Inertial frame where rods have same length

[allyouneedislove]

Homework Statement

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 $l_1^2(c-v) < l_2^2(c+v)$.

Homework Equations

$\Delta x' = \gamma(\Delta x) - v'\Delta t)$
$u_1' = \frac{u_1 - v}{1 - \frac{u1 v}{c^2}}$
$L = \frac{L_0}{\gamma}$

The Attempt at a Solution

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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?

 

[Orodruin]

Now, if we were to subtract these two values and make a LT into another frame where their lengths are equal -

This is not a correct approach. You must take into account that the second rod moves in the rest frame of the first and therefore the coordinate of the end of the second rod will depend on time (and therefore $\Delta t$). Furthermore, in order to compare the lengths in the new system you must make sure that you are looking at events that are simultaneous in that system, not in any other system such as the rest frame of one of the rods.

Does the Minkowski diagram derivation make this result trivial?

Pretty much. It is a geometrical construction.

 

[allyouneedislove]

Okay, after some more careful thought and your insight I think I understand it a little better. I do want to try the geometrical construction at some point, but I would want to solve it algebraically first. Would I be solving the system $l_2 \gamma(v'-v) = l_1 \gamma(v')$ for $v'$? I'm thinking that since both rods and the unique frame lie along the same axis, $l_2$ would be contracted by a smaller value than would $l_1$. Their lengths would be measured simultaneously in $S'$ and we could just set them equal to each other and solve for $v'$?

 

[Orodruin]

You seem to be assuming that velocities add "normally". This is not the case, you need to use relativistic velocity addition if you are going to take that approach.

 

[allyouneedislove]

Okay, so I managed to derive it algebraically after seeing my error of not transforming the relative velocity. Do you have any hints for deriving it using the Minkowski diagram?