When I search on this question online, I get conflicting answers. Most sites will tell you that the amplitude is not affected by the doppler shift, but in Einstein’s 1905 publication on special relativity (§7) he shows that the amplitude of an electromagnetic wave transforms with the same factor as the frequency. How can it be that the amplitude of some waves is affected, while others are not?

  • $\begingroup$ Could you link a source saying the amplitude doesnt change? Also, is it possible that some sources were talking about different kinds of waves like sound waves (doppler shift works quite differently on sound waves)? $\endgroup$
    – AXensen
    Commented Apr 17, 2023 at 11:27
  • $\begingroup$ You got very nice theoretical answers below. Think about a simple handwaving argument: if you are stretching a wave packet out, but leave the amplitude the same, what happens to the total energy? It increases, right? Why? because the power (amplitude squared) of the wave packet would stay the same, but the duration would increase. This is clearly not possible, hence the amplitude has to decrease. $\endgroup$ Commented Apr 17, 2023 at 18:05
  • $\begingroup$ @FlatterMann why "not possible"? After all, the energy does increase in forward direction, and decrease backward etc. $\endgroup$ Commented Apr 17, 2023 at 18:30
  • $\begingroup$ @AndrewSteane You are correct, I was talking about a red shifted Doppler. You can make a similar argument for the blue shifted case. $\endgroup$ Commented Apr 17, 2023 at 18:33
  • $\begingroup$ @FlatterMann Yes that is true for the EM field wave where $energy \propto A^2$. There is a proof of it in this preprint. osf.io/wn3br $\endgroup$ Commented Apr 18, 2023 at 7:03

6 Answers 6


This is not related to "relativity". Note that the Doppler effect as usually taught assumes that the signal is a pure sinusoid. When the signal has a finite bandwidth then the Doppler effect is to be applied to every frequency component individually. When the signal is really wide bandwidth, say $TB > k_v \frac{c}{v}$, where $T$ and $B$ are the length and bandwidth of the pulse, and $v$ is the relative velocity between the emitter and receiver the Doppler shift across the pulse envelope becomes so significant that it affects its shape, stretches/compresses the envelope but without changing its energy. If phase errors across the pulse are to be limited to about $\pi/2$ then the numerical constant $k_v\approx 0.1$ and radars limit their bandwidth to $TB < 0.1 \frac{c}{v}$ to avoid this pulse shape distortion. Similar constraint can be established for the pule shape changing effect of acceleration $T^2B > k_a\frac{c}{a}$. For details see Rihaczek: Principles of high-resolution radar, Section 3.3.


It is correct that the amplitude of waves like water waves and sound waves are not affected by the doppler effect. It is also correct that the amplitude of an electromagnetic wave is affected by the doppler effect as Einstein shows. In this case

\begin{equation} \frac{A´}{A}=\frac{f´}{f} \end{equation}

To show that the amplitude of a water wave is not affected is relatively simple. If the wave propagates along the x-axis, and the amplitude is along the y-axis, a relativistic transformation along the x-axis will yield $y’=y$. Hence the y-coordinate of a $H_2 O$ molecule at the top of the wave will not change, and the amplitude stays constant.

So why is this different in case of an electromagnetic wave?

Because the electric field measures the gradient (rate of change) of the electromagnetic potential. Not the voltage of potential itself. If the electric field oscillates so will the potential, and since the amplitude of the potential is unaffected by the doppler effect, the slope of the wave will change with the frequency. In other words, the amplitude of the gradient will change with the frequency.

If we choose the vector potential of an electromagnetic wave to be $ \mathbf{A} =A c \sin(kz-\omega t) \hat{\mathbf{x}}$, and use the Gibbs gauge with scalar potential $V=0$, the electric field will be \begin{equation} \mathbf{E}=-\nabla V- \frac{1}{c} \frac{d \mathbf{A}}{dt}= -Ac \frac{d}{dt}{\sin(kz-\omega t) \hat{\bf{x}}} \end{equation}

\begin{equation} = \omega A {\cos(kz-\omega t) \hat{\bf{x}}} \end{equation}

If we now consider a doppler shifted wave with identical ampliude, $A$, and doppler shift factor $\Gamma$, so that $\omega'=\Gamma \omega$, we get

\begin{equation} \mathbf{E}'=-\nabla V'- \frac{1}{c} \frac{d \mathbf{A'}}{dt}= -Ac \frac{d}{dt}{\sin(kz-\Gamma \omega t) \hat{\bf{x}}} \end{equation}

\begin{equation} = \Gamma \omega A {\cos(kz-\Gamma \omega t) \hat{\bf{x}}} \end{equation}

We see that the amplitude of the doppler shifted electric wave has changed with the same factor, $\Gamma$, as the frequency.

In the same way, if we measure the amplitude of a sound wave in pascal (SI unit of pressure) the amplitude will not be affected by the doppler effect. But if we instead measure the pressure gradient off the same wave, $\frac{dP}{dx}$, the amplitude will be Doppler shifted.


