Is it Possible to Resolve the SR Time Dilation Problem? Help Needed!

[Curious3141]

Please help me resolve this problem.

Suppose we have 3 objects traveling 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.

 

[Doc Al]

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

(To properly relate measurements in two frames you must consider [itex]\Delta x[/itex] and [itex]\Delta t[/itex] together.)

 

[Curious3141]

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

(To properly relate measurements in two frames you must consider [itex]\Delta x[/itex] and [itex]\Delta t[/itex] together.)

Thanks for the reply, Doc. But could you please amplify further. [itex]\Delta x[/itex] 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.

 

[Doc Al]

But could you please amplify further. [itex]\Delta x[/itex] has dimensions of distance but exactly what is it. Can you give me more detail or point to a page with this explanation ?

Just like two events can be separated in time, they can be separated in distance. When you made the statement

"A particular time interval tA is recorded by A"

you are talking about the time interval [itex]\Delta t'[/itex] between two events. But what about the distance ([itex]\Delta x'[/itex]) 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:
[tex]\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)[/tex]

[tex]\Delta x = \gamma(\Delta x' + v\Delta t')[/tex]

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

So, in general, your statement:

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.

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.

 

[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. :)