 ## The Twins Problem, With Acceleration

In the most common SR version of the "twins paradox", acceleration is neglected, and the traveling twin is assumed to change velocity in negligible time when he starts the trip and when he turns around to come home.

On this page, I will demonstrate why that assumption is made, by dispensing with it.  This makes the problem much messier.  In fact, it is so messy that I will derive the solution from the point of view of the "stationary" twin, and just deduce the point of view of the traveling twin from that.  Working the problem directly in the moving twin's FoR appears to be harder.  I will take full advantage of any other shortcuts which present themselves, as well.

### Contents of the Page

Statement of the Problem
The Traveler's Acceleration
The Traveler's Velocity
The Traveler's Location
Time as a Function of Distance
The Traveler's Elapsed Time
Putting it Together: How Long does the Trip Take?
From the Traveler's Frame of Reference
Earth Time, in the Traveler's MCRF, blastoff to turnover
Earth Time, in the Traveler's MCRF, from turnover to arrival
The Porthole View
Some Graphs

### Statement of the Problem

All distances will be measured in lightyears, and all times will be measured in years.

The traveling twin, T, will travel to Alpha Centaurus, which is assumed to be 4 LY away along the X axis, and will then return.  As measured in the frame of the stay-at-home twin, H, the traveler will accelerate for the first 2 LY of the trip, then turn over and decelerate for the remaining 2 LY, arriving at AC with zero velocity.  Upon arrival, the traveler will immediately start upon the return journey, or in other words he will just keep accelerating Earth-wards, as he has been doing for the entire second half of the trip out.  At the 2 lightyear point, he will again reverse his acceleration, and will eventually arrive at Earth with velocity = 0.

The acceleration of the traveler will be a fixed value of 1 lightyear/year2, which is, by one of those amazing cosmological coincidences, almost exactly 1 Earth gravity.  That is his acceleration from his own point of view -- a spring-balance accelerometer in his ship will read 1g.  From the point of view of the Earth twin, T's acceleration decreases as his velocity approaches c.

There are many questions we could ask about this trip, but the primary question we are going to try answer here is how long does the trip take?  We wish to know how long it takes as viewed by H and as viewed by T -- and, consequently, what the difference in their ages is when T returns home.

Our first task -- and our hardest task -- is to find the equations of motion of T as viewed by H.  Once we have those, we will know how long the trip took from H's point of view, and we can integrate T's 4-velocity along his path, as viewed by H, in order to find T's elapsed (proper) time, which is how long the trip took from T's point of view.

With the equations of motion in hand, we can also find the "porthole view":  if each twin sends telemetry data to the other, at what rate does each see the other's clock ticking at each point on the trip, and how old does the other twin seem to be at each moment?  This is likely to be quite different from the "MCRF" view, since the signals received by each twin are necessarily delayed by the distances between them.

### The Traveler's (3-space) Acceleration

In the traveler's MCRF -- the inertial frame which is moving at the same velocity as the traveler at this moment -- his acceleration is a constant value:  a' = ak

We'll represent the traveler's MCRF as the "primed" frame.  The velocity of the MCRF, viewed from the Earth twin's (unprimed) frame, is v.  Immediately we have: Since the velocity v is 0 in the MCRF, we also have: and, using (2c) with composition of velocities, we have: and we now can find the acceleration in H's FoR:  ### The Traveler's (3-space) Velocity

Integrate the acceleration, eqn (4c), to get the velocity as a function of time:  At this point we can find γ as a function of time, which will be useful later on:  At this point, it's interesting to pause and look at three special cases: the traveler has just started on the trip, the traveler has been traveling just long enough to reach C in a Newtonian world, and the traveler has been on the way for a long, long time: In plain English, when the traveler has been under way only a short time, his velocity is approximately at, just as we would expect.  When he's been under way long enough to reach C in a Newtonian world, he'll actually only be traveling at about 2/3 C.  And when he's been traveling for a long long time, his velocity will approach C, but remain forever smaller than it by 1/(2(at)2).

### The Traveler's Location

And integrate the velocity, eqn. (9), to get the position:   Two other forms of the same equation may provide additional insight into what's going on:  It's not immediately obvious from (13), (14), or (15) (to me, at least) that this formula makes sense or reduces to the Newtonian formula for small values of akt.  From (14) and (15), however, we can see that the limits for large and small values of akt may be interesting.  We find, For small accelerations or short amounts of time, (13) reduces to (16a), which is just the Newtonian limit of the position given a fixed rate of acceleration, as we expected.  For akt = 1, Newtonian mechanics predicts x = 0.5 / ak.  The value in (16b) is a  little smaller than that:  about 0.4/ak.  So the traveler won't have gone quite as far as a Newtonian object would have gone at that point.

