# Rods of two different lengths have same length in unique inertial frame

I was re-reading my old Relativity book (by Rindler) and taking a look at some of the problems. He asks: Using a Minkowski diagram to establish the following result:

Given two rods of rest lengths $l_1$ and $l_2 (l_2 < l_1)$, moving along a common line with relativity velocity $v$, there exists a unique inertial frame $S'$ moving alone the same line with velocity $c^2 [l_1 - l_2/\gamma(v)]/(l_1v)$ relative to the longer rod, in which the two rods have equal lengths, provided $l_1^2(c-v) < l_2^2(c+v)$.

I'm generally confused on how to use a Minkowski diagram to derive this mathematical result, but I'm more confused on conceptual grounds. In the frame $S$ of $l_1$ you have the the lengths $l_1$ and $\frac{l_2}{\gamma(v)}$ for the two rods. Now, if we were to subtract these two values and make a LT into another frame where their lengths are equal - $\Delta x' = \gamma((l_1 - l_2/\gamma(v)) - v'\Delta t) = 0$ we can solve for $v'$. But $\Delta t$ should be $0$ since you're measuring these lengths simultaneously. Since you'll measure these lengths simultaneously in $S'$ as well $\Delta t'$ will be zero too. So you have three coordinate differences that are zero, which doesn't seem to make sense for this type of problem. Am I just approaching this all wrong? Does the Minkowski diagram derivation make this result trivial? Also, assuming that all both rods will contract by the same factor in this inertial frame, I don't see how it's possible for $l_1$ to ever be equal to $l_2$.

• @hijasonno Sorry, using Minkowski diagrams isn't something I know much about. The simplest solution is to equate the two lengths measured in the observer's frame : $l_1/\gamma(v_1) = l_2/\gamma(v_2)$. – sammy gerbil Feb 23 '18 at 14:35