Relativistic Doppler effect

If a source and a receiver move relative to each other, the frequency of the light received changes, exactly as the whistle of an approaching train sounds higher-pitched than when it recedes. In special relativity the formula is derived by reasoning event by event, using only the constancy of $c$ and time dilation.

Setup. A scientist at rest in SRA sends two brief flashes of light towards an object at rest in SRB, which moves at speed $v$ relative to SRA (positive if receding). We define four events:

• $E_1$: departure of the first flash from the scientist;
• $E_2$: departure of the second flash;
• $E_3$: the first flash reaches the object;
• $E_4$: the second flash reaches the object.

We take $E_1$ as the synchronisation event between SRA and SRB: $t_{1(A)} = t_{1(B)} = 0$ and $x_{1(A)} = x_{1(B)} = 0$. The scientist emits the flashes with interval $\Delta t_{(A)} = t_{2(A)} - t_{1(A)}$, measured on their own clock.

Step 1: dilation between cospatial events. $E_1$ and $E_2$ both occur at the point where the scientist stands: they are cospatial in SRA. So in SRB the interval is dilated:

$\Delta t_{(B)} = t_{2(B)} - t_{1(B)} = \gamma\,\Delta t_{(A)}$

Step 2: equations of motion in SRB. In SRB the object is at rest at the origin, while the scientist recedes at speed $-v$. The first flash leaves $x = 0$ at instant $t = 0$ and travels at speed $+c$: law $x = ct$. The second flash leaves $x_0 = -v\,\Delta t_{(B)}$ (the scientist has moved in the meantime) at instant $t_0 = \Delta t_{(B)}$, again at $+c$: law $x = -v\,\Delta t_{(B)} + c\,(t - \Delta t_{(B)})$. Imposing $x = 0$ (the object is at the origin of SRB) gives $t_{3(B)} = 0$ and:

$t_{4(B)} = \Delta t_{(B)}\,(1 + v/c)$

Step 3: the reception interval in SRB. The difference $\Delta t'_{(B)} = t_{4(B)} - t_{3(B)}$ is, using the identity $\gamma(1+v/c) = \sqrt{(1+v/c)/(1-v/c)}$:

$\Delta t'_{(B)} = \Delta t_{(B)}\,(1 + v/c) = \gamma\,(1 + v/c)\,\Delta t_{(A)} = \Delta t_{(A)}\,\sqrt{\frac{1 + v/c}{1 - v/c}}$

Relativistic Doppler effect

If the source sends signals with proper interval $\Delta t_{(A)}$ towards an object moving at speed $v$ (positive if receding), the object receives them with proper interval $\ev{\Delta t'_{(B)} = \Delta t_{(A)}\sqrt{\frac{1 + v/c}{1 - v/c}}}$ and, for frequencies $f = 1/\Delta t$, $\ev{f' = f\,\sqrt{\frac{1 - v/c}{1 + v/c}}}$ With $v > 0$ (receding) we have $f' < f$: redshift. With $v < 0$ (approaching) we have blueshift.

Why square roots?

In the classical Doppler effect (for sound) a single reflection of the signal would give the factor $(1+v/c)/(1-v/c)$, without a square root. The relativistic effect “adds” the time dilation between cospatial events of SRA and SRB: the overall factor becomes $\gamma(1+v/c) = \sqrt{(1+v/c)/(1-v/c)}$. Square roots are relativity’s trademark: they appear every time a time interval is “dragged” from one frame to another through the dilation formula.

Cosmological redshift

The relativistic Doppler effect has a spectacular application in astrophysics: distant galaxies emit light that reaches us with a redshift (shift towards red) that is greater the farther away they are. It is the proof that the universe is expanding. A galaxy with redshift $z=1$ has $f' = f/2$, which implies a recession speed $v \approx 0.6\,c$. By observing ever larger redshifts, astronomers manage to peer at the universe as it was billions of years ago.

Links

Topics: Special relativity Concepts: Doppler effect Skills: Changing reference frame Methods: Minkowski diagram

Related exercises: Problem — A red traffic light seen as green · Problem — Composition via Doppler · Problem — Redshift of a galaxy