Doppler Effect: What is formula for Rate of Change of Period observed by accelerating receiver?

Question

(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.

EXPERIMENT

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 $\dot{P}$.

QUESTION

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

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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.

Answer 1

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 !

Answer 2

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.