Show these two equations are equivalent (involves sines and cosines)

  • #1
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Homework Statement


Show that
[tex]sin(\alpha)-sin(\beta)=-\frac{v}{c}sin(\alpha+\beta)[/tex]

is equivalent to

[tex]cos(\beta)=\frac{-2\frac{v}{c}+(1+\frac{v^2}{c^2})cos(\alpha)}{1-2\frac{v}{c}cos(\alpha)+\frac{v^2}{c^2}}[/tex]

Homework Equations


This is the last step in the derivation of the relativistic reflection law. The first equation was derived using the principle of least time and the second is from Einstein's paper on special relativity.
[itex]\alpha[/itex] = angle of incidence
[itex]\beta[/itex] = angle of reflection
[itex]v[/itex] = velocity of mirror away from light
[itex]c[/itex] = speed of light

link to diagram

The Attempt at a Solution


I used the identity [itex]sin(\alpha+\beta = sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)[/itex] on the RHS, squared both sides, changed by the squared sines into cosines squared - some more algebra and this is the best I've got at the moment:

[tex]2c^2(1-sin(\alpha)sin(\beta))-(c^2-v^2)(cos^2(\alpha)+cos^2(\beta))=-2v^2cos^2(\alpha)cos^2(\beta)+2sin(\alpha)cos(\beta)sin(\beta)cos( \alpha)[/tex]

How do I get rid of the sines?
 

Answers and Replies

  • #2
haruspex
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Is this supposed to be true without any further relationship between v/c and the two angles? Doesn't seem to be to me.
I tried writing v/c as tan(phi). The second eqn becomes
##\sin(2\phi) = \frac{\cos(\alpha)-\cos(\beta)}{1-\cos(\alpha)\cos(\beta)}##
Plugging in 0.5 and 1 for the two angles did not give the same phi as given by the first eqn.
 
  • #3
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The only relations between v, c and the angles are the ones on the diagram

[tex]cos( \alpha) = \frac{d_0 + vt_a}{ct_a} = \frac{d_0+vt_a}{ \sqrt{(d_0+vt_a)^2+x^2)}}[/tex]
[tex]cos( \beta) = \frac{d_0 + vt_a}{ct_b} = \frac{d_0+vt_a}{\sqrt{(d_0+vt_a)^2+(l-x)^2)}}[/tex]
 
  • #4
haruspex
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The only relations between v, c and the angles are the ones on the diagram

[tex]cos( \alpha) = \frac{d_0 + vt_a}{ct_a} = \frac{d_0+vt_a}{ \sqrt{(d_0+vt_a)^2+x^2)}}[/tex]
[tex]cos( \beta) = \frac{d_0 + vt_a}{ct_b} = \frac{d_0+vt_a}{\sqrt{(d_0+vt_a)^2+(l-x)^2)}}[/tex]

From that, it would appear that there are 2 degrees of freedom for the variables, so effectively no constraints beyond the eqn in the OP. I would try plugging in some numbers for the variables and check that the equations really are equivalent.
 

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