Unable to Reconcile my Understanding of Length Contraction with the Lorentz Transformation
Hello Stack Exchange,
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 $\frac{1}{\gamma}5 * 0.85 = 4\space years$. 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 $(x', ct') = ( [\gamma (x - \frac{u}{c}ct)],\space[\gamma (ct -\frac{u}{c}x]))$. 275
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? 272