My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.

In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.

I told her that:

Doppler effect is about alteration of the perceived frequency of a signal produced by the relative movement between transmitter and receiver. The quoted video is about relativity, which is a "deeper" effect. Maybe doppler effect can be understood as the effect of relativity on a wave phenomena.

Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.

Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?


As pointed out in earlier answers, Doppler effect and relativistic effects are independent.

There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.

However, it's interesting to note that mathematically there are parallels in the descriptions of the two.

In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.

General wave equation in one dimension:

$$ \frac{\partial^2 \phi }{\partial x^2} = \frac{1}{u^2} \frac{\partial^2 \phi}{\partial t^2} $$

(u = propagation speed of the wave)

As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.

Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.

Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.

That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.

Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)

The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.

  • $\begingroup$ Thank you for addressing the Lorentz transformation relation to both theories! To clarify: with sound being a wave phenomena, could we say that similar maths would apply if he was studying doppler on sound waves (not the relativistic part, but the Doppler to quasi Lorentz derivation)? $\endgroup$ – jjmontes Mar 19 at 15:08
  • 1
    $\begingroup$ @jjmontes Well, Kevin Brown describes specifically that Voigt was exploring a general wave equation, not a specific equation tied to a specific case. The general equation describes an idealized case. To a first approximation this idealized equation applies for sound waves, but of course actual sound propagation is not quite the idealized case. For instance, the higher the frequency of a sound, the faster it dissipates to heat. As far as we know the Lorentz transformations are exhaustively correct. For actual waves (any kind): the idealized case is an approximation. $\endgroup$ – Cleonis Mar 19 at 20:24

The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.

The formula for the Doppler effect is $$f_o = \frac{v - v_o}{v - v_s} f_s$$ where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have $$\gamma_0 f_0 = \frac{v - v_o}{v - v_s} \gamma_s f_s.$$ This is a very small correction assuming the speeds are small.

When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have $$f_o = \frac{c - v_r}{c} f_s$$ where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r \neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source, $$f_o = \frac{c - v_r}{c} \gamma_s f_s = \sqrt{\frac{1 - v_r/c}{1 + v_r/c}} f_s.$$ There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r \ll c$.

  • $\begingroup$ the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards. $\endgroup$ – craq Mar 18 at 2:32

There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.

However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.

If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.


Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis. If $\Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$\Delta t+\frac{\beta \Delta t}{1-\beta}=\frac{\Delta t}{1-\beta}=\Delta T$$. From the invariance of spacetime interval $$c^2\Delta \tau^2=c^2(\Delta T)^2(1-\beta^2)$$$$\Delta \tau=\Delta T\sqrt{1-\beta^2}$$$$\Delta \tau=\Delta t \sqrt{\frac{1+\beta}{1-\beta}}$$


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