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Migrate to KaTeX

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Indigo5684
2025-09-30 13:19:34 -05:00
parent e3b95ba3b3
commit 6ed3c08d70
27 changed files with 369 additions and 391 deletions
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2
.vscode/settings.json vendored
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@@ -8,11 +8,13 @@
"indeterminates",
"infimum",
"infty",
"Injective",
"macrostate",
"macrostates",
"Magnetostatic",
"Magnetostatics",
"mathbb",
"mathbf",
"mathcal",
"monic",
"Noetherian",

10
docs/javascripts/katex.js Normal file
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@@ -0,0 +1,10 @@
document$.subscribe(({ body }) => {
renderMathInElement(body, {
delimiters: [
{ left: "$$", right: "$$", display: true },
{ left: "$", right: "$", display: false },
{ left: "\\(", right: "\\)", display: false },
{ left: "\\[", right: "\\]", display: true }
],
})
})

26
docs/javascripts/mathjax.js
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@@ -1,26 +0,0 @@
window.MathJax = {
loader: {load: ['[tex]/physics', '[tex]/gensymb']},
tex: {
inlineMath: [["\\(", "\\)"], ["\\$", "\\$"]],
displayMath: [["\\[", "\\]"]],
processEscapes: true,
processEnvironments: true,
packages: {'[+]': ['physics', 'gensymb']},
macros: {
"Hom": ["\\operatorname\{Hom\}"],
"Ann": ["\\operatorname\{Ann\}"],
"Tor": ["\\operatorname\{Tor\}"]
}
},
options: {
ignoreHtmlClass: ".*|",
processHtmlClass: "arithmatex"
}
};
document$.subscribe(() => {
MathJax.startup.output.clearCache()
MathJax.typesetClear()
MathJax.texReset()
MathJax.typesetPromise()
})

4
docs/math/abstract-algebra/16-rings.md
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@@ -155,13 +155,13 @@ $$
\end{align}
$$
**Theorem**. Chineese Remainer Theorem. Let $n_1, \ldots, n_k \in \mathbb{N}$ be given such that $\gcd(n_i, n_j) = 1$. Then, for any integers $a_1, \ldots, a_k$, the system
**Theorem**. Chinese Remainder Theorem. Let $n_1, \ldots, n_k \in \mathbb{N}$ be given such that $\gcd(n_i, n_j) = 1$. Then, for any integers $a_1, \ldots, a_k$, the system
$$
\begin{align}
x &\equiv a_1 \pmod{n_1} \\
x &\equiv a_2 \pmod{n_2} \\
\vdots
\ldots
x &\equiv a_k \pmod{n_k}
\end{align}
$$

2
docs/math/abstract-algebra/18-integral-domains.md
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@@ -53,7 +53,7 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
**Lemma**. 18.14: Let $D$ be a PID. Let $I_1 \subseteq I_2 \subseteq \ldots$. Then, there exists some integer $N$ such that $I_n = I_N$ for all $n > N$. That is, any chain of ideals converges.
**Definition**. Any commutative ring that satisfies the above condition (the *ascending chain condition*), even if it's not a PID, is called a *Noetherien ring*.
**Definition**. Any commutative ring that satisfies the above condition (the *ascending chain condition*), even if it's not a PID, is called a *Noetherian ring*.
**Theorem**. 18.15: Every PID is a UFD. Note that the converse is not true.

18
docs/math/abstract-algebra/DF-10-modules.md
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@@ -5,7 +5,7 @@
**Definition**. Let $R$ be a ring. A *left $R$-module* or a *left module over $R$* is a nonempty set $M$ together with
1. A binary operation $+$ on $M$ under which $M$ is an abelian group
2. An action $\cross$ of $R$ on $M$, that is, a map or function $R \cross M \rightarrow M$, denoted $rm$, that for all $r, s \in R, m, n \in M$ satisfies
2. An action $\times$ of $R$ on $M$, that is, a map or function $R \times M \rightarrow M$, denoted $rm$, that for all $r, s \in R, m, n \in M$ satisfies
- $(r + s)m = rm + sm$
- $(rs)m = r(sm)$
- $r(m + n) = rm + rn$
@@ -21,12 +21,12 @@
---
**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformation from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \cross V \rightarrow V$.
**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformation from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \times V \rightarrow V$.
Now, for a given linear transformation $T$, consider some polynomial $p(x) = a_n x^n + \ldots + a_0$ and some $v \in V$. We define $p(x) \cross v$ by$
Now, for a given linear transformation $T$, consider some polynomial $p(x) = a_n x^n + \ldots + a_0$ and some $v \in V$. We define $p(x) \times v$ by$
$$
p(x) \cross v = a_n T^n(v) + a_{n-1} T^{n-1}(v) + \ldots + a_0 v
p(x) \times v = a_n T^n(v) + a_{n-1} T^{n-1}(v) + \ldots + a_0 v
$$
with $T^n$ being defined as applying $T$ a total of $n$ times.
@@ -50,20 +50,20 @@ with $T^n$ being defined as applying $T$ a total of $n$ times.
**Theorem**. An $R$-module homomorphism is an *isomorphism* if it is 1-1 and onto, and said modules are *isomorphic*.
**Definition**. Let $M, N$ be $R$-modules. The set $\Hom_R(M, N)$ is the set of all homomorphisms from $M$ to $N$.
**Definition**. Let $M, N$ be $R$-modules. The set $\text{Hom}_R(M, N)$ is the set of all homomorphisms from $M$ to $N$.
**Proposition**. Let $M$, $N$, and $L$ be $R$-modules. Then,
1. A function $\varphi: M \rightarrow N$ is an $R$-module homomorphism if and only if $\varphi(rx + y) = r\varphi(x) + \varphi(y)$ for all $x, y \in M$ and $r \in R$.
2. Let $\varphi, \psi \in \Hom_R(M, N)$. Then, define $\varphi + \psi$ as
2. Let $\varphi, \psi \in \text{Hom}_R(M, N)$. Then, define $\varphi + \psi$ as
$$
(\varphi + \psi)(m) = \varphi(m) + \psi(m)
$$
Then, $\varphi + \psi \in \Hom_R(M, N)$. Additionally, if $R$ is commutative, with $(r\varphi)(m) = r(\varphi(m))$, then $r\varphi \in \Hom_R(M,N)$
3. If $\varphi \in \Hom_R(L, M)$ and $\psi \in \Hom_R(M, N)$, then $\psi \circ \varphi \in \Hom_R(L, N)$
4. $\Hom_R(M, M)$ is a ring with identity. With $R$ being commutative, $\Hom_R(M, M)$ is an $R$-algebra.
Then, $\varphi + \psi \in \text{Hom}_R(M, N)$. Additionally, if $R$ is commutative, with $(r\varphi)(m) = r(\varphi(m))$, then $r\varphi \in \text{Hom}_R(M,N)$
3. If $\varphi \in \text{Hom}_R(L, M)$ and $\psi \in \text{Hom}_R(M, N)$, then $\psi \circ \varphi \in \text{Hom}_R(L, N)$
4. $\text{Hom}_R(M, M)$ is a ring with identity. With $R$ being commutative, $\text{Hom}_R(M, M)$ is an $R$-algebra.
**Proposition**. Let $R$ be a ring, $M$ an $R$-module, and $N \subseteq M$ an $R$-submodule. then, $M/N$ can be made into an $R$-module by defining addition. With $r \in R$ and $x + N \in M/N$,

14
docs/math/abstract-algebra/DF-12-modules-pids.md
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@@ -25,22 +25,22 @@ there exists some $k \in \mathbb{N}$ such that given any $n \in \mathbb{N}$ with
**Definition**. Given $R$ an integral domain and $M$ an $R$-module,
$$
\Tor(M) = \{ x \in M | rx = 0 \text{ for any } r \neq 0 \}
\text{Tor}(M) = \{ x \in M | rx = 0 \text{ for any } r \neq 0 \}
$$
This is the *torsion submodule* of $M$. If $\Tor(M)$ is empty, then $M$ is *torsion-free*.
This is the *torsion submodule* of $M$. If $\text{Tor}(M)$ is empty, then $M$ is *torsion-free*.
**Definition** Let $R$ be an integral domain and $M$ be an $R$-module. Then, given a submodule $N$,
$$
\Ann_R(N) = \{r \in R | rn = 0 \text{ for all } n \in N \}
\text{Ann}_R(N) = \{r \in R | rn = 0 \text{ for all } n \in N \}
$$
This ideal of $R$ is the *annihilator of $N$*. That is, $\Ann(N)$ is the set of elements of $R$ such that $(r)N = \{ 0 \}$.
This ideal of $R$ is the *annihilator of $N$*. That is, $\text{Ann}(N)$ is the set of elements of $R$ such that $(r)N = \{ 0 \}$.
Note that if $N$ is not a torsion submodule of $M$, then $\Ann(N) = (0)R$. Additionally, given $N, L$ are submodules of $M$ with $N \subseteq L$, then $\Ann(N) \subseteq \Ann(L)$.
Note that if $N$ is not a torsion submodule of $M$, then $\text{Ann}(N) = (0)R$. Additionally, given $N, L$ are submodules of $M$ with $N \subseteq L$, then $\text{Ann}(N) \subseteq \text{Ann}(L)$.
Additionally, if $R$ is a PID, as $\Ann_R(N)$ is an ideal, $\Ann(N) = (n)R$ and $\Ann(L) = (l)R$ for some $n, l \in R$ such that $n | l$.
Additionally, if $R$ is a PID, as $\text{Ann}_R(N)$ is an ideal, $\text{Ann}(N) = (n)R$ and $\text{Ann}(L) = (l)R$ for some $n, l \in R$ such that $n | l$.
**Definition**. Given any integral domain $R$, the *rank* of an $R$-module $M$ is the maximum number of $R$-linearly independent elements of M.
@@ -64,7 +64,7 @@ $$
- Note that
$$
\Tor{M} \cong \frac{R}{(a_1)R} \oplus \frac{R}{(a_2)R} \oplus \ldots \oplus \frac{R}{(a_m)R}
\text{Tor}{M} \cong \frac{R}{(a_1)R} \oplus \frac{R}{(a_2)R} \oplus \ldots \oplus \frac{R}{(a_m)R}
$$
As a consequence, $M$ is a torsion module if and only if $r = 0$.

