An angled photon (special relativity)

[philnow]

Homework Statement

A photon moves at an angle theta with respect to the x' axis in the frame S'. Frame S' moves with speed v with respect to frame S (along the x' axis). Calculate the components of the photon's velocity in S and verify that it's speed is c.

The Attempt at a Solution

I break down the photon speed in S' into Ccos(theta) for X and Csin(theta) for y. There is no change from the transformation of frames with respect to the y-axis, only the x-axis. So the speed of the horizontal component of the photon in S frame is then

(Ccos(theta) + V) / (1 + (Ccos(theta)*V/C^2))

by velocity addition.

I would like to show that this squared, plus Csin(theta) squared should be equal to C^2. The problem is the algebra... whatever I do, I end up with cosines to odd and even powers that just won't simplify, so here I am. Any hints?

 

[tauon]

well, this part is right

$$X' : \ c_x= c\cos\theta$$
$$Y' : \ c_y= c\sin\theta$$

but there is a change in both vector components when you transform from $$S'$$ to $$S$$
(there is no change from the transformation of frames with respect to the y-axis only when $$\theta=0 \ or \ \pi$$)

as for what you're trying to do, obviously:

$$(\frac{c\cos\theta + v}{1+\frac{vc\cos\theta}{c^2}})^2 + (c\sin\theta)^2 \neq c^2 \ (i)$$

because

$$(\frac{c\cos\theta + v}{1+\frac{vc\cos\theta}{c^2}})^2 + (c\sin\theta)^2=\frac{c^2\cos^2\theta +v^2+2vc\cos\theta+c^2\sin^2\theta+2vc\cos\theta\sin^2\theta+v^2\cos^2\theta\sin^2\theta}{\frac{c^2+2vc\cos\theta+v^2\cos^2\theta}{c^2}}=$$

$$=c^2\frac{c^2+v^2+2vc\cos\theta(1+\sin^2\theta)+v^2\cos^2\theta\sin^2\theta}{c^2+2vc\cos\theta+v^2\cos^2\theta}$$

$${not} (i) \ \Leftrightarrow \ (v=0) \vee (\cos\theta=1\Leftrightarrow \theta\in \{0,\pi \})$$

which means that either $$S'$$ is not moving relative to $$S$$ either the velocity vector of the photon is parallel with the x-axis (and both cases are contradictory with the premises of the exercise).

you must go back to square one. you probably forgot how to calculate the photon's velocity vector components in S. you should have that in your textbook. good luck.

 

[philnow]

Ok that makes sense. But how does the Y component of the photon's velocity change from S' to S?

 

[tauon]

Ok that makes sense. But how does the Y component of the photon's velocity change from S' to S?

well , (you should have this in your textbook but...) the Y component from S is:

$$\displaystyle \frac{c_y}{\lambda (1+\frac{vc_x}{c^2})}$$, where $$\lambda$$ is the Lorentz factor...

 

[philnow]

Hi again. I'm assuming you meant without the lorentz factor, I think the that is only for S to S', and not vice versa. So using this, I'm getting

Ux = C*sin / (1 + cv*cos/c^2)

and Uy = C*cos / (1 + cv*cos/c^2)

Once again, my algebra defeats me. This time it simplifies much better, though... I'm getting:

C^2*(c^2 + v^2 + 2vc*cos)/(c^2 + cos^2*v^2 + 2vc*cos)

I can't get rid of that damn cos^2 in front of the v in the denominator. I realize this is trivial algebra, but I can't figure out the problem! I checked with my professor and we have the same Ux and Uy.

 

[philnow]

Ux^2 + Uy^2... should equal C^2

What's the problem folks :(