A carrot-slicing machine consists of eight parallel blades spaced 5 cm apart, held together in a framework

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A carrot-slicing machine consists of eight parallel blades spaced 5 cm apart, held together in a framework that allows all the blades to descend at once upon an unsuspecting carrot laid out horizontally in the machine. The result is several carrot pieces of length \(5 \mathrm{~cm}\), plus random bits left over at each end. Now suppose that a carrot is made to move lengthwise at speed \(4 / 5 c\) into the machine just as the blades descend. The Lorentz contraction ensures that the carrot will be shorter in the machine frame than in its rest frame, so there will be fewer carrot pieces. Each of these non-end pieces will still have length \(5 \mathrm{~cm}\) in the machine frame because that is the spacing of the blades, so it appears they must be longer than \(5 \mathrm{~cm}\) when finally brought to rest: In fact, each should have restlength \(5 \mathrm{~cm} /(3 / 5 c)=81 / 3 \mathrm{~cm}\). Now view the exact same procedure in the rest-frame of the carrot. Then it is the slicing machine that moves at \(4 / 5 c\), so it contracts as a whole, and the distance between blades is Lorentz-contracted to \(5 \mathrm{~cm} \sqrt{1-(4 / 5)^{2}}=3 \mathrm{~cm}\). That is, it seems that it produces carrot pieces \(3 \mathrm{~cm}\) long in their rest-frame. These conclusions \((81 / 3 \mathrm{~cm}\) and \(3 \mathrm{~cm})\) cannot both be correct, since it is the same carrot that was involved in both sets of reasoning. Which is the correct answer (if either) and why is the other answer or answers wrong?

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Modern Classical Mechanics

ISBN: 9781108834971

1st Edition

Authors: T. M. Helliwell, V. V. Sahakian

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