ph1bA_08_hw3s

ph1bA_08_hw3s - Ph1b(analytic track Solution to Homework 3...

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Ph1b (analytic track): Solution to Homework 3 Chefung Chan ([email protected]), Winter 2008 Problem 7-5 (a) (2.5 points) Recall that proper time is related to lab time (see sec. 5.8) by: = 1 γ dt . Then, it is straightforward to see that m dt = γmv dx 0 + γm dt 0 = γvm dx 0 + γ ( γ 0 m ) E = γvp 0 x + γE 0 . Where we have used Eq. (7-5) in the last setup: E = m dt = γm , p x = m dx , E 0 = γ 0 m, p 0 x = m dx 0 . Similarly, m dx = dx 0 + dt 0 p x = γp 0 x + γvE 0 m dy = m dy 0 p y = p 0 y m dz = m dz 0 p z = p 0 z . (b) (2.5 points) The above procedure can be repeated to give the inverse trans- formations. Or, one can obtain the inverse transformations from the “direct” transformations above by changing the sign of v (i.e., v 7→− v ) and interchang- ing lab and rocket labels (primed unprimed coordinates): E 0 = x + p 0 x = x p 0 y = p y p 0 z = p z . 1
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Problem 7-6 (a) (2 points) Given K =47GeV=47 , 000 MeV, the energy gain (which we’re told is linear) is: Energy gain = 47 , 000 MeV 3 , 000 m =16MeV/m . The Newtonian kinetic energy for an electron traveling at speed c is: K Newt ( v = c )= 1 2 mc 2 = 1 2 0 . 5MeV=0 . 25 MeV . Thus, the distance to be traveled is distance to travel = 0 . 25 MeV 16 MeV/m =0 . 016 m = 16 mm . (b) (2 points) First, we need to ±nd what γ is for these electrons (recall that γ (1 v 2 ) 1 / 2 ). We also need that the kinetic energy K
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ph1bA_08_hw3s - Ph1b(analytic track Solution to Homework 3...

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