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Special Relativity: Angles in different Inertial Systems

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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.
 

Answers and Replies

  • #2
BruceW
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hi, nice work so far. uh, for part c) you are meant to find the angle from part b), i.e. you are meant to find the angle theta in this equation: ##\cos(\theta)=v/c## And you are meant to assume v is close to c. So you are still considering a source which emits light, except now they are saying the source itself is going at close to the speed of light. It is not too difficult from here, as long as you start in the right place. You have two equations:
##\gamma=\frac{1}{1-v^2/c^2}##
##\cos(\theta)=v/c##
And you are told to get an equation which relates some function of ##\theta## to ##\gamma## (i.e. no ##v## involved). What is the simplest way to do this?
 
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Ah, yes. I see now. You just insert [itex] \cos^2 \theta [/itex] in place of [itex] v^2/c^2 [/itex] in the Gamma function, and reduce for the angle, while using the trigonometric identities given in the problem. A large part of my problem was that I did not even realise what angle to look for. Thanks a bunch!
 
  • #4
BruceW
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ah yeah, no problem. yes, I agree the hardest bit of part c) was seeing what it is they wanted. (which can often be the hardest bit of a lot of physics problems!)
 

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