Why E/c = p ?

  • #1
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Moved from a technical forum, so homework template missing
Hi people, I studying electromagnetic waves (intermediate) and
I don't understand how the expression for linear momentum of a wave is obtained, if the wave doesn't carry any mass.
In particular, I have to explain why the radiation pressure on a perfect absorber is half that on a perfect reflector

So, I do this:

P_rad= pressure
\vec{p}= momentum of wave
A= transversal area
Volumen= ctA
p_den= density of momentum =p/vol
P_rad=\frac{Δp}{Δt *A} =\frac{ p_den *(c *t*A)}{ t*A} = p_den *c
but i the books're saying that E/c =p ???? Why?
Finally, p_den *c = I *c/ c² = I/c to case of absorbent surface.
 
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Answers and Replies

  • #2
DrClaude
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In a relativistic context, which you must use when considering light, the energy is given by
$$
E^2 = p^2 c^2 + m^2 c^4
$$
Therefore, a massless particle has momentum ##p = E/c##.

As for the second question, think about the difference in the before and after pictures of the collision and conservation of momentum.
 
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