lys04
- 144
- 5
- Homework Statement
- ^^As title suggests
- Relevant Equations
- Lorentz transformation matrix, current 4-vector
A charge distribution stationary in its own frame S’ has a static charge density ##\rho ’##, a total charge of 0 and a net electric dipole ##\vec{p'}##. An observer in frame S sees the charge distribution moving with constant velocity ##\vec{v}=c \vec{\beta}## What is the magnetic dipole moment in S?
Split the electric dipole moment into components perpendicular and parallel ##\vec{\beta}##, i.e ##\vec{p'}_\parallel## and ##\vec{p'}_\perp##
Then the current density in frame S is given by ##\vec{j} = c \gamma \rho ' \vec{\beta}##
Now using the definition of magnetic dipole moment: ##\vec{m} = \frac{1}{2} \int_V \vec{x} \times \vec{j} dV##
Split $\vec{x}$ into components perpendicular and parallel to ##\vec{\beta}##, i.e ##\vec{x}_\parallel## and ##\vec{x}_\perp##
##\vec{m} = \frac{c \gamma}{2} \int_V ( \vec{x}_\parallel + \vec{x}_\perp) \times (\rho ' \vec{\beta}) dV##
##\vec{m} = \frac{c \gamma}{2} \int_V \rho ' ( \vec{x}_\parallel + \vec{x}_\perp) \times \vec{\beta} dV##
##\vec{m} = \frac{c \gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{\beta} dV##
##\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{v} dV##
##\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp dV \times \vec{v} ##
##\implies \vec{m} = \frac{\gamma}{2} \vec{p'}_\perp \times \vec{v}##
Is this correct?
Split the electric dipole moment into components perpendicular and parallel ##\vec{\beta}##, i.e ##\vec{p'}_\parallel## and ##\vec{p'}_\perp##
Then the current density in frame S is given by ##\vec{j} = c \gamma \rho ' \vec{\beta}##
Now using the definition of magnetic dipole moment: ##\vec{m} = \frac{1}{2} \int_V \vec{x} \times \vec{j} dV##
Split $\vec{x}$ into components perpendicular and parallel to ##\vec{\beta}##, i.e ##\vec{x}_\parallel## and ##\vec{x}_\perp##
##\vec{m} = \frac{c \gamma}{2} \int_V ( \vec{x}_\parallel + \vec{x}_\perp) \times (\rho ' \vec{\beta}) dV##
##\vec{m} = \frac{c \gamma}{2} \int_V \rho ' ( \vec{x}_\parallel + \vec{x}_\perp) \times \vec{\beta} dV##
##\vec{m} = \frac{c \gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{\beta} dV##
##\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{v} dV##
##\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp dV \times \vec{v} ##
##\implies \vec{m} = \frac{\gamma}{2} \vec{p'}_\perp \times \vec{v}##
Is this correct?