4
docs/math/abstract-algebra/DF-13-fields.md
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@@ -1,8 +1,8 @@
# Dummit & Foote Chapter 12 - Field Theory
# Dummit & Foote Chapter 13 - Field Theory
## Section 13.1 Basic Theory of Field Extensions
**Definition**. The *charactaristic* of a field $F$ is the smallest positive integer $p$ such that $1_F * p = 0$. It follows that $p$ is $0$ or prime, and $p \alpha = 0$ for any $\alpha \in F$.
**Definition**. The *characteristic* of a field $F$ is the smallest positive integer $p$ such that $1_F * p = 0$. It follows that $p$ is $0$ or prime, and $p \alpha = 0$ for any $\alpha \in F$.
**Definition**. If $K, F$ are fields such that $F \subseteq K$, then $K$ is an *extension field* or *extension* of $F$, denoted $K / F$.

2
docs/math/diffeq/2-1st-order.md
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@@ -103,7 +103,7 @@ $$
Since $y$ is really $y(x)$, we can make the following substitution:
$$
u = y(x) \text{ and } du = y'(x)dx = \frac{dy}{dx}{dx}
u = y(x) \text{ and } du = y'(x)dx = \frac{dy}{dx} dx
$$
This reduces the integral to the following:

20
docs/math/real-analysis/11-metric-spaces.md
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@@ -24,19 +24,19 @@ $$
---
**Definition**. Let $(S, d)$ be a metric space. Then, for each $\epsilon > 0$, the *$\epsilon$-neighborhood* or *$\epsilon$-ball* of a point $a \in S$ is the set
**Definition**. Let $(S, d)$ be a metric space. Then, for each $\varepsilon > 0$, the *$\varepsilon$-neighborhood* or *$\varepsilon$-ball* of a point $a \in S$ is the set
$$
V_\epsilon(a) = {x \in S | d(a, x) < \epsilon}
V_\varepsilon(a) = {x \in S | d(a, x) < \varepsilon}
$$
**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *open* if for each $x \in G$, there exists some $\epsilon > 0$ so that $V_\epsilon(x) \subseteq G$.
**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *open* if for each $x \in G$, there exists some $\varepsilon > 0$ so that $V_\varepsilon(x) \subseteq G$.
**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *closed* if its complement $C(G) = S - G = S \ F$ is closed.
**Definition**. Let $(S, d)$ be a metric space. A point $c \in S$ is a *cluster point$ of a set $A \subseteq S$ if every $\epsilon$-neighborhood of $c$ contains some point $a \in A$ such that $a \neq c$.
**Definition**. Let $(S, d)$ be a metric space. A point $c \in S$ is a *cluster point$ of a set $A \subseteq S$ if every $\varepsilon$-neighborhood of $c$ contains some point $a \in A$ such that $a \neq c$.
**Theorem**. Every $\epsilon$-neighborhood of a point is an open set.
**Theorem**. Every $\varepsilon$-neighborhood of a point is an open set.
**Theorem**. The union of an arbitrary collection of open sets is open.
@@ -50,22 +50,22 @@ $$
---
**Definition**. A *sequence* $(x_n)$ in a metric space $(S, d)$ converges to a point $x \in S$ if given any $\epsilon > 0$, there exists a $K \in \mathbb{N}$ such that given $n \in \mathbb{N}$,
**Definition**. A *sequence* $(x_n)$ in a metric space $(S, d)$ converges to a point $x \in S$ if given any $\varepsilon > 0$, there exists a $K \in \mathbb{N}$ such that given $n \in \mathbb{N}$,
$$
n \geq K \Rightarrow d(x_n, x) \leq \epsilon
n \geq K \Rightarrow d(x_n, x) \leq \varepsilon
$$
**Theorem**. Let $(x_n)$ be a sequence in metric space $(S, d)$. Then,
- $(x_n)$ converges to $x$ if and only if every $\epsilon$-neighborhood of $x$ contains all but finitely many terms of $(x_n)$.
- $(x_n)$ converges to $x$ if and only if every $\varepsilon$-neighborhood of $x$ contains all but finitely many terms of $(x_n)$.
- If $(x_n) \rightarrow x$ and $(x_n) \rightarrow x'$, then $x = x'$.
- If $(x_n)$ converges, then $(x_n)$ is bound.
**Definition**. A sequence $(x_n)$ in metric space $(S, d)$ is a *Cauchy sequence* if for every $\epsilon > 0$, there exists some $H \in \mathbb{N}$ such that for any $m, n \in \mathbb{N}$,
**Definition**. A sequence $(x_n)$ in metric space $(S, d)$ is a *Cauchy sequence* if for every $\varepsilon > 0$, there exists some $H \in \mathbb{N}$ such that for any $m, n \in \mathbb{N}$,
$$
m, n \geq H \Rightarrow d(x_n, x_m) < \epsilon
m, n \geq H \Rightarrow d(x_n, x_m) < \varepsilon
$$
**Definition**. A metric space in which every Cauchy sequence converges is said to be *complete*.

20
docs/math/real-analysis/2-reals.md
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@@ -56,32 +56,32 @@
2. $|ab| = |a||b|$
3. $|a + b| \leq |a| + |b|$
**Corollary**. Given $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
**Corollary**. Given $a, b \in \mathbb{R}$, then $||a| - |b|| \leq |a - b|$.
**Remark**. Every field has at least one absolute value function.
**Theorem**. In an ordered field $F$, for any $r > 0$, we know that
1. $\abs{x = r}$ if and only if $x = r$ or $x = -r$
2. $\abs{x < r}$ if and only if $-r < x < r$
3. $\abs{x > r}$ if either $x > r$ or $x < -r$
1. $|x = r$ if and only if $x = r$ or $x = -r$
2. $|x < r$ if and only if $-r < x < r$
3. $|x > r$ if either $x > r$ or $x < -r$
---
**Definition**. The *standard distance function* or *metric* on the real numbers $\mathbb{R}$ given $a, b$ is $\abs{a - b}$.
**Definition**. The *standard distance function* or *metric* on the real numbers $\mathbb{R}$ given $a, b$ is $|a - b|$.
**Theorem**. For any real numbers $a, b, c$,
1. $\abs{a - b} > 0$ if and only if $a \neq b$ and $\abs{a - b} = 0$ if and only if $a = b$
2. $\abs{a - b} = \abs{b - a}$
3. $\abs{a - c} \leq \abs{a - b} + \abs{b + c}$
1. $|a - b| > 0$ if and only if $a \neq b$ and $|a - b| = 0$ if and only if $a = b$
2. $|a - b| = |b - a|$
3. $|a - c| \leq |a - b| + |b + c|$
**Definition** A set together with a function satisfying these three properties is known as a *metric space*.
**Definition** The $\epsilon$-neighborhood of $a \in \mathbb{R}$, denoted $V_\epsilon(a)$ is the set of all real numbers $x \in \mathbb{R}$ such that $\abs{x - a} < \epsilon$. That is,
**Definition** The $\varepsilon$-neighborhood of $a \in \mathbb{R}$, denoted $V_\varepsilon(a)$ is the set of all real numbers $x \in \mathbb{R}$ such that $|x - a| < \varepsilon$. That is,
$$
V_\epsilon(a) = (a - \epsilon, a + \epsilon)
V_\varepsilon(a) = (a - \varepsilon, a + \varepsilon)
$$
---

22
docs/math/real-analysis/3-sequences-series.md
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@@ -26,10 +26,10 @@ $$
---
**Definition**. A sequence $X = (x_n)$ is said to *converge* to a number $x \in \mathbb{R}$ if when given any $\epsilon > 0$, there exists some $K \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ with $n \geq K$,
**Definition**. A sequence $X = (x_n)$ is said to *converge* to a number $x \in \mathbb{R}$ if when given any $\varepsilon > 0$, there exists some $K \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ with $n \geq K$,
$$
\abs{x_n - x} < \epsilon
|x_n - x| < \varepsilon
$$
If this is the case, we say that $X$ converges to $x$, and $x$ is a *limit* of X. This can be written as
@@ -66,9 +66,9 @@ $$
**Theorem**. Suppose $(x_n)$ is a sequence if real numbers. Then,
1. If $x_n \rightarrow x$, then $\abs{x_n} \rightarrow \abs{x}$
2. If $\abs{x_n} \rightarrow 0$, then $x_n \rightarrow 0$
3. $x_n \rightarrow x$ if and only if $\abs{x_n - n} \rightarrow 0$
1. If $x_n \rightarrow x$, then $|x_n| \rightarrow |x|$
2. If $|x_n| \rightarrow 0$, then $x_n \rightarrow 0$
3. $x_n \rightarrow x$ if and only if $|x_n - n| \rightarrow 0$
**Theorem**. Suppose $(x_n)$ is a sequence if real numbers, with each $x_n \geq 0$. Then, given some $k \in \mathbb{N}$, if $x_n \rightarrow x$, then $\sqrt[k]{x_n} \rightarrow \sqrt[k]{x}$.
@@ -100,11 +100,9 @@ is a *subsequence* of $X$,
## Section 3.5 - The Cauchy Criterion
**Definition**. A sequence $(x_n)$ is said to be a *Cauchy sequence* such that for any given $\epsilon$, there exists a natural number $H$ such that all natural numbers $m, n \geq H$, then
**Definition**. A sequence $(x_n)$ is said to be a *Cauchy sequence* such that for any given $\varepsilon$, there exists a natural number $H$ such that all natural numbers $m, n \geq H$, then
$$
\abs{x_m - x_n} \leq \epsilon
$$
$$|x_m - x_n \leq \varepsilon$$
**Theorem**. If $(x_n)$ is a Cauchy sequence, then $(x_n)$ is convergent.
@@ -130,15 +128,15 @@ $$
In particular, $s_n - s_{n - 1} = x^n$. Thus, the Cauchy criteria takes the form
**Theorem**. Cauchy Criteria for Series. The series $\sum x_n$ converges if and only if, for a given $\epsilon$, there exists some natural number $H$ such that when $m > n > H$,
**Theorem**. Cauchy Criteria for Series. The series $\sum x_n$ converges if and only if, for a given $\varepsilon$, there exists some natural number $H$ such that when $m > n > H$,
$$
\abs{s_m - s_n} = \abs{\sum_{i = m + 1}^n x_i} < \epsilon
|s_m - s_n| = |\sum_{i = m + 1}^n x_i| < \varepsilon
$$
**Corollary**. $n$-th Term Test. If $\sum x_n$ converges, then $x_n \rightarrow 0$.
**Corollary**. Absolute Convergence Test. If $\sum \abs{x_n}$ converges, then $\sum x_n$ converges.
**Corollary**. Absolute Convergence Test. If $\sum |x_n|$ converges, then $\sum x_n$ converges.
---

