to be d = v t / sqrt(1 - (v/c)^2). When the stationary observer's clock reads 1 second, the stationary observer views the traveling observer's clock to read .5 second. If clocks are placed everywhere within the stationary frame, then a clock that directly coincides with the traveling observer will read 1 second while the traveling observer's clock reads .5 second, so a direct comparison between the two clocks reveals the traveling observer's clock to be ticking at half the rate according to the stationary observer's clocks. As for what the traveling observer measures for the time dilation of the stationary observer's clock, the stationary observer's clock reads 1 second while the clock at the back of the ship which directly coincides with the stationary observer will read

t' = sqrt(1 - (v/c)^2) t + d v / c^2

= sqrt(1 - (v/c)^2) t + [v t / sqrt(1 - (v/c)^2)] v / c^2

= .5 t + (v/c)^2 t / .5

= .5 t + (3 / 4) t / .5

= 2 t

So the clock at the back of the ship that coincides with the stationary observer's clock reads 2 seconds while the stationary observer's clock reads 1 second, and the traveling observer says that half the time has passed for the stationary observer's clock than for the clocks in his own frame, so with a time dilation of .5 as well. We only make a direct comparison of what clocks in the same place will measure, a time dilation of .5 in this case, so we don't keep switching that time dilation back and forth since we compare the clocks directly once. Now, one might wonder, if the stationary frame's clock that coincides directly with the traveling observer's clock reads twice as great as that of the traveling observer's clock, wouldn't the traveling observer say that with the direct comparison, the stationary observer's clocks are ticking faster? The answer is no. The traveling observer can only compare the clocks within its own frame to the single clock of the stationary observer that it timed after passing at T=0, and says that 1 second has passed upon that clock as compared to 2 seconds within his own frame, while the clock that the traveling observer now coincides with was never initially timed, so the rates of time cannot be directly compared until after passing and comparing the elapsed times, in which case the same time dilation of .5 will be measured for that clock as well. In the same way that the stationary observer views a simultaneity difference of a greater time per distance along the length of the traveller's ship, the traveller will view a greater time along the stationary frame's line of travel as well, so says that is why the clock that now directly coincides with the traveling observer reads a greater time, but that is not an elapsed time which is necessary to find the time dilation.