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- Thread starter fireball3004
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If you define momentum as the ability to deliver an impulse, then light has it. This is experimentally demonstrable. See Compton scattering.

It is a tenet of special relativity that no mass can be accelerated so as to break the light barrier in any frame of reference, therefore the photon is assigned zero rest mass on principle. Experiments to test this set a very low possible mass for the photon.

See the FAQ on the Experimental Basis of Special Relativity.

It may sound contradictory to some, but that's the way it seems to be.

It is a tenet of special relativity that no mass can be accelerated so as to break the light barrier in any frame of reference, therefore the photon is assigned zero rest mass on principle. Experiments to test this set a very low possible mass for the photon.

See the FAQ on the Experimental Basis of Special Relativity.

It may sound contradictory to some, but that's the way it seems to be.

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jtbell

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[tex]E^2 = (pc)^2 + (m_0 c)^2[/tex]

If [itex]m_0 = 0[/itex], then E = pc.

It's possible to show that

[tex]\frac{v}{c} = \frac{pc}{E}[/tex]

so if E = pc, then v = c.

Reversing this logic, we can see that if v = c, then [itex]m_0 = 0[/itex].

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jtbell

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If you're talking about "relativistic mass" then you can combine [itex]E = mc^2[/itex] and [itex]E = pc[/itex] to get [itex]p = mc[/itex].

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As the momentum is defined as the velocity times relativistic mass ( not rest mass ) in special relativity, p=mc in case of photon ( a quantum particle of light ) where m is the relatvistic mass and c the speed of light.

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Its more simple than that since relativist mass, m, is

If you're talking about "relativistic mass" then you can combine [itex]E = mc^2[/itex] and [itex]E = pc[/itex] to get [itex]p = mc[/itex].

Pete

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...I think I've got the gist of it now, thanks guys

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