# Relativistic Doppler Effect from source’s frame

Could someone please help me understand the derivation of the relativistic Doppler effect of light for objects moving away as derived from the source's reference frame? There’s no problem for me to understand it from the receiver's reference frame but for some reason I can’t comprehend it while trying to do so from the source's reference frame. The problem I'm having is before considering relativistic effects such as time dilatation or length contraction.

Thanks in advance!

• – Frobenius Jun 23 '19 at 11:40

## 4 Answers

Lets consider that the source is at rest and the receiver is moving away from the source at some velocity $$v$$ and $$v$$ is taken to be positive.Lets use the subscript '$$s$$' for source and '$$r$$' for receiver.

Lets say at time $$t_s = 0$$ and $$x_s =0$$ the first pulse was received by the receiver. Then at $$t_s = 0$$ the second pulse is at a distance $$\lambda_s$$ from the receiver, $$\lambda_s$$ being the wavelength of light pulses as observed from the frame of reference of the source. Now let $$t_{r s}$$ be the time at which the receiver receives the second pulse as seen from the reference frame of the source. Then $$\lambda_s + vt_{rs}=ct_{rs}\tag{1}$$

That is the in time $$t_{rs}$$ the source moved a distance $$vt_{rs}$$ and light pulse travelled a distance $$ct_{rs}$$. As light and the receiver is at the same position at $$t_{rs}$$, the distance light travelled should be as in equation (1). Now we know that $$\lambda = \frac{c}{f_s}$$

$$\begin{eqnarray} \lambda_s=(c-v)t_{rs} \\ \frac{c}{f_s} = (c-v)t_{rs}\\ \frac{1}{t_{rs}} = (1-\frac{v}{c})f_s \tag{2} \end{eqnarray}$$

Now we need to change $$t_{rs}$$ to $$t_{r}$$, the time at which the observer recieves the second pulse. Using Lorentz transformations. $$\begin{eqnarray} t_r = \frac{1}{\gamma}(t_{rs}-\frac{v}{c^2}\Delta x) \end{eqnarray}$$ $$\gamma = \frac{1}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}$$ and $$\Delta x$$ being the distance between the two events i.e receiving the first pulse and then the second pulse. But $$\Delta x = vt_{rs}$$. $$\begin{eqnarray} t_r = \gamma(1-\frac{v^2}{c^2})t_{rs}\\ t_r = \frac{1}{\gamma} t_{rs} \tag{3} \end{eqnarray}$$ Substituting (3) in (1) we get $$\begin{eqnarray} \frac{1}{ \gamma t_{r}} = (1-\frac{v}{c})f_s \\ f_{r} = \gamma(1-\frac{v}{c})f_s\\ f_{r} = \frac{1}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}(1-\frac{v}{c})f_s\\ f_{r} = \left(\frac{1-\frac{v}{c}}{1+\frac{v}{c}} \right)^{\frac{1}{2}}f_s \end{eqnarray}$$

• First of all, thanks a lot for your great explanation. Secondly, wikipedia stated that in the first part of your first equation, you sould use t,s instead of t,r,s. That what got me all confused and actually your explanation makes sense. So wikipedia is wrong? – G.Bar Jun 24 '19 at 4:47
• @G.Bar Frankly, their explanation makes no sense to me. Its probably a mistake. – Kolandiolaka Jun 24 '19 at 6:06
• Thanks, good to know I'm not alone there... :) – G.Bar Jun 28 '19 at 16:43

$$\let\l=\lambda \def\tE{t_{\rm E}} \def\xE{x_{\rm E}}$$ I'm going to offer you a derivation making no use of time dilation, length contraction, Lorentz transformations.

You have two (inertial) frames: Earth's one, say E, and spaceship's one, say S. It's assumed that spaceship is receding from Earth at a velocity $$v$$ - the same can be said of Earth in spaceship's frame.

Source in S is emitting a monochromatic wave of period $$T$$. You want to know how to find received period $$T'$$ on Earth, i.e. in E frame. Instead of thinking of the full wave system it's useful to single out e.g. wave crests as distinct objects. Two consecutive crests are separated in time by period $$T$$ and in space by wavelength $$\l=c\,T$$. In all frames crests propagate with velocity $$c$$.

