(I have looked at other questions relating to Doppler Effect with acceleration but none seem to provide the formula in question.) Edit: I wish to avoid using the Einsteinian relativity model. So consider the pulses as sound pulses moving in a body of water at rest in the IRF of the source.


A fixed source (at position $x=0$) emits brief spike pulses at a regular time interval between pulses with period $P_e$. The pulses travel away from the source at constant radial speed $s$ (positive). A distant receiver (at large $x+$) accelerates directly towards the source at constant rate of acceleration $a$ (negative). The receiver velocity $V$ (initially zero at some undefined time in the past) is always directed towards the source (the sign of $V$ is negative). The time interval between received pulses is $P$. Assume that the receiver remains distant from the source during the experiment (i.e. it doesn't arrive at or go past the source). Also the receiver velocity $V$ remains very small compared to $s$. Clearly the value of $P$ will decrease as time passes. The rate of change of $P$ with time is $\dot{P}$.


What is the formula for the rate of change of period in terms of $P_e, a, s, V$?

MY 1st ATTEMPT (shown to be incorrect by Sofia's answer).

The (wavelength) distance between pulses is $\lambda=P_e s$. The period between pulse receptions is $P=\lambda/(s-V)=P_es/(s-V)$. Consider the periods $P_1$,$P_2$ at times $t_1,t_2$ where $P_2=P_1+\Delta P$ and $t_2=t_1+\Delta t$. So $$ \Delta P = P_2-P_1 = P_e s \left( \frac{1}{s-V_2}-\frac{1}{s-V_1} \right) = P_e s \left( \frac{(s-V_1)- (s-V_2)}{s^2-sV_1-sV_2+V_1V_2} \right) $$

$$ = P_e s \left( \frac{V_2-V_1}{s^2-sV_1-s(V_1 + a\Delta t)+V_1(V_1 + a\Delta t)} \right) $$

$$ = P_e s \left( \frac{a\Delta t}{s^2 -2sV_1 - s a\Delta t + V_1^2 +V_1 a\Delta t} \right) $$

$$ = \frac{P_e s}{s^2} \left( \frac{a\Delta t}{1 -2V_1/s - a\Delta t/s + V_1^2/s^2 +V_1 a\Delta t/s^2} \right) $$ Thus $$ \frac{\Delta P}{\Delta t}= \frac{P_e }{s} \left( \frac{a}{1 -2V_1/s - a\Delta t/s + V_1^2/s^2 +V_1 a\Delta t/s^2} \right) $$

and as $\Delta t \rightarrow 0$ we get

$$ \dot P =\frac{\mathrm{d} P}{\mathrm{d} t}= \frac{P_e }{s} \left( \frac{a}{1 -2V/s + V^2/s^2 } \right). $$

Using $P = P_e s/(s-V) \longrightarrow P(s-V)/s = P_e$ we can express $\dot P$ in terms of $P$ rather than $P_e$ as follows

$$ \dot P = \frac { P(s-V) }{s^2} \left( \frac{a}{1 -2V/s + V^2/s^2 } \right) $$

and if $V << s$ then

$$ \dot P \approx \frac { P a }{s} . $$

(Thanks to Sofia for pointing out the error of this 1st solution).

My 2nd Attempt is contained in a separate Answer on this page.

  • $\begingroup$ This answers your question if you were talking about light pulses. physics.stackexchange.com/questions/137387/… Or at least it can easily be answered by differentiating the expression for doppler shiftedfrequency. $\endgroup$
    – ProfRob
    Jul 2, 2015 at 7:21
  • $\begingroup$ @Rob Jeffries. Thanks I didn't indicate as much in my question but I wanted to avoid the (Einsteinian) relativity model. $\endgroup$
    – steveOw
    Jul 7, 2015 at 20:27

2 Answers 2


Your calculus of P is not correct, because the velocity varies even during a period, Also, labeling each shortened period as $P_1$, $P_2$, ..., etc., note that $P_2$ doesn't begin after $P_e$, but after $P_1$, and $P_3$ after $P_2$, and so on.

