The Reading of an Accelerated Clock

G.R.Dixon, July 10, 2008

In this note an accelerating clock’s readings are plotted against its initial rest frame’s clocks. The accelerated clock’s time is denoted as t’’, and the initial rest frame’s clock times are denoted as t. Prior to t’’ = t = 0, the clock is at rest in inertial frame K. At time t’’ = t = 0 the clock acquires a constant acceleration, a, in the positive x direction. At some time tfinal < c/a, the acceleration drops back to zero.

It is a well-tested fact that a "clock," moving at a constant speed, runs slowly by the factor (1-v2/c2)1/2. And there is also reason to believe that the same holds when the clock accelerates. Thus in general, if dt’’ is an increment of time on the accelerating clock, and dt is the corresponding increment of time in the initial rest frame, then

. (1)

In the present case (where v = at) we have

. (2)

Or, rearranging,

. (3)

Now mathematically,

. (4)

And

(5)

Thus

. (6)

Fig. 1 plots t’’ vs. t when a value of a = 10m/sec2 is used. Note that the curve’s slope decreases as t (and hence v = at) increases. (Were v to go to c, the slope would go to zero and the moving clock would be "frozen.")

Figure 1

t’’ vs. t

The Visual Basic Routine that produces the above graph follows.

Private Sub cmdPlot_Click()

Const N As Long = 285

Const c As Double = 300000000#

Const FinalV As Double = 0.95 * c

Const accel As Double = 10

Const FinalT As Double = FinalV / accel

Const deltat As Double = FinalT / N

Dim t(N), tpp(N) As Double

Dim Counter As Long

Open "c:\WINMCADC\Physics\Twins.PRN" For Output As #1

For Counter = 0 To N - 1

t(Counter) = Counter * deltat

tpp(Counter) = accel * t(Counter) / (2 * c) * Sqr(c ^ 2 / accel ^ 2 - t(Counter) ^ 2) + c / (2 * accel) * Atn((accel * t(Counter) / c) / Sqr(1 - accel ^ 2 * t(Counter) ^ 2 / c ^ 2))

Next Counter

For Counter = 0 To N - 1

Write #1, t(Counter), tpp(Counter)

Next Counter

Close

MsgBox ("Plot")

Stop

End Sub