__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 twin’s 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 twin’s velocity reverses and he returns to the origin. The clocks of K
have advanced 2T, but the moving twin’s onboard clock has advanced only (1-v^{2}/c^{2})^{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-v^{2}/c^{2})^{1/2}.
Why is it that the moving twin will have aged more slowly at the trip’s 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 pod’s
thruster engages and exerts a constant force, F, on the pod. The pod accelerates
in the positive x direction until it reaches some speed v_{max}<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 v_{max} 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 pod’s mass depends on its speed as follows:

. (1)

(In Eq. 1, m_{o} is
the pod’s rest mass.) Newton’s 2^{nd} law
applies, provided the dependence of inertial mass on speed is taken into
account. The equation of motion is thus

. (2)

Therefore

. (3)

But

. (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)

and

. (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)

or

. (11)

Substituting from Eq. 7,

. (12)

Rearranging terms produces

. (13)

But

(14)

Presumably the onboard clock reads t’’=0 when the thruster engages. Then integrating both sides of Eq. 13 produces

, (15)

and

. (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
v_{max} 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.

Program

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.")

Stop

End Sub