Calculating Time Dilation in Relativity: A Solution for Clocks in Motion

[aura]

Hi,

I am a new member of this forum...

I am stuck with a problem.Could anybody provide me the solution for this problem?

Two clocks, separated by 5 light hours, are synchronized according to an observer at rest relative to the clocks.How much are they out of synchronization according to a spaceship observer who is in motion at v=0.8c from clock 1 to clock2.

post any explanation that could prove useful..

thanks in advance...

 

[aura]

use of lorentz transformation is preferable though four vector is also welcome...

 

[StatusX]

Imagine there is a lightbulb exactly halfway between the two clocks, and it's switched on at some instant. In the rest frame, the light will hit the clocks at the same time. But in the moving frame, clock 2 is moving toward the lightbulb and clock 1 is moving away, but the light still starts in the middle and moves at c. Calculate the difference in the time the light arrives at each. Then remember the clocks are moving in this frame, and will be ticking slow, so divide this difference by a factor of gamma to get the difference between the times read on the clocks.

edit: you'll also have to account for lorentz contraction. On second thought, this probably isn't the best way to do it, but it's sort of instructive.

 

[OlderDan]

use of lorentz transformation is preferable though four vector is also welcome...

Another alternative- You know that when the spaceship observer arrives at the second clock it has to read the "at rest" time it took him to get there, and you know that while he is moving that distant clock runs slow compared to his clock. You can use Lorentz transformations to find the times involved in both reference frames and deduce the synchronization difference that way.

 

[aura]

Imagine there is a lightbulb exactly halfway between the two clocks, and it's switched on at some instant. In the rest frame, the light will hit the clocks at the same time. But in the moving frame, clock 2 is moving toward the lightbulb and clock 1 is moving away, but the light still starts in the middle and moves at c. Calculate the difference in the time the light arrives at each. Then remember the clocks are moving in this frame, and will be ticking slow, so divide this difference by a factor of gamma to get the difference between the times read on the clocks.

edit: you'll also have to account for lorentz contraction. On second thought, this probably isn't the best way to do it, but it's sort of instructive.

first of all 1 know that two clocks are synchronized wrt an observer but i need to find the time diff wrt the other observer who is also in motion.Now that means I will need to consider 4 reference frames ...first find out lorentz eqn for each of the clocks wrt the 1st observer (here time will be same)then between 1st observer and the spaceship...equate those to get the eqn between the spaceship and the clocks...this eqn should yield the result but its becoming too complex to solve for 4 ref frames due to gamma modification in each case!is there any other way??

 

[aura]

Another alternative- You know that when the spaceship observer arrives at the second clock it has to read the "at rest" time it took him to get there, and you know that while he is moving that distant clock runs slow compared to his clock. You can use Lorentz transformations to find the times involved in both reference frames and deduce the synchronization difference that way.

these are mere descriptions but i want the method of solution.it can't be too cumbersome but i am making it complex and more complex with solution far away and probably will commit mistake somewhere before i get the solution...gamma factor has become complex for the 3rd ref frame itself...

 

[StatusX]

first of all 1 know that two clocks are synchronized wrt an observer but i need to find the time diff wrt the other observer who is also in motion.Now that means I will need to consider 4 reference frames ...first find out lorentz eqn for each of the clocks wrt the 1st observer (here time will be same)then between 1st observer and the spaceship...equate those to get the eqn between the spaceship and the clocks...this eqn should yield the result but its becoming too complex to solve for 4 ref frames due to gamma modification in each case!is there any other way??

A reference frame covers all of space and time for a certain velocity. You don't need a different frame for each point, that defeats the whole purpose. If you want to use reference frames for this problem, there is one in which the clocks are at rest and one where the spaceship is. Say the position of the clocks in the clock frame are (0,t) and (5,t), for all t. Use the transformation equation to get these coordinates in the spaceship frame, and see what the difference in t is for each clock at a fixed t', where t' is the time in the spaceship frame. You'll need the equation with x, t and t', since those are the variables you have or want.

 

[OlderDan]

these are mere descriptions but i want the method of solution.it can't be too cumbersome but i am making it complex and more complex with solution far away and probably will commit mistake somewhere before i get the solution...gamma factor has become complex for the 3rd ref frame itself...

If by "mere descriptions" you mean I have not done the calculations for you, you are correct. You do not need 4 reference frames to do this problem. The method suggested by StatusX requires two reference frames, and the method I sugggeted requires two reference frames. The light pulse method is more fundamental, using the fact that all observers see light traveling at the same speed. The method I proposed uses the fact that when the space observer gets to the location of the second clock, it can only read one time. Both observers must agree on what the clock reads.

Light pulse method outline: I will call the observer who sees the clocks as syncronized the Earth observer, and the one who does not the space observer. The Earth observer synchronizes his clocks by placing a light source half way between them and sends out a pulse. That pulse arrives at both clocks at the same time, according to the Earth observer. That pulse triggers the clocks to start running.