Finally, for large values of akt, achieved with either very high acceleration or after the traveler has been in flight for a very long time, according to (16c) the traveler will move from a position near x=t back to a position near x=t-(1/a).  What does this mean?  It means that a photon which starts out at the same time as the traveler, and which is always located at xp = t, will initially pull ahead.  As the traveler's velocity approaches C, the rate at which the photon is leaving him behind will decrease.  The traveler will continue to fall behind, ever more slowly, and will eventually fall to about 1/a units behind the photon, where he will remain ... if he continues accelerating at ak indefinitely.

### Time as a function of distance

We still need the traveler's time of turnover (the halfway point) and time of arrival.  For that we need time as a function of distance.  From (13):   ### The Traveler's Elapsed (Proper) Time

We finally have all our ducks in a row, and we can integrate to obtain the traveler's elapsed time.  We need to integrate We can use the value of dτ/dt in the MCRF of the traveler.  We know that is and, again in the MCRF of the traveler, we know that Plugging in equation (9) for the velocity, we see  And now the integral of the change in proper time with respect to time, taken over coordinate time, becomes which is If t0 = 0, then this is We can use eqn (19) to convert this to a function of x.  Assuming that x0 = 0, we obtain  Taking limits in eqn (28) we see that, Equation (29a) is the Newtonian limit for small velocities: elapsed time goes up as the square root of the distance.  Equation (29b) says that, for a long trip, the elapsed time of the traveler only increases as the log of the distance -- a long trip at constant acceleration takes far less subjective time in a relativistic universe than a purely Newtonian universe.

### Putting It Together:  How Long does the Trip Take?

At this point, we can determine how long the first quarter of the trip takes: the time from Earth blastoff to turnover.

Ideally we should re-run the integrals for a negative acceleration to obtain the elapsed time from turnover to arrival at Alpha Centaurus, but it's pretty clear even before we begin that everything's symmetric and the integral of the proper time will work out the same way.  So we can just double the proper time to obtain the total proper time on the trip from Earth to Alpha Centaurus.

Since the direction of travel didn't enter into the calculations in any interesting way, it's also clear that the elapsed times on the trip home will be the same as the elapsed times on the trip out.  So, we can just double everything again to obtain the round trip times.

Plugging in a distance to turnover of 2 ly, and an acceleration of 1 ly/y2, we obtain the elapsed Earth-time from equation (19) and we can then obtain the elapsed time for the traveler either by plugging the elapsed Earth time into eqn (26) or by plugging the distance into eqn (28).  Here we use (26): Multiplying each of these times by 4 to cover the whole trip, we obtain, ### From the Traveler's Frame of Reference

We worked the problem out in the inertial Earth frame.  We could also switch to coordinates in which the traveler is at rest, and compute the integrals there.

The hardest and most interesting part of this, however, still seems to be finding the equations of motion of the traveler in the frame of the Earth, which we needed to do in flat space.  Without that, we don't know when turnover occurs and we're at a loss as to how to map the traveler's coordinates to the Earth frame coordinates.  That's been done.  To switch to the traveler's PoV we'd just define the coordinates, compute the metric in the new coordinates (which would not be a Lorentz frame, and so would have some different metric), and then integrate the proper time for both parties once more.

That seems like a lot of work to obtain the same answer all over again, so I'm not going to go through it here.

### Earth Time, In the Traveler's MCRF, from blastoff to turnover

What time is it on Earth at each moment during the traveler's flight?

In other words, if at some particular moment, the traveler looks out his window and sees someone who is not accelerating, but who happens to be moving at the same speed he's moving at, and there happens to be another observer moving at the same speed as the two of them who is just passing Earth at the same time according to his own clock, and the clocks of the two observers who are comoving with the traveler are synchronized in their own rest frame, then what time will the observer who is passing Earth see when he looks at the Earth clocks?

This is about as simple as we can make the question while casting it in English and still retaining a sensible meaning.

