- #1

allyouneedislove

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## 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?