Curious3141
Dec23-04, 08:37 AM
Please help me resolve this problem.
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.
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.