# Doppler shift for a uniformly accelerating observer

This was given in textbook as an example.

An observer on a spaceship with a four velocity $u$ is approaching from $x = +\infty$ a star at rest in the reference frame $S$ while undergoing constant proper acceleration $a > 0$. Its distance of closest approach is $a^{-1}$. The star emits light of frequency $\omega_{star}$. The observed Doppler shifted frequency of the light from the star is $\omega(\tau) = \omega_{star}e^{-a\tau}$

Now how did they get that as the frequency? I've tried looking back over the text and for a more elaborate example but that's it. I know the equation for Doppler-shifted frequency is $$v_{obs} = v_{source}\sqrt\frac{1+\beta}{1-\beta}.$$ I just don't know how the distance comes into play to get the example answer. • What textbook is this?
– BMS
Sep 26, 2014 at 17:07
• @BMS this is a textbook written by lecturer for special relativity Sep 27, 2014 at 10:22
• @Rob Jeffries the symbol is not alpha but 'a' standing for proper acceleration and yes tau is proper time Sep 27, 2014 at 10:24
• The observed doppler shift will depend on distance to the source if the spaceship is not aimed directly at the source, or if it is accelerating. What system of units are you using where $a \tau$ is unitless? Sep 27, 2014 at 10:46
• @Rob Jeffries Umm...I suppose standard? To be honest, I don't know as we don't really talk units in class. For the distance there was something given in class. The worldline of the ship is $x^2 - t^2 = a^{-2}$. I've edited my original post with an image from my notes. Sep 27, 2014 at 11:43

Use the standard relationship between acceleration in the two frames of reference.

i.e. the proper acceleration $$a$$ is given by $$a = \gamma^3 \frac{dv}{dt}, \ \ \ \ \ \ \ \ \gamma = \frac{dt}{d\tau}$$ $$\frac{dv}{d\tau} = \frac{dv}{dt} \frac{dt}{d\tau} = \gamma^{-2} a = (1-v^2)a$$

This can be integrated to give $$v$$ and hence $$\gamma$$ as a function of $$\tau$$. $$\int \frac{dv}{1-v^2} = \int a\ d\tau$$ Let $$v = \tanh(x)$$ and use the identity $$1 - \tanh^2(x) = 1/\cosh^2(x)$$ $$\frac{dv}{dx} = \frac{\cosh^2(x) - \sinh^2(x)}{\cosh^2(x)} = \frac{1}{\cosh^2(x)}$$ and so the integral becomes $$\int dx = \int a\ d\tau$$ $$\tanh^{-1} (v) = a\tau + A,$$ where $$A$$ is a constant determined by the initial velocity.

Let $$v=v_0$$ when $$\tau=0,$$hence: $$v = \tanh[a\tau + \tanh^{-1}(v_0)]$$

The doppler shift can be written as: $$\omega = \omega_0 (1-v)\gamma$$

NB: This expression comes from here, with the source at rest, but is I think only strictly valid when the velocity of the observer does not change significantly between wavefronts. For optical light, this requires that (expressing $$a$$ in SI units for a moment) $$a \ll 10^{24}$$ ms$$^{-2}$$ - which is probably ok for a spaceship!

$$\omega = \omega_0\left[1 - \tanh[a\tau + \tanh^{-1}(v_0)]\right]\left[1 - \tanh^2[a\tau + \tanh^{-1}(v_0)]\right]^{-1/2}$$ $$\omega = \omega_0 \left[1 - \tanh[a\tau + \tanh^{-1}(v_0)]\right]\cosh[a\tau + \tanh^{-1}(v_0)]$$.

This is the general expression. For the specific case addressed by the OP, we have $$v_0=0$$. In this case: $$\omega = \omega_0[1 - \tanh(a\tau)]\cosh(a\tau)$$ $$\omega = \omega_0\left[\frac{\cosh(a\tau) - \sinh(a\tau)}{\cosh(a\tau)}\right] \cosh(a\tau)$$ Expressing the hyperbolic functions in terms of exponentials: $$\omega = \frac{\omega_0}{2}[\exp(a\tau) + \exp(-a\tau) - \exp(a\tau) + \exp(-a\tau)] = \omega_0 \exp(-a\tau)$$ as required.

A similar treatment is provided by Cochran 1989 (section II), leading to the same result.

A more useful result is obtained by noting that a coordinate transform of the form $$\tau^{\prime} = \tau + \frac{\tanh^{-1}(v_0)}{a}$$ can make life more easy for general cases, since this also leads to the result $$\omega = \omega_0 \exp(-a\tau^{\prime})$$

This does make life easier - for instance we can show that we recover the standard doppler shift when $$a=0$$, since $$a\tau^{\prime} = \tanh^{-1}(v_0)$$ and so $$\omega = \omega_0 \exp[-\tanh^{-1}(v_0)] = \omega_0 \exp\left[-\frac{1}{2}\ln \left(\frac{1+v_0}{1-v_0}\right)\right]$$ $$\omega = \omega_0\left( \frac{1+v_0}{1-v_0}\right)^{-1/2} = \omega_0(1-v_0)[(1+v_0)(1-v_0)]^{-1/2} = \omega_0(1-v_0)\gamma\ .$$

• Rob Jeffries: "The doppler shift can be written as: $$\omega = \omega_0~(1 - v)~\gamma.$$ Do you claim that this relation holds exactly, even if $$a \gg c~\omega_0$$? Otherwise, please discuss the approximation you're using; or (even better) use an exact expression for "_the doppler shift_" (in terms of $\omega_0$, $v$ and $a$). Jul 9, 2015 at 21:59
• @user12262 I think I know what you are getting at - perhaps you could expand? I believe the expression I have used is ok so long as the speed of the observer does not change significantly during the time between wavefronts. Practically, this means that $a \omega_0^{-1} \ll c$ (NB: In my answer, the unit system is such that $c=1$). So for optical light with $\omega_0 \sim 3\times 10^{15}$ s$^{-1}$, it means that the uniform acceleration should be much less than $10^{24}$ ms$^{-1}$, which seems reasonable for a spaceship! Jul 9, 2015 at 23:24
• Rob Jeffries: "[...] ok so long as the speed of the observer does not change significantly during the time between wavefronts." -- Well ... I believe that "your expression" $$\omega = \omega_0~(1 - v)~\gamma$$ is not perfectly correct even for uniform motion, even with $$(\Delta \mathbf r \cdot \mathbf v)^2 = (\Delta \mathbf r)^2~(\mathbf v)^2 ;$$ but even then only if "$\Delta \mathbf r \cdot \mathbf v$ doesn't invert its sign between wavefronts". (It's of course a separate "technician's problem" to decide what to consider still "probably ok" in some circumstance, or the other.) Jul 10, 2015 at 5:37
• Rob Jeffries: "NB: In my answer, the unit system is such that $c=1$" -- That's alright, I suppose; mostly since $1$ is explicitly different from $0$; and as long as you (can) steer clear of expression that would "mix apples and oranges" (i.e. not to use expressions such as "$(a / \omega_0) - (\omega_0 / a)$", or somesuch). But in any case, to be consistent, you should therefore remove any reference to the unit "$\text m$" from your answer, too. Jul 10, 2015 at 5:38