[Openal-devel] Re: [Openal] Speed of sound... does it incur propagation delays?
bartramf at acm.org
Mon Dec 5 13:59:25 PST 2005
On Dec 5, 2005, at 04:14 PM, Stephen J Baker wrote:
> However, this makes me wonder (again) about the asymmetry between
> listener motion and source motion in the doppler math. I was
> about this at the outset - and now, I'm even more sceptical. Since
Yes! because the the Doppler math IS wrong!
The OpenAL doppler shift is the classical form with the addition of a
In this case there IS an asymmetry between listener and source.
f' = f * (VD - DF*vl) / (VD + DF*vs)
f'/f = (VD - DF*vl) / (VD + DF*vs)
Letting VD = 1 and DF = 1
f'/f = (1 - vl) / (1 + vs).
For a stationary listener, vl = 0 and vs = v, this becomes
f'/f = 1 / (1 + v)
The difference in the two cases can be seen clearly by expanding in
f'/f = 1 / (1 + v) = 1 - v + v^2 - v^3 +....
For a stationary source, vs = 0 and vl = v we get
f'/f = (1 - v).
So it is easy to see that the two cases are not the same. => (1-v)
1 - v + v^2 - v^3 +...
I am not a physicist and I didn't sleep in a hotel last night but
this an example is one of the differences between the classical
and the relativistic (Lorentzian) one in which the source and
listener velocities are indistinguishable.
f'/f = sqrt((1-u)/(1+u))
f' = f sqrt((1-u)/(1+u)) <==
where u is a composition of the source and listener velocities
u = (vl+vs)/(1 + (vl*vs)/VD^2).
If the Listener is stationary then u = vs.
If the Source is stationary then u = vl;
So in either case if we let vs = v or vl = v, expanding in series form,
f'/f = 1 - v + (1/2)v^2 - (1/2) v^3 + (3/8)v^4 - (3/8)v^5 ...
which differs from both classical cases and IS symmetrical with
respect to the source and listener.
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