Congratulations!

Right now she is traveling from her homeplanet Homey to another planet Destiny, located 200 light years away. In her spaceship, Apollo-Out, she travels with a velocity \(v=0.99c\).

We know that the planets do not move with respect to each other in order to have them in the same frame of reference, so that we only have two frames of reference:

• Unprimed \((x,t)\): The planets, with Homey at the origin and Destiny 200 light years away.
• Primed \((x',t')\): Apollo-Out, always at the origin in this frame.

We have two events:

Let's get the basics before we begin. Using equations of movement we have that the trip from Homey to Destiny should take \(\Delta t = \frac{\Delta x}{v}=202\) years from the Homey frame of reference. Using time dilation, \(\Delta t = \frac{\Delta t'}{\sqrt{1-v^2/c^2}}\), we see that Lisa will measure \(\Delta t' = \Delta t \sqrt{1-v^2/c^2}=28.5\) years on her clock (natural units!). Using the same arguments we will see that her trip back takes the same amount of time, meaning the total journey takes \(\Delta t = 404\) years in the planet frame and \(\Delta t' = 57\) years in Lisa's frame.

That all made sense, yeah? So where is the paradox? Let's keep going, friend, maybe we will find it along the way.

Now, let's for a hot minute switch frames, and we should start to see some spicy numbers. We use the same laws of physics in both frames of reference, so this should be fine. The situation now is that Lisa remains at the origin in the unmarked system, and the planets move (with velocity \(v=0.99c\)) in the marked system.

We still have that from the time it takes Homey to leave Lisa, and Destiny to arrive, she measures \(\Delta t = 28.5\) years on her watch. If we use time dilation to measure the planet frame of reference, we then see that the time measured from the planet is \(\Delta t' = \Delta t \sqrt{1-v^2/c^2}=4\) years! The time it takes for Homey to reach Lisa again can be found in the same way, resulting in the total travel time being \(\Delta t' = 8\) years in the planet frame.

Let's sum up what we have so far: In the original (unmarked) planet frame it takes Lisa \(404\) years to make the entire journey, spending the same time each way. In Lisa's (marked) frame it takes her \(57\) years, and she also measures that she spends the same time each way. Finally, when we switch frames of reference, we see that the (now marked) planets measure that the whole journey takes only \(8\) years, but the time interval remains the same each way. All of these were found using only time dilation and equations of movement!

This means that Lisa aged 50 years during the trip, while her friends aged 404 years (and are kinda dead.. bummer). But, what about when we switched frames? Then we get that her friends only aged \(8\) years? They can't both be true, can they?

That, my friend, is the paradox!

So, what, if anything, went wrong here? We have assumed that the planet frame and the Apollo-Out frame are equal, and that we can chose whether we consider the marked system as either the planet frame or the Apollo-Out frame as we wish, but are these identical?

One difference that should come to mind, is that in our first roles, Lisa changes direction, while the planets do not. After we switched, the planets change directions. Go outside and run 60 meters one way, and then back exactly the way you came from. Did you have to stop? I hope so. And when you stopped running (and started again the other way) you had accelerate!

The thing about Homey is that the planet (or the people on the planet, if you will) remains in the same frame of reference the whole time, while Lisa changes her frame when she accelerates in order to change direction and return to Homey. The planet does not, no matter if we put it in the laboratory frame or in the moving frame.

When we derived the expression for time dillation, \(\Delta t = \gamma \Delta t'\), we did it for constant velocity. The expression is not valid anymore, and we need to rethink how we want to calculate the actual time it takes.

We've added another planet, Beyond, located 400 lightyears from Homey along the positive x-axis. We've also added a second spaceship, Apollo-In, traveling from Beyond with velocity \(v=-0.99c\), with astronaut Peter.

Now, to better understand what's going on, imagine one infinite train traveling from Homey towards Destiny, with the same velocity and in the same direction. The trains are sentient, like on thomas the tank engine, and capable of observation. We have the same situation from Beyond to Destiny. Lisa and Peter are still in their respective spaceships (or trains, in this case).

