diff --git a/book/02_maxwell_eq.md b/book/02_maxwell_eq.md index 1822e79..062061f 100644 --- a/book/02_maxwell_eq.md +++ b/book/02_maxwell_eq.md @@ -1,23 +1,43 @@ # Maxwell's Equations +## Definitions + +Variable | Name | Units +---|---|--- +$\mathbf{E}$ | Electric field strength | $V/m$ +$\mathbf{D}$ | Electric flux density | $C/m^2$ +$\mathbf{H}$ | Magnetic field strength | $A/m$ +$\mathbf{B}$ | Magnetic flux density | $T$ +$\mathbf{J}$ | Current density | $A/m^2$ +$\rho$ | Charge density | $C/m^3$ +$Q$ | Charge | $C$ +$I$ | Current | $A$ +$V$ | Voltage | $V$ + +## Maxwell's Equations + +$\mathbf{D}$ and $\mathbf{B}$ are auxilary fields which capture the impact of atomic dipole moments, magnetization, and more on a macroscopic scale. + $$ \begin{align} -D &= \epsilon E \\ -B &= \mu H +\mathbf{D} &= \epsilon \mathbf{E} \\ +\mathbf{B} &= \mu \mathbf{H} \end{align} $$ -Where $\epsilon$ is the electric permittivity and $\mu$ is the magnetic permeability. In prose I will sometimes refer to both $D$ and $E$ as the electic field and $B$ and $H$ as the magnetic field so don't get tripped up on that. Technically $D$ is the electric displacement field and $B$ is magnetic flux density. +Some texts will keep everything in $\mathbf{E}$ and $\mathbf{H}$ microscopic fields by adding additional polarization ($\mathbf{P}$) and magnetization ($\mathbf{M}$) fields. There may be some benefits to such a representation when looking at semiconductor physics or material science but that is beyond my area of expertise and I will be using the macro fields. + +Note that in prose I will sometimes refer to both $\mathbf{D}$ and $\mathbf{E}$ as the electic field and $\mathbf{B}$ and $\mathbf{H}$ as the magnetic field so don't get tripped up on that. There are two ways to look at Maxwell's Equations: derivative form and integral form. Both can be useful however I think the derivative form tends to be a more concise way to look at things so we'll start with that. Don't worry about understanding all of these yet, we'll go through them one at a time. $$ \begin{align} -\nabla \cdot D &= \rho \\ -\nabla \cdot B &= 0 \\ -\nabla \times E &= - \frac{\delta B}{\delta t} \\ -\nabla \times H &= J + \frac{\delta D}{\delta t} +\nabla \cdot \mathbf{D} &= \rho \\ +\nabla \cdot \mathbf{B} &= 0 \\ +\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\ +\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t} \end{align} $$ @@ -25,14 +45,14 @@ While I like the derivative forms for conceptual understanding, often it makes m $$ \begin{align} -\oiint D \cdot dS &= \iiint \rho dV \\ -\oiint B \cdot dS &= 0 \\ -\oint E \cdot d\ell &= - \frac{d}{dt}\iint B \cdot dS \\ -\oint H \cdot d\ell &= \iint J \cdot dS + \frac{d}{dt} \iint D \cdot dS +\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\ +\oiint \mathbf{B} \cdot d\mathbf{s} &= 0 \\ +\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s} \\ +\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s} \end{align} $$ -$dS$ is directed outward. +$dS$ is directed outward for closed surfaces and uses the right hand rule when looped by a closed path. That's all there is to it. Isn't everything clear now? We'll break down each of these equations to understand their implications. @@ -41,20 +61,20 @@ We'll break down each of these equations to understand their implications. $$ \begin{align} -\nabla \cdot D &= \rho \\ -\oiint D \cdot dS &= \iiint \rho dV \\ +\nabla \cdot \mathbf{D} &= \rho \\ +\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\ &= Q \end{align} $$ -In plain terms, this says that charge