The E and B of electromagnetism are components of the electromagnetic field tensor $F^{\mu\nu}$.

$$ F^{\mu\nu}=\begin{bmatrix} 0 & cB_z & -cB_y & E_x \cr -cB_z & 0 & cB_x & E_y \cr cB_y & -cB_x & 0 & E_z \cr E_x & E_y & E_z & 0 \end{bmatrix} $$

The EM wave is going in the z direction so $E_z=B_z=0$. We then transform the EM field tensor F to a frame boosted in the -z direction.

\begin{align} F_{New} & = \begin{bmatrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & \cosh{\lambda} &-\sinh{\lambda} \cr 0 & 0 &-\sinh{\lambda} & \cosh{\lambda} \end{bmatrix} \begin{bmatrix} 0 & 0 & -cB_y & E_x \cr 0 & 0 & cB_x & E_y \cr cB_y & -cB_x & 0 & 0 \cr E_x & E_y & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & \cosh{\lambda} & \sinh{\lambda} \cr 0 & 0 & \sinh{\lambda} & \cosh{\lambda} \end{bmatrix}\\ &=\begin{bmatrix} 0 & 0 &-\gamma(cB_y-\beta E_x) &\gamma(E_x-\beta cB_y) \cr 0 & 0 & \gamma(cB_x+\beta E_y) &\gamma(E_y+\beta cB_x) \cr \gamma(cB_y -\beta E_x) &-\gamma(cB_x+\beta E_y) & 0 & 0\cr \gamma(E_x -\beta cB_y) & \gamma(E_y+\beta cB_x) &0 & 0 \end{bmatrix}\end{align}

where the substitutions $\gamma=\cosh{\lambda}$ and $\beta\gamma=\sinh{\lambda}$ were made. The $\lambda$ is the Lorentz Boost parameter (also known as rapidity), and where $\beta=\frac{v}{c}=\tanh{\lambda}$.

You can see that the new E and B are different. For example $E^{New}_{x}=\gamma(E_x-\beta cB_y)$. As you quoted Einstein, the same Doppler factor $\gamma$ appears.

For the boost being in the opposite direction the EM wave is travelling, the frequency of the EM wave seen by the boosted observer is Doppler shifted to: $$ \nu_{new}=e^{\lambda}\nu $$

  • 1
    $\begingroup$ Out of curiosity: how comes you choose $x^3 = c t$ instead of $x^0$? Is there a context where that is the usual convention? $\endgroup$
    – Albert
    Commented Feb 16 at 20:05
  • $\begingroup$ @Albert: I label the components of a 4-vector as 1,2,3,4 with $x^4=ct$, instead of your convention 0,1,2,3 with $x^0=ct$. I just got used to thinking of t as the fourth component. $\endgroup$ Commented Feb 18 at 19:45

When an EM-wave pulse collides with an antenna of a still standing spaceship, the kinetic energy of the pulse becomes EM-energy.

When an EM-wave pulse collides with an antenna of a spaceship moving away at speed about 0.5 c, half of the kinetic energy of the pulse becomes EM-energy, other half becomes kinetic energy of the ship.

When an EM-wave pulse collides with an antenna of a spaceship moving towards the pulse at speed about 0.5 c, all of the kinetic energy of the pulse becomes EM-energy, and equal amount of kinetic energy of the spaceship is converted to EM-energy.

When an sound-wave pulse collides with a diaphragm of a still standing microphone, the very small kinetic energy of the pulse becomes sound energy.

So sound-waves and EM-waves work the same way.

But it's quite difficult to move towards a sound wave at relativistic speed, and then measure the amplitude. And the Doppler effect would be very large. Except if we have sound wave that moves very fast in some special material, then it's more like EM-wave.


So much noise about nothing.

For those, that abhorr Maxwells tensor analysis: A transversal harmonic e.m. plane wave in the x-t-plane of $\mathbb {R}^4$ with wave vector $(\omega,k,0,0)$ and electric amplitude $E \mathbb {d}y$ has the 1-form

$$\mathit{A} = \frac{E}{\omega}\ \sin( k\ x - \omega \ t) ) \ \mathbb{ d}y $$

The field is the exterior derivative (4-d curl)

$$\mathbb{d} \mathit{A} =\mathit{F} = \left(\frac{E \ k }{\omega} \ \ \mathbb{d} x \wedge \mathbb{ d}y + E \ \mathbb{ d}t \wedge \mathbb{d}y \right) \quad \cos ( k (x - c t) ) .$$

The exterior second derivative is zero by $\mathbb{d}^2=0$, thats the homogenuous part of Maxwells vacuum equations.

