The standard equation for the doppler effect from a frame of reference with constant velocity is

$f_0$ initial frequency;

$v$ is the velocity of waves in the medium;

$v_r$ is the velocity of the receiver relative to the medium;

$v_s$ is the velocity of the source relative to the medium;

$$f = \left(\frac{v + v_r}{v + v_s}\right)f_0$$

Now I'd like to know what's the equivalent for an accelerating frame, I have

$$f = \left(\frac{v'\text{dt} + v_r}{v'\text{dt} + v_s}\right)f_0$$

But it just doesn't seem correct.


Mea Culpa - I was thinking of the signal being emitted from an accelerating train B as being the frame of reference.

So, second attempt.

1)You are standing at a platform.

2)The wind is stationary.

3)A train is accelerating towards you - blowing its whistle.

What is the formula for the frequency received at the station? Where the standard formula is

$$f = \left(\frac{v }{v + v_s}\right)f_0$$

rewriting for an accelerating frame of reference for the source $$f = \left(\frac{v }{v + (dv_s/dt)dt}\right)f_0$$

but I don't think this goes anywhere, and yes a non-linear solution is probably the answer but it's like a three legged black dog with rabies - it's better than no dog at all.

  • $\begingroup$ There's a good clue that something's gone wrong with your formula: the units don't balance. $v'\,dt$ has units of length, while $v_r$ has units of length/time. So you can't add them. $\endgroup$
    – Ted Bunn
    Commented Jul 1, 2011 at 14:02
  • $\begingroup$ read dv/dt for v' $\endgroup$
    – metzgeer
    Commented Jul 1, 2011 at 14:12
  • $\begingroup$ Sorry, that was stupid of me. I'm not used to primes as time derivatives, but in context I should've figured it out. Never mind! $\endgroup$
    – Ted Bunn
    Commented Jul 1, 2011 at 14:18
  • $\begingroup$ Some details related to "Doppler Effect for Non-uniformly Moving Source, Observer, and Mirror" may be found in this paper. $\endgroup$
    – izogfif
    Commented Mar 10, 2023 at 12:29

4 Answers 4


In the classical Doppler effect, you should think of $v_s$ and $v_r$ to be the velocities of the source and receiver relative to the medium, at the time when the signal was sent and received. In an accelerated reference frame, it is essentially the medium that is moving. Remember that the Doppler shift concerns the frequency sent by the source compared to the frequency seen by the receiver. So it does not depend on which reference frame you, as a third party, is in, but only depends on the intrinsic wave speed of the medium $v$ (which we assume to be constant, when measured in a frame in which the medium is at rest), and the velocities relative to the medium of the source and receivers.

With that said, that means you can compute relative to the unaccelerated reference frame by doing a coordinate change.

Suppose you are in a reference frame in which the medium itself is moving with variable velocity $\tilde{v}(t)$, and you measure the source and receivers to be moving with velocity $\tilde{v}_s(t)$ and $\tilde{v}_r(t)$. Then the Doppler effect can be computed by looking at the reference frame in which the medium is not moving (so the corresponding velocities are $v_s(t) = \tilde{v}_s(t) - \tilde{v}(t)$ and similarly $\tilde{v}_r(t)$). If in this reference frame $v_s$ and $v_r$ are not constant in time, then given a time $t_r$ when the signal was received, you will have to compute (using that the wave travels with constant speed $v$ and using the trajectory of the receiver versus the source) the time $t_s$ at which the signal was sent, and plug those values into your Doppler shift expression (with $v_s \to v_s(t_s)$ and $v_r \to v_r(t_r)$).

If the wave speed itself $v$ depends on the macroscopic of the entire medium, then you will have to solve the very complicated system for elastodynamics/acoustic in the medium, for which the final answer would depend on how you model the medium.

Edit for the new clarification

In the particular case, assume you, the receiver at the platform is measuring the signal at time $t_0 = 0$. And choose coordinate system so that you are at the origin. Let $x(t)$ denote the position of the train that is approaching the station. (or leaving the station, you just need to adjust the sign appropriately.) The signal you observe will have been emitted by the train at time $t_s < 0$, when $|x(t_s)| = v \cdot (0-t_s)$ where $v$ is the speed of sound in air. (For the time being we'll assume you have a non-supersonic train, so there is a unique time $t_s$ that satisfies the requirements.) Then the expression for the doppler shift would be

$$ f = \left( \frac{v}{v + \dot{x}(t_s)}\right) f_0 $$

Unfortunately, solving for $t_s$ is generally hard. But we can get a good estimate on $f$ if we assume that our train is reasonably close to the station when you are making the observation, and if the train is not accelerating too much, by using Newton's method.

