Why does a photon transfer only partial energy in Compton scattering?

[Orman]

The Compton's Apparatus is not good enough of an explanation. I would like to know at a fundamental level the geometry of both the entire atom and photon and what causes a photon to give off only some energy as kinetic.

 

[malawi_glenn]

Light pressure is just momentum and energy conservation, and is a mesoscopic concept.

An atom don't have a 'geometry', it is a quantum object, and a photon is as we know pointlike, i.e no spatial extension.

 

[Orman]

Real objects take up space, so how does a photon take up space? An infinite number of photons?

 

[malawi_glenn]

Who has said that a photon must occpuy a volume or anything like that?
Do you want me to point you to a source which goes throug radiation pressure?

 

[Orman]

A beam of light requires space. No space would mean no reference to anything not even to itself. Electron configurations are shown for atoms, so why isn't there a photon configuration as well?

 

[malawi_glenn]

Electrons are pointlike particles aswell...

Ever heard of delta function? The form factor for electrons is a delta function. So there is no distance between the point x=2.402397 and x=2.9027643 on the x-axis even though the points themselves occupy zero width?

My ultimate point is that if you want to understand radiation pressure, you only need the concept of momentum and energy conservation, you don't need to bother about the quantum nature of atoms and photons since radiation pressure is a mesoscopic concept.

 

[Orman]

It would help to know what happens at that level because creating light propulsion would be ideal for all forms of travel. Better materials could be created in maximizing energy conversion to kinetic energy rather than having most energy lost as heat or in reflection.

 

[malawi_glenn]

It would help to know what happens at that level because creating light propulsion would be ideal for all forms of travel. Better materials could be created in maximizing energy conversion to kinetic energy rather than having most energy lost as heat or in reflection.

Then you are in the branch of Solid State and Atmoic Physics, specially optical properties of solids. In this forum we are dealing with particle and nuclear physics.

But the maximum light pressure is obtained for a perfect reflecting surface, so you want a good reflecting material. Why? Well since:

$$P = |\frac{d\vec{F}}{dA}|$$

and

$$\vec{F} = \frac{d\vec{p}}{dt}$$

Maximum change in momentum is when the momentum vector is totally reflected: $$\vec{p} \rightarrow - \vec{p}$$

voila

 

[Orman]

I've read about the solar sails developed by nasa, interestingly they just plan to use a lot of surface area to achieve a feasible propulsion.

So a photon in a chamber of perfectly reflective surfaces would lose all of its energy eventually by momentum transfer solely? Still, why can't the photon lose all its energy as kinetic?

 

[russ_watters]

A beam of light requires space. No space would mean no reference to anything not even to itself.

While true, that doesn't have anything to do with whether a photon or electron has dimensions.

Electron configurations are shown for atoms, so why isn't there a photon configuration as well?

That is a meaningless combination of concepts.

 

[russ_watters]

So a photon in a chamber of perfectly reflective surfaces would lose all of its energy eventually by momentum transfer solely? Still, why can't the photon lose all its energy as kinetic?

I think you're missing a critical concept here: the energy of the photon does not and cannot change. And the value of the momentum doesn't change either, only the sign. This is easy enough to see using the equations for the two concepts:

e=1/2 mv^2 -- notice that with v being squared, direction is irrelevant to energy
M1(light)+mv1(object)=M2(light)+mv2(object) -- notice that if the sign of the momentum of the light is reversed as the light is reflected, the final momentum of the object is 2M, where M is the momentum of the light.