- #1

Guybrush

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## Homework Statement

a) A light-source at rest in an inertial system S' is emitting light with angle θ' relative to the x'-axis. The System S' is moving at speed v (along the x-axis) relative to the laboratory system S. Show that the emitted light is making an angle θ with the x-axis of:

[itex] \cos \theta = \frac{c \cos \theta' + v}{c + v\cos \theta'} [/itex]

b) Now assume that the light-source is emitting its light isotropically in its own reference system. Show, that the part of the light being emitted into the forward-going hemisphere is located within a cone with half angle:

[itex] \cos \theta = v/c [/itex]

c) Show that this angle, for speeds close to c, can be approximated with [itex] \gamma^{-1} [/itex], where [itex] \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} [/itex]

(Tip given in the text: [itex] \sin^2 x + \cos^2 x = 1 [/itex], and [itex] \sin x \approx 1, x << 1 [/itex])

## Homework Equations

The relevant equations are the equations for transforming speeds in different inertial systems

[itex] u_x = \frac{u'_x + v}{1 + u'_xv/c^2} [/itex]

[itex] u_{y,z} = \frac{u'_{y,z}}{\gamma (1 + u'_xv/c^2)} [/itex]

## The Attempt at a Solution

Parts a), and b) I have solved with no problem. In part a) you just realise that the speed of light is constant in all reference frames so that [itex] u = u' = c [/itex]. Using standard trigonometry you get that [itex] u'_x = c \cos \theta' [/itex]. You get [itex] u_x [/itex] by inserting in the equation above. Then [itex] \cos \theta = u_x/u [/itex], which reduces to the correct answer.

In part b) you just use the limiting case where θ' = 90°, which reduces to the correct equation.

In part c) I am not sure what to do. In large part because the problem text seems very ambiguous (what angle am I supposed to find? What speed is close to the speed of light, is it the relative speed between the inertial systems or have we ceased looking at light, and instead some other particle emitting at speeds close to the speed of light?). I have tried taylor expanding the above result in v/c around 1, but the math gets so convoluted, that I doubt this is the right approach.

I have translated the problem text above from another language into english, but I hope it is still understandable. Any help would be much appreciated.