- #1

fisher garry

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## Summary:

- from the theory in Griffiths. Calculate the energy in an EM wave by starting with time derivative of energy density is equal to divergence of poynting vector

You dont have to read all this theory to answer my question. I added it just in case.

Above they use the definition work energy theorem in vacuum to get to (8.12). Since it is in vacuum I would guess that one could use the equations for B and E field from EM-waves so that magnitude poynting vector due to orthogonality of B and E field of EM-waves this becomes:

##|\frac{1}{\mu_0} \pmb{E}\times \pmb{B} |=\frac{1}{\mu_0} \pmb{E} \pmb{B}##

by using

## \pmb{E}=c \pmb{B}##

##|\frac{1}{\mu_0} \pmb{E}\times \pmb{B} |=\frac{1}{\mu_0} c \pmb{B^2}##

Is it possible to expand this so that you can calculate the energy of a single EM-wave? By integrating (8.12) or something? It should be equal to E=hf if I am not wrong.

Above they use the definition work energy theorem in vacuum to get to (8.12). Since it is in vacuum I would guess that one could use the equations for B and E field from EM-waves so that magnitude poynting vector due to orthogonality of B and E field of EM-waves this becomes:

##|\frac{1}{\mu_0} \pmb{E}\times \pmb{B} |=\frac{1}{\mu_0} \pmb{E} \pmb{B}##

by using

## \pmb{E}=c \pmb{B}##

##|\frac{1}{\mu_0} \pmb{E}\times \pmb{B} |=\frac{1}{\mu_0} c \pmb{B^2}##

Is it possible to expand this so that you can calculate the energy of a single EM-wave? By integrating (8.12) or something? It should be equal to E=hf if I am not wrong.

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