So, let's do the calculus together: for simplicity I deprive V of sign. So, in the first period $P_1$ we have

s$P_1$ + a$P_1$^2 / 2 = s$P_e$ .

The second period begins after $P_1$, s.t. we have in total

s($P_1$ + $P_2$) + a($P_1$ + $P_2$)^2 / 2 = 2s$P_e$ ,

and in general,

(1) s∑$P_i$ + a(∑$P_i$)^2 / 2 = ns$P_e$ .

where the sum is from i = 1, to i = n. Writing this equality for n-1, we get

(2) s∑$P_i$ + a(∑$P_i$)^2 / 2 = (n-1)s$P_e$ ,

where the summation is from i = 1, to i = n-1. Subtracting (2) from (1), and denoting ΔP = $P_n$ - $P_{n-1}$

(3) s$P_n$ + a($P_n$^2 - 2 $P_n$ ΔP)/2 = s$P_e$

From this equation I isolate

(4) ΔP/$P_e$ = (s/a)[1/$P_n$ - 1/$P_e$] - $P_n$/$P_e$ .

You can consider $P_e$ as the unit time. What remains is only to express $P_n$ in s, V, a, $P_e$. And this is simple. We can write,

(5) s$P_n$ + (*V*$P_n$ + a$P_n$^2 / 2) = s$P_e$,

where V is the velocity of the observer when $P_n$ begins. The space travelled by the observer during $P_n$ is what is written between the round parentheses. This is an equation of 2nd degree in $P_n$, and its solution is interms of s, V, a, $P_e$.

Good luck !

  • $\begingroup$ Thanks. I see I got it wrong. But in your equation (3) instead of ($-2P_n \Delta P$) I get ($+2P_n\sum_i^{N-1} P_i$). Please can you explain how you get ($-2P_n \Delta P$) in (3)? $\endgroup$
    – steveOw
    Dec 10, 2014 at 10:50
  • $\begingroup$ Following your very helpful lead I have compiled a second solution (presented in the question above). $\endgroup$
    – steveOw
    Dec 10, 2014 at 22:06
  • $\begingroup$ Moved my solution to a separate answer. $\endgroup$
    – steveOw
    Jul 7, 2015 at 20:36

2nd ATTEMPT (after taking advice from Sofia's answer)

Let us consider the experiment in the inertial rest frame of the pulses as they move along the $x$-axis in the $x+$ direction. In this frame the source is moving along the $x$-axis in the $x-$ direction at constant velocity $-s$. At all times $t_*$ the receiver is accelerating towards $x-$ with acceleration $-a$ (which is frame independent). The distance between pulses is constant $L = s.P_e$.

At $t_0$ let the receiver be moving in the same direction as the source, with the same velocity $V_0 = -s$. Also at $t_0$ let the receiver "collide" with pulse $K_0$.

The next collision is with pulse $K_1$ which occurs at time $t_1$. We then have period $P_1 = t_1 - t_0$. In the time interval $P_1$ (with all pulses being stationary in their common rest frame) the receiver has moved distance $L$. From the well-known formula for distance travelled under uniform linear acceleration we obtain $-L = -s.t_1 - 0.5a.t_1^2$.

The next collision is with pulse $K_2$ which occurs at time $t_2$. We then have period $P_2 = t_2 - t_1$. In the time interval $P_1 + P_2$ the receiver has moved distance $2L$. The total distance travelled since $t_0$ is $-2L = -s.t_2 - 0.5a.t_2^2$.

The general formula for distance travelled (since $T_0$) at time of collision with pulse $N$ is thus $$ -N.L = -s.t_N-0.5a.t_N^2 \qquad [1]$$

and the distance travelled (since $T_0$) at collision with pulse $K_{N+1}$ is thus $$ -(N-1).L = -s.t_{N-1}-0.5a.t_{N-1}^2 \qquad [2].$$