With Earth coordinates of (t,x) and MCRF coordinates of (τ,ξ), we move the origins to:

(MC-1a)   (t, x(t))  in the Earth FoR
(MC-1b)   (τ(t,x(t)), ξ(t,x(t))) in the traveler's MCRF frame,

where x(t) is given by equation (13) and τ(t) is given by equation (26b), and the velocity is given by (9).  Then the transformation from these translated Earth coordinates to the translated MCRF coordinates is given by the Lorentz transform, The spaceship is located at τ=0 in these coordinates, and we need to find the Earth coordinates which map to a time of τ=0 also.  An easy way to do this is to observe that Earth's distance from the traveler in the traveler's frame is contracted by 1/γ, and we can easily map that coordinate to the Earth frame using the inverse of (MC-3). To find the actual Earth time corresponding to this, we need to translate the origin back to Earth by adding back the offset we subtracted from the time, and we need to plug in (9) and (13) to expand v and x in terms of time: And so, for the first quarter of the trip (acceleration phase, ending at turnover), we can plug in (9.4) to obtain: From the point of view of an observer in the MCRF of the traveler, the clocks on Earth read progressively farther and farther behind the clocks in the Earth's FoR which the traveler is currently passing.  Strange, perhaps, but not unexpected; just another consequence relativity of simultaneity.

We would like to know the rate of the clocks on Earth relative to the traveler's clock: Using the value of γ we found in (9.4), we have, for the part of the trip up to turnover: If the twin stopped accelerating, the formula would become 1/γ, which is just "time dilation" of Earth time, viewed from the traveling twin's point of view.  At the start of the trip, the value was 1 -- the twin's clock was going at the same rate as the Earth clock.  The clock rate ratio reaches a maximum at turnover.

### Earth Time, in the Traveler's MCRF, from turnover to arrival

From the turnover point to arrival at Alpha Centaurus, the traveler is decelerating.  It's pretty clear from the equations that the motion will be the mirror image of the motion during acceleration, so I will just assume that is the case without any further effort at proving it.

With that assumption and a few additional definitions we can immediately write the equations we need for the second half of the outbound trip.  Here are the definitions we'll need: We will assume the following equations are correct: By plugging in the formulas we found earlier for position, velocity, and gamma during the first half  (equations (13),  (9), and (9.4)), we obtain: As we did earlier, we will move the origin to the location of the ship, and then compute the time on an Earth clock as seen by an observer in the MCRF for the traveler whose clock is in sync with the traveler's clock.  We will again use the fact that the distance to Earth is contracted by γ for the traveler.  Mapping time τ=0 and the contracted distance back from the traveler's frame to the Earth frame, and adding back the offset to the origin, we obtain: Plugging (MC-16) and (MC-17) into (MC-23), and replacing t with tf - tc, we obtain: Plugging in (MC-21) this reduces a little more, to As the twin proceeds from turnover to the end of the journey, he slows down, and time on Earth approaches the local time in the Earth's FoR.  When T arrives at the end of the journey, Earth time matches the time at Alpha Centaurus, as we would expect.

Differentiating (MC-26) we obtain the relative rate of clocks on Earth, compared with the traveler's clock, as viewed from the traveler's MCRF.  (Note that dtc/dt = -1.) Finally, substituting γ from (MC-21) again, we obtain This is strange!  At the moment when the twin arrives at Alpha Centaurus, with zero velocity, the Earth clocks are still ticking at a different rate from T's clock!  The local clocks on Alpha Centaurus are ticking in sync with T's clock (though there's now a large offset, due to the "shorter path" T followed to get to AC).  But the Earth clocks are not.  Plugging in real numbers, the acceleration is 1, final distance is 4, final gamma is 1, and the Earth clocks appear to be running 5 times faster than T's clock.

Why is this?

It's because T is still accelerating, and Earth is far, far away.  As T accelerates, the transform we use to obtain values in his frame changes.  The effects of the change are larger as we view things farther and farther away from T.

The culprit here is the cross term in the Lorentz transform for time.  To obtain the Earth time corresponding to a particular point in T's frame, we add the value vγx to the (transformed) time from T's frame.  Gamma is 1 in this case, but x is 4, and v is changing linearly at dv/dt = 1.

### The Porthole View from Earth, during the Outbound Trip

If H, staying on Earth, points a telescope at the traveler and reads the face of his clock, what will it say?

Equivalently, if H and T send out telemetry signals throughout the trip, with one beep being emitted each second, how many will each of them receive at each point in the trip?

There are several similar questions here, but to start with we'll answer just one of them.  During the outbound trip, we'll assume a photon leaves the traveler's ship at time τ.  We will find the time at which it arrives at Earth, and we'll find the rate of change of the arrival time with respect to a change in the traveler's time.  This will provide us with the "porthole view" from Earth during the outbound trip:  it will tell us what time H sees looking through a telescope at T, and it will tell us how fast T's clock appears to be going when H looks at it.

