1
$\begingroup$

I've been thinking about the doppler effect, where electromagnetic radiation experiences a redshift if the radiation source is relatively moving away from the observation point, and a blueshift if it's moving closer.

If I have a light source, a sensor which is stationary relative to the light source, and a mirror either moving towards or away from the light source while reflecting the light to the sensor — the sensor would pick up the electromagnetic waves as shifted, no? And thus photons (if they are being reflected) basically lose energy to or gain energy from an object (with which they interact) in accordance to the object's momentum relative to the vector of the wave?

I believe that should be the case as I know solar sails are "pushed" by light. Thus I assume light in a way should be able to be "pushed back" and gain energy (which causes the frequency shift).

I'm sorry if this is an obvious concept already explained by the doppler effect. I'm not formally learned in physics and didn't find this specific phenomenon explained in the texts I read.

$\endgroup$
0
$\begingroup$

You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is $$ f = \frac{1 - v/c}{1 + v/c} f_0 $$ where $f_0$ is the original frequency and $v$ is the speed at which the mirror is receding from the source.

There are various ways this can be derived. This is in fact the classical equation for the Doppler shift when the source is at rest with respect to the "propagation medium." This works because the speed of light is $c$ in the reference frame of the source of the waves, and there is no time dilation since waves are being emitted and received by the source and detector which are at rest with respect to each other. You can also show this result using Maxwell's equations, without using the idea of photons, or even explicitly using special relativity (Maxwell's equations are consistent with special relativity). Finally, as you argue in your question, conservation of energy can also be used to arrive at this result.

$\endgroup$
0
$\begingroup$

The collision between an object and a photon can be modelled like any other collision in physics; you take a reference frame (I recommend the rest frame of the object being struck) and give the photon an energy $h\nu$ and a momentum $\frac{h\nu}{c}$. If the photon is absorbed, then the object gains velocity $\frac{h\nu}{M}$, or if it is reflected then the velocity $v$ and reflected/transmitted frequency $\nu'$ solve the equations $\frac{1}{2}Mv^2+\frac{h\nu'}{c}=\frac{h\nu}{c}$ and $Mv\pm h\nu' = h\nu$. This can lead to the photon losing energy and increasing its wavelength.

$\endgroup$
  • $\begingroup$ And if the object being struck is not at rest, will it give energy to the photons if it's moving towards it (quickly enough) and perhaps remove more energy from the photon if it's moving in the same direction (so "away" from it)? $\endgroup$ – TLSO Jul 27 '19 at 22:58
  • $\begingroup$ The best way to approach this is to enter the rest frame of the object and alter the frequency of the photon accordingly using redshift: $\nu=1+(\frac{u}{c})\nu_0$, so an object moving towards the photon increases it's pre-collision energy, and an object moving away form it would decrease the pre-collision energy $\endgroup$ – Samuel Gunatilleke Jul 27 '19 at 23:19
  • $\begingroup$ @SamuelGunatilleke: The frequency shift upon a Lorentz boost is different from what you gave. If a photon has frequency $\nu_0$ in some frame, then in another frame Lorentz boosted by velocity $v$ along the direction the photon is moving in, the frequency of this photon is $\nu = \sqrt{\frac{1-v/c}{1+v/c}} \nu_0$. $v$ is negative if this boosted frame is moving in the opposite direction to the velocity of the photon. $\endgroup$ – Puk Jul 27 '19 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.