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- Thread starter Orman
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malawi_glenn

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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.

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

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malawi_glenn

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Do you want me to point you to a source which goes throug radiation pressure?

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malawi_glenn

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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.

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malawi_glenn

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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:

[tex]P = |\frac{d\vec{F}}{dA}|[/tex]

and

[tex]\vec{F} = \frac{d\vec{p}}{dt}[/tex]

Maximum change in momentum is when the momentum vector is totaly reflected: [tex] \vec{p} \rightarrow - \vec{p} [/tex]

voila

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So a photon in a chamber of perfectly reflective surfaces would lose all of its energy eventually by momentum transfer solely? Still, why cant the photon lose all its energy as kinetic?

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russ_watters

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While true, that doesn't have anything to do with whether a photon or electron has dimensions.A beam of light requires space. No space would mean no reference to anything not even to itself.

That is a meaningless combination of concepts.Electron configurations are shown for atoms, so why isn't there a photon configuration as well?

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russ_watters

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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:So a photon in a chamber of perfectly reflective surfaces would lose all of its energy eventually by momentum transfer solely? Still, why cant the photon lose all its energy as kinetic?

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.

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