The Twins Problem, With Acceleration |

On this page, I will demonstrate

The Traveler's Acceleration

The Traveler's Velocity

The Traveler's Location

Time as a Function of Distance

The Traveler's Elapsed Time

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

The traveling twin,

The acceleration of the traveler will be a fixed value of 1 lightyear/year

There are many questions we could ask about this trip, but the primary question we are going to try answer here is

Our first task -- and our hardest task -- is to find the equations of motion of

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.

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

Since the velocity

and, using (2c) with composition of velocities, we have:

and we now can find the acceleration in H's FoR:

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

In plain English, when the traveler has been under way only a short time, his velocity is approximately

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

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 a

Finally, for large values of a

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 t

We can use eqn (19) to convert this to a function of x. Assuming that x

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

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/y

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,

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

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.

In other words, if at some particular moment, the traveler looks out his window and sees someone who is

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

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

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.

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

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

Plugging (MC-16) and (MC-17) into (MC-23), and replacing t with t

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. (

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

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.

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

From blastoff to turnover, we can apply eqn (26b) to find τ

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

And we can differentiate that to obtain the relative rate of the Earth clock compared with the time observed in the telescope,

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, τ

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.

We continue to define x

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 t

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 t

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

Finally, to obtain the ratio of clock rates

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 x

We also note that (PH-28) is easily integrable, and so we finally get τ

After turnover, we once again work in terms of y

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 t

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 τ

Taking the integral from turnover to time

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 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

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

Finally, the

This is the key to the "special relativity" explanation of the twins "paradox": the relative clock rates of

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.