Twins Paradox with specialized clocks.

This is the standard twins problem, where there are two identical twins in an inertial reference frame. One twin starts moving at velocity V relative to the initial inertial reference frame. In this variant of the problem, I used specialized clocks to see if the clocks in the other frame run slower. Let V = sqrt(3)/2 * c. This gives a factor of two as the time dilation.

Now the stay at home twin wants to measure if the moving twins clock runs at half the rate as his own. So the stay at home twin makes a clock that runs at twice the rate of a standard clock. He gives this to the moving twin to take on his journey. Similarly, the moving twin wants to verify that his clock is running at twice the rate of the stay at home twin's clock. So the moving twin makes a clock that runs at half the rate of a standard clock and takes that with him on his journey. Obviously, both of the modified clocks traveling with the moving twin do not show the same time since one is running at four times the rate of the other. The clock running at twice the standard rate doesn't show the same elapsed time as the clock running at half the standard rate since they are in the same inertial reference frame (the moving twin's frame), yet the two specialized clocks supposedly each measure exactly how the stay at home twin's clock is running. One view has to be incorrect. Any insights?

David Seppala

Bastrop TX