I'm wondering if one can arrive at [itex]E=mc^2[/itex] using only the physics of the late 19th century, in the following way:(adsbygoogle = window.adsbygoogle || []).push({});

As light waves pass over an electrically charged particle, they push it in the direction of the wave motion, transferring both (kinetic) energy and momentum to the particle. Let's call these quantities [itex]\Delta KE [/itex] and [itex]\Delta p [/itex]. So after the waves have passed over the particle, they've lost [itex]\Delta[/itex]KE of energy and [itex]\Delta[/itex]p of momentum.

As far as I know, the physics of the late 19th century regarded light waves as massless. But what if we plug in the classical Newtonian definition of momentum p=mv and — since the light waves won't change speed — say that the waves have "lost mass" according to [itex]\Delta m = \Delta p / v [/itex] where v is the speed of light [itex] c=1/ \sqrt{\epsilon_o \mu_o} [/itex].

Do the quantities [itex]\Delta KE[/itex] and [itex]\Delta m [/itex] work out such that [itex]\Delta KE = c^2 \Delta m [/itex]? I'm guessing yes but frankly don't have the experience working out the mathematics of electromagnetism to find out for myself.

Thanks.

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# I Can one get E=mc^2 using classical EM theory?

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