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Doppler effect for light  how can it be explained ? 
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#19
Feb512, 01:03 PM

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If one observer measures an object's momentum and energy as p and E, then another observer who is moving with velocity v with respect to the first observer measures the energy and momentum as
[tex]p^\prime = \gamma \left( p  \frac{vE}{c^2} \right)[/tex] [tex]E^\prime = \gamma (E  vp)[/tex] where [itex]\gamma = \frac{1}{\sqrt{1  v^2/c^2}}[/itex] This is the Lorentz transformation for momentum and energy, similar to the L.t. for position and time which you may have seen if you've studied relativity. For a photon, these two equations are equivalent, as you can see by substituting E = pc and E' = p'c. 


#20
Feb512, 01:48 PM

P: 3

Aaah I see. Its what I thought but just didnt have a basis to explain it on. Thanks a lot! :)



#21
Feb512, 03:47 PM

P: 3,187




#22
Feb512, 03:58 PM

P: 3,187

In laymen's terms, just as with the sound of the horn of a train passing by, also the colour of the light that you receive is affected by the velocity of the source as well as by your own velocity. And the tweaks of relativity make that it is always as if there is a light medium that is in rest with respect to the reference frame that you use; only the relative velocity matters for the Doppler calculation. 


#23
Feb512, 04:17 PM

P: 67

Edit: I should mention that the analogy applies to frequency and wavelength. The energy of the light photons is the frequency times Plank's constant. Why so, is a question to be answered elsewhere. 


#24
Feb512, 04:36 PM

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