Ball hitting a wall with relativistic effects

[Figaro]

Homework Statement

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?

Homework Equations

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

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?

 

[Figaro]

Anyone?

 

[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.

 

[Figaro]

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.

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).

 

[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.

 

[Matt King]

Don't forget that the wall is at a different angle to the x-axis and y-axis in the S frame than it is to the x' and y' axes in the S' frame too (due to Lorentz contraction along the y' axis).

You can work out the wall angles to the x-axis and y-axis as follows:

If the wall contains points (t',x',y') = (0,1,-1), (0,0,0) and (0,-1,1) in the primed frame, you can transform these (the origin stays the same), but now the transformed origin is NOT at the same time as the other two points, so you have to work out where the middle of the wall is at these two points in time ( x=0, y = v * t) and compute the angles, giving

tan (Angle wall to x axis) = 1/gamma which would be 1 under Galilean transform
tan (Angle wall to y axis) = gamma which would also be 1 under Galilean transform

This just means we've Lorentz contracted the y' co-ordinate. Remember tan ( (pi/2) - x ) = 1/ tan (x) and this is still a straight wall; it's just at a different angle.

where gamma = (1 - (v/c)^2)^-0.5 as usual.

For the velocities, let the ball start with initial velocity I' = (dx'/dt' , dy'/dt') = (-A', -B') and end with final velocity F' = (B',A') in the S' frame, A' and B' positive, colliding with wall at (t=0,x=0,y=0) = (t'=0, x'=0 , y'=0) (both frames; co-incident origin).

Then using the Lorentz transforms for velocity (which are easy to derive from the Lorentz transforms for position; just let x' -> dx' and t -> dt' and divide top and bottom by dt') gives

I = ( -A' / ( gamma * ( 1- (vB'/(c^2)))) , (v - B') / (1 - (vB'/(c^2))) ) which would be ( -A' , v - B') under Galilean transform
F = ( B' / (gamma * (1 + (vA'/(c^2)))) , (A' + v) / (1 + (vA'/(c^2))) ) which would be (B' , A' + v) under Galilean transform

Note: DON'T try subtracting out the v from the y component of velocity to try to work out the relative velocity of the ball to the wall in the S frame, which one would do in classical mechanics when using the equal angle rule to work out the final velocities in this frame; this is mixing Galilean and Lorentz transforms and NOT OK. If one was to do this properly one would just get the original I' and F' back of course. I made that mistake.

You can then work out the angles of the transformed velocities:

tan (Angle incoming velocity to x axis) = gamma * (B' - v) / A' which would (B' - v) / A' under Galilean transform.
tan (Angle outgoing velocity to y axis) = B' / (gamma * (A' + v) ) which would be B' / (A' + v) under Galilean transform.

and then use the standard formula tan (k + l) = tan (k) + tan (l) / (1 - tan(k)*tan(l)) for all k,l to add the velocity angles to the wall angles:

tan (incident angle, S frame) = ((gamma * ((B' - v) / A')) + (1/gamma) ) / (1 - ((B' - v) /A')) which would be ( ((B'-v)/A') + 1 ) / (1 - ((B' -v) / A')) under Galilean transform.

and

tan(reflected angle, S frame) = ((B' / (gamma* (A'+v))) + gamma ) / (1 - (B/(A+v))) which would be ( (B / (A+v)) + 1) / (1 - (B/(A+v))) under Galilean transform.

which is a terrible mess. The question is a little vague; it asks 'From the point of view of S, what is the relationship between the incident angle and the reflected angle?'. If there's a way of expressing this relationship that is cleaner than the above equations I'd LOVE to know what it is.

As for the relationship between the velocities, in the prime frame if you pre-multiply I' by the matrix

(0 -1)
(-1 0)

you get F'.

If there's a similar matrix in the unprimed frame I can't see what it is. Technically the above formulae for I and F do give the relationship.

Can anyone do better? I'd love to know what the author was expecting to see.

I hope the rest of them aren't this hard... that's the first exercise in the book!

 

[Matt King]

...all the As and Bs should be A' and B' in the last line where I forgot to prime them. It wouldn't let me fix it afterwards - I did try. Sorry about that.

 

[George Keeling]

I have an answer at
https://drive.google.com/open?id=178ahu6VD3AzHl7N6ks07i3PlKlNn8ptg
It's in a pdf file and also refers to another answer by Petra Axolotl, which agrees with mine as far as it goes.

One thing I found fascinating was that the reflected angle in S is often greater than 90°, so it might seem that the ball goes through the wall. I was able to show that it does not, even thought the angle is greater than 90°.

There is also a spreadsheet which draws all the lines.

 

[Ray Vickson]

I have an answer at
https://drive.google.com/open?id=178ahu6VD3AzHl7N6ks07i3PlKlNn8ptg
It's in a pdf file and also refers to another answer by Petra Axolotl, which agrees with mine as far as it goes.

One thing I found fascinating was that the reflected angle in S is often greater than 90°, so it might seem that the ball goes through the wall. I was able to show that it does not, even thought the angle is greater than 90°.

There is also a spreadsheet which draws all the lines.

Could you please just attach the file herein? I cannot get my Google account to work, so you document is not accessible to me, and perhaps not to others as well.

 

[George Keeling]

Sorry Ray, it works for me even when I am not logged into Google. I'm new here.
I have uploaded the pdf and the Excel file. Also available: Microsoft Word and Powerpoint.

Here's the direct reference again for those lucky enough to be able to use it
https://drive.google.com/open?id=178ahu6VD3AzHl7N6ks07i3PlKlNn8ptg

 

[Ray Vickson]

Sorry Ray, it works for me even when I am not logged into Google. I'm new here.
I have uploaded the pdf and the Excel file. Also available: Microsoft Word and Powerpoint.

Here's the direct reference again for those lucky enough to be able to use it
https://drive.google.com/open?id=178ahu6VD3AzHl7N6ks07i3PlKlNn8ptg

Thank you for doing that; the files open perfectly now.