Ibix said:
Not without extra information. Your velocity with respect to the medium and the original frequency, off the top of my head.
Exactly there is no difference between a 440Hz sound with both observer and source stationary and a 440Hz sound that is produced by relative motion. According to what I have read hear you would not need any extra information because just from the sound you can measure the frequency and wavelength. If the wave velocity ##v=f*\lambda ## is different from a source that is stationary then the source is moving. Don't take the last sentences out of the context of this paragraph.
Ibix said:
Where did you get your expressions for the Doppler effect (link, or your working)? They aren't trivially related to the expressions I derive, and they don't match (for example)
https://en.wikipedia.org/wiki/Relat...#Systematic_derivation_for_inertial_observers. Also, they have the implausible property you describe, that the speed of sound is invariant in all frames. If that were the case, an aircraft traveling at Mach 1 would be paradoxical.
In any fundamentals of physics book that derives the Doppler effect you can get the equations from there. the standard equations that is used to convert from frequency shift to wavelength shift is ##f=\frac{\nu}{\lambda} ## and ##f'=\frac{\nu}{\lambda'} ##. That is how the wavelength equation is derived for the relativistic equation. here is a link that describes Doppler wavelength change
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html#c2. Under the Doppler wavelength change section there are these two equations.
Source receding
##\lambda'=\frac{\nu +v_{s}}{\nu }*\lambda_{s} ##
Source approaching
##\lambda'=\frac{\nu -v_{s}}{\nu }*\lambda_{s} ##
Which can be combined to say if only the source is moving
##\lambda'=\frac{\nu \pm v_{s}}{\nu }*\lambda_{s} ##
The equation is not complete because they did not address a moving observer.
The definition of wave speed is ##\v=\lambda*f ## which applies to all waves.
so the equation for speed of wave with source approaching the observer is
##v=\lambda' * f' = \frac{\nu -v_{s}}{\nu }*\lambda_{s} *\frac{\nu}{\nu-v_{s}}*f_{s} = \lambda{s} * f_{s} ##
Which means that the source would measure the same wave speed as the observer.
As for the derivation for inertial observers it proves my point to.
##f'=\gamma (1-\frac{v_o}{\nu })*f ##
##\lambda'=\frac{1}{\gamma} (\frac{1}{1-\frac{v_o}{\nu }})*\lambda ##
##f'*\lambda'=f*\lambda ##
Waves do not have mass. Waves transport energy not mass.
Can i see the expressions that you derive. Why would an aircraft traveling at Mach 1 be paradoxical