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PHYSICS 131 hw-3-soln

# PHYSICS 131 hw-3-soln - HOMEWORK SET 3 PHYS 131 Instructor...

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HOMEWORK SET 3, PHYS. 131 Instructor: Anthony Zee Email: [email protected] TA: Matt Pillsbury Email: [email protected] Handout Problem: Consider a free particle in Fat spacetime with coordinates y α moving according to d 2 y α 2 = 0 (1) with the proper time 2 = - η αβ dy α dy β . Now transform to another frame with coordinates x μ , a frame that may be accelerating, curvilinear, whatever. Simply plugging y a ( x μ ) into the equation above and using the rules of calculus show that x μ satis±es the equation the motion d 2 x μ 2 + Γ μ νλ dx ν dx λ = 0 (2) where the quantity Γ μ could be calculated knowing y α ( x μ ) . Using the coordinate transformation y α ( x μ ) (which will, in general, have a well-de±ned inverse x μ ( y α )), we can use the chain rule to ±nd dx μ = ∂x μ ∂y α dy α (3) dy α = μ α dy α (4) Di²erentiating (3) with respect to τ then yields d 2 x μ 2 = μ α d 2 y α 2 + 2 x μ α β dy α dy β The ±rst term on the right vanishes due to (1), and (4) gives = ± μ α d 2 y α 2 α ν β λ ² dx ν dx λ If we identify Γ μ = - μ α d 2 y α 2 α ν β λ then (2) is satis±ed. 1

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6.7. (a) An accelerated laboratory has a bottom at x ± = 0 and a top at x ± = h , both with extent in the y ± - and z ± -direction. Use the line element derived in part (a) of Problem 6.6 to show that the height of the laboratory remains constant in time, i.e., the laboratory moves rigidly. The line element in question is ds 2 = - (1 + gx ± ) 2 ( dt ± ) 2 +( dx ± ) 2 dy ± ) 2 dz ± ) 2 Note a displacement entirely in the x ± -direction will have no dependence on t ± , so the distance from the bottom to the top of the lab will not change with time. (b) Compute the acceleration a ( a · a ) 1 / 2 , where a α = d 2 x α /dτ 2 , and show that it is diFerent at the top and the bottom of the laboratory. Using the results of part (c) of Problem 6.6, at x ± = 0, we have = dt ± . If we’re standing still in the lab’s frame, we’ll have dx ± /dt ± = 0 everywhere, and d 2 t 2 = 2 t ∂t ± 2 = g sinh( gt ± ) , d 2 x 2 = 2 x ± 2 = g cosh( ± ) and ( a · a ) 1 / 2 = - ( a t ) 2 a x ) 2 = g .
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PHYSICS 131 hw-3-soln - HOMEWORK SET 3 PHYS 131 Instructor...

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