10
docs/math/real-analysis/4-limits.md
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@@ -18,10 +18,10 @@
---
**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if given any $\epsilon > 0$, there exists some $\delta > 0$ such that
**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if given any $\varepsilon > 0$, there exists some $\delta > 0$ such that
$$
0 < |x-c| < \delta \Rightarrow |f(x) - L| < \epsilon
0 < |x-c| < \delta \Rightarrow |f(x) - L| < \varepsilon
$$
**Theorem**. For a given function and cluster point, there can be at most one limit at said point.
@@ -32,12 +32,12 @@ $$
**Definition**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set with supremum and infimum. Note that this set is no longer a field.
**Definition**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$, there exists some $V_\delta(c)$ such that for all $x \in V_\epsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
**Definition**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$, there exists some $V_\delta(c)$ such that for all $x \in V_\varepsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
**Definition**. The limit of a function at infinity is defined if for a given $\epsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,
**Definition**. The limit of a function at infinity is defined if for a given $\varepsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,
$$
x > \alpha \Rightarrow |f(x) - L| < \epsilon
x > \alpha \Rightarrow |f(x) - L| < \varepsilon
$$
## Section 4.2 - Limit Theorems

12
docs/math/real-analysis/5-continuity.md
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@@ -2,10 +2,10 @@
## Section 5.1 - Continuous Functions
**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
$$
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon
$$
Note that if $a$ is an *isolated point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
@@ -20,10 +20,10 @@ Note that a function cannot be continuous at a point outside of its domain, even
---
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
$$
d_S(x, a) < \delta \Rightarrow d_T(f(x), f(a)) < \epsilon
d_S(x, a) < \delta \Rightarrow d_T(f(x), f(a)) < \varepsilon
$$
**Theorem**. A function $f: S \rightarrow T$ is continuous at a point $a \in A$ if and only if given some neighborhood $V(f(a)) \in B$, there exists some $U(a) \in A$ such that $f(U) \subseteq V$.
@@ -63,10 +63,10 @@ $$
---
**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\epsilon > 0$, there exists some $\delta > 0$ depending only on $\epsilon$ such that for any $x, y \in A$,
**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\varepsilon > 0$, there exists some $\delta > 0$ depending only on $\varepsilon$ such that for any $x, y \in A$,
$$
|x - y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon
|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon
$$
Note that if $f$ is uniformly continuous, it must be continuous on $A$.

