Twin Watches

This problem came from Richard P. Feynman in the 1960s while one of us (F. P.), an undergraduate, was with him in his car on the way to Malibu, California, where he gave weekly physics lectures. The third person in the car, B. Winstein, then a graduate student in physics, contributed to a discussion that became quite involved!

Charlotte holds two identical ideal watches at the same height, one in each hand. She holds one steady in her left hand and tosses the other into the air straight up. At the instant the upward-moving watch is alongside the other at the same height above the ground, she sees that the two watches are synchronized, with the exact same readout value. Later, on its downward free fall path, she reads the time on both watches when they are again alongside each other and at the same height above the ground. Assuming that the moving watch is always in free fall, what would you predict for the two watch readings?

Answer

A watch ticks at its fastest rate when at rest and when there is no gravitational field. So there are two effects to consider: (1) from the special theory of relativity (STR), the motion of the watch with respect to the laboratory frame affects the ticking rate; and (2) the change in gravitational potential according to the general theory of relativity (GTR) affects the clock ticking rate. For a watch in free fall, the two effects are exactly opposite and cancel! The two watches agree again when she takes the second reading. Now for the details. First, are there any symmetry considerations that would simplify the calculation? Yes; the two parts of the journey for the moving watch the upward and the downward parts are time reflections of each other, and these two parts require the same elapsed time in the laboratory frame and in the moving watch frame. Pick the laboratory frame of reference. As the watch goes upward in the lab frame at its maximum velocity initially, the STR makes the watch tick faster as the velocity decreases, and the GTR makes the watch tick faster as greater height is achieved. On the downward journey, the watch ticks slower and slower by both STR and GTR effects. So we need only calculate the changes in the tick rate when the watch has gone upward by a small amount Δh, say.

From the STR, the time interval T between ticks at velocity v is given by T = T ′/ , where T ′ is the time interval between ticks of the watch in its own reference frame. At two different heights h₁ and h₂ = h₁ + Δh the time intervals between ticks are T₁ = T ′/ and T₂ = T′/ ,respectively, because the velocities will be different at the two heights. Since v << c, and assuming a uniform acceleration approximation for free fall, by the third golden rule of kinematics, v²₂ = v²₁ – 2 g Δh. Substitute the velocities squared into the watch’s time interval relations and expand the square roots in the denominators by the Taylor series expansion  One calculates T₂ ~ T₁ – T′ g Δh/c², a quantity proportional to the change in height. From the GTR, the time interval T between clock ticks at radial distance R outside of a body of mass M is given by T = T′ . In the limit of very large R, the clock ticks at its fastest rate. By definition, g = GM/R² at the surface of Earth. Substitute the above heights for the two distances from the massive body and take the difference. One calculates that T₂ ~ T₁ – T′ g Δh/c², a quantity proportional to the change in height and a quantity from the GTR that changes as fast as the quantity from the STR.

So the total change in the tick rate going upward is canceled by the total change in the tick rate coming downward, to make no net change when they are once again at the same height. If this argument has any flaws, do not blame either of my colleagues, Richard P. Feynman (deceased) or B. Winstein, for they know not what they had wrought!

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