sol01 - 1 University of California Berkeley Physics H7C...

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1 University of California, Berkeley Physics H7C Spring 2011 ( Yury Kolomensky ) SOLUTION TO PROBLEM SET 1 Special Relativity Maximum score: 200 points Due: January 28, 2011 Composed and formatted by E.A. Baltz, M. Strovink, Yu. Kolomensky, S. Oliver, and T.R. Beals 1. (20 points) Is it consistent with the principles of the Special Relativity to think of an electron as a uniform rotating sphere of mass m = 9 . 11 · 10 31 kg (electron mass) and radius R = 2 . 82 · 10 15 m (classical electron radius) if its spin (internal angular momentum) is S = 9 . 1 · 10 35 kg · m 2 / s ? 1. Solution One of the main principles of Special Relativity is that no physical object (or point on a physical object, for that matter) can move faster than a speed of light. It is clear that of electron were a solid rotating sphere, a point on its equator would be moving with linear veloc- ity v = ωR = SR I where I = 2 5 mR 2 is the moment of inertia of a uniform solid sphere. Plug- ging in numbers, v = 5 S 2 mR = 8 . 9 · 10 10 m / s c which is unphysical. Electron does have an internal an- gular momentum degree of freedom (spin), but it can- not be described by any mechanical equivalent. More on this in the quantum mechanics part of H7C. .. 2. (30 points) Taylor and Wheeler problem 27: the twin paradox (a part thereof) On their twenty-first birthday, Peter leaves his twin Paul behind on the earth and goes off in the x direction for seven years of his time at 24/25 the speed of light, then reverses direction and in another seven years of his time returns at the same speed. [In this most elementary ver- sion of the problem, we assume that the necessary peri- ods of acceleration are infinitesimal in duration, requir- ing Peter’s acceleration to be infinite. Nonetheless, our plucky twin remains uninjured.] (a) Make a spacetime diagram ( ct vs. x ) showing Peter’s motion. Indicate on it the x and ct co- ordinates of the turn-around point and the point of reunion. For simplicity idealize the earth as an inertial frame, adopt this inertial frame in the construction of the diagram, and take the origin to be the event of departure. (b) How old is Paul at the moment of reunion? 2. Solution On a spacetime ( ct vs. x ) diagram in Paul’s (unprimed) frame, Peter begins at (0,0) and proceeds with slope β 1 = 25 24 for a time interval c Δ t = γc Δ t + γβ x = 0) = γc Δ t = r 1 1 ( 24 25 ) 2 c Δ t
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2 = 25 7 c Δ t = 25 lightyr At Peter’s point of maximum excursion, ( ct = 25 , x = 24) lightyr. Peter then returns with slope β 1 = 25 24 , reaching x = 0 at ct = 50 lightyr where he reunites with Paul. Peter has aged only 14 years, while Paul has aged 50 years (and has reached the age of 71). 3.
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sol01 - 1 University of California Berkeley Physics H7C...

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