Momentum of electromagnetic wave

Derivation of momentum of electromagnetic wave:

Maxwell's had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. Let a particle which has mass $m$ moving with velocity then the momentum of a particle,

$\overrightarrow{P}=m\overrightarrow{v} \qquad(1)$

According to mass-energy relation


Here $U$ - Total energy of the particle

$m=\frac{U}{c^ {2}} \qquad(2)$

From equation $(1)$ and equation $(2)$

$\overrightarrow{P}=\frac{U}{c^{2}} \overrightarrow{v} \qquad(3)$

If the electromagnetic wave is propagating along the x-axis then

$\overrightarrow{v}=c \hat{i}$

Put this value in the above equation $(3)$

$\overrightarrow{P}=\frac{U}{c} \hat{i} \qquad(4)$

We know that the equation of energy flow in electromagnetic wave

$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{n}$

Here wave is propagating along x-axis i.e 


$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$

The energy density in plane electromagnetic wave in free space:

$U=\epsilon_{0} E^{2}$

Where $E$ - Magnitude of electric field

$E^{2}=\frac{U}{\epsilon_{0}} \qquad(6)$

Now substitute the value of $E^{2}$ in equation$(5)$

$\overrightarrow{S}= \frac{1}{\mu_{0} c} \frac{U}{\epsilon_{0}} \hat{i} $

$\overrightarrow{S}= \frac{c^{2}}{c} U \hat{i} \qquad (\because \frac{1}{\sqrt{ \mu_{0} \epsilon_{0}}}=c) $

$\overrightarrow{S}= c U \hat{i} $

$U \hat{i}=\frac{\overrightarrow{S}}{c} \qquad(7)$

Now substitute the value of $ U \hat{i} $ in equation $(4)$. Then


$\overrightarrow{P}=\frac{(\overrightarrow{E} \times \overrightarrow{B})}{ \mu_{0}c^{2}}$

$\overrightarrow{P}=\epsilon_{0}(\overrightarrow{E} \times \overrightarrow{B})$

This is the equation of "Momentum of electromagnetic wave"