Unable to Reconcile my Understanding of Length Contraction with the Lorentz Transformation

Question

I'm currently taking an undergraduate course in Special Relativity. I thought that I understood the material, but reviewing for the test has made my more confused than when I began. Particularly we were given the following problem:

A spaceman is chasing an alien space ship. As seen from a stationary observer on Earth, the aliens are going at a constant speed of $0.4c$ and the spaceman is going at a constant speed of $0.6c$. Initially they are at $1\space lightyear$ apart. In the spaceman's reference frame how long does it take for the aliens to be caught?

Preface

The correct answer is given as 4 years.

They arrived at this answer by calculating how long it would take in the Earth frame: $\frac{1 \space light year}{0.6c - 0.4c} = 5\space years.$ Then the professor noted that the astronauts time runs more slowly because it is moving relative to the earth at $0.6c$, so the duration of the chase in the astronaut's frame is $\frac{1}{\gamma}*5 * 0.8*5 = 4\space years$. This solution made sense to me AFTER reading it.

I was able to solve it in another way. I parameterized the spacemans's worldline in the Earth frame as $$\vec r_S = (0.6\lambda, \lambda)$$ In the earth's frame, the alien's worldline is $$\vec r_A = (0.4\lambda + 1, \lambda).$$ NOTE, these both are written in terms of $(x, ct)$. If we transform into the spaceman's frame, then according to the Lorentz Transform, the new form of the world lines will be $(x', ct') = ( [\gamma (x - \frac{u}{c}ct)],\space[\gamma (ct -\frac{u}{c}x]))$. In this primed frame, $$\vec r'_S = (0, 8\lambda)$$ $$\vec r'_A = (1.25 - 0.25\lambda, 0.95\lambda - 1.75)$$ To find the time of the collision, I found where $x'_S = x'_A$. Skipping the algebra, this occurs at the coordinates $$\lambda = 5: (x', ct') = (0, 4)$$ -- just as expected.

My question

I initially got this problem wrong because I tried to find the solution using length contraction in the spaceman's frame. I reasoned that if the spaceman is moving relative to Earth, then his measurement of distances in the x-direction will be contracted and shorter by a factor of $1/\gamma$. When we discussed the twin paradox in class, we talked about how the Earth-twin will "see" Rocket-twin's clock tick more slowly, which explains why Rocket-twin thinks the trip to the star is shorter. Then we talked about how Rocket-twin will see the length of their trip contracted, because "1 m" in their frame is shorter than a meterstick in the Earth's frame, and the distance was originally measured in the Earth's frame. See this question for another instance of this logic.

This reasoning failed. If the distance between the man and the aliens measured on Earth is contracted in the man's frame by $1/\gamma \mid \beta=0.6$, then he will see the aliens as only $0.8\space lightyears$ away initially. Additionally, we look at the slope of $r'_A$ in the primed frame, we see that to the spaceman the aliens are moving toward him at $0.263c$. So, shouldn't he see the chase take only $\frac{0.8}{0.263} = 3.04\space years$?

Even ignoring this incorrect solution. If the alien's speed is $0.263c$ toward him, then shouldn't he see them travel $4 * 0.263c = 1.052\space ly$ in his frame?

1.) Why can't I use length contraction and relative velocity to solve this problem?

2.) Why do I calculate the alien's moving 1.052 light years in the spaceman's frame? MORE than in the Earth's frame?

Answer 1

I'm working with c=1 and vectors in the form $(t,x)$

Length contraction is really a statement about parallel lines.

You take worldlines $(t,x_0)$ and $(t,x_0+\ell)$, apply a Lorentz transformation, and reparameterize, to get world lines $(t', x_0'+\beta t')$ and $(t', x_0'+\ell/\gamma+\beta t')$. The distance between the two lines changes to $\ell/\gamma$, and that's your length contraction.

If you can't phrase the length contraction you're interested in in terms of parallel lines, then it needn't exist, or you can get distances becoming larger! This might sound like a contradiction on face value (if stuff contracts and other things expand, won't they run into each other at some point?) but the Lorentz transformation is invertible - there can be no contradiction.

The facts of the matter are this:

Earth coordinates $(t,x)$:

• Event where the ship leaves: $(0,0)$.
• Event where the aliens start running: $(0,1)$
• Event where the ship catches the aliens: $(5,3)$

Ship coordinates $(t',x')$:

• Event where the ship leaves: $(0,0)$.
• Event where the aliens start running: $(-0.75, 1.25)$ (simultaneity is broken)
• Event where the ship catches the aliens: $(4,0)$
• Place where the aliens were at at $t'=0$: $(0., 1.0526)$

Those facts follow directly from Lorentz transformation. The general answer for the position of the aliens at $t'=0$, for spaceship moving at speed $\beta$, alien moving at speed $\alpha$, initially a distance $\ell$ from earth (all statements at t=0 in Earth's coordinate system), is $\frac{1}{\gamma}\frac{\ell}{1-\alpha \beta}$, with $\gamma=1/\sqrt{1-\beta^2}$. THIS is the quantity you can divide by the velocity of the aliens in the spaceship's frame, to get the time taken as measured by the spaceship.

That looks eerily similar to length contraction and velocity addition, so maybe there's a nice solution using those.

[edit] found one! Proceed as follows: Imagine a line parallel to the alien's world line and passing through the origin, in the Earth's frame. Go to a frame where both world lines are vertical: the horizontal distance between the lines expands by a factor of $1/\sqrt{1-\alpha^2}$ and the spaceship is moving at speed $\beta'=\frac{\beta-\alpha}{1-\beta \alpha}$. Now go into the frame where the spaceship is at rest: the new distance shrinks by a factor of $\sqrt{1-\beta'^2}$. So the answer is:

$$\ell''=\ell \sqrt{1-\beta'^2}/\sqrt{1-\alpha^2}$$

Chucking this into mathematica (I'm too lazy to actually expand $\beta'^2$!) shows that indeed:

$$\ell_0'=\ell\sqrt{1-\beta'^2}/\sqrt{1-\alpha^2}=\ell\frac{\sqrt{1-\beta^2}}{1-\alpha\beta}$$

Dividing by the absolute value of velocity $\frac{\beta-\alpha}{1-\beta \alpha}$ gives:

$$t'=\ell\frac{\sqrt{1-\beta^2}}{|\alpha-\beta|}$$

Plugging in $\ell=1$, $\beta=0.6$, $\alpha=0.4$ gives $t'=4.0$. So $\ell_0'/v'$ really gives the correct result.