Doppler effect
General

In classical physics, where the speeds of source and the receiver relative to the medium are lower than the speed of waves in the medium, the relationship between observed frequency f {\displaystyle f} and emitted frequency f 0 {\displaystyle f_{\text{0}}} is given by: f = (c ± v r c ∓ v s) f 0 {\displaystyle f=\left({\frac {c\pm v_{\text{r}}}{c\mp v_{\text{s}}}}\right)f_{0}} where

Note this relationship predicts that the frequency will decrease if either source or receiver is moving away from the other.

Equivalently, under the assumption that the source is either directly approaching or receding from the observer: f v w r = f 0 v w s = 1 λ {\displaystyle {\frac {f}{v_{wr}}}={\frac {f_{0}}{v_{ws}}}={\frac {1}{\lambda }}} where

If the source approaches the observer at an angle (but still with a constant speed), the observed frequency that is first heard is higher than the object's emitted frequency. Thereafter, there is a monotonic decrease in the observed frequency as it gets closer to the observer, through equality when it is coming from a direction perpendicular to the relative motion (and was emitted at the point of closest approach; but when the wave is received, the source and observer will no longer be at their closest), and a continued monotonic decrease as it recedes from the observer. When the observer is very close to the path of the object, the transition from high to low frequency is very abrupt. When the observer is far from the path of the object, the transition from high to low frequency is gradual.

If the speeds v s {\displaystyle v_{\text{s}}} and v r {\displaystyle v_{\text{r}}\,} are small compared to the speed of the wave, the relationship between observed frequency f {\displaystyle f} and emitted frequency f 0 {\displaystyle f_{\text{0}}} is approximately

where

Given f = (c + v r c + v s) f 0 {\displaystyle f=\left({\frac {c+v_{\text{r}}}{c+v_{\text{s}}}}\right)f_{0}}

we divide for c {\displaystyle c} f = (1 + v r c 1 + v s c) f 0 = (1 + v r c) (1 1 + v s c) f 0 {\displaystyle f=\left({\frac {1+{\frac {v_{\text{r}}}{c}}}{1+{\frac {v_{\text{s}}}{c}}}}\right)f_{0}=\left(1+{\frac {v_{\text{r}}}{c}}\right)\left({\frac {1}{1+{\frac {v_{\text{s}}}{c}}}}\right)f_{0}}

Since v s c ≪ 1 {\displaystyle {\frac {v_{\text{s}}}{c}}\ll 1} we can substitute using the Taylor's series expansion of 1 1 + x {\displaystyle {\frac {1}{1+x}}} truncating all x 2 {\displaystyle x^{2}} and higher terms: 1 1 + v s c ≈ 1 − v s c {\displaystyle {\frac {1}{1+{\frac {v_{\text{s}}}{c}}}}\approx 1-{\frac {v_{\text{s}}}{c}}}
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