Spec Rel: Proof concerning IF through Minkowski diag

[phz]

Homework Statement

This problem is from "Relativity" by Rindler, second edition, problem 3.4:
Use a Minkowski diagram to establish the following result: Given two rods of equal length $$l_1$$ and $$l_2$$ ($$l_2 < l_1$$), moving along a common line with relative velocity $$v$$, there exists a unique inertial frame $$S'$$ moving along the same line with velocity $$c^2 \frac{l_1-l_2/\gamma (v)}{l_1 v}$$ relative to the longer rod, in which the two rods have equal lengths, provided $$l^2_1 (c-v) < l^2_2(c+v)$$.

Homework Equations

$$\tan \Theta = \frac{v}{c}$$ : (1)

The Attempt at a Solution

I have denoted the velocity of $$S'$$ relative the rod $$l_1$$ as $$v'$$. By use of (1) I identify that $$tan \Theta = \frac{v'}{c} = c \frac{l_1-l_2/\gamma (v)}{l_1 v}$$ : (2). When I saw this I thought I should be able to find a trigonometric relation in my diagram satisfying that relation.

I drew the diagram with $$l_1$$ along the principal x-axis, identified a side with length $$l_1-l_2/\gamma(v)$$ in the figure, found through similarity that $$\Theta$$ was in that triangle and calculated through Pyth. theorem the cathetus remaining (denoted $$X$$ in the figure) to be $$X = l_2 \sqrt{ 1 - \frac{1}{\gamma (v) ^ 2} } = l_2 \sqrt{ 1-(1-v^2/c^2) } = l_2 \frac{v}{c}$$. Had it been $$l_1 \frac{v}{c}$$ I would have been done I believe since that would have shown (2). What am I missing?

I also don't see the what effect the last constraint $$l^2_1 (c-v) < l^2_2(c+v)$$ has on the problem.

EDIT: Right now I'm thinking that I am using the Minkowski diagram incorrectly. When trying to find the length of $$l_2$$ on the principal x-axis I shouldn't draw a line perpendicular to the x-axis but instead parallell to the time axis corresponding to the inertial frame in which $$l_2$$ is at rest, right? That makes my geometry wrong, and more complicated I guess.

EDIT 2: But when I'm looking at the figures in the book it seems like I interpretated the diagrams correctly the first time, and I'm back at my previous problem again. Or more precisely, I'm more lost than before. Is it the calibrating hyperbolas I am missing? How do they affect my drawing?

EDIT 3: If I calculate my $$X$$ another way, by denoting the angle between $$l_1$$ and $$l_2$$ as $$\alpha$$ and using that $$\tan \alpha = \frac{v}{c}$$ I get $$X = \frac{v l_2}{c \gamma(v)}$$ which gives $$v' = c^2 ( l_1 - \frac{l_2}{\gamma(v)} ) \frac{\gamma(v)}{v l_2 }$$, a different result. In this case, if $$\frac{l_2 }{\gamma (v) } = l_1$$ the given expression for $$v'$$ would have been fulfilled. The fact that I get different results depending on which relation I use to calculate $$X$$ clearly indicates that I have misunderstood something.

 

[phz]

I used the Minkowski diagrams incorrectly. The projection of $$l_2$$ onto $$l_1$$ wasn't perpendicular but with an angle corresponding to the time-axis for $$l_2$$, not included in my image above. At first I thought this would lead to a contradiction when I tried to project $$l_2$$ onto $$x'$$, but that was because I didn't consider the calibrating hyperbola.

I believe the last constraint is just an expression that says that the relativistic speed limit should be obeyed.