To the zero'th order you'd get $t_s = 0$. To the first order in Newton's method, you can solve and get $t_s ~ x(0) / (v - \dot{x}(0))$. Then we can approximate the velocity $\dot{x}(t_s) \sim \dot{x}(0) + \ddot{x}(0) t_s$ which leads to a first order correction to the doppler shift formula

$$ f \sim \left( \frac{v}{v+ \dot{x}(0) + \frac{\ddot{x}(0) x(0)}{v - \dot{x}(0)}}\right) f_0 $$

  • $\begingroup$ nope doesn't help, I accept its a nonlinear eqn, but I cant see how changing the frame of ref is relevant. Just assume the velocity of the media is constant to the observer $\endgroup$
    – metzgeer
    Commented Jul 1, 2011 at 14:21
  • $\begingroup$ @Metzgeer: now I am confused. What Exactly do you mean by "accelerated frame"? Do you mean that the signal receiver is accelerating? But that doesn't matter, since Doppler effect only depends on the instantaneous speed of the receiver. Do you mean that everything is moving around? Then the above shows that you can do a change of coordinates to a frame where the medium is not moving. Who is the observer in your comment? $\endgroup$ Commented Jul 1, 2011 at 17:44
  • $\begingroup$ Ah yes Willie, this is what I was looking for. Obvious now I see it. All the talk about Gen-Rel and covariant frames must have arisen because I framed the question poorly. Thank you. $\endgroup$
    – metzgeer
    Commented Jul 3, 2011 at 4:19

Your first formula can be rewritten as $\frac{f_r}{c_r+v_r}=\frac{f_s}{c_s+v_s}$, where the equivalences with your notation are $f_s=f_0$, $f_r=f$, and $c_s=v=c_r$ are the velocities of the medium at the sender and at the receiver. This equation tells us that the number of wave crests per unit length is the same for the sender and for the receiver. We should preserve this equality in an expression for the number of wave crests per unit length in the accelerating frame of reference. I take it we're working in 1+1 dimensions.

In the accelerating frame of reference, we can take $v_s$ and $v_r$ to be constant, but the velocities $c_r$ and $c_s$ to be functions of time, $c_r(t)=c_r+at$, $c_s(t)=c_s-at$, say, with the difference of sign because the direction in which the sender is sending waves is opposite to the direction in which the receiver receives (this classically, not SR, of course). Presuming that the wavelength is short enough relative to the characteristic lengths introduced by the ratios of the acceleration relative to the velocities in the problem, the equation for the frequency relationship would be, approximately, $\frac{f_r}{c_r(t)+v_r}=\frac{f_s}{c_s(t)+v_s}$. You have to get the relative signs of the acceleration and of the various velocities right, of course.

This is so far from a generally covariant description that there doesn't seem much point to consider the question of accelerated coordinates, but whatever. The elementary idea of counting wave crests, for example, does not extend to a general situation in which there is a (continuous or discrete) distribution of energy at different wave numbers.


The observed "frequency" in an accelerated reference frame should be equal at least to the "frequency" in an inertial co-moving reference frame. It means replacing the RF velocity with a time-dependent velocity $v_r (t)$.



non relativistic, i.e. all speeds are small compared to the speed of the waves in the medium, the receiver stay still in the medium, and the problem of taking measures at distance is not important.

Be $\frac{f}{f_{0}}=\frac{v+v_{r}}{v+v_{s}}$, lets use dimensionless units, be $v=1,\: v_{s}\leftarrow v_{s}/v,\: v_{r}\leftarrow0$ (the platform is still).

$\frac{f}{f_{0}}=\frac{1}{1+v_{s}(t_{E})}$, where $v_{s}(t_{E})$ is the speed of the emitter at the time of the emission, measured by the receiver.

The speed of the emitter can be represented by f(t), measured by the receiver.

Say for instance $f(t)=t^{2}$ (dimensionless), as if the emitter is moving subjected to a central force like gravity.

$\frac{f}{f_{0}}=\frac{1}{1\pm t_{E}^{2}}$, and the $\pm$ sign account for the relative direction of movement.

A complete study of Doppler effect, like no other, is needed to understand light properties under the CMB referential, is presented here: Cosmological Principle and Relativity - Part I.

EDIT add:
the motion of the emitter is along the direction emitter-receiver (angle = 0)


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