• Unprimed \((x,t)\): The planets, with Homey at the origin, Destiny 200 light years away, and Beyond 400 light years away.
• Primed \((x',t')\): Apollo-Out, traveling from Homey to Destiny, always at the origin. These are the outgoing trains!
• Double primed \((x'',t'')\): Apollo-In, traveling from Beyond to Destiny, always at the origin. These are the returning trains!

Our events are now:

Event B' is pictured below:

From equations of movement we can say that Lisa arrives at planet Destiny at a time \(t_B = L_0/v\). We can then use the Lorentz transformations, \(t'_B = t_B/\gamma\), in order to find an expression for the time when Lisa arrives at Destiny on her own watch, \(t'_B=28.5\) years.

From these expressions we find an expression for \(t_{B'}\):

\(t_{B'} = v\gamma x'+\gamma t' = L_0(\frac{1}{v}-v)=4.1\)years.

This time is the time when the observer in the outgoing train reads the time at planet Homey clocks at the same time (in his frame) as Lisa arrives at Destiny. This means that the people on planet Homey know that Lisa has reached Destiny in her own frame. In the planet frame, they still need to wait another 173.4 years before they see her reach the destination. Event B' still happens after 28.5 years in Lisa's frame, though.

When arriving at Destiny, Lisa launches herself frm the yellw to a red spaceship (Apollo-In), and returns again to Homey. In the planet frame we see the blue light signal after 4 years, and the people of Homey know that Lisa has reached Destiny in her own frame of reference.

Let's see how this looks to spectators in Lisa's frame of reference:

We know that the blink that Homey sees coencided with Lisa's arrival in her own frame of reference. For Lisa, only 4 years have passed on Homey when she arrives at Destiny, even if the people of Homey have to wait 202 years to see her arrive for themselves! The observation of Homey's clocks from the blinking spaceship and Lisa's arrival at Destiny is not simultaneous in the planet frame! What is happening???

As we've stated, the secret lies in the changing of frames of reference. When Lisa jumps over to the red spaceship, she changes her frame of reference B-->B'. In order to do this, she has to accelerate, going from \(\vec{v}=0.99c\) to \(\vec{v}=-0.99c\). Special relativity only works for constant acceleration! Time is still relative for accelerating systems and we need to account for this. Things are getting relative!!!

Okay, okay, okay. But how am I supposed to understand all this, how can Lisa spend 404 years on the entire journey, but only 8 of them are spent traveling between planets, you ask? Let's add some more events, you know, to make things less confusing and all.

We cannot use Lorentz transformation because the clocks in the double primed reference system is not correctly synchronized with Homey clocks. Therefore we use the spacetime interval, \(\Delta s^2 = \Delta s'^2 = \Delta s''^2\), which is independent of the intertial frame of reference.

We can already tell that the spacecrafts from event A and D use equal amounts of time to event B, as they both have equal velocity and do not accelerate.

We see that the time of arrival of the red spaceship is 28.5 years, as we expected. Let's check if we can find the time \(t''_B\) at Peter's wrist watch when he arrives at Destiny.

\(\Delta t_{BD}=L_0/v\)

\(\Delta x_{BD}=L_0\)

\(\Delta t_{BD}''=t''_B\)

\(\Delta x_{BD}''=0\)

Invariance of the spacetime interval gives \(\frac{L_0^2}{v^2}-L_0^2=(t_B'')^2\), which gives \(t_B''=L_0/(v\gamma)\). We know that \(t'_B=28.5\), and as we expected \(t'_B=t''_B\), in words, Lisa and Peter spend the same amount of time traveling from their home planets to Destiny.

Let's summarize in a table:

What about the time on Homey that the person in the blinking train observes when Peter arrives at Destiny (event B'')? Is this the same as for Lisa (4 years)? In other words, when do the people of Homey think that Lisa has been transferred to Peter? Again we want to use a spacetime interval, \(\Delta s_{DB''}\).

We can find that:

\(\Delta x_{DB''}=0-2L_0=2L_0\) (Removed the negative sign as length is what matters) 

\(\Delta t_{DB''}=t_{B''}-0=t_{B''}\) (We do not know this yet)

\(\Delta x''_{DB''}=x''_{B''}-0=\frac{L_0}{\gamma}\)

\(\Delta t''_{DB''}=t''_B-0=\frac{L_0}{v\gamma}\)

We can again use invariance of the spacetime interval of event D and B'' in order to find \(t_{B''}=\frac{L_0}{v}+L_0v=400\) years.