Assume a photon leaves T at time t in the frame of reference of H.  The ship is at x(t) distance from Earth at that point, so the photon travels for a duration of x in H's frame to get to Earth.  Call the time of arrival at Earth ta, and call the time of emission in the T's frame of reference τe.  Then we have From blastoff to turnover, we can apply eqn (26b) to find τe: And we can obtain the photon's time of arrival at Earth by plugging equation (14) into (PH-1b): Substituting (PH-3) into (PH-5) we obtain the time of arrival as a function of T's proper time of emission, for all times before turnover: And we can differentiate that to obtain the relative rate of the Earth clock compared with the time observed in the telescope, before turnover: Substituting (PH-3) and (9.4) into (PH-7), we get, and finally, substituting the value of v(t) from equation (9) into (PH-7b) we see This is just the (inverted) usual formula for redshift from a source moving directly away from the observer  (which leads us to think the foregoing derivation may actually be correct!).

After turnover we need to turn the formulas around, since the ship is decelerating.  We start by finding the post-turnover proper time, τp, as a function of tc, by using equation (26b): To find the arrival time, we add the distance to earth, from (MC-19), to the emission time in H's frame of reference: Substitute (PH-10) into this, to obtain and differentiate it to obtain the relative clock rates, Substituting (PH-10) and (9.4) into (PH-16a) we obtain and again substituting equation (9) into (PH-16b) we finally see This is identical to (PH-7c) and is again just the (inverted) relativistic redshift formula which is usually obtained by more mundane and less lengthy means.

### The Porthole View from Earth, during the Return Trip

For the return trip, we translate the time origin so that t = τ = 0 at turnaround.  We leave the spatial origin on Earth, however.

We continue to define xf = 4 = location of Alpha Centaurus.

We need a few extra definitions.

From blastoff from Alpha Centaurus to turnover (the acceleration phase) we define the position and velocity of the ship to be, and from turnover until arrival back on Earth, we define the position and velocity functions to be Note that we are just assuming that the velocity and position curves will have the same shapes on the trip back as the trip out.  The additional definitions are just needed for convenience in the calculations.

As above, we let ta be the arrival time of a photon on Earth, and τe be the emission time in the traveler's frame of reference.  Unsubscripted t is the emission time in the Earth frame.  Then we have and, from eqn (14), From equation (26b) and the assumption that proper time is the same function of Earth time on the trip back as it was on the trip out, we have Equations PH-20 and PH-21 together tell us, implicitly, what the watcher on Earth will see in the telescope.  They are not easy to interpret through casual inspection.  We'll solve for one in terms of the other a little later.  First, though, we want the apparent rate of the ship's clock when viewed from Earth, and for that we need to take the derivative of ta with respect to τe.  We'll start by taking differentials of (PH-20) and (PH-21). Dividing (PH-25) into (PH-23),  we obtain Simplifying and plugging in eqn (9.4) for gamma, Substituting the value of v(t) from (9) we obtain which is just the (inverted) relativistic blueshift formula for a source approaching at v.

Finally, to obtain the ratio of clock rates from blastoff to turnover as a function of the observer's time, we substitute (PH-20) into (PH-27a) and fiddle a little to obtain: At blastoff from AC, t = 0, time of arrival of the photon at Earth will be equal to the distance from Earth to AC, which is just xf, and PH-28 will be 1.  The ratio will increase until turnover, at which point time photon's arrival time will be half of xf later than Earth time at the moment of emission.  Earth time of turnover will be about 2.3, and PH-28 will evaluate to about 1.3.

We also note that (PH-28) is easily integrable, and so we finally get τe as a function of ta, again valid from blastoff to turnover: After turnover, we once again work in terms of yR, wR, and tc, as most of the formulas turn out to be nearly identical to the previous ones but with time to arrival used in place of time since takeoff.  We start by finding time of arrival of the photon in the Earth frame and time of emission in the traveler's frame.    We've now got the porthole view implicitly, in terms of the Earth time coordinate of the ship when the photon is emitted.  We  again find the derivative of the arrival time versus the emission time by differentiating the two equations and dividing:  We divide (PH-37) into (PH-35) to obtain which, upon multiplying out and simplifying, turns into the far less alarming, Substituting the value for velocity as a function of tc, equation (9), we get which is the (inverted) usual relativistic blueshift formula for an approaching emitter.

To obtain the ratio as a function of the observer's time, we substitute eqn (PH-31) into (PH-39a), This can be integrated to obtain τe as a function of ta: Taking the integral from turnover to time t, where time proper time at turnover is τf/2 and Earth time is tf/2, ### The Porthole View from the Traveler's Ship, during the Outbound Trip

Not done yet -- watch this space!

### The Porthole View from the Traveler's Ship, during the Return Trip

Not done yet -- watch this space!

### Some Graphs

In this section, we show the results in a form which will -- hopefully! -- mean a little more than the raw equations.  The Gnuplot file for each of the graphs is linked to from the caption.

