[size=150]The Twin Paradox:[/size]

Space: the Final Frontier

Second Edition*

The Setup:

Two Twins, Kirk I and Kirk II, board the identical spaceships Enterprise I and Enterprise II respectively. Since Vulcans live longer than Humans, the overwhelming majority of Physicists have voted that Spock should stay on the (inertial) space station. Additionally, the clocks are calibrated such that each ship leaves at time t = t’ = 0. Spock’s clock will also read 0 when the ships depart.

The rest mass of each ship, m0, = 1000 kilograms. The acceleration rate, relative to Spock will be a constant A = 10 meters per second per second (roughly the same as the gravity on the Earth’s surface).

In order to avoid problems with declining rest masses, we will assume some alternate propulsion (perhaps electromagnetic). In addition, we will set the experiment up in the inter galactic space, half way between Andromeda and the Milky Way, as viewed from our galaxy, to ensure that the experiment is conducted with an absolute minimum of outside gravitational influences.

As is customary in the analysis of this problem, we will ignore any gravitational affects among the spaceships and space station.

We will have prior agreements that the flights will take place in a single dimension, x, and each ship will accelerate to a speed of .5c relative to Spock.

Finally, Spock will send a light signal, calculated to reach each craft at the end of 25 years of inertial flight, which will notify Kirk I and Kirk II when it is time to turn around.

The following diagram, as viewed from Spock’s perspective, should show the situation described:

Let the Kobayashi Maru Test begin.

This humble rubber stamp manufacturer, marooned in the desolate frost bitten far north, finds himself at war with the Klingons - err nearly the whole of the evil empire of Physics.

There are four cases to study.

Case I: Spock is at the origin.

This is the standard argument made by professional physicists. Spock is in an inertial frame and Kirk II is moving relative to Spock. It should be noted that, due to the symmetry of the setup, the analysis of the motion of Kirk I, with respect to Spock should yield the same results.

Case II: Kirk I is at the origin and Spock is moving.

Here we look at the story from the point of view of Kirk I. This view is generally unreported by most Physicists as they dismiss it for various reasons. e.g. It is not an inertial system; the acceleration yields a discontinuity with respect to the first or second derivatives; et al. It was also the most difficult for me to construct.

The obvious problem, with objections to this case, is that Kirk I is real, (well OK he’s not real, but you get the point) and all the talking in the world will not change the fact that he has a perspective. Furthermore, it is the whole point of general relativity that all reference systems are equivalent.

Case III: Kirk I is at the origin and Kirk II is moving.

I imagine this to be the first part of the Twin paradox as viewed by the original author of the paradox. I don’t know who he was. It is interesting because of the relativistic addition of velocities and multiple gravitational forces.

Case IV: Kirk II is at the origin and Kirk I is moving.

This is simply the complementary view required to complete the analysis.

The Goal:

My general goal is to show that each period of acceleration is bounded not only in t, as measured by the clock at the appropriate origin, but also in t’, the clock on the vessel perceived to be moving relative to the origin.

This leaves the travel times t and t’, in the inertial systems, as independent quantities. In turn, if the travel times in the inertial systems are large compared with the total elapsed time in then accelerated systems, then the total elapsed time in t’ will appear much less than the total elapsed time in t. Since this is true from both reference frames, the famous twin paradox will appear to be valid.

Analysis:

Case I: Spock is at the origin.

This is the standard analysis. Kirk II undergoes a constant acceleration, A, until he reaches a speed of .5c. (All motion is in the x dimension so I will not need to deal with vectors. Additionally, I will use u, instead of the customary v, to designate velocity until we deal with Case III).

The equation for relativistic force, where the force is a constant, A, is given by:

Since the initial time t[size=85]0[/size] is 0, and after dividing by A we get:

This means that the acceleration cutoff time, t[size=85]cutoff[/size] at u[size=85]max[/size] is

Substituting 1000 for m0, 10 for A, and .5c for u[size=85]max[/size] we get:

t[size=85]cutoff[/size] ~ .55 years

We observe here that Kirk II is moving in an accelerated reference frame, relative to Spock. From the Equivalence Principal we can deduce that t’ is less than t, because clocks in a gravitational field, relative to an inertial field, run slow.

So the initial acceleration time intervals in Spock’s reference frame are given by:

After the acceleration phase we have the inertial travel phase, which we will deem to be 25 years. Thus we will have:

But in two inertial systems we have:

Now we can write that the first inertial travel times are given by:

The elapsed time for deceleration and acceleration, due to the symmetry from Spock’s point of view, are the same.

Similarly the second acceleration period or turn around period, as viewed by Spock, will be the same as the first acceleration period in both t and t’. Thus:

During the inertial part of the trip back we will again have:

and again we have:

The reader might notice that u[size=85]max[/size] is actually – u[size=85]max[/size] but since umax is squared the difference in sign does not play any role.