The Hodge dual, wedge contraction with the volume element, yielding the stress tensor $D,H $ is

$$\star \mathit F = \left< \mathbb{d}t \wedge \mathbb{d}x\wedge \mathbb{d}y\wedge \mathbb{d}z, \mathit F\right> = \left(E k \mathbb{d}t \wedge \mathbb{d}z + E \ \mathbb{d}x \wedge \mathbb{d}z\right)\ \cos (k x - \omega t) $$

simply by exchange of the two forms with their counterparts in the volume form, signs in the exterior algebra of the Lorentz metric automatically correctly set, with vanishing differential 3-form, the dual of the current density vector.

$$\mathbb {d}\star \mathit{F} = 0$$

By inspection of a Lorentz transformation in the $t-x$-plane $$\mathbb{d} {t'} =\mathbb{d} \left(t \cosh u + \frac{x}{c} \sinh u \right),\mathbb{d} {x'} =\mathbb{d} \left(x \cosh u + c t \sinh u \right) $$ we see without mor calculation, that both components of $\mathit F$ aquire a factor $\cosh u =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ by its dominant differential form.

In other words: $E_y = F_{t,y}, B_z=F_{x,y}$ both have one of the indifferent indices $y,z$ and on of the active indices $t,x$, such that both transform as vector components by a boost in the longitudinal direction of the wave.

It comes a bit of a surprise, that the transversal components $(E,B)$ grow by $\cosh u$ if measured in a system moving in the $k$-direction, while longitudinal components wrt to motion direction are the same in all systems.

On the other hand, this fact yields a constant longitudinal E field for the accelerated charged observer, yielding the motion on a hyperbola with constant proper acceleration, and, in a sense, this fact conserves charge, transversality and helicity during acceleration.


Short Answer:

Some material waves such as a wave in a string have a restoring force that is proportional to frequency. Therefore the deflection can remain the same yet perform the same thing as a corresponding increase in the E field.

Long Answer:

A way to look at whether or not the electric field will increase or decrease depending on frequency change due to doppler shifts is to remember that the strength of the electric field can be represented by how many electric field lines are within a certain distance. If due to being in an inertial frame of reference whereby the wave becomes twice frequency of another inertial frame of reference, the E field will double. The power will be 4 times as great relative to the other observer, since power is proportional to ExB according to Poynting's theorem, B (magnetic field amplitude) being doubled as well. Four times the power makes sense since there are twice as many photons per time (two cycles per an interval of time compared to one cycle in the other frame; no change in photons per cycle), and each photon is doubled in frequency, and thus double in energy per photon E = hf (Planck's constant times frequency). The energy per cycle in this new frame has doubled, just as the E field has doubled.

Material waves in general have restoring force proportion to there displacement. For a wave in a string the displacement is traverse from the direction of travel. The energy per cycle of a wave in a string is proportional to the frequency in the x-direction even though its displacement is constant. For the wave in a string the effect of doubling the amplitude is achieve by a double in the restoring force for a twice frequency wave as measured in the x-direction. Since traveling at the high speeds in the x-direction to get any kind of contraction of distance for special relativity is not practical one can not easily change the frequency in the x-direction. Therefore, the restoring force will not change as a function of x along the string in any practical manner. Changing the frequency in the x-direction would have to be induced by the source of the wave, and not a doppler change. In the time direction, it is a much different story. One can readily move at different speed and change the time frequency. However, changing the time frequency is not going to help to change the restoring force as a function of time frequency. So for all practical purposes, the restoring force remains the same for different speeds and different frequencies. This simply means one will not get four times the power when changing from 1 cycle per second to say two cycles per second. One will only achieve a double of power received. There will be no increase in the energy per cycle. This is a doppler effect in time, but it no longer is similar to that of the doppler effect for the EM wave in that doubling the frequency in time by by moving faster towards the source achieves 4 times the power for doubling the frequency.

The EM wave is unique too in that its power cycle is a sinusoid, whereas these material waves have constant energy density. The difference is that the potential energy of the material wave is offset from its kinetic energy by 90 degrees in phase along the direction wave is traveling. On the other hand, the B field is rotated from the E field by 90 degrees relative to the direction of travel of the wave using the right-hand rule.

  • $\begingroup$ I don't think I ever saw a drawing af an EM wave using field lines. Usually vectors are used. But I think you are right that it can be viewed that way. $\endgroup$ Commented Feb 14 at 13:48
  • $\begingroup$ As I recall, the electric field lines usually start at positive charges and end at a negative charge. A charge by itself will have field lines that go radially in or out of a charge. For traveling electromagnetic fields, the field lines will eventually close onto themselves. As a em wave becomes more planer, the field lines will eventually go close to themselves but in opposite directions. Faraday liked to use them as I recall. $\endgroup$
    – Wantabe
    Commented Feb 14 at 23:15
  • $\begingroup$ Feyman says these lines are "only an approximation, and it will require, in general, that new lines sometimes start up in order to keep the number to the strength of the field" $\endgroup$
    – Wantabe
    Commented Feb 15 at 4:34
  • $\begingroup$ Greetings! We've noticed you made a large number of small edits to this answer. Be aware that each edit you make "bumps" this question to the top of the front page, where it displaces a post made by someone who wrote what they intended the first time. Please try to make you edits substantive, rather than repeated minor corrections. If your answer is still evolving, you might edit it on your own computer, or using a service like stackedit.io $\endgroup$
    – rob
    Commented Feb 18 at 1:20
  • $\begingroup$ Will do. Thank you for letting me know. $\endgroup$
    – Wantabe
    Commented Feb 18 at 16:06

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