Frequency of Reflected light in special relativity

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The discussion centers on the application of the Doppler effect in special relativity, specifically regarding the frequency of reflected light in different frames of reference. There is confusion about why the Doppler effect is applied twice in frame K and whether the signs should be switched since the photon is moving away from the source. Clarification is provided that the correct frequency in frame K can be determined by analyzing the left- and right-going light pulses in the mirror frame. The conversation emphasizes the importance of understanding how frequencies transform between different inertial frames. Overall, the discussion seeks to resolve misunderstandings related to the Doppler effect in the context of special relativity.
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Homework Statement
In a reference frame K a photon of frequency f falls normally on a mirror approaching it with relativistic velocity v. Find the momentum imparted to the mirror during the reflection of the photon (a) in the reference frame fixed to the mirror; (b) in the frame K.
Relevant Equations
f'/f = sqrt((1+(v/c))/(1-(v/c)))
In the. solution attached I'm not too sure why in frame K, we apply the doppler effect twice. Also, since the photon is moving away from the source, shouldn't the signs be switched? Thanks
Screenshot 2024-09-24 at 10.21.05 PM.png
 
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Your image is hard to read, but you seem to have the right answer in the mirror frame. So in that frame you have left- and right-going light of frequency ##f'##. What frequency will those pulses have in ##K##?
 
Last edited:
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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