54
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@@ -5,34 +5,34 @@
Consider an empty space. Then, it is evident that
$$\begin{align}
\div \vb{E} &= 0 \\
\div \vb{H} &= 0
\nabla \cdot \mathbf{E} &= 0 \\
\nabla \cdot \mathbf{H} &= 0
\end{align}$$
$$\begin{align}
\curl \vb{E} + \frac{\partial \vb{B}}{\partial t} &= 0 \\
\curl \vb{H} - \frac{\partial \vb{D}}{\partial t} &= 0
\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} &= 0 \\
\nabla \times \mathbf{H} - \frac{\partial \mathbf{D}}{\partial t} &= 0
\end{align}$$
As $\vb{B} = \mu_0 \vb{H}$ and $\vb{D} = \varepsilon_0 \vb{E}$ in a vacuum, the third and fourth equations can be rewritten as
As $\mathbf{B} = \mu_0 \mathbf{H}$ and $\mathbf{D} = \varepsilon_0 \mathbf{E}$ in a vacuum, the third and fourth equations can be rewritten as
$$\begin{align}
\curl \vb{E} + \mu_0 \frac{\partial \vb{H}}{\partial t} &= 0 \\
\curl \vb{H} - \varepsilon_0 \frac{\partial \vb{E}}{\partial t} &= 0
\nabla \times \mathbf{E} + \mu_0 \frac{\partial \mathbf{H}}{\partial t} &= 0 \\
\nabla \times \mathbf{H} - \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} &= 0
\end{align}$$
We can take the curl of both equations and then substitute to see that
$$\begin{align}
\curl \curl \vb{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \\
\curl \curl \vb{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0
\nabla \times \nabla \times \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \\
\nabla \times \nabla \times \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0
\end{align}$$
We can apply a vector identity to see
$$\begin{align}
-\nabla^2 \vb{E} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \\
-\nabla^2 \vb{H} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0
-\nabla^2 \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \\
-\nabla^2 \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0
\end{align}$$
### Section 10.1.1 - The Wave Equation and Plane Waves
@@ -43,34 +43,34 @@ Consider some function $f(s)$. If $s = x - vt$ or $x + vt$, it is trivial to see
**Definition**. A *plane wave* is a solution to the Laplacian form of the last two Maxwell equations for empty space that also satisfy the one-dimensional wave equation. However, these solutions may not be valid electromagnetic waves as they are not guaranteed to satisfy the first two Maxwell equations.
Notably, the functions for $\vb{E} = \vb{E}_0 f(s)$ and $\vb{H} = \vb{H}_0 g(s)$ do not have to be equal. However, $v = c$.
Notably, the functions for $\mathbf{E} = \mathbf{E}_0 f(s)$ and $\mathbf{H} = \mathbf{H}_0 g(s)$ do not have to be equal. However, $v = c$.
**Definition**. A *plane electromagnetic wave* is a plane wave which satisfies the first two Maxwell equations. The divergence equations restrict $\vb{E}_0$ and $\vb{H}_0$ to be in the plane normal to the direction of motion. That is, electomagnetic plane waves are transverse, not longitudinal.
**Definition**. A *plane electromagnetic wave* is a plane wave which satisfies the first two Maxwell equations. The divergence equations restrict $\mathbf{E}_0$ and $\mathbf{H}_0$ to be in the plane normal to the direction of motion. That is, electomagnetic plane waves are transverse, not longitudinal.
Additionally, the curl equations force $f(s) = g(s)$, such that $H_0 = E_0 \sqrt{\frac{\varepsilon_0}{\mu_0}}$.
**Definition**. The quantity $Y_0 = \sqrt{\frac{\varepsilon_0}{\mu_0}}$ is the *vacuum admittance* and its inverse, $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}$ is the *vacuum impedance*.
If we assume the direction of propagation can be written as $\vu{k}$, we can write $f(s) = f(\vu{k} \vdot \vb{r} - vt)$, such that $\vb{E}(\vb{r}, t) = \vb{E}_0 f(\vu{k}\vdot\vb{r} - vt)$, where $\vu{k}\vdot\vb{E}_0 = 0$.
If we assume the direction of propagation can be written as $\hat{\mathbf{k}}$, we can write $f(s) = f(\hat{\mathbf{k}} \cdot \mathbf{r} - vt)$, such that $\mathbf{E}(\mathbf{r}, t) = \mathbf{E}_0 f(\hat{\mathbf{k}}\cdot\mathbf{r} - vt)$, where $\hat{\mathbf{k}}\cdot\mathbf{E}_0 = 0$.
From this, we can see that $\vb{H}(\vb{r}, t) = \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 f(\vu{k} \vdot \vb{r} - vt)$. Similarly, $\vu{k} \vdot \vb{H} = 0$.
From this, we can see that $\mathbf{H}(\mathbf{r}, t) = \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 f(\hat{\mathbf{k}} \cdot \mathbf{r} - vt)$. Similarly, $\hat{\mathbf{k}} \cdot \mathbf{H} = 0$.
Additionally, we can compute $\vb{S} = \vb{E} \times \vb{H} = c \epsilon_0 E_0^2 f^2(\vu{k} \vdot \vb{r} - vt) \vu{k}$. We can also see that $\epsilon_0 E^2 = \mu_0 H^2$ at any given time.
Additionally, we can compute $\mathbf{S} = \mathbf{E} \times \mathbf{H} = c \varepsilon_0 E_0^2 f^2(\hat{\mathbf{k}} \cdot \mathbf{r} - vt) \hat{\mathbf{k}}$. We can also see that $\varepsilon_0 E^2 = \mu_0 H^2$ at any given time.
### Section 10.1.2 - Monochromatic Plane Waves
In any simple material, we like to say that $\vb{D} = \epsilon \vb{E}$ and $\vb{B} = \mu \vb{H}$. However, this only holds true at a fixed frequency $\omega$. For multiple frequencies, we see that $\vb{D}(\omega) = \epsilon{\omega}\vb{E}(\omega)$ and $\vb{B}(\omega) = \mu(\omega)\vb{H}(\omega)$. This causes problems. As such, we will want to consider waves that are only composed of one frequency under Fourier decomposition.
In any simple material, we like to say that $\mathbf{D} = \varepsilon \mathbf{E}$ and $\mathbf{B} = \mu \mathbf{H}$. However, this only holds true at a fixed frequency $\omega$. For multiple frequencies, we see that $\mathbf{D}(\omega) = \varepsilon(\omega)\mathbf{E}(\omega)$ and $\mathbf{B}(\omega) = \mu(\omega)\mathbf{H}(\omega)$. This causes problems. As such, we will want to consider waves that are only composed of one frequency under Fourier decomposition.
**Definition**. A *monochromatic* plane wave is a plane wave in which the full Fourier series of $f(x)$ has only one term. That is, $f(x)$ is $\sin(x)$ or $\cos(x)$. We furthermore define a *wave vector* $\vb{k}$ as $\vb{k} = k \vu{k}$, so that $\omega = kc$. Then,
**Definition**. A *monochromatic* plane wave is a plane wave in which the full Fourier series of $f(x)$ has only one term. That is, $f(x)$ is $\sin(x)$ or $\cos(x)$. We furthermore define a *wave vector* $\mathbf{k}$ as $\mathbf{k} = k \hat{\mathbf{k}}$, so that $\omega = kc$. Then,
$$\begin{align}
\vb{E}(\vb{r}, t) &= \vb{E_0} \cos(\vb{k} \vdot \vb{r} - \omega t) \\
\vb{H}(\vb{r}, t) &= \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 \cos(\vb{k} \vdot \vb{r} - \omega t)
\mathbf{E}(\mathbf{r}, t) &= \mathbf{E_0} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t) \\
\mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)
\end{align}$$
Notably, the frequency, or number of cycles per second, is $f = \frac{\omega}{2\pi}$, and wavelength $\lambda = \frac{2\pi}{k}$.
We can calculate the energy density $u$, energy current density $\vb{S}$, momentum density $\vb{g}$, and momentum current density $-\overleftrightarrow{\vb{T}}$
We can calculate the energy density $u$, energy current density $\mathbf{S}$, momentum density $\mathbf{g}$, and momentum current density $-\overleftrightarrow{\mathbf{T}}$
### Section 10.1.3 - Monochromatic Plane Waves in a Linear Model
@@ -78,13 +78,13 @@ Monochromatic plane waves with frequency $\omega$ in a simple linear material ar
### Section 10.1.4 - Polarization of Monochromatic Plane Waves
Any plane wave described in such a way that $\vb{E} = \vb{E}_0 f(\vb{k} \vdot \vb{r} - ct)$ is linearly polarized in the direction of $\vb{E}_0$. That is, the direction of polarization is the direction of $\vb{E}$, and if that direction is unchanging, the wave is linearly polarized.
Any plane wave described in such a way that $\mathbf{E} = \mathbf{E}_0 f(\mathbf{k} \cdot \mathbf{r} - ct)$ is linearly polarized in the direction of $\mathbf{E}_0$. That is, the direction of polarization is the direction of $\mathbf{E}$, and if that direction is unchanging, the wave is linearly polarized.
Notably, an elliptically polarized wave can be described as follows:
$$\begin{align}
\vb{E}(\vb{r}, t) &= E_{x0} \vu{x} \cos(kz - \omega t) + E_{y0} \vu{y} \sin(kz - \omega t) \\
\vb{H}(\vb{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \vu{y} \cos(kz - \omega t) - E_{y0} \vu{x} \sin(kz - \omega t))
\mathbf{E}(\mathbf{r}, t) &= E_{x0} \hat{\mathbf{x}} \cos(kz - \omega t) + E_{y0} \hat{\mathbf{y}} \sin(kz - \omega t) \\
\mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \hat{\mathbf{y}} \cos(kz - \omega t) - E_{y0} \hat{\mathbf{x}} \sin(kz - \omega t))
\end{align}$$
If $E_{x0} = E_{y0}$, the wave is said to be circularly polarized.
@@ -93,8 +93,8 @@ If $E_{x0} = E_{y0}$, the wave is said to be circularly polarized.
This section will focus on plane monochromatic waves incident from material 1 onto material 2, where both materials are homogenous insulators and the surface between the two materials is smooth (on the scale of the wavelength).
In this case, we must re-consider Maxwell's equations. We know from previous sections that $\div \vb{E} = \frac{\vb{\rho_e}}{\varepsilon_0}$ and $\div \vb{H} = \frac{\vb{\rho_m}}{\mu_0}$. We also know that $\div \vb{D} = \rho_{ef}$ and $\div \vb{B} = \rho_{mf}$.
In this case, we must re-consider Maxwell's equations. We know from previous sections that $\nabla \cdot \mathbf{E} = \frac{\mathbf{\rho_e}}{\varepsilon_0}$ and $\nabla \cdot \mathbf{H} = \frac{\mathbf{\rho_m}}{\mu_0}$. We also know that $\nabla \cdot \mathbf{D} = \rho_{ef}$ and $\nabla \cdot \mathbf{B} = \rho_{mf}$.
Consider the boundary between the two materials. If we consider $\div \vb{D}$, and take the integral over a Gaussian pillbox on the boundary, we can apply divergence theorem to see that $\int_V \div \vb{D} dV = \int_{SofV} D \vdot \vu{n} dS = \rho_{efree}$. If we assume the materials are insulating, we do not expect to find any electrical charge, so $\rho_{efree} = 0$. Thus, we can say that $\int_{SofV} D \vdot \vu{n} = 0$, so $\vb{D}_1 \vdot \vu{n} + \vb{D}_2 \vdot \vu{n} = \vb{D_1} \vdot \vu{z} + \vb{D}_1 \vdot (-\vu{n}) = 0$. Then, we can say that $\vb{D}_1 \vdot{z} = \vb{D}_2 \vdot{z}$, or in simpler terms, $\vdot{D}_1^\perp = \vdot{D}_2^\perp$.
Consider the boundary between the two materials. If we consider $\nabla \cdot \mathbf{D}$, and take the integral over a Gaussian pillbox on the boundary, we can apply divergence theorem to see that $\int_V \nabla \cdot \mathbf{D} dV = \int_{SofV} \mathbf{D} \cdot \hat{\mathbf{n}} dS = \rho_{efree}$. If we assume the materials are insulating, we do not expect to find any electrical charge, so $\rho_{efree} = 0$. Thus, we can say that $\int_{SofV} D \cdot \hat{\mathbf{n}} = 0$, so $\mathbf{D}_1 \cdot \hat{\mathbf{n}} + \mathbf{D}_2 \cdot \hat{\mathbf{n}} = \mathbf{D_1} \cdot \hat{\mathbf{z}} + \mathbf{D}_1 \cdot (-\hat{\mathbf{n}}) = 0$. Then, we can say that $\mathbf{D}_1 \cdot \hat{\mathbf{n}} = \mathbf{D}_2 \cdot \hat{\mathbf{n}}$, or in simpler terms, $\mathbf{D}_1^\perp = \mathbf{D}_2^\perp$.
Applying the same logic to $\vb{B}$, we see that $\vdot{B}_1^\perp = \vdot{B}_2^\perp$. Note that due to the existence of polarization and magnetization, we cannot say the same regarding $\mathbf{E}$ or $\mathbf{H}$.
Applying the same logic to $\mathbf{B}$, we see that $\mathbf{B}_1^\perp = \mathbf{B}_2^\perp$. Note that due to the existence of polarization and magnetization, we cannot say the same regarding $\mathbf{E}$ or $\mathbf{H}$.