We will have the lapsed time of the inertial return trip, exactly the same as before:

Finally we have the end of the trip, which we will designate as:

Again based on the symmetry the basic laws should apply relative to Spock’s frame of reference. Therefore:

Now we need to compare the length of the elapsed times, from Spock’s reference frame. The total elapsed time should be:

Substituting the unprimed equations from above, and using the inequalities we get:

Note the strict inequality.

From Spock’s reference system:

Experiments have been conducted and verified to a high accuracy that the time contraction matches the theoretical results.

Case II:

This case is profoundly different from Case 1. In Case I Spock is in an inertial reference frame. Here Kirk I experiences 3 periods of acceleration including one immediately upon the separation of the spaceship and space station. This means that Kirk I is experiencing a gravitational force, and we will be required to deal with the general theory of relativity.

There will again be six time intervals to consider:

During Spock’s travels, Kirk I will experience three periods of gravitational pull.

Now we consider:

An intuitive approach:

Before we start though we should review the relationship between t’ and t in an inertial reference frame.

Drawing 1:

The relationship between the primed interval and the unprimed interval is given by:

Things to notice:

becomes nearly 0, which will make the time interval very large.

The angle of this line will never pass 45 degrees.

Now let’s look at some examples of a general relationship

between t’ and t.

Here there are chunks of time that are missing on the curve. If we write t’ as a function f of t, we would say that f is not continuous. One of the criteria of f from the point of view of General Relativity is that the relationship must be continuous.

Another example is the following:

Here we see that as t’ increases t will at first increase and then decrease. We, intuitively would like a curve that never goes backward in time. Edit 10/24/09 A We will describe such a function as non-decreasing. End Edit 10/24/09 A

The following is an example of a function f where t is always going forward, but t’ will at first move forward and later move backward in time. Again we would intuitively like to see t’ not go backwards.

So our preference would be one of the two following types of curves.

Graph 1:

Graph 2:

One of the striking things about Graph 1 is that the tangent curve, the red line, appears to be inclined at more than 45 degrees.

This makes us question if this is possible. The answer is no!

A formal approach:

We know that there exists two functions, f1 and f2 such that f1(t, x) = t’ and f2(t, x) = x’, because uniform acceleration is a resolved problem in general relativity.

These functions exhibit certain properties. For example if w is an arbitrary parameter then the line defined by {f1(t(w)),x(w)), f2(t(w),x(w))} will be some arbitrary curve on the t’ - x’ surface and its’ arc length s will be defined by:

Edit 9/04/2009 A:

I forgot to add a negitive sign to the right side of the second equation.

End Edit 9/04/2009 A

The relationship of the tangent plane to the t’-x’ surface will be defined by the partial derivatives:

This means that the partial derivative of f1 with respect to t will lie in the tangent plane of t’-x’, and that f1 defines, within a constant, the relationship, f, between t’ and t. (It is the blue line in the drawings above).

From our definition of f we have:

Consider the following graph:

The equation for the blue line is given by t’ = mt + b, where m is the slope and b is the t’ intercept.

And since this equation must be valid at (f(t[size=85]0[/size]), t[size=85]0[/size]), we get:

Solving for b we get:

So the equation for the line would be:

Since the slope of the line is

Since t[size=85]0[/size] is arbitrary, this means that during the initial acceleration

Edit 4/10/2010: The reader should note that this line, even if it were entirely in the tangent space, would not be straight. Luckily all we need is that the tangent at t[size=85]0[/size] is in the tangent space, which in fact is the case. End Edit.

The Bounded Reference Frame Theorem:

If [A, B]’ is a time interval in an inertial 2 dimensional reference frame, that corresponds to a time interval [A, B] in a gravitational field, then B’ - A’ < B - A

In our particular case:

There is no interval [A, B] during the period that Kirk I is accelerating, where

Assume Not.

Consider the following graph:

Graph 3:

Since f(B) is greater than C we must have f(B) – f(A) > B – A. Additionally, we have assigned N to be the amount that f(B) – f(A) is greater than B – A.

Now let’s consider the following graph:

Graph 4:

Here we have a curve that is approximated by the red segmented lines between each t[size=85]i[/size].

As in Graph 3, C is the point where C- f(A) = B – A.

Let t[size=85]i[/size] be the partition of [A, B] where t[size=85]0[/size] = A and t[size=85]n[/size] = B and t[size=85]i[/size] = t[size=85]i-1[/size] + (B – A)/n.

Then we have:

Since df(t)/dt exists for all t in this interval, we must formally have:

Edit 09/04/2009 B:

Since M and L depend on t[size=85]i-1[/size], we must write M[size=85]i[/size] and L[size=85]i[/size]. Therefore consider M and L to be M[size=85]i[/size] and L[size=85]i[/size] through out the following.

