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Imagine a photon hits an object, then bounces off and travels in the other direction, all while retaining it’s wavelength.

If the object is slightly accelerated in the direction of the incident photon’s travel, then it seems like conservation of energy is violated, because the photon lost no energy and the object gained kinetic energy.

If the object is not accelerated, it seems like conservation of momentum is violated, because the net velocity of the entire system (the object and the photon) has changed; the object doesn’t move but the photon changed direction.

Is it even possible for a photon to change directions without changing wavelength? If so, why is this paradox invalid?

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Imagine that the object with mass M gains momentum $p = \frac{2h}{\lambda}$ and now has energy $E = \frac{p^2}{2M}$. Let's get the new wavelength from the new photon energy, taking off the energy stolen by the object.

$\frac{hc}{\lambda'} = E' = \frac{hc}{\lambda} - \frac{p^2}{2M } = \frac{1}{2}pc - \frac{p^{2}}{2M} = \frac{1}{2}pc \left[1 - \frac{p}{Mc} \right] = \frac{hc}{\lambda} \left[ 1 - \frac{Mv}{Mc}\right]$

Hopefully it's clear from this approximation that the wavelength has changed by a factor of

$1 - \frac{v}{c}$

where v is the velocity of the object. When $v <<c $ this justifies the assumption I made at the beginning, where the collision is a near perfect reflection and the photon is able to provide nearly twice the momentum, because the new wavelength is very close to the old wavelength.

For large mirrors, this is why the color of photons does not perceptibly change upon reflection.

For tiny particles, this is why you get Compton scattering, for example.

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When a photon bounces of a mass, there are two things to remember:

  1. the momentum transfer is orthogonal to the mass's four momentum.
  2. there is always a frame in which the energy transfer is zero.

In the rest-frame of the mass:

$$ k^{\mu} = (k,0,0,k) $$ $$ p^{\mu} =(M,0,0,0) $$

and for 180 degree scattering:

$$ k'^{\mu} = (k',0,0,-k')$$ $$ p'^{\mu} =(\sqrt{M^2+p'^2},0,0,p') $$

and the 4-momentum transfer of the photon is:

$$ q^{\mu}=k'^{\mu}-k^{\mu}=(k'-k,0,0,k'+k)$$

From here, you can use:

$$q^{\mu}p_{\mu}=0$$

to derive the relativistic radar-Doppler shift (with recoil). It's also a matter of algebra to find the Breit-Frame,in which:

$$q_{\rm Breit}^{\mu}=(0,0,0,2k_{\rm Breit})$$

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