54
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@@ -44,11 +44,11 @@ $$\cos \phi = \frac{R}{\sqrt{R^2 + X_C^2}}$$
### Section 8.1.2 - Quasi-Static Error for a Parallel Plate Capacitor
Consider a parallel-plate capacitor. We know that within the capacitor, the electric flux is $\vb{D}(t) = \varepsilon_0 \vb{E}(t)$. With the charge on a plate given by $Q(t)$, we can say that $\vb{D}(t) = \frac{Q(t)}{\pi R^2}$.
Consider a parallel-plate capacitor. We know that within the capacitor, the electric flux is $\mathbf{D}(t) = \varepsilon_0 \mathbf{E}(t)$. With the charge on a plate given by $Q(t)$, we can say that $\mathbf{D}(t) = \frac{Q(t)}{\pi R^2}$.
We also know by Ampere's law that $\curl \vb{H} = \frac{\partial}{\partial t} \vb{D}$. From this, given circular platFrom this, applying Stokes to Ampere's Law, we see that
We also know by Ampere's law that $\nabla \times \mathbf{H} = \frac{\partial}{\partial t} \mathbf{D}$. From this, given circular platFrom this, applying Stokes to Ampere's Law, we see that
$$\vb{H} = \frac{\partial Q}{\partial t} \frac{s}{2 \pi R^2} \vu{\varphi}$$
$$\mathbf{H} = \frac{\partial Q}{\partial t} \frac{s}{2 \pi R^2} \hat{\mathbf{\varphi}}$$
This continues on in this manner, however, I've opted to skip most of the math.
@@ -60,11 +60,11 @@ For an inductor, the voltage drop across an inductor is directly proportional to
$$\Delta V_L = L \frac{d^2Q}{dt^2} = L\frac{dI}{dt}$$
Consider a circular current loop with a voltage source in the $x-y$ plane. Then, applying Faraday's law, $\int_{circle} = (\curl \vb{E}) \cdot \vu{z} dS = -\frac{\partial}{\partial t} \int_{circle} \vb{B} \cdot \vu{z} dS$.
Consider a circular current loop with a voltage source in the $x-y$ plane. Then, applying Faraday's law, $\int_{circle} = (\nabla \times \mathbf{E}) \cdot \hat{\mathbf{z}} dS = -\frac{\partial}{\partial t} \int_{circle} \mathbf{B} \cdot \hat{\mathbf{z}} dS$.
Apply Stokes' law to the left hand side to see that $\int_{circumference} \vb{E} \cdot d\vb{l} = -\frac{\partial \Phi_B}{\partial t}$.
Apply Stokes' law to the left hand side to see that $\int_{circumference} \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial \Phi_B}{\partial t}$.
With $\Phi_B = LI$, we see the "back EMF" opposing the increasing current will be $\vb{\mathcal{E}} = -L \frac{\partial I}{\partial t}$.
With $\Phi_B = LI$, we see the "back EMF" opposing the increasing current will be $\mathbf{\mathcal{E}} = -L \frac{\partial I}{\partial t}$.
Consider a circuit with a voltage source, a switch, an inductor, and a resistor in series. Then, by KVL, $V_{cell} = L \frac{dI}{dt} + IR$.
@@ -84,9 +84,9 @@ $$\sin \phi = -\frac{X_L}{Z}; \cos \phi = \frac{R}{Z}; I_0 = \frac{V_0}{Z}$$
### Section 8.1.4 - Calculation of Inductance
Recall that the back EMF $\vb{\mathcal{E}} = -L \frac{\partial I}{\partial T}$. We can calculate the work done by this force as follows.
Recall that the back EMF $\mathbf{\mathcal{E}} = -L \frac{\partial I}{\partial T}$. We can calculate the work done by this force as follows.
$$\frac{dW}{dt} = I(\vb{\mathcal{E}}) = IL \frac{dI}{dt} = \frac{1}{2} L \frac{d}{dt}(I^2)$$
$$\frac{dW}{dt} = I(\mathbf{\mathcal{E}}) = IL \frac{dI}{dt} = \frac{1}{2} L \frac{d}{dt}(I^2)$$
Then, integrating both sides, we see that
@@ -96,9 +96,9 @@ We also know from Section 3.2 that the work needed to create a magnetic field is
$$W = \frac{1}{2} \mu_0 \int_V H^2 dV$$
Now, consider a long air-filled solenoid with $n$ turns per unit length and cross-sectional area $A$. Then, we know the flux through a cross-section of the solenoid will be $\Phi = BA = \mu_0 n I A$. Then, the back-EMF for one loop of the solenoid can be given by $\vb{\mathcal{E}}_{1 loop} = - \frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$. Then, the total induced EMF will be $\vb{\mathcal{E}} = nl\vb{\mathcal{E}}_{1 loop}$, where $l$ is the length of the solenoid. Thus, by the definition of back-EMF,
Now, consider a long air-filled solenoid with $n$ turns per unit length and cross-sectional area $A$. Then, we know the flux through a cross-section of the solenoid will be $\Phi = BA = \mu_0 n I A$. Then, the back-EMF for one loop of the solenoid can be given by $\mathbf{\mathcal{E}}_{1 loop} = - \frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$. Then, the total induced EMF will be $\mathbf{\mathcal{E}} = nl\mathbf{\mathcal{E}}_{1 loop}$, where $l$ is the length of the solenoid. Thus, by the definition of back-EMF,
$$\vb{\mathcal{E}} = -nl \mu_0 nA \frac{dI}{dt} = -L\frac{dI}{dt}$$
$$\mathbf{\mathcal{E}} = -nl \mu_0 nA \frac{dI}{dt} = -L\frac{dI}{dt}$$
So, $L = \mu_0 n^2 Al$
@@ -106,15 +106,15 @@ We can also compute thsi by energy. We know that $W = \frac{1}{2} \mu_0 \int_V H
Now, consider a coaxial cable. That is, consider a solid cylinder of radius $a$ that conducts current in the $+z$ direction. The circuit is completed by a thin cylindical shell outside of the conductor yet still with radius $a$. We also assume that current density is uniform within the cylindrical conductor.
Recall that from Ampere's Law, for $s \in (0, a)$, we have $\vb{H}(s) = \frac{I_enc}{2\pi s}\vu{\varphi}$, with $I_enc = \frac{I\pi s^2}{\pi a^2}$. Thus, $\vb{H} = \frac{Is}{2\pi a^2}\vu{\varphi}$. Then,
Recall that from Ampere's Law, for $s \in (0, a)$, we have $\mathbf{H}(s) = \frac{I_enc}{2\pi s}\hat{\mathbf{\varphi}}$, with $I_enc = \frac{I\pi s^2}{\pi a^2}$. Thus, $\mathbf{H} = \frac{Is}{2\pi a^2}\hat{\mathbf{\varphi}}$. Then,
$$W = \frac{1}{2} \mu_0 \int_V H^2 dV = \frac{1}{2} \mu_0 l \int_0^a 2\pi s ds (\frac{Is}{2\pi a^2})^2 = \frac{I^2}{2} \frac{\mu_0 l}{2\pi} \frac{1}{4}$$
This implies that $L = \frac{\mu_0 l}{8\pi}$.
We can also solve this via flux. We know that $\vb{B}(s) = \frac{u_0 Is}{2\pi a^2}\vu{\varphi}$. Then,
We can also solve this via flux. We know that $\mathbf{B}(s) = \frac{u_0 Is}{2\pi a^2}\hat{\mathbf{\varphi}}$. Then,
$$\Phi = \int_0^a \vb{B} \vdot \vu{n} dS = \int_0^a \frac{\mu_0 Is}{2\pi a^2} l ds = \frac{\mu_0 I l}{2\pi} \frac{1}{2}$$
$$\Phi = \int_0^a \mathbf{B} \cdot \hat{\mathbf{n}} dS = \int_0^a \frac{\mu_0 Is}{2\pi a^2} l ds = \frac{\mu_0 I l}{2\pi} \frac{1}{2}$$
This is off by a factor of $2$. Instead, multiply by a fator of $f(s) = \frac{s^2}{a^2}$ to see
@@ -126,9 +126,9 @@ This becomes complicated, and is thus omitted.
### Section 8.1.5 - Quasi-static Error for a Solenoidal Inductor
In section 5.2, we learned that in a long solenoid, $\vb{H} = nI \vb{z}$, where $I$ is the current and $n$ the number of turns per uniut length. This, however, was dependent of the current being constant. Now, let current be represented as $I(t) = I_0 \cos (\omega t + \phi)$. Now, $\vb{H}(t) = nI(t) \vu{z}$.
In section 5.2, we learned that in a long solenoid, $\mathbf{H} = nI \mathbf{z}$, where $I$ is the current and $n$ the number of turns per uniut length. This, however, was dependent of the current being constant. Now, let current be represented as $I(t) = I_0 \cos (\omega t + \phi)$. Now, $\mathbf{H}(t) = nI(t) \hat{\mathbf{z}}$.
This implies magnetic flux with a density of $\vb{B} = \mu_0 n I(t) \vu{z}$. Given the inductor with radius $s$, this will cause changing flux $\Phi_B(t) = \vb{B} * A = \mu_0 n I(t) \pi s^2$.
This implies magnetic flux with a density of $\mathbf{B} = \mu_0 n I(t) \hat{\mathbf{z}}$. Given the inductor with radius $s$, this will cause changing flux $\Phi_B(t) = \mathbf{B} * A = \mu_0 n I(t) \pi s^2$.
By Faraday's Law,
@@ -136,13 +136,13 @@ $$E_{induced} = -N \frac{\partial \Phi_B}{\partial{t}}$$
This can be rearranged to see
$$2\pi s \vb{E}_{induced}(t) = -\frac{\partial \Phi_b(t)}{\partial t} = -\mu_0 n \pi s^2 \frac{\partial}{\partial t} I(t)$$
$$2\pi s \mathbf{E}_{induced}(t) = -\frac{\partial \Phi_b(t)}{\partial t} = -\mu_0 n \pi s^2 \frac{\partial}{\partial t} I(t)$$
This can be used to find $\vb{E} = -\mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \vu{\varphi}$, equivalent to a flux density $\vb{D}(t) = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \vu{\varphi}$. We can then apply Ampere's law to see that
This can be used to find $\mathbf{E} = -\mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \hat{\mathbf{\varphi}}$, equivalent to a flux density $\mathbf{D}(t) = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \hat{\mathbf{\varphi}}$. We can then apply Ampere's law to see that
$$\curl \vb{H} = \frac{\partial \vb{D}(t)}{\partial t} = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial^2}{\partial t^2}I(t) \vu{\varphi}$$
$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}(t)}{\partial t} = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial^2}{\partial t^2}I(t) \hat{\mathbf{\varphi}}$$
We can work backwards to find that $\vb{H}(s, t) = (1 - \frac{\mu_0 \varepsilon_0 s^2 \omega^2}{4}) H_{z0}(t)\vu{z}$. Then, the same conditions should apply as in 8.1.2. That is, $\omega$ should be low enough or the d evice small enough that light can easily propagate across the device during one period of oscillation.
We can work backwards to find that $\mathbf{H}(s, t) = (1 - \frac{\mu_0 \varepsilon_0 s^2 \omega^2}{4}) H_{z0}(t)\hat{\mathbf{z}}$. Then, the same conditions should apply as in 8.1.2. That is, $\omega$ should be low enough or the d evice small enough that light can easily propagate across the device during one period of oscillation.
## Section 8.2 - Circuits with Resistance, Capacitance and Inductance and a Sinusoidal EMF
@@ -213,24 +213,24 @@ $$u = \frac{1}{2}(\varepsilon_0 e^2 + \mu_u H^2)$$
In a capacitor, we cannot directly state the energy. However, we can state its rate of change:
$$\frac{du_E}{dt} = \vb{E} \vdot \frac{d\vb{D}}{dt}$$
$$\frac{du_E}{dt} = \mathbf{E} \cdot \frac{d\mathbf{D}}{dt}$$
In the case where $\vb{D} = \varepsilon \vb{E}$, that is, in the presence of a simple dielectric,
In the case where $\mathbf{D} = \varepsilon \mathbf{E}$, that is, in the presence of a simple dielectric,
$$u_E = \frac{1}{2} \varepsilon E^2 = \frac{1}{2}\vb{E} \vdot \vb{D}$$
$$u_E = \frac{1}{2} \varepsilon E^2 = \frac{1}{2}\mathbf{E} \cdot \mathbf{D}$$
This is mirrored in solenoids / inductors, where
$$\frac{du_M}{dt} = \vb{H} \vdot \frac{d\vb{B}}{dt}$$
$$\frac{du_M}{dt} = \mathbf{H} \cdot \frac{d\mathbf{B}}{dt}$$
In the case where $\vb{B} = \mu \vb{H}$, that is, in the presence of a simple magnetic material,
In the case where $\mathbf{B} = \mu \mathbf{H}$, that is, in the presence of a simple magnetic material,
$$u_M = \frac{1}{2}\vb{H} \vdot \vb{B}$$
$$u_M = \frac{1}{2}\mathbf{H} \cdot \mathbf{B}$$
Lastly, this continues for electromagnetic fields.
$$\frac{du_{EM}}{dt} = \vb{E} \vdot \frac{d\vb{D}}{dt} + \vb{H} \vdot \frac{d\vb{B}}{dt}$$
$$\frac{du_{EM}}{dt} = \mathbf{E} \cdot \frac{d\mathbf{D}}{dt} + \mathbf{H} \cdot \frac{d\mathbf{B}}{dt}$$
In simple electromagnetic materials, we see that
$$u_{EM} = \frac{1}{2}(\vb{E} \vdot \vb{D} + \vb{H} \vdot{B}) = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$$
$$u_{EM} = \frac{1}{2}(\mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B}) = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$$

