Help on Relativity: Showing y' Acceleration in xy Plane of S

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The discussion focuses on deriving the y' component of acceleration for a particle in the xy plane of reference frame S. It clarifies that the xy planes of S and S' are parallel, but the axes can be rotated if the particle's motion is not aligned with the axes. Participants address a formatting issue with the gamma symbol in the equations. One user confirms they understand the derivation and seeks verification of their modified formula for y' acceleration. The conversation emphasizes the importance of correctly interpreting the relativistic effects on acceleration components.
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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?
 
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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 said:
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
 
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