I Classical Aberration Formula: Understanding & Application

Click For Summary
The discussion centers on the classical aberration formula for light, specifically the non-relativistic version and its physical relevance. Participants explore whether this formula can apply to real-world scenarios, such as the perception of rain while driving, questioning the significance of the emission point in defining angles. It is clarified that the angles involved are related to the velocity of the particle and the coordinate axes, rather than the emission point. The non-relativistic expression is deemed adequate due to the minimal impact of stellar aberration, with a simpler measurable result provided. Overall, the conversation emphasizes the formula's applicability and the relationship between angles and velocity.
Kairos
Messages
182
Reaction score
16
I believe I have understood the formula of aberration of light ##\tan \theta' = \dfrac{\sin \theta}{\beta + \cos \theta} \sqrt{1-\beta^{2}} ##

but I wonder if the non-relativistic formula ## \tan \theta' = \dfrac{\sin \theta}{\beta + \cos \theta} ## has a physical relevance. Does this formula apply to a real physical situation (for sound ?) or is it just an approximation for small ## \beta ##?
 
Physics news on Phys.org
Did you ever drive a car in rain? Did the rain seem to be coming from ahead?
 
  • Like
Likes vanhees71
Yes I did. But as the point of emission of the rain is not well defined, I have difficulty in defining ## \sin \theta ## and ## \sin \theta' ##
 
Kairos said:
Yes I did. But as the point of emission of the rain is not well defined, I have difficulty in defining ## \sin \theta ## and ## \sin \theta' ##
The emission point is irrelevant. The only relevant thing is the angle of the rain relative to the velocity of the car.
 
  • Like
Likes vanhees71
Does this mean that this formula is valid for an emission by a plane but not for a point source?
 
Kairos said:
But as the point of emission of the rain is not well defined, I have difficulty in defining ## \sin \theta ## and ## \sin \theta' ##
##\theta## and ##\theta'## have nothing a priori to do with the emission point, they're related to the angle between the velocity of the particle and the coordinate axes i.e. ##\mathbf{v} = (v\cos{\theta}, v\sin{\theta})## and ##\mathbf{v}' = (v'\cos{\theta'}, v'\sin{\theta'})##.
 
thank you
 
Stellar aberration is so small that the the NR expression, derived and observed in 1729 by James Bradley,
is sufficient. A simpler measurable result is ##\theta -\theta'=\beta \sin\theta.##