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@@ -6,70 +6,70 @@ The laws of physics exhibit temporal, spatial, and angular symmetry. That is, th
The conservation of chargee is another symmetry-based conservation law, derived from "gauge invariance". In this section, this is conservation law is assumed to be valid.
Recall the *continuity equation*, that is, $\frac{\partial \rho(\vb{r})}{\partial t} = - \div \vb{J}(\vb{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\vb{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \vb{J}(\vb{r}, t) \vdot \vu{n} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
Recall the *continuity equation*, that is, $\frac{\partial \rho(\mathbf{r})}{\partial t} = - \nabla \cdot \mathbf{J}(\mathbf{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\mathbf{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \mathbf{J}(\mathbf{r}, t) \cdot \hat{\mathbf{n}} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
## Section 9.2 - Conservation of Electomagnetic Energy
Consider a volume $V$ with surface $\partial V$, that encloses some magnetic and electric point charges. Then, for any electric charge $q_{ei}$ or magnetic charge $q_{mj}$, the force on each charge is
$$\begin{align}
\vb{F}_i &= q_{ei} (\vb{E}_i + \vb{v}_i \times \vb{B}_i) \\
\vb{F}_j &= q_{mi} (\vb{H}_j - \vb{v}_j \times \vb{D}_j)
\mathbf{F}_i &= q_{ei} (\mathbf{E}_i + \mathbf{v}_i \times \mathbf{B}_i) \\
\mathbf{F}_j &= q_{mi} (\mathbf{H}_j - \mathbf{v}_j \times \mathbf{D}_j)
\end{align}$$
We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \vb{F}_i \vdot \vb{v}_i$ and $\frac{dw_m}{dt} = \vb{F}_j \vdot \vb{v}_j$. This allows us to conclude that at any point $\vb{r}$ inside the volume, the rate at which the mechanical energy density changes is
We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \mathbf{F}_i \cdot \mathbf{v}_i$ and $\frac{dw_m}{dt} = \mathbf{F}_j \cdot \mathbf{v}_j$. This allows us to conclude that at any point $\mathbf{r}$ inside the volume, the rate at which the mechanical energy density changes is
$$\frac{du_{mech}}{dt} = \sum_i \delta(\vb{r} - \vb{r}_i) \vb{F}_i \vdot \vb{v}_i + \delta(\vb{r} - \vb{r}_j) \vb{F}_j \vdot \vb{v}_j$$
$$\frac{du_{mech}}{dt} = \sum_i \delta(\mathbf{r} - \mathbf{r}_i) \mathbf{F}_i \cdot \mathbf{v}_i + \delta(\mathbf{r} - \mathbf{r}_j) \mathbf{F}_j \cdot \mathbf{v}_j$$
Since $(\vb{v}_i \times \vb{B}_i) \vdot \vb{v}_i = 0$ and $(\vb{v}_j \times \vb{D}_j) \vdot \vb{v}_j = 0$, and the current densities are given as $\vb{J}_e(\vb{r}) = \sum_i \vb{v}_i q_{ei}\delta(\vb{r} - \vb{r}_i)$ and $\vb{J}_m(\vb{r}) = \sum_j \vb{v}_j q_{mj}\delta(\vb{r} - \vb{r}_j)$, we can rewrite this as
Since $(\mathbf{v}_i \times \mathbf{B}_i) \cdot \mathbf{v}_i = 0$ and $(\mathbf{v}_j \times \mathbf{D}_j) \cdot \mathbf{v}_j = 0$, and the current densities are given as $\mathbf{J}_e(\mathbf{r}) = \sum_i \mathbf{v}_i q_{ei}\delta(\mathbf{r} - \mathbf{r}_i)$ and $\mathbf{J}_m(\mathbf{r}) = \sum_j \mathbf{v}_j q_{mj}\delta(\mathbf{r} - \mathbf{r}_j)$, we can rewrite this as
$$\frac{du_{mech}}{dt} = \vb{J}_e(\vb{r}) \vdot \vb{E}(\vb{r}) + \vb{J}_m(\vb{r}) \vdot \vb{H}(\vb{r})$$
$$\frac{du_{mech}}{dt} = \mathbf{J}_e(\mathbf{r}) \cdot \mathbf{E}(\mathbf{r}) + \mathbf{J}_m(\mathbf{r}) \cdot \mathbf{H}(\mathbf{r})$$
Combine this with the Maxwell equations to remove current densities, we see that
$$\frac{du_{mech}}{dt} = (\curl \vb{H} - \frac{\partial \vb{H}}{\partial t}) \vdot \vb{E} + (-\curl \vb{E} - \frac{\partial \vb{B}}{\partial t}) \vdot \vb{H}$$
$$\frac{du_{mech}}{dt} = (\nabla \times \mathbf{H} - \frac{\partial \mathbf{H}}{\partial t}) \cdot \mathbf{E} + (-\nabla \times \mathbf{E} - \frac{\partial \mathbf{B}}{\partial t}) \cdot \mathbf{H}$$
With a vector identity, this simplifies to
$$\frac{du_{mech}}{dt} = -\curl (\vb{E} \times \vb{H})$$
$$\frac{du_{mech}}{dt} = -\nabla \times (\mathbf{E} \times \mathbf{H})$$
**Definition**. We call $\vb{S} = \vb{E} \times \vb{H}$ the *Poynting vector*.
**Definition**. We call $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ the *Poynting vector*.
With this vector, we can define
$$\frac{du_{mech}}{dt} = -\curl \vb{S}$$
$$\frac{du_{mech}}{dt} = -\nabla \times \mathbf{S}$$
### Section 9.2.2 - Energy Density for Linear Materials
For a simple material, that is, one in which $\vb{D} = \epsilon \vb{E}$ and $\vb{B} = \mu \vb{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\vb{E} \vdot \vb{D} + \vb{H} \vdot \vb{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\epsilon E^2 + \mu H^2)$.
For a simple material, that is, one in which $\mathbf{D} = \varepsilon \mathbf{E}$ and $\mathbf{B} = \mu \mathbf{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$.
**Definition**. A material is said to be an *anisotropic linear material* if $\vb{D} = \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} = \overleftrightarrow{\vb{\mu}} \vdot \vb{H}$. Then, the internal electromagnetic energy becones
**Definition**. A material is said to be an *anisotropic linear material* if $\mathbf{D} = \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}$ and $\mathbf{B} = \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}$. Then, the internal electromagnetic energy becomes
$$u_{em} = \frac{1}{2}(\vb{E} \vdot \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} + \vb{H} + \overleftrightarrow{\vb{\mu}} \vdot \vb{H}) = \frac{1}{2}\sum_{i,j}(\epsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
$$u_{em} = \frac{1}{2}(\mathbf{E} \cdot \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}+ \mathbf{H} + \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}) = \frac{1}{2}\sum_{i,j}(\varepsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
If the dyadics are symmetric, the energy functions uniquely specify the energy in terms of fields. In general, the polarization or magnetization of a material may depend on its past history and on time, and as such, the energy density for such materials cannot be expressed entirely in terms of the fields.
### Section 9.2.3 - Poynting Vector Examples
Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\vb{H} = \frac{R}{2\pi s R}\vb{\varphi}$. Additionally, we know that
Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\mathbf{H} = \frac{R}{2\pi s R}\mathbf{\varphi}$. Additionally, we know that
$$V=\int_a^b \vb{E} \vdot d\vb{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
$$V=\int_a^b \mathbf{E} \cdot d\mathbf{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
This then implies that $\vb{E} = \frac{V}{\ln(\frac{b}{a})s} \vu(s)$. We can then solve for both the energy density and Poynting vector, as well as $\vb{v} = \vb{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
This then implies that $\mathbf{E} = \frac{V}{\ln(\frac{b}{a})s} \hat{\mathbf{s}}$. We can then solve for both the energy density and Poynting vector, as well as $\mathbf{v} = \mathbf{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
$$\vb{v} = \frac{2c \vu{z}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
$$\mathbf{v} = \frac{2c \hat{\mathbf{z}}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
---
Consider a long cylindrical ohmic wire of radius $a$, length $L$, and resistivity $\rho$ along the $z$-axis. If this wire is carrying a constant current $I$, we know that inside the wire, $E_z = \rho J = \rho \frac{I}{\pi a^2}$. From Ampere's Law, $H_\phi = \frac{s}{2\pi a^2} I$. Then, inside the wire,
$$\vb{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \vu{s}$$
$$\mathbf{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \hat{\mathbf{s}}$$
Outside the wire, we know that $\vb{E} = \frac{\rho I}{\pi a^2} \vu{z}$ and $\vb{H} = \frac{I}{2\pi s}\vu{\varphi}$, so
Outside the wire, we know that $\mathbf{E} = \frac{\rho I}{\pi a^2} \hat{\mathbf{z}}$ and $\mathbf{H} = \frac{I}{2\pi s}\hat{\mathbf{\varphi}}$, so
$$\vb{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \vu{s}$$
$$\mathbf{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \hat{\mathbf{s}}$$
Notably, inside the wire, $\div \vb{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
Notably, inside the wire, $\nabla \cdot \mathbf{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
---
@@ -79,50 +79,50 @@ Further examples are present but omitted.
We know that the rate of change of momentum for any given particle is simply the force acting on it. To calculate this, recall the force density:
$$\vb{f} = \sum_i \vb{F}_i\delta(\vb{r}-\vb{r}_i) + \sum_j \vb{F}_j\delta(\vb{r}-\vb{r}_j) = \sum_i q_{ei}\delta(\vb{r}-\vb{r}_i)(\vb{E} + \vb{v}_i \cross \vb{B}) + \sum_j q_{ej} \delta(\vb{r}-\vb{r}_i) (\vb{H} - \vb{v}_j \times \vb{D})$$
$$\mathbf{f} = \sum_i \mathbf{F}_i\delta(\mathbf{r}-\mathbf{r}_i) + \sum_j \mathbf{F}_j\delta(\mathbf{r}-\mathbf{r}_j) = \sum_i q_{ei}\delta(\mathbf{r}-\mathbf{r}_i)(\mathbf{E} + \mathbf{v}_i \times \mathbf{B}) + \sum_j q_{ej} \delta(\mathbf{r}-\mathbf{r}_i) (\mathbf{H} - \mathbf{v}_j \times \mathbf{D})$$
Converting to currents, we see that
$$\vb{f}(\vb{r}) = \rho_e(\vb{r})\vb{E}(\vb{r}) + \vb{J}_e(\vb{r}) \times \vb{B}(\vb{r}) + \rho_m(\vb{r}) + \vb{H}(\vb{r}) - \vb{J}_m(\vb{r}) \times \vb{D}(\vb{r})$$
$$\mathbf{f}(\mathbf{r}) = \rho_e(\mathbf{r})\mathbf{E}(\mathbf{r}) + \mathbf{J}_e(\mathbf{r}) \times \mathbf{B}(\mathbf{r}) + \rho_m(\mathbf{r}) + \mathbf{H}(\mathbf{r}) - \mathbf{J}_m(\mathbf{r}) \times \mathbf{D}(\mathbf{r})$$
Substituting in Maxwell's Equations, we see that
$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \varepsilon_0 (\div \vb{E})\vb{E} + (\curl \vb{E})\times\vb{D} + \mu_0(\div \vb{H})\vb{H} + (\curl \vb{H})\times\vb{B}$$
$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \varepsilon_0 (\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times\mathbf{D} + \mu_0(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times\mathbf{B}$$
Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\vb{T}}$, so that
Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\mathbf{T}}$, so that
$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \div \overleftrightarrow{\vb{T}}$$
$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\div \overleftrightarrow{\vb{T}} = \div \overleftrightarrow{\vb{T}}_e + \div \overleftrightarrow{\vb{T}}_m$. Then, we can state
This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\nabla \cdot \overleftrightarrow{\mathbf{T}} = \nabla \cdot \overleftrightarrow{\mathbf{T}}_e + \nabla \cdot \overleftrightarrow{\mathbf{T}}_m$. Then, we can state
$$\begin{align}
\div \overleftrightarrow{\vb{T}}_e &= \varepsilon_0 [(\div \vb{E})\vb{E} + (\curl \vb{E})\times \vb{E}] \\
\div \overleftrightarrow{\vb{T}}_m &= \varepsilon_0 [(\div \vb{H})\vb{H} + (\curl \vb{H})\times \vb{H}]
\nabla \cdot \overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 [(\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times \mathbf{E}] \\
\nabla \cdot \overleftrightarrow{\mathbf{T}}_m &= \varepsilon_0 [(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times \mathbf{H}]
\end{align}$$
We know that $\div(\vb{EE}) = (\div \vb{E})\vb{E} + (\vb{E} \vdot \nabla)\vb{E}$ and $\div(\overleftrightarrow{\vb{I}}f) = \grad f$. If we let $f = \frac{1}{2}\vb{E} \vdot \vb{E}$, we see that $\grad(\frac{1}{2}\vb{E} \vdot \vb{E}) = (\vb{E} \vdot \nabla)\vb{E} + (\curl \vb{E})\vb{E}$. Then, we see that
We know that $\nabla \cdot(\mathbf{EE}) = (\nabla \cdot \mathbf{E})\mathbf{E} + (\mathbf{E} \cdot \nabla)\mathbf{E}$ and $\nabla \cdot(\overleftrightarrow{\mathbf{I}}f) = \nabla f$. If we let $f = \frac{1}{2}\mathbf{E} \cdot \mathbf{E}$, we see that $\nabla(\frac{1}{2}\mathbf{E} \cdot \mathbf{E}) = (\mathbf{E} \cdot \nabla)\mathbf{E} + (\nabla \times \mathbf{E})\mathbf{E}$. Then, we see that
$$\begin{align}
\overleftrightarrow{\vb{T}}_e &= \varepsilon_0 \vb{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\vb{I}}(\vb{E} \vdot \vb{E}) \\
\overleftrightarrow{\vb{T}}_m &= \mu_0 \vb{HH} - \frac{\mu_0}{2} \overleftrightarrow{\vb{I}}(\vb{H} \vdot \vb{H})
\overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 \mathbf{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{E} \cdot \mathbf{E}) \\
\overleftrightarrow{\mathbf{T}}_m &= \mu_0 \mathbf{HH} - \frac{\mu_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{H} \cdot \mathbf{H})
\end{align}$$
Knowing that $\overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}}_e + \overleftrightarrow{\vb{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
Knowing that $\overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}}_e + \overleftrightarrow{\mathbf{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
$$\overleftrightarrow{\vb{T}} = \varepsilon_0 \vb{EE} + \mu_0 \vb{HH} - \overleftrightarrow{\vb{I}}u$$
$$\overleftrightarrow{\mathbf{T}} = \varepsilon_0 \mathbf{EE} + \mu_0 \mathbf{HH} - \overleftrightarrow{\mathbf{I}}u$$
Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\vb{g}(\vb{r}) = \vb{D}(\vb{r}) \times \vb{B}(\vb{r})$. Thus,
Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\mathbf{g}(\mathbf{r}) = \mathbf{D}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})$. Thus,
$$\vb{f}(\vb{r}) = \frac{\partial}{\partial t}\vb{g} = \div \overleftrightarrow{\vb{T}}$$
$$\mathbf{f}(\mathbf{r}) = \frac{\partial}{\partial t}\mathbf{g} = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
---
The Divergence Theorem states that $\int_V(\div \overleftrightarrow{\vb{T}}) dV = \int_{SofV} dS \vu{n} \vdot \overleftrightarrow{\vb{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\vb{T}}$ is symmetric, $\vu{n} \vdot \overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}} \vdot \vu{n}$.
The Divergence Theorem states that $\int_V(\nabla \cdot \overleftrightarrow{\mathbf{T}}) dV = \int_{SofV} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\mathbf{T}}$ is symmetric, $\hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}}$.
Given a static field, the momentum does not change in time. That is,
$$\frac{\partial}{\partial t} \vb{g} = \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = 0$$
$$\frac{\partial}{\partial t} \mathbf{g} = \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = 0$$
Then, we can see that $\vb{f}(\vb{r}) = \div \overleftrightarrow{\vb{T}}$. We can thus integrate over the volume to find force on an object.
Then, we can see that $\mathbf{f}(\mathbf{r}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$. We can thus integrate over the volume to find force on an object.
$$\vb{F} = \int_V \vb{f}(\vb{r}) dV = \int_V \div \overleftrightarrow{\vb{T}} dV = \int_{\partial V} dS \vu{n} \vdot \overleftrightarrow{\vb{T}} = \int_{\partial V} \overleftrightarrow{\vb{T}} \vdot \vu{n} dS$$
$$\mathbf{F} = \int_V \mathbf{f}(\mathbf{r}) dV = \int_V \nabla \cdot \overleftrightarrow{\mathbf{T}} dV = \int_{\partial V} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \int_{\partial V} \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}} dS$$

