I have read and understood most of chapter 1. I am slightly stuck on the final section about Classical Field Theory because I have forgotten or was never taught much about this at university. So I decided to do a few of the exercises. The first one was Exercise 1.01.

In fact, the wall is moving upwards so fast that it stays ahead of the ball.

Read my full answer here.

In order to check my answer I used an equation on Wikipedia for relativistic velocity transformations here. But I first had to slightly improve it by showing that

$$ \large \frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}=\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right)$$which I proved in the complete answer above and here.

The question was also discussed on Physics Forums here. I added my answer to the thread.

## Question

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 wall (travelling in the x-y plane) and the reflected angle? What about the velocity before and after?## Answer

I found this the most difficult of the first seven exercises and gave up and came back to it. I think I have the right answer now. Sometimes the reflected ball has an angle of reflection > 90°, which means it seems to go through the wall, as illustrated below.In fact, the wall is moving upwards so fast that it stays ahead of the ball.

Read my full answer here.

In order to check my answer I used an equation on Wikipedia for relativistic velocity transformations here. But I first had to slightly improve it by showing that

$$ \large \frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}=\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right)$$which I proved in the complete answer above and here.

The question was also discussed on Physics Forums here. I added my answer to the thread.

Funny! I am re-learning GR and this time I am lazy and opted to search for solutions online, instead of solving them myself. To my surprise and amusement, I found links to my answers from many years ago on your blog.

ReplyDeleteNice to see you back! I'm sure my answer could be better. It was my very first ...

DeleteBTW, I scanned through your solution. I think it can be made much simpler if you use 4-velocities v=(\cosh \alpha, \sinh \alpha \cos \beta, \sinh \alpha \sin \beta, 0) for the ball, and use matrix multiplications to transform 4-velocities back and forth between the two frames. It should take no more than a few lines of calculations.

ReplyDeleteIn the primed frame, you assumed that after reflection, the velocity acquires a negative sign. Isn't that only valid when the ball's hitting the wall head on? I.e. incident angle is zero.

ReplyDeleteI assume you're referring to equation (1) which is 'calculated' from the diagram. On that the a's are clearly positive and the b's clearly negative. So no.

DeleteAlright I read the subscripts wrong, it's not just acquiring a negative sign but the indices also flipped

Delete