# SR time dilation problem

1. Dec 23, 2004

### Curious3141

Suppose we have 3 objects travelling at constant velocity, and hence 3 inertial frames of reference. Call them object A, B and O respectively. O is at rest with respect to the observer.

Let the relative velocity of A from the point of view of B be speed u in a "rightward" direction.

Let the relative velocity of B from the point of view of O be speed v in a "leftward" direction.

Let the magnitude (speed) |u| < |v|.

A particular time interval tA is recorded by A. The equivalent interval recorded by B is tB and is given by :

tA = tB*sqrt (1 - u^2/c^2) = tB*L1, where L1 is the Lorentz correction between B and A.

Similarly as B records a time of tB, O records a time of tO, where :

tB = tO*sqrt (1 - v^2/c^2) = tO*L2.

Putting those two equations together, we get :

tA = tO*L1*L2 ---(eq 1)

meaning the Lorentz correction between A and O should be given by L1*L2.

But another way of looking at it is like this :

Let the relative velocity of A as seen by O be given by w.

|w| = (v - u)/[1 - (uv/c^2)] by relativistic addition of the velocities.

and the direction of w is "leftward" from the point of view of O.

It is obvious that |w| < |v|

tA can also be given by :

tA = tO*sqrt (1 - w^2/c^2) = tO*L3 -- (eq 2)

Since |w| < |v|, L3 > L2.

But by comparing eq 1 and eq 2, we see that L3 = L1*L2

Since L1 is necessarily less than one, L3 < L2

So we have L3 > L2 and L3 < L2, which is a contradiction, creating a paradox.

I got this from trying to figure out one of the accounts of the Hafale-Keating experiment, where a clock on a plane and a clock on the earth surface were compared. I found the method described strange, because the only proper time was considered to be at a hypothetical clock at the center of the earth, and everything was calculated relative to the center of the earth. I thought it should be fine to use the surface of the earth as inertial, even though there is a centripetal acceleration, there is no change in speed of the surface clock in a tangential direction. To figure out the mathematical approach I would use, I considered 3 idealised inertial reference frames and tried to compute relativistic intervals between them, expecting everything to come out OK. But I ran into the above problem that I can't resolve.

This looks like a fairly simple problem that must have been addressed before, and I think I'm making some horrible mistake somewhere. Any ideas ? Thanks.

2. Dec 23, 2004

### Staff: Mentor

One problem is that you are assuming that time intervals in two frames transform like this:
$$\Delta t = \gamma(\Delta t')$$
But this is only true when $\Delta x' = 0$. (The special case of a moving clock.) The general Lorentz transformation for time is:
$$\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$$

(To properly relate measurements in two frames you must consider $\Delta x$ and $\Delta t$ together.)

3. Dec 23, 2004

### Curious3141

Thanks for the reply, Doc. But could you please amplify further. $\Delta x$ has dimensions of distance but exactly what is it. Can you give me more detail or point to a page with this explanation ? Thanks so much.

4. Dec 24, 2004

### Staff: Mentor

Just like two events can be separated in time, they can be separated in distance. When you made the statement
you are talking about the time interval $\Delta t'$ between two events. But what about the distance ($\Delta x'$) between those events? (Here I am calling the A frame the primed frame.)

To figure out what another frame (the B frame) would measure as the time and distance between those same two events, one must apply the Lorentz transformations:
$$\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$$

$$\Delta x = \gamma(\Delta x' + v\Delta t')$$

Where v is the speed of the A frame with respect to the B frame, and $\gamma = 1/\sqrt{1- v^2/c^2}$. Note that the time measured in one frame ($\Delta t$) depends on both the time and distance measured in the other frame.

is not true.

If you wish to transform measurements made in the A frame to those made in the O frame, you are free to do so in two steps (A measurements ==> B measurements ==> O measurements) or in one (A measurements ==> O measurements) as long as you use the correct relative speeds and the correct Lorentz transformations. Either way gives the same answer.

5. Dec 24, 2004

### Curious3141

^Thanks, Doc for all your help. I worked through the algebra and I have verified that it comes out the same way.

Happy Holidays. :)