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@@ -5,69 +5,69 @@
**Definition**. A *dyadic* is a representation of two-ish vectors.
$$
\stackrel{\leftrightarrow}{\vb{D}} = \begin{matrix}
D_{xx} \vu{x}\vu{x} &+ D_{xy} \vu{x}\vu{y} &+ D{xz} \vu{x}\vu{z} \\
+ D_{yx} \vu{y}\vu{x} &+ D_{yy} \vu{y}\vu{y} &+ D{yz} \vu{y}\vu{z} \\
+ D_{zx} \vu{z}\vu{x} &+ D_{zy} \vu{z}\vu{y} &+ D{zz} \vu{z}\vu{z}
\stackrel{\leftrightarrow}{\mathbf{D}} = \begin{matrix}
D_{xx} \hat{\mathbf{x}}\hat{\mathbf{x}} &+ D_{xy} \hat{\mathbf{x}}\hat{\mathbf{y}} &+ D{xz} \hat{\mathbf{x}}\hat{\mathbf{z}} \\
+ D_{yx} \hat{\mathbf{y}}\hat{\mathbf{x}} &+ D_{yy} \hat{\mathbf{y}}\hat{\mathbf{y}} &+ D{yz} \hat{\mathbf{y}}\hat{\mathbf{z}} \\
+ D_{zx} \hat{\mathbf{z}}\hat{\mathbf{x}} &+ D_{zy} \hat{\mathbf{z}}\hat{\mathbf{y}} &+ D{zz} \hat{\mathbf{z}}\hat{\mathbf{z}}
\end{matrix}
$$
**Definition**. If a dyadic can be written as a composition of two vectors $\vb{A}$ and $\vb{B}$, it is called a *dyad*.
**Definition**. If a dyadic can be written as a composition of two vectors $\mathbf{A}$ and $\mathbf{B}$, it is called a *dyad*.
$$
\vb{AB} = \begin{matrix}
A_x B_x \vu{x}\vu{x} &+ A_x B_y \vu{x}\vu{y} &+ A_x B_z \vu{x}\vu{z} \\
+ A_y B_x \vu{y}\vu{x} &+ A_y B_y \vu{y}\vu{y} &+ A_y B_z \vu{y}\vu{z} \\
+ A_z B_x \vu{z}\vu{x} &+ A_z B_y \vu{z}\vu{y} &+ A_z B_z \vu{z}\vu{z}
\mathbf{AB} = \begin{matrix}
A_x B_x \hat{\mathbf{x}}\hat{\mathbf{x}} &+ A_x B_y \hat{\mathbf{x}}\hat{\mathbf{y}} &+ A_x B_z \hat{\mathbf{x}}\hat{\mathbf{z}} \\
+ A_y B_x \hat{\mathbf{y}}\hat{\mathbf{x}} &+ A_y B_y \hat{\mathbf{y}}\hat{\mathbf{y}} &+ A_y B_z \hat{\mathbf{y}}\hat{\mathbf{z}} \\
+ A_z B_x \hat{\mathbf{z}}\hat{\mathbf{x}} &+ A_z B_y \hat{\mathbf{z}}\hat{\mathbf{y}} &+ A_z B_z \hat{\mathbf{z}}\hat{\mathbf{z}}
\end{matrix}
$$
The dot product of a dyad $\stackrel{\leftrightarrow}{\vb{D}} = \vb{AB}$ and vector $\va{v}$ can be written as follows:
The dot product of a dyad $\stackrel{\leftrightarrow}{\mathbf{D}} = \mathbf{AB}$ and vector $\mathbf{v}$ can be written as follows:
$$
(\vb{AB}) \vdot \va{v} = \vb{A} (\vb{B} \vdot \va{v})
(\mathbf{AB}) \cdot \mathbf{v} = \mathbf{A} (\mathbf{B} \cdot \mathbf{v})
$$
**Definition**. A *symmetric/antisymmetric* dyadic is defined the same way that a matrix is.
**Definition**. The *identity dyadic* is $\stackrel{\leftrightarrow}{\vb{I}} = \vu{x}\vu{x} + \vu{y}\vu{y} + \vu{z}\vu{z}$.
**Definition**. The *identity dyadic* is $\stackrel{\leftrightarrow}{\mathbf{I}} = \hat{\mathbf{x}}\hat{\mathbf{x}} + \hat{\mathbf{y}}\hat{\mathbf{y}} + \hat{\mathbf{z}}\hat{\mathbf{z}}$.
**Definition**. FOr a *tensor*, with coordinates $u^i$, we have two sets of basis vectors:
$$
\vb{e}_i = \pdv{\vb{r}}{u^i}
\mathbf{e}_i = \pdv{\mathbf{r}}{u^i}
$$
$$
\vb{e}^i = \grad{u^i}
\mathbf{e}^i = \nabla{u^i}
$$
## 1.9 - Helmholtz Theorem
Given an arbitrary vector field $\vb{F}(\vb(r))$, we can write said field as a composition of a curl-free component $\vb{\Phi}(\vb{r})$ and a divergence-free component $\vb{A}(\vb{r})$ as follows:
Given an arbitrary vector field $\mathbf{F}(\mathbf(r))$, we can write said field as a composition of a curl-free component $\mathbf{\Phi}(\mathbf{r})$ and a divergence-free component $\mathbf{A}(\mathbf{r})$ as follows:
$$
\vb{F}(\vb{r}) = - \grad{\vb{\Phi}(\vb{r})} + \curl{\vb{A}(\vb{r})}
\mathbf{F}(\mathbf{r}) = - \nabla{\mathbf{\Phi}(\mathbf{r})} + \nabla \times{\mathbf{A}(\mathbf{r})}
$$
**Definition**. Here, the gradient of the scalar potential is $\grad{\vb{\Phi}(\vb{r})}$ and the curl of the vector potential is $\curl{\vb{A}(\vb{r})}$. Thus, the scalar potential is $\vb{\Phi}(\vb{r})$ and the vector potential is $\vb{A}(\vb{r})$.
**Definition**. Here, the gradient of the scalar potential is $\nabla{\mathbf{\Phi}(\mathbf{r})}$ and the curl of the vector potential is $\nabla \times{\mathbf{A}(\mathbf{r})}$. Thus, the scalar potential is $\mathbf{\Phi}(\mathbf{r})$ and the vector potential is $\mathbf{A}(\mathbf{r})$.
Letting said field be over bounded volume $V$ with closed surface $\partial V$, and the functions $\vb{C}(\vb{r}) = \curl{\vb{F}(\vb{r})}$ and $\vb{D}(\vb{r}) = \div{\vb{F}(\vb{r})}$ are known, we can say that
Letting said field be over bounded volume $V$ with closed surface $\partial V$, and the functions $\mathbf{C}(\mathbf{r}) = \nabla \times{\mathbf{F}(\mathbf{r})}$ and $\mathbf{D}(\mathbf{r}) = \nabla \cdot \mathbf{F}(\mathbf{r})$ are known, we can say that
$$
\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \frac{\vb{F}(\vb{r}') \vdot \va{n}'}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
\mathbf{\Phi}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{D(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'} - \frac{1}{4 \pi} \int_{\partial V} \frac{\mathbf{F}(\mathbf{r}') \cdot \mathbf{n}'}{|{\mathbf{r}-\mathbf{r}'}|} d{S'}
$$
$$
\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \va{n}' \cross \frac{\vb{F}(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
\mathbf{A}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{C(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'} - \frac{1}{4 \pi} \int_{\partial V} \mathbf{n}' \times \frac{\mathbf{F}(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{S'}
$$
Now, assume that $\lim(\frac{\vb{F}(\vb{r})}{\vb{r}}) = 0$ as $\vb{r} \rightarrow \infty$, with a large enough volume, we see that the second terms vanish.
Now, assume that $\lim(\frac{\mathbf{F}(\mathbf{r})}{\mathbf{r}}) = 0$ as $\mathbf{r} \rightarrow \infty$, with a large enough volume, we see that the second terms vanish.
$$
\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
\mathbf{\Phi}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{D(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'}
$$
$$
\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
\mathbf{A}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{C(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'}
$$