As we've mentioned, the paradox seems to happen when Lisa changes her frame of reference on destiny (when she accelerates). On Homey only 4 years pass untill she arrives, but somehow she then spends 396 years on Destiny, changing her frame, in a move that is instantanious in her frame of reference, because special relativity does not account for changes in acceleration.When she is accelerated, she experiences fictive forces, but this does not happen to Homey.

Recall how we switched the frames earlier? This cannot be done when only one frame experiences acceleration! They are not interchangable anymore.

So far we've looked at a case where Lisa travels between planets Homey and Destiny, and a case where we switched the frames so that the planets traveled away from/towards Lisa. This is where we first saw how the planets only spend 4 years on a journey that took Lisa 404 years (in the planet frame). We also saw that Lisa spend 57 years in her own frame. We then added another planet, Beyond, and an astronaut Peter who travels from Beyond to Homey, and imagined an infinite amount of trains moving between the planets, two of which send a signal to Homey, one when Lisa makes it to Destiny, another when Peter makes it (the infinite trains make it easier to see how the blinking trains are still in the frame of their respective astronaut, even though they are far away).

From this we saw how, if we use Lorentz geometry and transformations, we will get strange answers when looking at the time Lisa spends traveling. Lisa's watch will show 28.5 years each way, a total of 57 years. Homey's clocks will show that Lisa arrives after 4 years (in her own frame of reference). However, when Lisa arrives back on Homey and into the planets frame, 404 years will have passed, even though she (in her own frame) only sees 4 years passing each way, until she steps onto Homey.

We've discovered that in relativity we can think of acceleration as being in a gravitational field!

If you thought we went deep into relativity before, you'll want to hang on tight for this! You may be able to accept that acceleration (and the subsequent 'forces' acting on Lisa) make things wonky, but what actually happens with time and space during the accelerated phase? If we want to know this, we need to step into gravity and general relativity, so let's explore the accelerated phase in detail.

Earlier we found that when Lisa arrives on Destiny, 4 years have passed on Homey, using the Lorentz transformation. We can also do this using the invariance of the spacetime interval \(\Delta s_{BB'}\), and we get that \(t_{B'}=L_0/v-L_0v=L_0/(v\gamma^2)=4\). So the time i takes her to get to Destiny is the same, but 396 years pass on Homey between this point and when she begins the journey back to Homey.

Acceleration is key, time on Homey made a huge jump when shifting from one system of reference to the other. So what if we now assume that Apollo-Out, starting at Destiny, starts deaccelerating, \(a = g\), measured in the planet frame. At a time, \(t_{turn}\), measured on planet system clocks, the spaceship has reached the turning point where the velocity has reached 0 and will start returning to Destiny. We can already say a few things about this accelerating period. First, in order to find the time of the turning point, \(t_{turn}\), we can use the fact that time is given by velocity over acceleration and we see that \(t_{turn}=t_B-v_0/g=202+94=296years\). So at the turning point, 296 years have passed in the planet frame. Second, we know that at the point when the spaceship reached Destiny it had a velocity, \(\vec{v}_0=-0.99c\). After this point it has a constant acceleration of \(\vec{a} = -g\), meaning that once it reaches Destiny the second time around, it is only logical that it will have a velocity \(\vec{v}_0=0.99c\), the same value as before, in the opposite direction.

We will now calculate how time on Homey runs for the space ship system during the accelerated phase! This should tell us something about why Lisa spends such a long time on her journey.

We assume that during the accelerated phase, all the trains are 'chooing', in a series of events we call \(Y\) (at \(t_Y\) at the current position of the spaceship), and \(Y'\) (simultaneously with event Y in the spaceship frame at Homey). These are very similar to event B and B' from before!

Now, say that the gravity is given by \(g=-0.1m/s^2\) in the planet frame. We do not want to use these seconds in our relativity, this is a natural units only zone! We instead divide it by \(c^2\), in order to get rid of these seconds, and we get that \(g/c^2 = -1.1\cdot 10^{-18}m^{-1}\). Much better!

