# 3.3 Electromagnetism in three dimensions

A First Course in String Theory

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(a) Find the reduced Maxwell equations in three dimensions by starting with Maxwell’s equations and the force law in four dimensions, using the ansatz (3.11), and assuming that no field can depend on the $z$ direction.

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\displaystyle{\begin{aligned} \nabla \cdot \mathbf {E} &= \rho \\ \nabla \cdot \mathbf {B} &= 0 \\ \nabla \times \mathbf {E} &= - \frac{1}{c} {\frac {\partial \mathbf {B} }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}

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Eq. (3.11):

\displaystyle{\begin{aligned} E_z &= 0 \\ B_x &= 0 \\ B_y &= 0 \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ \frac{\partial B_z}{\partial z} &= 0 \\ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= - \frac{1}{c} {\frac {\partial B_z }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{\partial B_z }{\partial y} - \frac{\partial B_y }{\partial z} &= \frac{1}{c} j_x + \frac{1}{c} {\frac {\partial E_x }{\partial t}} \\ \frac{\partial B_x }{\partial z} - \frac{\partial B_z }{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} {\frac {\partial E_y }{\partial t}} \\ \frac{\partial B_y }{\partial x} - \frac{\partial B_x }{\partial y} &= \frac{1}{c} j_z + \frac{1}{c} {\frac {\partial E_z }{\partial t}} \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{\partial B_z }{\partial y} &= \frac{1}{c} j_x + \frac{1}{c} {\frac {\partial E_x }{\partial t}} \\ - \frac{\partial B_z }{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} {\frac {\partial E_y }{\partial t}} \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{d \vec p}{dt} &= q \left( \vec E + \frac{\vec v}{c} \times \vec B \right) \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} (v_y B_z - v_z B_y) \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} (v_x B_z - v_z B_x) \right) \\ \frac{d p_z}{dt} &= q \left( E_z + \frac{1}{c} (v_x B_y - v_y B_x) \right) \\ \end{aligned}}

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\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} v_y B_z \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} v_x B_z \right) \\ \frac{d p_z}{dt} &= 0 \\ \end{aligned}}

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— Me@2022-10-22 04:17:10 PM

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