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
$U=mc^{2}$
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
$\hat{n}=\hat{i}$
$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$
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
|
$\overrightarrow{P}=\frac{\overrightarrow{S}}{c}$
|
$\overrightarrow{P}=\frac{(\overrightarrow{E} \times \overrightarrow{B})}{ \mu_{0}c^{2}}$
|
$\overrightarrow{P}=\epsilon_{0}(\overrightarrow{E} \times \overrightarrow{B})$
|
This is the equation of "
Momentum of electromagnetic wave"