Let us do the subtraction [1]-[2] , noting that $t_{N-1}=t_N-P_N$, to produce $$ -L = -s.P_{N} - 0.5a(t_N^2 - (t_{N}-P_N)^2 ) \qquad [3.1]$$ $$ -L = -s.P_{N} - 0.5a(t_N^2 - (t_N^2 -2.t_{N}P_N + P_N^2 ) \qquad [3.2]$$ $$ -L = -s.P_{N} - 0.5a(2.t_{N}P_N - P_N^2 ) \qquad [3.3]$$ $$ -L = -s.P_{N} - a.t_{N}P_N + 0.5.a.P_N^2 \qquad [3.4].$$

This is a quadratic equation in $P_N$, $$ 0 = [L] - [s + a.t_N]P_N + [0.5.a]P_N^2 \qquad [4]$$

using the standard formula for solving quadratic equations we get $$P_N = \frac{(s + a.t_N) \pm \sqrt{(s + a.t_N)^2 - 4*0.5*a.L}}{a}$$

$$P_N = \frac{s + a.t_N \pm \sqrt{s^2 + 2s.a.t_N + a^2.t_N^2 - 2*a.L}}{a}$$

$$P_N = (s/a) + t_N \pm (s/a)\left[1 + 2a.t_N/s + (a^2.t_N^2)/s^2 - (2*a.L)/s^2 \right]^{1/2}$$

For $x<<1$ we can use the rough binomial approximation for the square root $(1+x)^{1/2} \approx (1+x/2)$ so for very large pulse velocity $s$ (relative to $(a^2.t_N^2)$ and $(2*a.L)$) we can approximate $P_N$ by

$$P_N \approx (s/a) + t_N \pm (s/a)\left(1 + a.t_N/s + (a^2.t_N^2)/2s^2 - (a.L)/s^2 \right)$$

$$P_N \approx (s/a) + t_N \pm \left((s/a) + t_N + (a.t_N^2)/2s - L/s \right)$$

by inspection the $"\pm"$ becomes a $"-"$ leading to

$$P_N \approx L/s -(a.t_N^2)/2s$$

using $L=s.P_e$ we obtain

$$P_N \approx P_e -(a.t_N^2)/2s $$

and so $$P_{N-1} \approx P_e -(a.t_{N-1}^2)/2s $$

We find $\Delta P$ for the time interval $\Delta t = t_N - t_{N-1} = P_N$ by the subtraction $P_N - P_{N-1}$ thus

$$\Delta P \approx P_N - P_{N-1} \approx -(a/2s)\left( t_N^2 -t_{N-1}^2 \right) $$ $$\Delta P \approx -(a/2s)(t_N^2 -t_{N-1}^2) \approx -(a/2s)\left( t_N^2 -(t_N-P_N)^2 \right)$$ $$\Delta P \approx -(a/2s)\left( t_N^2 -t_N^2 -P_N^2 +2t_N.P_N \right) \approx -(a/2s)\left( -P_N^2 +2t_N.P_N \right)$$

so $$\frac{\Delta P}{\Delta t} = \frac{\Delta P}{P_N} \approx -(a/2s)\left( -P_N +2t_N \right)$$

Substituting for $P_N$ we get $$\frac{\Delta P}{P_N}\approx -(a/2s)\left( -P_e +(a.t_N^2)/2s +2 t_N \right)$$

$$\frac{\Delta P}{P_N} \approx +a.P_e/2s -(a^2.t_N^2)/4s^2 - a.t_N/s $$

For constant $a,s$ and $k=a/s$ we obtain:- $$\frac{\Delta P}{P_N} \approx k.P_e/2 - k.t_N -k^2.t_N^2/4 $$

We expect $\frac{\Delta P}{P_N}$ to be negative because the receiver is accelerating towards the source and so the period should get smaller with time.

After $t_N$ has increased past $P_e/2$ so $\frac{\Delta P}{P_N}$ becomes increasingly negative. With $a<<s$ so $k<<1$, at small $t_N$ the term $-k.t_N$ will dominate. At large $t_N$ the term $-k^2.t_N^2/4$ will dominate. The terms will be equal in magnitude when $k.t_N = 4$.

Note that $\Delta P$ indicates the change in the time interval $P$ between one collision and the next and that $\Delta P/P_N \approx dP/dT$.

In the case where the receiver is accelerating away from the source the period will increase (and the frequency will decrease) over time.


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