# Ball hitting a wall with relativistic effects

Tags:
1. Sep 13, 2016

### Figaro

1. The problem statement, all variables and given/known data
I have encountered a problem in Sean Carroll's GR book, exercise 1.1

Consider an inertial frame S with coordinates $~x^μ = (t, x, y ,z)~$, and a frame S' with coordinates $x^{μ'}$ related to S by a boost with velocity parameter $v$ along the y-axis. Imagine we have a wall at rest in S', lying along the line x' = -y'. From the point of view of S, what is the relationship between the incident angle of a ball hitting the mirror (traveling in the x-y plane) and the reflected angle? What about the velocity before and after?

2. Relevant equations
Lorentz Transformation
Let w be the initial velocity and u be the final velocity ( v is the velocity parameter )
$β = \frac{v^2}{c^2}$

3. The attempt at a solution
If I were to let the wall be parallel to the x-axis first and calculate the relationship between the incident angle and reflected angle as seen in the frame S, I would get,

Initial velocity can be broken down to:
$w_y = \frac{ w'_y + v}{ 1 + \frac{v w'_y}{c^2} }~~, ~~w_x = \frac{dx}{dt} = \frac{dx'}{ γ(dt' + \frac{v dy'}{c^2}) } = \frac{w'_x}{ γ(1 + \frac{v w'_y}{c^2}) }$

Final velocity can be broken down to:
$u_y = \frac{ -u'_y + v}{ 1 - \frac{v u'_y}{c^2} }~~, ~~u_x = \frac{dx}{dt} = \frac{dx'}{ γ(dt' - \frac{v dy'}{c^2}) } = \frac{u'_x}{ γ(1 - \frac{v u'_y}{c^2}) }$

Thus,
$tanθ_i = \frac{w'_x (1 - β^2)^½}{ w'_y + v } = \frac{ tanθ'_i (1 - β^2)^½}{ 1 + \frac{v}{w'_y} }$
$tanθ_f = \frac{u'_x (1 - β^2)^½}{ -u'_y + v } = \frac{ tanθ'_f (1 - β^2)^½}{ -1 + \frac{v}{u'_y} }$

Since $~θ'_i = θ'_f~$, the relationship between $~θ_i~$ and $~θ_f~$ is,
$tanθ_i (1 + \frac{v}{w'_y} ) = tanθ_f (-1 + \frac{v}{u'_y} )$

Can somebody comment on my solution if it is correct?

Now what I'm not sure about is if the wall is inclined, I'm not sure on how should the ball approach the wall, because as I'm thinking, if it hits the wall at some incident angle, the y-component final velocity can be positive or negative depending on the size of the incident angle, if it is too big (with respect to the normal of course) then the ball will have a positive y-component final velocity, while if it is small then it can be negative. So how do I approach this problem?

2. Sep 14, 2016

### Figaro

Anyone?

3. Sep 15, 2016

### kuruman

Without having done the problem, this is what I would try.
1. Find the "before" and "after" velocities in the primed frame (trivial)
2. Transform the velocities into the unprimed frame. Instead of deriving these, I would use the inverse of the expressions in the wikipedia article
https://en.wikipedia.org/wiki/Lorentz_transformation#Transformation_of_velocities, section 4.4, Transformation of velocities.

I hope this helps.

4. Sep 15, 2016

### Figaro

I understand what you are saying but what I'm wondering about is how should the ball approach the inclined wall, since I think the answer will be different if the ball approach the wall at different incident angles (as I have stated in my post#1).

5. Sep 15, 2016

### kuruman

In the primed frame I think it is fair to assume that the angle of incidence is the same as the angle of reflection. Pick an angle θ and write the incident and reflected velocities as vectors. Transform them into the unprimed frame and you will have two new vectors. From their components you should be able to figure out what the angles of incidence and reflection look like in the unprimed frame.