A survey network of buoys defines the inertial frame ; a second inertial frame moves…

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Special relativity: spacetime and kinematics · Special relativity: spacetime and kinematics

Relativity and Gravitation, Ch Minkowski spacetime, four-vectors and causal structure

Problem

A survey network of buoys defines the inertial frame SS; a second inertial frame S′S' moves along the common +x+x axis in the standard configuration with rapidity ψ=1.200\psi = 1.200, so that the boost reads cΔt′=cΔtcosh⁡ψ−Δxsinh⁡ψc\Delta t' = c\Delta t\cosh\psi - \Delta x\sinh\psi and Δx′=Δxcosh⁡ψ−cΔtsinh⁡ψ\Delta x' = \Delta x\cosh\psi - c\Delta t\sinh\psi. Gravity is ignored. Two flashes occur in SS separated by Δt=2.001 μs\Delta t = 2.001\ \mu\mathrm{s} and Δx=1000.0 m\Delta x = 1000.0\ \mathrm{m}, with Δy=Δz=0\Delta y = \Delta z = 0. ((a)) Using only cosh⁡2ψ−sinh⁡2ψ=1\cosh^2\psi - \sinh^2\psi = 1, substitute the boost above into −c2Δt′2+Δx′2-c^2\Delta t'^2 + \Delta x'^2 and show that it equals −c2Δt2+Δx2-c^2\Delta t^2 + \Delta x^2 for any ψ\psi. ((b)) Evaluate cΔt′c\Delta t' and Δx′\Delta x' for this pair of flashes, and show numerically that Δs′2=Δs2\Delta s'^2 = \Delta s^2. ((c)) A probe moves along the +x+x axis of SS with rapidity ψ=1.200\psi = 1.200. Show that its four-velocity is U=(ccosh⁡ψ, csinh⁡ψ, 0, 0)\mathbf{U} = (c\cosh\psi,\ c\sinh\psi,\ 0,\ 0), evaluate the two components, and compute U⋅U\mathbf{U}\cdot\mathbf{U}. Take c=2.998×108 m s−1c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}, cosh⁡1.200=1.81066\cosh 1.200 = 1.81066, sinh⁡1.200=1.50946\sinh 1.200 = 1.50946.

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