End Edit 09/04/2009

Edit 8/18/2009:

To make this logically correct I need to add:

End Edit 8/18/2009:

Edit 10/24/09 B

Since f(t[size=85]i[/size]) => f(t[size=85]i-1[/size]), because f is monotone increasing non-decreasing End Edit 10/24/09 B, t[size=85]i[/size] > t[size=85]i-1[/size] by definition and L > 0 because Kirk I must travel at sub light, we can write:

Adding L to each side of the inequality and multiplying by (t[size=85]i[/size] – t[size=85]i-1[/size]) we get:

Since L is the derivative of f(t), Lemma 1 implies that L < 1. Therefore

By choosing n large enough so that t[size=85]i [/size]– t[size=85]i-1[/size] < 1 and

Substituting we get:

f(B) – f(A) < B – A + N/2

Since we know that f(B) – f(A) = B – A + N, by substitution we get B – A + N < B – A + N/2 Or N < N/2,

which is a contradiction.

Therefore, there is no interval [A, B] during the period that Kirk I is accelerating, where

And The Bounded Reference Frame Theorem is hereby proven.

This means that for any partition P[size=85]i[/size] of time, in which Kirk I is accelerating,

f(P[size=85]i[/size] ) < P[size=85]i[/size] therefore:

Now we will examine the inertial flight time from Kirk I’s perspective.

Again we have:

We know that Spock will send a signal to arrive at the Enterprise I after 25 years of inertial travel. Kirk I will receive that signal ~ 28.868 years latter.

So now we have:

The elapsed time for Spock’s deceleration and acceleration, which Kirk I experiences under a gravitational force, will be assumed to be twice the initial acceleration time. This assumption is due to the symmetries of the physical events.

Likewise the acceleration in periods should be the same

The Spock’s inertial flight back will need to be the same length of time as his flight out so we can write:

Finally the docking time should equal the time for Spock to accelerate out, so we can write:

To complete this analysis of Case II we sum the component times to get:

Substituting our values for t’ we get:

The analysis for Kirk II will yield the same results.

So for the same physical event, as viewed by Spock, Kirk II will have aged at least 6.7 years less than Spock; and as viewed by Kirk II, Spock will have aged at least 7.7 years less than Kirk II.

The Twin Paradox appears to be true in this pair of cases.

Case III:

Here we assume that Kirk II is moving relative to Kirk I.

Again there are six time intervals to consider:

As in Case II, the first period of acceleration is the most complex. Here we have both Kirk I and Kirk II in gravitational fields.

Here we observe that we can establish t*= f(t, x) for Kirk I to Spock from Case II, and we find t’ = g(t*, x) from Spock to Kirk II. This means that we can write h from t to t’ by looking at h(t, x) = g(f(t, x)).

Now we consider d(h(t, x))/dt by applying the chain rule:

dh/dt = (df/dt)(dg/df)

Since each of the derivatives df/dt and dg/df are less than 1 that means that dh/dt <1, and from Case II we can deduce:

During the inertial flight times, we will need to determine the velocity v of Kirk II to Kirk I.

We know that if u is the velocity of a moving reference frame and u’ is the velocity of a second reference frame relative to the first moving frame then:

Here u = u’ = .5c and we get:

v = c/(1 + .25) = .8c

So during the inertial flight the time, from Case II, will be 28.868 years and we get:

To complete this analysis of Case III we sum wthe component times to get:

Similarly for t’ we get:

Case IV:

This is exactly the same as Case III, except that Kirk II is the reference frame.

I think that the true spirit of the paradox can best be addressed by comparing Case III and Case IV.

Case III yielded Kirk II to be at least 23 years younger than Kirk I, and Case IV will yield Kirk I to be at least 23 years younger that Kirk II.

Again, the Twin Paradox appears to be true.

Summary:

This is just a cursorary view of the matter. There are other views that include a Doppler analysis and an examination of the problems occurring at points where the differentials are not continuous.

Fundamentally, I have broken the analysis into six different segments:

where i varies by the reference frame being considered.

in both the t and t’ systems.

However, even if this symmetry is not valid, each of the intervals in t’ are less than the corresponding intervals in t.

It should also be noted, and I stressed this point, that each of the intervals of acceleration are of fixed length, depending only on the acceleration. The inertial intervals can be as long as required to make the acceleration intervals seem insignificant.

All of which means:

In Spock’s system, Kirk I and Kirk II will age less than Spock. Kirk I’s system, Spock and Kirk II will age less than Kirk I. Finally, in Kirk II’s system, Kirk I and Spock will age less than Kirk II.

As a final word on the subject, I thought that I might note that each of the labeled transit intervals should be treated, mathematically, as open intervals (there are six of them). Each of the transit points between the intervals, and at the beginning and end points (there should be seven of them) have no measure because there is no passage of time at these points.

Basically this subject matter is discussed at length under the branch of Mathematics called Measure Theory.

However, you should get the same results from a physical point of view, because time gets “grainy” at intervals of about 10^-44 seconds.

On references:

- First Edition located at:

phpbb/viewtopic.php?f=4&t=168317&start=0

The special relativity formulae come from my college text book “Introduction To Special Relativity” by Robert Resnick.

However, I did have to improvise in the general relativity sections of this problem and the Bounded Reference Frame Theorem is my own creation.