Calculations For The Twin Paradox Problem In Special Relativity

Introduction

This page provides the equations, and their calculations, for the of special relativity. The final equations are demonstrated in the Twin Paradox Graphs And Equations page.

Constant Velocity Situation

Scenario

Consider an astronaut floating in space with a flashlight and a mirror at a constant distance x. She briefly shines a flashlight at the mirror, and measures the elapsed time for the light to travel twice the distance x (away and back). Next, imagine placing a planet just under the astronaut and mirror, with the astronaut and mirror moving at some velocity v relative to the planet (or, depending on your point of view, the planet moving at some velocity -v relative to the astronaut and mirror). In this case, the burst of light would no longer travel away and back along the same path to a scientist standing on the planet, but instead appear to travel away along a diagonal path, reflect off the mirror, and then travel back along a different diagonal path (see picture below). Both diagonal paths are longer than the original separation x. Somehow, in the same amount of time, the light has traveled a farther distance according to the scientist! How could this happen? Did the light travel at two different speeds?

Reference Frame

With the information we have so far, there is nothing to tell us which person, astronaut or scientist, is moving. Depending on your point of view, either one is moving, or even both if your point of view is from some meteoroid flying by. However, we will select the choice that is more familiar, and say the astronaut and mirror are moving, and the scientist is at rest. The reference frame moving with the astronaut will therefore be called M (=moving), and the reference frame attached to the planet will therefore be called R (=resting). As a bonus, this definition of reference frame M still works later on when we apply acceleration to the astronaut, in which case the astronaut even feels herself moving.

Solving the Equations

At this point, we do not know which of the time, distance and velocity (t, x and v) measurements is different. Our daily experience no longer applies to situations where velocities are close to the speed of light, as we shall see. So, our instincts are no longer reliable, and any assumptions we make may turn out to be wrong.

Since t, x and v measurements may be different for the train passenger and observer, we will label them with subscripts: tM, dM, vM, tR, xR and vR. We will solve, at the start, our equations for the general case in which all of these quantities can have separate values, because at this point we do not know any better.

We will follow Einstein's proposal that the speed of light c is the same for both people, and we remember that vM 0 was a starting assumption. If our calculations work out, then these assumptions are valid (and they will).

Moving Train Passenger

The moving train passenger sees the light travel away a distance xM, and then return the same distance xM. Using the well known relationship

For this geometry

Observer At Rest

The stationary observer sees the light travel diagonally away from the train, and then diagonally back to the train (although not easy, one could put up sensors along the path). The distance traveled diagonally now depends on x, and the distance traveled (v*t). According to the Pythagorean theorem:

Applying our diagonal distance to equation (3)

First Try With t, x, v and c The Same

Let's assume, quite reasonably, that t, x, v and c are the same for both. From equation (3) we get

From equation (7) we get

Substituting in x from (8)

which is only true when v = 0, which violates our starting assumption, and/or t=0 which tells us something is wrong when t>0 (and maybe even when t=0, but the point is that t is not allowed to be greater than 0).

Second Try With t, v and c The Same But x Different

Let's try again, assuming x is different, but t, v and c are the same for both

which means xR=xM at t=0, but before and after that xR>xR. This means t=0 is special, but it really should not be special - we never said that t=0 had any significance.

Third Try With x, t and c The Same But v Different

Let's try again, assuming v is different, but x, t and c are the same for both

This does not help - vM does not even show up so whether or not it is different has no effect.

Final Try With x, v and c the Same But t Different

Let's try again, assuming the times are different, but x, v, and c are the same

This is the "time dilation equation". Although this result may not seem obvious, this makes sense. If v is much less than c then tR is extremely close to tM which is what we are used to in our normal, never-near-the-speed-of-light, lives (does your wristwatch slow down on the highway? yes, but not enough to make a difference). In fact, only when v is close to c does tR differ from tM very much. If v=c then the equation blows up, which suggests that v must stay less than c. So, because this result works so well we will assume from now on that x, v and c are the same but tR and tM are different according to equation (22).

It is often useful to simplify equations like the time dilation equation with the gamma factor

Note that we have not proved beyond a doubt that the time dilation equation is the final answer for the constant velocity case. However, the time dilation equation seems to work well in terms of not having unfixable paradoxes, and explaining various experimental measurements made over the years.

Success!

Constant Acceleration

In this case, there is a person at rest and a moving person. The moving person experiences a constant acceleration. How does special relativity affect the time, position, velocity and acceleration of the moving person in the stationary reference frame?

First Try, Assuming aR Is Constant

Let's try to solve for velocity, assuming aR=constant

Since aR=dvR/dtR, then dvR=aR*dtR. Integrating both sides gives

This says that if we wait long enough, then vR>c. But that conflicts with the requirement that nothing can go faster than the speed of light, so aR is not a constant.

Final Try, With Acceleration Adjusted By Gamma

At this point, we are forced to turn to external sources for information that will let us go further (Wikipedia), in this case because that information involves forces which is a whole topic in itself. Specifically, in special relativity, the nonrelativistic equation dv/dt=a becomes

Note that equation (25) is well approximated by the nonrelativistic equation dv/dt = a when v is much less than c, since gamma is close to 1.

Solving For Velocity vR

There is a vR hiding in gamma

Solving for vR we get

Although this result probably does not seem obvious, this makes sense. Initially, tR=0 which causes gamma0 to disappear so vR=vR(0) when tR=0, which was our starting assumption. Also, as t approaches infinity vR approaches c but never exceeds it. This is consistent with our result in the first section which says the math prevents velocities from exceeding the speed of light.

Solving For Acceleration aR

Given vR(tR), we can differentiate to get aR(tR)

when the math is worked out we get

Although this result probably does not seem obvious, this makes sense. Initially when tR=0 and vR(0)=0 we get aR(0)=aM(0) which means the acceleration is initially the same in both systems when the initial velocity is zero. Also, as t approaches infinity, aR approaches 0, which means the effective acceleration gets smaller and smaller, which is consistent with the velocity growing over time but never exceeding c.

Solving For Position xR

Given vR(tR), we can integrate to get xR(tR)

After trying several different substitutions that do not work, we find one that does:

Although this result probably does not seem obvious, it makes sense. When tR is small and vR(0)=0, the radical is approximately (using a Taylor series expansion and dropping higher order terms)

which means the relativistic equation has about the same value as the nonrelativistic equation when tR is very small and vR(0)=0.

When tR is very big, the radical is approximately

which is consistent with the velocity eventually getting very close to c, and the effective acceleration getting very close to 0, so after enough time the relatively finite acceleration period contributes relatively little to the position, and the relatively long constant-velocity period contributes very much (constant-velocity is described by distance=rate*time).

Solving For Time tM

How about the time in the observer's frame? We can start with equation (23) and apply it to the time differentials. How do we know we can do this? At this point, we don't know for sure. However, we will try and see what happens.

Replacing vR and rearranging gives

After maybe trying some substitutions that do not work, we find one that does:

So we can integrate to get

There is relatively unknown function called arsinh (not to be confused with arcsinh) that is defined as

So tR becomes

Although this result probably does not seem obvious, it makes sense. When tM is zero, arsinh is zero, so tR(0) is also zero (tR and tM start out at the same value which is what we assumed in the beginning). When tM goes to infinity, tR is pproximately (c/aM) log (2 * aM*tM/c), which is not very informative and does not appeal to us as a result that simply "feels right" (although it is monotonically increasing at least which is consistent with the idea that time always goes in one direction). So, at this point our assumption of equation [18] is maybe looking weak. However, various experimentalists have tested the equations of special relativity, and the experimental results agree well with theory.

Success!

Summary