As usual, we'll begin by listing the time and place of each event:

where the equations in blue are those we needed to find.

We already know that at the turningpoint, when \(v=0\), the planet and spaceship do not have any velocity with respect to each other, we know that the time of this event in the spaceship frame must be equal to the time of the event in the planet frame, \(t_{Y'}=t_{Y}\). 

The expression for the time of event Y' can be found using invariance of the spacetime interval \(\Delta_{YY'}\),which yields the answer, \(t_{Y'}=t_Y\pm X_Yv\),. We can then use that event Y is simply a generalization of event B, meaning that we use the negative sign. At the time of the event at the turning point in the spaceship frame, v=0, and we can see the time on Homey:

\(t_{Y'}=t_Y-( L_0-(v_0(t_Y-t_B)+\frac{1}{2}g(t_Y-t_B)^2))v=296years\).

The same as in the planet frame!

During the deacceleration frame, the spaceship sees the clocks of Homey run much faster! At this point, the spaceship is 'catching up' with Homey, sort of, and it will seem as if time on Homey flies by in the blink of an eye. This makes sense, as when v=0, the spaceship is in the same frame as Homey, and cannot see things differently. Then, as we speed back up, time will go slower until we are at full speed back.

By adding the events (imagine a train blowing its whistle, or any short event you like) along the way, we can more easily see how the acceleration affects time. At a time when the spaceship is turning, its velocity is zero in Homeys frame, and so the events happen simultaneously!

We have that the time spent accelerating is 292 years each way! If we add the travel time from Homey to Destiny, 4 years, we get that the total time passed before Lisa again reaches Destiny (for the second time) is \(4+292+292=588\) years! We see how time on Homey has gone from 4 years, to 588 years during the accelerated phase.

Finally, we can find Lisa's age when she reaches Homey. We know she aged 57 years during the constant acceleration phase, and we only have to add the accelerated phase. We make this a little easier restarting our clocks and calculating this for the returning phase, at the moment when the velocity is 0, which we will call event E. As before, we assume the infinite trains heading back towards Destiny with different velocities from 0 to v₀. We imagine Lisa's acceleration by having her jump from train to train at interval \(\Delta t'\).

After Lisa turns around (event E), and jumps on the returning train, until she has reached the original velocity, v₀, the short intervals that she spends on each train (as she is always jumping forward/accelerating) corresponds to a short time interval \(\Delta T\) on planet clocks. By using the usual formula for time dilation\(\Delta t = \Delta t' /\gamma \leftrightarrow \Delta t' = \sqrt{1-v^2}\Delta t\), keeping in mind that \(v=g\Delta t\), we can make an expression for the relation between a time interval in the spaceship frame and the planet frame:

\(\Delta T' = \sqrt{1-g^2t^2}\Delta T\)

where we use T rather than t to keep track of the fact that this is after we've reset the time to 0 at event E.

We have now looked at the acceleration as an infinite sum of reference frames with different constant velocities, and therefore we can integrate the equation for the time intervals with respect to T, from 0 to \(v_0/g\). When we do this, we get the time Lisa ages from the point when she jumps on the first train, all the way till the final train that travels with constant velocity from Destiny to Homey. We won't do this integration here, but when we solve it, and insert our values, we see that the total time Lisa has aged when returning to Destiny is  \(74.5\) years. By using symmetry, we see that over the full acceleration phase, Lisa has aged \(149\) years. Adding the 57 years she spends traveling with constant velocity, she has aged a total of

206 years!

Let's put it all in a table:

So while the individual times are different, the total time is the same for the Planet frame and what is read from Homey clocks in Lisa's frame.

But Lisa still doesn't age as much as the people of Homey, why is this? If we think back to the principle of maximal aging, we get our answer. In spacetime distance, the distance in space is subtracted from the time. In other words, any observer in the lab system will measure a longer time between events than the observer in the rest system, as a way to make up for the lost spacetime difference from the position. A greater distance in space means that the time will go faster. The more 'ticks' on the clock between two events mean that the time is faster, and Homey measures more 'ticks' than Lisa does in her frame.

And that is the twin paradox!