Assume a given crest #1 is emitted form source in $$x=0$$ at time $$t=0$$. (Actually this is no assumption - we simply choose origins of space and time coordinates in S accordingly.) We may think of a crest as having a world-line in spacetime: its equation is $$x_1 = c\,t_1.\tag1$$

The next crest #2 is emitted in $$x_2=0$$ at time $$t_2=T$$, so its world-line equation is $$x_2 = c\,(t_2 - T).\tag2$$

A third step is to write Earth's world-line equation in S. It is $$\xE = a + v\,\tE.\tag3$$ The constant $$a$$ is unknown but you'll see it disappears in the following.

And now we have to consider arrival events $$E_1$$, $$E_2$$ when both crests reach the Earth. As to $$E_1$$ this happens when $$t_1 = \tE \qquad x_1 = \xE.$$ Using (1) and (3) we find $$t_1 = {a \over c - v} \qquad x_1 = {a\,c \over c - v}.\tag4$$ As to $$E_2$$ we have $$t_2 = \tE \qquad x_2 = \xE$$ and from (2), (3) we get $$t_2 = {a + c\,T \over c - v} \qquad x_2 = {c\,(a + v\,T) \over c - v}.\tag5$$

This far SR almost didn't enter - almost, because we assumed that light speed is $$c$$ in spaceship's frame and this frame has nothing special, so how can we assume that? Of course, thanks to SR's postulates.

But now the time has come to make a heavier use of SR. As I said in the beginning I'll use neither dilations nor contractions nor Lorentz. I'll only apply invariance of spacetime interval. In defining spacetime interval not a single event is involved but a couple of events, e.g. our $$E_1$$ and $$E_2$$. Given their coordinates in a frame, like S, the spacetime interval between $$E_1$$ and $$E_2$$ is defined as $$s = c^2 (t_2 - t_1)^2 - (x_2 - x_1)^2.$$ For the same events we can also compute interval in another frame, say E. If coordinates in E are $$(t_1',x_1')$$ for $$E_1$$, $$(t_2',x_2')$$ for $$E_2$$, the spacetime interval in E is $$s' = c^2 (t_2' - t_1')^2 - (x_2' - x_1')^2.$$ Invariance simply means $$s=s'$$ (actually, spacetime interval has the same value in every inertial frame).

Since we know coordinates in S (eqs. (4), (5)) we can compute $$s$$. But what about frame E? Although we don't know the $$x'$$-coordinates, we are certain they are the same for both events, since both happen on Earth. As to time coordinates, their difference $$t_2'-t_1'$$ is nothing but the received period $$T'$$. So $$s' = c^2 {T'}^2.\tag6$$

Let's compute $$s$$: $$s = c^2 {c^2 T^2 \over (c - v)^2} - {c^2 v^2 T^2 \over (c - v)^2} = c^2 T^2 {c^2 - v^2 \over (c - v)^2} = c^2 T^2\,{c + v \over c - v}.\tag7$$ By comparing (6) and (7) we obtain $$T'$$: $$T' = T\,\sqrt{c + v \over c - v}.$$ Once you know $$T'$$ it's easy to deduce $$\l'$$ and $$f'$$.

To the source sending out the light (let's say the spaceship is moving away from you and the crew were flashing a laser signal in green light), the light would seem to have its normal wavelength,but to you back on Earth it would be red-shifted. The amount of the redshift would of course depend on the speed of recession. The Doppler effect is not a relativistic effect and can be detected with light or sound emanating from objects moving at low speed. Even at speeds a considerable fraction of the speed of light, the crew aboard the spaceship don't observe any relativistic effects in their own frame of reference: no time dilation, length contraction or mass increase. These relativistic effects are only apparent to observers back on Earth when they observe the spaceship.

Source says: Each second I produce 3*10^9 meters of waves. Each second there are x meters more waves between me and the destination, where x is the speed of the destination in meters per second. Each second the destination receives 3*10^9 - x meters of waves. The ratio of produced waves and received waves is 3*10^9 / (3*10^9 - x) , which is the classical Doppler shift.

Then Source adds: The destination thinks that he received those waves that he received in one second in less than one second. That is the relativistic part of relativistic Doppler shift.