Plot 1 (plotfile lta_plot_1): In Plot 1 we show the basic information for a 1-way trip to AC.  It's largely self-explanatory.  The velocity (red line) rises close to 1 (which is C) but never quite reaches it..  Gamma (magenta line) increases almost linearly to a peak as the ship accelerates, then abruptly reverses course at "turnover".  Ship's time slows as gamma increases (green line curves downward) then speeds up again as gamma decreases (green line curves up again during second half of trip).

Plot 2 (plotfile lta_plot_2): Plot 2 is, in my opinion, more interesting.  It shows a number of things in the ship's frame of reference -- the inertial frame which is, for a brief instant, moving at the same speed as the ship.  Gamma and v are plotted once again, just for reference.  The most interesting line, however, is the dark blue line ("t_simultaneous"):  It shows the time on Earth, as shown by Earth clocks, at each moment in the ship's frame of reference.  It's not at all the same as the "Earth time" of the spaceship!  The latter, shown on the thin brown line ("t"),  is the time shown by a "stationary" clock which the spaceship passes.  Because of "relativity of simultaneity", from the traveler's point of view, the stationary clocks it passes do not appear to be in sync with the clocks on Earth.

The difference between "Earth time at the spaceship" and the time shown on the Earth clocks in the MCRF of the ship is shown on the magenta line ("ts_diff").  Of course, when the ship is stationary relative to Earth, the difference must be 0, and indeed it is.  At three points, at the start and end of the trip and at arrival at Alpha Centaurus, clocks are in sync everywhere and the line crosses zero.  In the second half of the trip out and the first half of the trip back, however, when the ship is far from Earth and is going quickly, the difference becomes very large.  Note that the difference is positive after the ship starts back from Alpha Centaurus -- in the ship's MCRF the clocks in Earth show a time which is in the future relative to an Earth-frame clock at the location of the ship.

Finally, the rate at which the clocks on Earth are ticking relative to the clocks on the ship, when viewed from the MCRF of the ship, is shown on the light blue line.  It starts at 1, of course, as we would expect -- the ship starts out on Earth with v=0, and at that moment its clock and the Earth clocks are in sync and going at the same rate.  The tick rate of the clocks on Earth then seems to decrease as viewed from the ship's frame -- that's "time dilation" and it's symmetric.  However, at turnover things start to get strange -- first, there's a hiccup in the graph as the acceleration changes suddenly, and then the relative tick rate of the Earth clocks starts increasing -- and just keeps increasing, until they're going about five times faster than the clock on the ship.  And that happens when the ship is stationary relative to Earth!  As the ship continues accelerating toward Earth on the trip home, the relative rate of the Earth clocks drops again, until by the homeward-bound turnover point the Earth clocks are once again ticking more slowly than the traveler's clock.

This is the key to the "special relativity" explanation of the twins "paradox":  the relative clock rates of distant clocks are affected by acceleration, because the time coordinate transform between the traveler and Earth depends on the product of the velocity and distance between the traveler and the Earth.  As the traveler accelerates, the coefficient on the distance term in the transform for time changes.  When the traveler is far away from Earth, acceleration toward Earth causes the Earth clocks to race ahead of the traveler's clock.  Keep in mind, however, that this happens in the MCRF of the traveler, where Earth and the traveler are separated by a spacelike interval.  The traveler cannot observe the effect.  Earth is safely hidden "in the magician's pocket" where it can't be seen while its clocks are racing ahead of the traveler's clock.

Plot 3 (plotfile lta_plot_3): After looking at Plot 2 it's natural to ask what the traveler and the Earth twin see if they look at each other.  Does the view look anything like the MCRF behavior?  The answer is "no".

Plot 3 shows the view of the ship, from Earth, for a full round trip.  Ship's time, shown on the red line, slows initially as the ship accelerates away from Earth, and then speeds up again as the ship slows to approach Alpha Centaurus, continues speeding up as the ship accelerates toward Earth, and finally slows again as the ship slows on approach to Earth.  This is all very much as we would expect.  The green line. showing the relative clock rate seen in a telescope, shows that there's redshift throughout the trip to AC.  Clock rates match exactly when the traveler arrives at AC (acceleration doesn't affect the "real" clock rate, please note).  Then on the trip back, there is blueshift the whole way, peaking sharply at turnover.  The plot of gamma (light blue) corresponds closely with the dt/dtau curve, with dt/dtau dropping below 1 for travel away from Earth and rising above 1 for travel toward Earth.

Porthole view from the ship hasn't been done but contains no new surprises.