Morin Problem 11.10

Physics · mechanics

Problem

A bunch of L.T.'s Using the results from the "Passing trains" examples in Sections 11.5.1 and 11.6, the relative speeds and the associated γ\gamma factors for the six pairs of frames are AB AC AD BC BD CD v 5c/13 4c/5 c/5\sqrt{5} 3c/5 c/5\sqrt{5} 5c/7\sqrt{7} γ\gamma 13/12 5/3 5/2\sqrt{5}/2 5/4 5/2\sqrt{5}/2 7/2\sqrt{7}/2 From the example in Section 11.6, the separations between the two events in the four frames are A B C D Δx −L L 5L 0 Δt 5L/c 5L/c 7L/c 262\sqrt{6}L/c The Lorentz transformations are Δx = γ\gamma(Δx' + vΔt'), (11.84) Δt = γ\gamma(Δt' + vΔx'/c²). For each of the six pairs, we'll transform from the faster frame to the slower frame. This means that the coordinates of the faster frame will be on the right-hand side of the L.T.'s. The sign on the right-hand side of the L.T.'s will therefore always be a "+." In the AB case, for example, we'll write "Frames B and A," in that order, to signify that the B coordinates are on the left-hand side and the A coordinates are on the right-hand side. We'll simply list the L.T.'s for the six cases, and you can check that they do indeed all work out. Frames B and A: L = 1312\frac{13}{12}(-L + 5c13\frac{5c}{13} · 5Lc\frac{5L}{c}), 5Lc\frac{5L}{c} = 1312\frac{13}{12}(5Lc\frac{5L}{c} + 5c(−L)13c2\frac{5c(-L)}{13c²}). Frames C and A: 5L = 53\frac{5}{3}(-L + 4c5\frac{4c}{5} · 5Lc\frac{5L}{c}), 7Lc\frac{7L}{c} = 53\frac{5}{3}(5Lc\frac{5L}{c} + 4c(−L)5c2\frac{4c(-L)}{5c²}). Frames D and A: 0 = 52\frac{\sqrt{5}}{2}(-L + c5\frac{c}{\sqrt{5}} · 5Lc\frac{5L}{c}), 26Lc\frac{2\sqrt{6}L}{c} = 52\frac{\sqrt{5}}{2}(5Lc\frac{5L}{c} + c(−L)5c2\frac{c(-L)}{\sqrt{5}c²}). Frames C and B: 5L = 54\frac{5}{4}(L + 3c5\frac{3c}{5} · 5Lc\frac{5L}{c}), (11.85) 7Lc\frac{7L}{c} = 54\frac{5}{4}(5Lc\frac{5L}{c} + 3cL5c2\frac{3cL}{5c²}). 11.13 Solutions 571 Frames B and D: L = 52\frac{\sqrt{5}}{2}(0 + c5\frac{c}{\sqrt{5}} · 26Lc\frac{2\sqrt{6}L}{c}), 5Lc\frac{5L}{c} = 52\frac{\sqrt{5}}{2}(26Lc\frac{2\sqrt{6}L}{c} + c(0)5c2\frac{c(0)}{\sqrt{5}c²}). Frames C and D: 5L = 72\frac{\sqrt{7}}{2}(0 + 5c7\frac{5c}{\sqrt{7}} · 26Lc\frac{2\sqrt{6}L}{c}), 7Lc\frac{7L}{c} = 72\frac{\sqrt{7}}{2}(26Lc\frac{2\sqrt{6}L}{c} + 5c(0)7c2\frac{5c(0)}{\sqrt{7}c²}).

Browse the problem library