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[Question] Frame $S'$ moves with velocity $V$ in the $x$-direction relative to frame $S$. A rod in the frame $S'$ lying on the $x'-y'$ plane makes an angle $\theta'$ with respect to the forward direction of motion.

What is the angle $\theta$ as measured in $S$?

[Personal Comments] -

Honestly, I have no idea how to properly solve this question. My best guess would be this involves Lorentz-transformations somehow. The only equation I recall and have jotted down on my lecture notes involving angles (besides the newly taught topic of rapidity) is:

$$\begin{cases} ct' = cosh(\phi)ct - sinh(\phi) x \\ x' = -sinh(\phi)ct + cosh(\phi)x \\ y' = y \\ z' = z\end{cases}$$

Other than this potential equation, I truly don't know how to begin solving for the question.

With that said, any help and guidance from the community to help me better understand how to solve this problem will be much appreciated. Thank you very much for reading through this as well as providing your insight and assistance here.

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  • $\begingroup$ It looks like problem 12.10 in the Griffith text book. The solution is available on-line. Just search Instructor’s Solution Manual Introduction to Electrodynamics griffith pdf, you will see multiple links. $\endgroup$ – verdelite Nov 22 '19 at 13:52
  • $\begingroup$ @verdelite , after viewing the different question from the recommended book, I arrived at the exact same answer where $\theta = arctan(\gamma tan(\theta '))$ once I went through the exact same answering process as the solution key. What do you think of the answer? Can the book's solution apply completely to this question? Perhaps, my biggest question is can I simply - and conveniently - throw out the $x'-y'$ part of information out of my solution? Thank you for your help and recommendation! $\endgroup$ – Athenian Nov 23 '19 at 3:34
  • $\begingroup$ I can not comment much on problems related to Special Relativity because I am not very familiar with it. I recognized your problem because I went through the problems in Griffith's book but I really can't say whether the solutions are correct or not. $\endgroup$ – verdelite Nov 23 '19 at 18:54
  • $\begingroup$ @verdelite , no problem. Regardless, thank you very much for helping me find the related sources to this question. I sincerely appreciate it! $\endgroup$ – Athenian Nov 24 '19 at 0:05

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