- #1
Adesh
- 735
- 191
Imagine that we have an electromagnetic wave or light propagating in [itex]x[/itex] direction, and [itex]\mathbf{E}[/itex] is oscillating in [itex]z[/itex] direction and [itex] \mathbf{B}[/itex] in [itex]y[/itex] direction. The picture looks something like this
Now, if there exists a charged particle [itex] q [/itex] on the xx axis at rest, then our B field can't do anything, but the electric field E will pull it upwards and as soon as it starts moving the magnetic field B comes into action and it goes like this $$ \mathbf{F_{mag}} = q (\mathbf{v} \times \mathbf{B}) $$ Since, E is going to cause a velocity in [itex] z [/itex] direction i.e. perpendicular to magnetic field B , therefore $$ F_{mag} = q~v~B$$
$$ F_{mag} = q~v~\frac{E}{c}$$
$$F_{mag} = v~\frac{(qE)}{c}$$
$$ F_{mag} = \frac{F_{electric}~v}{c}$$
$$ F_{mag} = \frac{dW_{electric}/dt}{c}$$ and after doing this he writes
That light carries energy we already know. We now understand that it also carries momentum, and further, that the momentum carried is always 1/c times the energy
After considering this line, I thought the last equation could be written as $$ \frac{dp}{dt} = \frac{1}{c} ~ \frac{dW}{dt}$$
$$ dp = \frac{dW}{c} $$
$$ p= \frac{E}{c}$$
But I find this thing sloppy because when I wrote [itex]F_{mag} = \frac{dp}{dt}[/itex] it was for the particle, that is force applied on the particle was equated with the rate of change of particle's momentum and not the light's momentum and similarly [itex]dW[/itex] is the energy imparted to the particle and not the energy of light itself.
So, these are my problems. I earnestly request you to please help me over here.
Now, if there exists a charged particle [itex] q [/itex] on the xx axis at rest, then our B field can't do anything, but the electric field E will pull it upwards and as soon as it starts moving the magnetic field B comes into action and it goes like this $$ \mathbf{F_{mag}} = q (\mathbf{v} \times \mathbf{B}) $$ Since, E is going to cause a velocity in [itex] z [/itex] direction i.e. perpendicular to magnetic field B , therefore $$ F_{mag} = q~v~B$$
$$ F_{mag} = q~v~\frac{E}{c}$$
$$F_{mag} = v~\frac{(qE)}{c}$$
$$ F_{mag} = \frac{F_{electric}~v}{c}$$
$$ F_{mag} = \frac{dW_{electric}/dt}{c}$$ and after doing this he writes
That light carries energy we already know. We now understand that it also carries momentum, and further, that the momentum carried is always 1/c times the energy
After considering this line, I thought the last equation could be written as $$ \frac{dp}{dt} = \frac{1}{c} ~ \frac{dW}{dt}$$
$$ dp = \frac{dW}{c} $$
$$ p= \frac{E}{c}$$
But I find this thing sloppy because when I wrote [itex]F_{mag} = \frac{dp}{dt}[/itex] it was for the particle, that is force applied on the particle was equated with the rate of change of particle's momentum and not the light's momentum and similarly [itex]dW[/itex] is the energy imparted to the particle and not the energy of light itself.
So, these are my problems. I earnestly request you to please help me over here.
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