- #1
Athenian
- 143
- 33
- 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 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!
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!
