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

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SUMMARY

The discussion focuses on deriving the y' component of acceleration for a particle moving in the xy plane of reference frame S. The formula presented is ay' = ay/?^2(1 - uxv/c^2) + axuxv/c^2/?^2(1 - uxv/c^2)^3. Participants clarify that the xy planes of S and S' are parallel, but the axes can be rotated if the particle does not move parallel to the x or y axes. The symbol "?" in the formulas is identified as the gamma factor (γ), which is crucial for relativistic calculations.

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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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