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Amy Truman.  Consider a luminous knot in a quasar jet that has been observed to move transversely across the sky with an apparent speed of beta = 3.6.

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Presentation on theme: "Amy Truman.  Consider a luminous knot in a quasar jet that has been observed to move transversely across the sky with an apparent speed of beta = 3.6."— Presentation transcript:

1 Amy Truman

2  Consider a luminous knot in a quasar jet that has been observed to move transversely across the sky with an apparent speed of beta = 3.6. ◦ Show that the actual speed of the material in the jet 0.95 < beta < 1. In doing so, be sure to include a carefully drawn and labeled (by you) side-view figure that clearly identifies and defines all the quantities you use and how you use them.

3 ◦ Assuming the slowest possible speed from the range in part 1, determine the angle between the jet and the line of sight. ◦ Plot the observed speed as a function of the angle between the jet and the line of sight and note the maximum angle for which superluminal motion can be observed.

4 A = Quasar B = Observer C = Jet Diagram

5  Jet emitted at t=0  We will see it emitted at t 1 ◦ t 1 =(D+vtcosθ)/c  We will see jet arrive at position C at t 2 ◦ t 2 =t+(D/c)  Time elapsed ◦ Δt=t 2 -t 1 =(t+D/c)-(D+vtcosθ)/c =t(1-βcosθ)

6  Angular Separation ◦ φ=vtsinθ/D  Infer a transverse velocity ◦ β app =v app /c =(1/c)(DΔφ/Δt) =vsinθ/c(1-βcosθ) =βsinθ/(1-βcosθ)

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8  When solving for cosθ ◦ cosθ=β  When solving for sinθ ◦ sinθ=(1-cos 2 θ) 1/2 =(1-β 2 ) 1/2 =1/γ  γ=(1-β 2 ) 1/2  Plugging cosθ and sinθ back into the original equation

9 BetaTheta  cosθ=β  θ=cos -1 β=15.5

10  Plug both theta and beta back into the apparent beta to check

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12  The actual speed of the jet is about 0.964c at an angle of 15.5°

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