The space observer sees the same events, but to him the two clocks are moving at .8c. One is moving toward the light source and one is moving away from it. The one moving toward the source is going to be triggered before the one that is moving away. It will then run slow because of time dilation until the second clock is triggered. The reading on the first clock when the second clock is triggered is the amount by which they are out of synchronization. To do the caclulation you need the relative velocities between the light pulse and the clock as seen by the space observer in his reference frame, which of course depends on how fast he sees the clocks moving. No other reference frames are needed.

The second approach uses the time the Earth observer calculates for the space observer to move from clock 1 to clock 2. When the space observer arrives and reads clock 2, he has to agree with the Earth observer about the reading. However, during his trip that clock has been running slow compared to his clock. If you calculate the time of the trip on his clock according to his point of view in which the distance is contracted, you can calculated the time that ticks off on the Earth clocks because they are running slow. The difference between what clock 2 reads when he gets there, and the amount that has ticked off during his trip is the amount by which he sees the two clocks out of synchronization. Again there are only two reference frames involved.

Choose your method and see if you can do the calculations, then check back and tell us what you came up with.

 

[aura]

A reference frame covers all of space and time for a certain velocity. You don't need a different frame for each point, that defeats the whole purpose. If you want to use reference frames for this problem, there is one in which the clocks are at rest and one where the spaceship is. Say the position of the clocks in the clock frame are (0,t) and (5,t), for all t. Use the transformation equation to get these coordinates in the spaceship frame, and see what the difference in t is for each clock at a fixed t', where t' is the time in the spaceship frame. You'll need the equation with x, t and t', since those are the variables you have or want.

thanks for the suggestion BUT in relativity u need to consider diff frames of references for objects moving with diff vels...and here none of the objects are mentioned to have same vel...in fact they are diff as evident from the problem...u can't consider a common ref frame in these problems as u need to solve for ref frames ONLY and NOT for objects seperately! from where u inferred that clocks are at rest wrt to some frame? they are synchronised that doesn't mean they are at rest!here we need to assume lot of variables...Once you find the common ref frame the problem is SOLVED...and to find that only the complexities are arising!U can't simply assume one as the common but will need to form an equation that holds good for all the other 3 frames..this is the problem..i hope u got my point

 

[StatusX]

You only need one, pretty simple equation, and I've even told you which one it is. The clocks are at rest wrt to each other. Read the problem again.

If the problem had said that the clocks were moving away from the observer at some velocity, then you would need 4 frames. But since they aren't, you can have three of the objects (the first observer and the clocks) at rest in a single frame.

 

[aura]

You only need one, pretty simple equation, and I've even told you which one it is. The clocks are at rest wrt to each other. Read the problem again.

If the problem had said that the clocks were moving away from the observer at some velocity, then you would need 4 frames. But since they are at rest, you can automatically include three of the objects (the first observer and the clocks) in a single frame.

one how? the observer is at rest wrt the clocks...but not the clocks ...they have a diff ref frame than the observer...also the observer views same time doesn't mean the time in respective frames of two clocks will be same...i,e if an observer(lets asuume there's one) in the ref frame of one clock views some time in his clock then another observer in the other ref frame having the 2nd clock need not view the same time in the 2nd/his clock as the 1st views in his clock...they are synchronised wrt a third observer in a third ref frame but not wrt each other...the 2nd observer is moving...i can't assume anything to be at rest specially when its not mentioned...

 

[StatusX]

First of all, if the clocks are at rest wrt the observer, then the observer is at rest wrt to the clocks. Think about that for a little while and you'll see why it wouldn't make sense otherwise. (eg, will they collide or not?) If you need to define a separate reference frame for each clock so that they're at x=0, go ahead, but it's a big waste of time. Since v is 0, the transformation equations would look something like: x=x'+5, t=t'.

Second, the relativistic effects like asynchronization and time dilation only occur when there are relative velocities involved. If everything is at rest, there are no such effects. If I'm at rest wrt to a clock on the other side of the galaxy, I will observe the same time on it as someone standing right next to it (of course, after accounting for the time it took the light to travel to me).

 

[panthera]

First of all, if the clocks are at rest wrt the observer, then the observer is at rest wrt to the clocks. Think about that for a little while and you'll see why it wouldn't make sense otherwise. (eg, will they collide or not?) If you need to define a separate reference frame for each clock so that they're at x=0, go ahead, but it's a big waste of time. Since v is 0, the transformation equations would look something like: x=x'+5, t=t'.

Second, the relativistic effects like asynchronization and time dilation only occur when there are relative velocities involved. If everything is at rest, there are no such effects. If I'm at rest wrt to a clock on the other side of the galaxy, I will observe the same time on it as someone standing right next to it (of course, after accounting for the time it took the light to travel to me).

okay yes u are correct. three are in same frame. two reference frames suffice.i concur.

 

[aura]

I got the point.that means the difference in time as seen by the 2nd observer will be due to the difference in position vector and motion of the spaceship wrt the clocks as relative motion of the clocks wrt each other is 0. i will solve it now using two ref frames ,one having the clocks and 1st observer and the other having the spaceship. thanks