Doppler shift derivation from Lorentz Transformations

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SUMMARY

The discussion focuses on deriving the Doppler transformation for the frequency of a receding source using Lorentz transformations, specifically the energy equation E = γ(E' + vp'x). The key equations involved include the relativistic transformations for momentum and energy, with β representing the ratio of velocity to the speed of light. The confusion arises from the distinction between β and the relative velocity v between reference frames. The correct observed frequency is given by γ' (1 - β), highlighting the importance of accurately applying Lorentz transformations in relativistic contexts.

PREREQUISITES
  • Understanding of Lorentz transformations in special relativity
  • Familiarity with the concepts of energy and momentum in relativistic physics
  • Knowledge of the relationship between frequency and energy (E = hν)
  • Basic grasp of the concept of relativistic velocity (β)
NEXT STEPS
  • Study the derivation of the Lorentz transformations in detail
  • Explore the implications of relativistic Doppler effect on electromagnetic radiation
  • Learn about the application of relativistic energy-momentum relations in particle physics
  • Investigate the differences between classical and relativistic Doppler shifts
USEFUL FOR

Students of physics, particularly those studying special relativity, as well as educators and researchers interested in the applications of Lorentz transformations and the Doppler effect in various fields of physics.

RyanP
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Homework Statement


Show how one can obtain the Doppler transformation for the frequency of a receding
source just using the Lorentz transformations for the energy (where E=h).

Homework Equations


Relativistic transformations for momentum and energy:
E = γ(E' + vp'x)
pc/E = v/c = β

The Attempt at a Solution


E = γ(E' + vp'x)
since p = Eβ/c, E = γ(E' + Eβ2) = γE' (1+β2).

Which then would imply that = γ'(1+β2). But I have in my notes that the observed frequency = γ' (1-β). Where did I go wrong?
 
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The speed ##\beta = \frac{pc}{E} = 1## for electromagnetic radiation. It's not the same as ##v##, the speed between the two reference frames.
 
vela said:
The speed ##\beta = \frac{pc}{E} = 1## for electromagnetic radiation. It's not the same as ##v##, the speed between the two reference frames.
That makes sense, thanks.
 

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