- #1

Athenian

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- Homework Statement
- 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##?

- Relevant Equations
- Refer to the solution below: ##\rightarrow##

Below is the attempted solution after researching the contents available on

To begin with, I defined the rod as having a length of ##l'## at rest in frame ##S'##. Thus, in frame ##S'##, the height of the rod is ##l' sin(\theta ')## and its horizontal projection will similarly be ##l' cos(\theta ')##.

While the height of the rod is not affected if seen by an observer from frame ##S##, the horizontal projection is - on the other hand - Lorentz contracted to ##\frac{1}{\gamma} l' cos(\theta ')##.

Thus, my conclusion is simple. The angle will therefore be:

$$tan \theta = \frac{l' sin(\theta ')}{\frac{1}{\gamma} l' cos(\theta')} = \gamma tan(\theta ') \Longrightarrow \theta = arctan(\gamma \tan(\theta '))$$

OR

$$tan \theta = \frac{tan (\theta ')}{\sqrt{1- v^2/c^2}} \Longrightarrow \theta = arctan(\frac{tan (\theta ')}{\sqrt{1- v^2/c^2}})$$

Thus, the above angle is the angle of the rod measured in frame ##S##. What does everybody think of the answer? What confuses me the most regarding the above-attempted solution is do I need to do anything about the ##x' -y'## part of the question? Or, can I conveniently disregard that piece of information? For some reason, I feel like the rod lying on the ##x'-y'## plane in the frame ##S'## plays a role in solving for the question.

Any help or guidance on the matter would be sincerely appreciated. Thank you all for your kind assistance!

*Introduction to Electrodynamics*by Griffith.To begin with, I defined the rod as having a length of ##l'## at rest in frame ##S'##. Thus, in frame ##S'##, the height of the rod is ##l' sin(\theta ')## and its horizontal projection will similarly be ##l' cos(\theta ')##.

While the height of the rod is not affected if seen by an observer from frame ##S##, the horizontal projection is - on the other hand - Lorentz contracted to ##\frac{1}{\gamma} l' cos(\theta ')##.

Thus, my conclusion is simple. The angle will therefore be:

$$tan \theta = \frac{l' sin(\theta ')}{\frac{1}{\gamma} l' cos(\theta')} = \gamma tan(\theta ') \Longrightarrow \theta = arctan(\gamma \tan(\theta '))$$

OR

$$tan \theta = \frac{tan (\theta ')}{\sqrt{1- v^2/c^2}} \Longrightarrow \theta = arctan(\frac{tan (\theta ')}{\sqrt{1- v^2/c^2}})$$

Thus, the above angle is the angle of the rod measured in frame ##S##. What does everybody think of the answer? What confuses me the most regarding the above-attempted solution is do I need to do anything about the ##x' -y'## part of the question? Or, can I conveniently disregard that piece of information? For some reason, I feel like the rod lying on the ##x'-y'## plane in the frame ##S'## plays a role in solving for the question.

Any help or guidance on the matter would be sincerely appreciated. Thank you all for your kind assistance!