Twin Paradox Redux

In this article a variation on the classic Twin Paradox is presented. By way of review, the classic paradox has one twin remaining at rest at the origin of an inertial frame (say frame K), and the other traveling out along the x axis at some constant speed v<c. Owing to time dilation, the moving twins onboard clock runs more slowly than the clocks in K. Indeed the moving twin ages more slowly than the stay-at-home twin. After some time, T, the moving twins velocity reverses and he returns to the origin. The clocks of K have advanced 2T, but the moving twins onboard clock has advanced only (1-v2/c2)1/2(2T). The paradox is of course that the moving twin is generally at rest in one inertial frame or another, and from the perspective of each of those frames the clocks in K run slowly by the factor (1-v2/c2)1/2. Why is it that the moving twin will have aged more slowly at the trips end? The answer lies in the change of mind that the moving twin has each time he changes inertial rest frames. For example, let us say that on the trip out along the x axis he is at rest in inertial frame K (which moves in the positive x direction of K at speed v). As long as he remains at rest in K he is perfectly justified in contending that it is the clocks of K which run slowly. (He notes the K clocks that whiz past him to be reading further and further ahead of his onboard clock, but attributes that to the "fact" that the K clocks are not synchronized with one another.) But as soon as he comes (momentarily) to rest in K, he reassesses the situation and concludes that it was his onboard clock, after all, that was running slowly. Similarly on the return trip.

In the present article we assume that the traveling twin is initially in a space pod at rest at the origin of K. At time t=0 the pods thruster engages and exerts a constant force, F, on the pod. The pod accelerates in the positive x direction until it reaches some speed vmax<c. This event occurs at time T. The thruster then reverses and returns the pod to a state of rest in K at time 2T. The thruster continues exerting a force in the negative x direction until the pod again reaches the speed vmax in the  negative x direction. This event occurs at time 3T. Finally the thruster reverses again and returns the pod to a state of rest at the origin at time 4T. We wish to determine how much the traveling twin has aged upon his return to the origin, and to compare this with how much the stay-at-home twin has aged.

The pods mass depends on its speed as follows:

        . (1)

(In Eq. 1, mo is the pods rest mass.) Newtons 2nd law applies, provided the dependence of inertial mass on speed is taken into account. The equation of motion is thus

        . (2)


        . (3)


        . (4)

Integrating both sides of Eq. 3, we have

        . (5)

Or more compactly

        . (6)

Squaring both sides and solving for v produces

        . (7)

In particular,

        , (8)


        . (9)

In order to calculate how much an onboard clock gains in the time T, we can exploit time dilation. Denoting an increment of onboard clock time as dt, and an increment of (resting) K clock time as dt, time dilation stipulates that

        , (10)


        . (11)

Substituting from Eq. 7,

        . (12)

Rearranging terms produces

        . (13)



Presumably the onboard clock reads t=0 when the thruster engages. Then integrating both sides of Eq. 13 produces

        , (15)


        . (16)

At time T, define T=t. Then

        . (17)

Finally, the onboard clock will have advanced 4T by the time the pod returns to the origin.

By way of an example, let us use the following values (where vmax is not really practical for space travel):

        , (18)

        , (19)

        . (20)

Then from Eq. 9,

        . (21)

And Eq. 17 indicates that

        . (22)

While his twin is away, the stay-at-home twin ages 4T=85 years. But upon returning home, the traveling twin will have aged only 4T=14.5 years.



Private Sub cmdConstantF_Click()

    Const c As Double = 300000000#

    Const vmax As Double = 0.999 * c

    Const m As Double = 1000#

    Const F As Double = 10000#

    Dim T1, T1pp As Double

    T1 = m * vmax * c / F / Sqr(c ^ 2 - vmax ^ 2)

    T1pp = m * c / F * (Log(T1 + Sqr(T1 ^ 2 + m ^ 2 * c ^ 2 / F ^ 2)) - Log(m * c / F))

    MsgBox ("T1 = " & T1 / 60 / 60 / 24 / 365 & " years, T1pp = " & T1pp / 60 / 60 / 24 / 365 & " years.")

    MsgBox ("Stay-at-home ages " & 4 * T1 / 60 / 60 / 24 / 365 & " years; Traveler ages " & 4 * T1pp / 60 / 60 / 24 / 365 & " years.")


End Sub