Do Photons Follow the Rules of Mechanics for Collisions?

[Jeebus]

If you do some simple manipulations of the equations E=hf, E=1/2mv^2, P=mv, and p=h/<lambda>, one can derive that the momentum is h/<lambda> or that it is h/(2<lambda>). I know that photons do not always follow normal mechanics, but do they follow the usual rules for mechanics for collisions?

I have heard of numerous situations in which photons have transferred their momentum to other particles or even larger masses, but it is unclear to me how exactly this works. It seems that one could build a device which would have an efficiency greater than 100 percent. In other words, it would not follow the law of conservation of energy.

Why can this not be done?

Is it because the photons transfer energy through various mechanisms, depending on energy?

 

[jcsd]

The momentum of a photon is h/&lambda; and yes they can be involved in collisions with free electrons (compton scattering), where their wavelength is depednet on their angle of scattering. Photons can be absorbed by electrons in bound states too.

 

[ahrkron]

Originally posted by Jeebus
I have heard of numerous situations in which photons have transferred their momentum to other particles or even larger masses

Is this why you say that...

...It seems that one could build a device which would have an efficiency greater than 100 percent.

??

The energy of a particle is given by

E² = m²c⁴ + p²c²

A photon, being massless, has an energy E = pc, which can be quite substantial, and it definitely can be transferred to massive particles.

 

[wawenspop]

a photon transfers its momentum or used to increase orbit enegy? 2 different cases or what? Exactly how does it transfer momemtum

 

[Drakkith]

In lower energy photons it usually results in heat. Higher energy photons, in about the UV range or greater, can ionize atoms by ejecting electrons. On the extreme end the photons can undergo Pair Production, where a photon interacts with a nucleus and its energy is used to create pairs of particles such as an electron and positron.