1
$\begingroup$

Does light reflecting from a material add kinetic energy to the material or does light only add kinetic energy when absorbed? And if reflection adds kinetic energy, where is this energy being subtracted from? The amplitude/frequency of light? Or is absorption statistically always possible, just highly unlikeley when having a material with high albedo?

EDIT: Ok, so I have mad a calculation for blue light photon reflecting from a 1kg body, this is what I got:

f' = E'/h

= (E - vBody * 1kg) / h

= (h * 650*10^12 - ( 2 * ((E / c^2 * c)/(1 + E / c^2)) ) ) / h

= (h * 650*10^12 - ( 2 * ((((4,29 * 10^-19) / c^2) * c)/(1 + (4,29 * 10^-19) / c^2)) ) ) / h

= ((6.62607004 × 10^-34) * 650*10^12 - ( 2 * ((((4.29 * 10^-19) / 299792458^2) * 299792458)/(1 + (4.29 * 10^-19) / 299792458^2)) ) ) / (6.62607004 × 10^-34)

= 6.499.. × 10^14 Hz

Is that correct?

$\endgroup$
  • 1
    $\begingroup$ Just make the fun calculation, a "blue" photon reflecting from 1 kg of mass. Conservation of momentum and energy. Assuming that the resulting kinetic energy of the 1 kg mass is coming from the photon, what is its red shift after reflection? $\endgroup$ – mikuszefski Sep 26 '16 at 9:28
  • $\begingroup$ I did, @mikuszefski. $\endgroup$ – Markus Appel Sep 26 '16 at 10:12
  • $\begingroup$ I am afraid, the calculation is wrong. The photon has momentum $p= h \nu/c$. The 1 Kg reflects so it gets basically $2p$ assuming that $p$ does not change too much. So kinetic enrgy of the body is $p^2/2m= \Delta E$. So $\Delta \nu=\Delta E/h=h \nu^2/(2 m c^2)$. You can rewrite it as $ \nu * h \nu/(m c^2)$ , i.e the relative change is $ \Delta \nu/ \nu = h \nu/(m c^2)$, the photon energy divided by the energy at rest of the body, which is ridiculously small. Nevertheless, check the photon propulsion $\endgroup$ – mikuszefski Sep 26 '16 at 12:08
  • $\begingroup$ Technically, you should consider that due to the decreased frequency, the momentum of the photon is smaller after reflection and the body does not get exactly $2p$ but the difference is so small that we forget about it. If however $m c^2 \approx h \nu$ things are different. $\endgroup$ – mikuszefski Sep 26 '16 at 12:13
  • $\begingroup$ By the way, the error in your calculation starts in line 2, where you mix up energy and momentum. (checking units is your friend.) $\endgroup$ – mikuszefski Sep 26 '16 at 12:15
1
$\begingroup$

With the assumption that the "material" is initially at rest and no complicating factors such as tricky coupling to other material you haven't yet mentioned - for example if it's a chunk of something floating in space and it is struck by a beam of light...

Does light reflecting from a material add kinetic energy to the material

yes

or does light only add kinetic energy when absorbed?

not only, it adds kinetic energy to that material in both cases.

And if reflection adds kinetic energy, where is this energy being subtracted from? The amplitude/frequency of light?

yep... the outgoing reflected photons will have a lower energy and therefore lower frequency/longer wavelength.

Or is absorption statistically always possible, just highly unlikeley when having a material with high albedo?

You can do the calculation without necessarily using statistics. Every photon with a frequency $\nu$ has a certain momentum $p=h \nu / c$. You can convert to wavelength $\lambda$ using $\nu = c/ \lambda$ . If it is absorbed, then the kinetic energy of the material of mass m is $p^2/2m$. If it is reflected, you calculate the same way you'd do elastic scattering. Assuming the mass of the material is much larger than the equivalent energy of the photon (works for visible light an even a single atom) then the energy loss of the photon is small ad the kinetic energy added is just double $p^2/m$ because you've changed both momenta by $2p$ assuming 180° reflection.

You can read more about radiation pressure.

$\endgroup$

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.