# Is the amplitude of a wave affected by the Doppler effect?

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?

• 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)? Apr 17 at 11:27
• 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. Apr 17 at 18:05
• @FlatterMann why "not possible"? After all, the energy does increase in forward direction, and decrease backward etc. Apr 17 at 18:30
• @AndrewSteane You are correct, I was talking about a red shifted Doppler. You can make a similar argument for the blue shifted case. Apr 17 at 18:33
• @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 Apr 18 at 7:03

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: $$\nu_{new}=e^{\lambda}\nu$$

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.