in-the-theory-of-relativity-the-lorentz-contraction-formula-l-l_0-sqrt1-v2c2-expresses-the-length-l-

Question

In the theory of relativity, the Lorentz contraction formula 
$$ L = L_0 \sqrt{1 - v^2/c^2} $$
expresses the length  $ L $ of an object as a function of its velocity $ v $ with respect to an observer, where $ L_0 $ is the length of the object at rest and $ c $ is the speed of light. Find $ \displaystyle \lim_{v 	o c}-L $ and interpret the result. Why is a left-hand limit necessary?

Daniel Jaimes · Accepted Answer

This is problem or fifty six of the Stuart Calculus eighth edition, section two point three. In the theory of relativity, killer ends contraction formula ends. Hell equals al subzero. Time is a square root. Uh, the quantity one minus b squared over C squared This formula expressive the length l of an object as a function of its velocity. Be with respect in it and observer, where all subzero is the length of a dob. Check that rest and C is the speed of light. Find the limit and be a purchase See from the left with the function l ah and interpret the results. Why is the last time limit necessary? So let&#x27;s answer the first question first. Well, we see that this functional is restricted by this square roots saying which it should be clear cannot take any negative numbers. It can&#x27;t be that in mind. The domain of the square in here or a quantity inside of this square root needs to be, I reckon, greater than or equal to zero. This means that one must figure than equal to this racial and privately C squared needs to be greater than equal to V squared and at this point, we see that see must be greater than equal to B and in this way has via purchasing. We notice Avi is always less than thie. So we definitely are approaching C from the left. Now find this Lim. We approximate what we may be as a week approach. See? Well, it&#x27;s used to excess institution one minus C squared over C squared. Excuse us. Screwed of one minus one. Which is, of course, zero. So the result is that and he doesn&#x27;t want seen. So as your velocity approaches the speed of light the length of the object purchase zero. This means that the object essentially disappears as you approach on the speed of light.

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If $ p $ is a polynomial, show that $ \displaysty…

Problem 56 Hard Difficulty

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