sources electric field. +In plain terms, this says that charge sources the electric field. ## Gauss's Law for Magnetism $$ \begin{align} -\nabla \cdot B &= 0 \\ -\oiint B \cdot dS &= 0 +\nabla \cdot \mathbf{B} &= 0 \\ +\oiint \mathbf{B} \cdot d\mathbf{s} &= 0 \end{align} $$ @@ -64,8 +84,8 @@ The divergence of the magnetic field is zero. This means that magnetic monopoles $$ \begin{align} -\nabla \times E &= - \frac{\delta B}{\delta t} \\ -\oint E \cdot d\ell &= - \frac{d}{dt}\iint B \cdot dS +\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\ +\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s} \end{align} $$ @@ -75,12 +95,13 @@ This means that a changing magnetic field will cause circulation in the electric $$ \begin{align} -\nabla \times H &= J + \frac{\delta D}{\delta t} \\ -\oint H \cdot d\ell &= \iint J \cdot dS + \frac{d}{dt} \iint D \cdot dS +\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t} \\ +\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s} \\ +&= I + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s} \end{align} $$ -There are two parts to this which are sometimes introduced individually since for slowly time varying fields $\delta D / \delta t$ is negligible. +There are two parts to this which are sometimes introduced individually since for slowly time varying fields $\delta \mathbf{D} / \delta t$ is negligible. The first part means that magnetic field will always circulate electric current. @@ -91,8 +112,8 @@ Near DC, a the $d/dt$ terms of Ampere's Law and Faraday's Law go to zero giving $$ \begin{align} -\nabla \times E = 0 \\ -\nabla \times H = J +\nabla \times \mathbf{E} = \mathbf{0} \\ +\nabla \times \mathbf{H} = \mathbf{J} \end{align} $$ @@ -102,7 +123,7 @@ Note that voltage is defined as: $$ \begin{align} -V = \int_p E \cdot d\ell +V = \int_p \mathbf{E} \cdot d\mathbf{\ell} \end{align} $$ @@ -111,4 +132,48 @@ At higher frequencies we have to worry about the path of integration. This is on ## Magnetic Hysteresis -## \ No newline at end of file +## Free Space Propagation + +In free space we can assume $J = 0$ since there are no charge carriers. +Similarly we can assume $\rho = 0$. While this is true in vacuum, it may not be valid in air, however superposition lets us ignore any constant fields when looking at propagation. + +This leaves us with a simplified set of equations + +$$ +\begin{align} +\nabla \cdot \mathbf{E} &= 0 \\ +\nabla \cdot \mathbf{H} &= 0 \\ +\nabla \times \mathbf{E} &= - \mu_0 \frac{\delta \mathbf{H}}{\delta t} \\ +\nabla \times \mathbf{H} &= \epsilon \frac{\delta \mathbf{E}}{\delta t} +\end{align} +$$ + +Note that $\epsilon$ has not been simplified to $\epsilon_0$ so this is valid in a dielectric as well as in true free space. + +We will need a fun identity to make this work + +$$ +\begin{align} +\nabla \cdot \left( A \times B \right) &= B \cdot \left( \nabla \times A \right) - A \cdot \left( \nabla \times B \right) +\end{align} +$$ + +Additionally, the following identity will come in handy +$$ +\begin{align} +\frac{d}{dt}(H \cdot H) &= \frac{d}{dt}(H^2) \\ +&= 2 H \frac{dH}{dt} \\ +\frac{d}{dt}(E \cdot E) &= 2 E \frac{dE}{dt} +\end{align} +$$ + +$$ +\begin{align} +\nabla \times E &= - \mu_0 \frac{\delta H}{\delta t} \\ +H \left( \nabla \times E \right) &= - \mu_0 H \cdot \frac{\delta H}{\delta t} \\ +\nabla \cdot (E \times H) + E \cdot (\nabla \times H) &= -\mu_0 H \cdot \frac{dH}{dt} \\ +\nabla \cdot (E \times H) + E \cdot \left(\epsilon \frac{\delta E}{\delta t}\right) &= -\mu_0 H \cdot \frac{dH}{dt} +\end{align} +$$ + +TODO: finish this part