Deriving the y' acceleration component in special relativity

[mntb]

For a particle moving in the xy plane of S, show that the y' component of the acceleration is given by ay'=ay/?^2(1-uxv/c^2)+axuxv/c^2/?^2(1-uxv/c^2)^3

I don't know if this means that ths S and S' xy plane aren't parrellel? If so, how do you get the angle, if not how come there are two terms?

 

[HallsofIvy]

No, the xy plane is exactly the same in S and S'. However, if the particle is not moving parallel to the x or y axis, the axes in that plane can be rotated (is that what you meant by "not parallel"?). By the way, there's a "?" showing up in you formulas on my reader. I presume that is some special character that is not being read correctly. Could you tell us what it is?

 

[mntb]

For a particle moving in the xy plane of S, show that the y' component of the acceleration is given by ay'=ay/?^2(1-uxv/c^2)+axuxv/c^2/?^2(1-uxv/c^2)^3

I don't know if this means that ths S and S' xy plane aren't parrellel? If so, how do you get the angle, if not how come there are two terms?

The ? are gamma, and actually I know how to get the answer now, but could you check for me to see if the answer is correct? I get ay'=ay/?^2(1-uxv/c^2)^2+axuxv/c^2/?^2(1-uxv/c^2)^3