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spaceofwaste
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Hi,
I'm working through an Excercise in Sean Carrol's spacetime and geometry book. The question asks you to consider an inertial frame S with coordinates x^\mu=(t,x,y,z) and a frame S' with primed coordinates. Which is related to S by a boost v in the y direction. Imagine a wall( or mirror) lying along the line x'=-y'. From the point of view of S, what is the relationship between the angle of incidence (assuming ball travels in x-y plane only) and angle of reflection? Also what is velocity before and after?
Unfortunately there are not even one word answers let alone solutions in this book, so I don't even know if what I have is correct.
The way I proceeded, was firstly to consider what angle the mirror would appear at in S. Since the y-direction will be Lorentz contracted, but the x -direction will remain unchanged, the mirror should appear to be at a greater angle than 45 to someone in the S frame. This angle would precisely be:
\theta_M=arctan(\gamma \frac{\Delta x'}{\Delta y'})=arctan(\gamma), the last equality following from the angle being 45 degrees in the primed frame meaning \Delta x'=\Delta y'
OK so that is my step 1. Now we need to work out how the incident velocity looks in S.
e.g. u_x(init)=\frac{\Delta x}{\Delta y}=\frac{\Delta x'}{\gamma (\Delta t'+v\Delta y')}= \frac{u_x'(init)}{\gamma(1+vu_y'(init))} Similarly, u_y(init)=\frac{u_y'(init)+v}{(1+vu_y'(init))}.
From this we can work out the angle of the incident ray (wrt -x-axis) in the S frame, this is therefore \alpha=\frac{u_y(init)}{u_x(init)} (which you can sub into from above). I work out then the actual incident angle, \theta_I=\alpha+\theta_M-90. (not 100% sure if this is correct)
Does this look like I'm going down the right path?
I'm working through an Excercise in Sean Carrol's spacetime and geometry book. The question asks you to consider an inertial frame S with coordinates x^\mu=(t,x,y,z) and a frame S' with primed coordinates. Which is related to S by a boost v in the y direction. Imagine a wall( or mirror) lying along the line x'=-y'. From the point of view of S, what is the relationship between the angle of incidence (assuming ball travels in x-y plane only) and angle of reflection? Also what is velocity before and after?
Unfortunately there are not even one word answers let alone solutions in this book, so I don't even know if what I have is correct.
The way I proceeded, was firstly to consider what angle the mirror would appear at in S. Since the y-direction will be Lorentz contracted, but the x -direction will remain unchanged, the mirror should appear to be at a greater angle than 45 to someone in the S frame. This angle would precisely be:
\theta_M=arctan(\gamma \frac{\Delta x'}{\Delta y'})=arctan(\gamma), the last equality following from the angle being 45 degrees in the primed frame meaning \Delta x'=\Delta y'
OK so that is my step 1. Now we need to work out how the incident velocity looks in S.
e.g. u_x(init)=\frac{\Delta x}{\Delta y}=\frac{\Delta x'}{\gamma (\Delta t'+v\Delta y')}= \frac{u_x'(init)}{\gamma(1+vu_y'(init))} Similarly, u_y(init)=\frac{u_y'(init)+v}{(1+vu_y'(init))}.
From this we can work out the angle of the incident ray (wrt -x-axis) in the S frame, this is therefore \alpha=\frac{u_y(init)}{u_x(init)} (which you can sub into from above). I work out then the actual incident angle, \theta_I=\alpha+\theta_M-90. (not 100% sure if this is correct)
Does this look like I'm going down the right path?
