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Let's say I drop a ball on the ground and it bounces up. According to classical physics, once the ball hits the ground it goes trough a short phase of deceleration until the velocity reaches zero, then it goes trough a quick phase of acceleration upwards. A shorter deceleration time implies a greater normal force acting on the ball.

From my understanding, a photon must be traveling at the speed of light at all times. This means that if it bounces off something, there is no deceleration phase, but rather it instantly changes it's momentum in another direction. If we think about the example above with the ball, in some sense the electron should experience an infinite normal force, but clearly this does not happen. Of course a hue difference is that the ball has mass while the photon does not, but the photon still has momentum. What is really going on to prevent this?

P.S. I am a layman, I understand mathematical equations and basic physics, but please try to avoid using highly specialized terminology.

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The main misconception in the question is treating the photon exactly like a point particle. Of course, your analysis is valid provided that photon is a point particle. But it is not exactly one. It is a quantum of the electromagnetic field, and therefore it exhibits both particle-like and wave like behaviour, when appropriate limits are taken. When a photon collides with, lets say a mirror, then it undergoes a scattering process with the electrically charged particles in the mirror. But you cannot model the photon exactly like a particle to treat this problem as a particle collision. For an exact treatment, you have to treat it as a quantum field, and calculate the scattering cross section with the help of Feynman rules, which might be tedious in this case. The concepts of acceleration/deceleration are valid in particle dynamics, however there is essentially no need for using these concepts in the framework for calculating cross sections for quantum fields, like the electromagnetic field of which the photon is a quanta.

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