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@@ -2,32 +2,32 @@
## Section 2.2 - Parallel Treatment of Electric and Magnetic Fields
Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\vb{r}$ relative to the origin.
Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\mathbf{r}$ relative to the origin.
Thus, according to Coulomb's Law,
$$
\begin{align}
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
F^e_{qQ}(\mathbf{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
F^m_{qQ}(\mathbf{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
\end{align}
$$
Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\vb{r}$.
Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\mathbf{r}$.
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
\end{align}
$$
Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}}$, and we see the following.
Now, let $Q$ be at point $\mathbf{r'}$. Then, the unit vector becomes $\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}}|$, and we see the following.
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
\end{align}
$$
@@ -35,97 +35,89 @@ With multiple charges, we can apply the *superposition principal* to see the fol
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
\end{align}
$$
We can convert this to an integral as $N$ goes to infinity.
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
\end{align}
$$
$$\begin{align}
E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \\
H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V'
\end{align}$$
## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
From a lot of advanced math, we know that
$$
\div{\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}} = 4 \pi \delta(\vb{r}-\vb{r'})
$$
$$\nabla \cdot \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = 4 \pi \delta(\mathbf{r}-\mathbf{r'})$$
Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnetostatic fields.
Now, apply the divergence operator over $\mathbf{r}$ to the electrostatic and magnetostatic fields.
$$
\begin{align}
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
\end{align}
$$
$$\begin{align}
\nabla \cdot E(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \\
\nabla \cdot H(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V')
\end{align}$$
As the divergence operator does not operate on $\vb{r'}$, we see that
As the divergence operator does not operate on $\mathbf{r'}$, we see that
$$
\begin{align}
\div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\
\div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_m(\vb{r})}{\mu_0}
\end{align}
$$
$$\begin{align}
\nabla \cdot E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
&= \frac{\rho_e(\mathbf{r})}{\varepsilon_0} \\
\nabla \cdot H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
&= \frac{\rho_m(\mathbf{r})}{\mu_0}
\end{align}$$
The curl of an electrostatic or magnetostatic is relatively simple.
$$
\begin{align}
\curl{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\nabla \times{E(\mathbf{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
\nabla \times{H(\mathbf{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
\end{align}
$$
Additionally, we know $\curl{f\vb{A}} = f \curl{\vb{A}} + \grad{f}\cross\vb{A}$. Thus,
Additionally, we know $\nabla \times{f\mathbf{A}} = f \nabla \times{\mathbf{A}} + \nabla{f}\times\mathbf{A}$. Thus,
$$
\begin{align}
\curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} &= \frac{1}{\abs{\vb{r}-\vb{r'}}^3} \curl{(\vb{r}-\vb{r'})} + (\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}}) \cross (\vb{r}-\vb{r'}) \\
\nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} &= \frac{1}{|\mathbf{r}-\mathbf{r'}|^3} \nabla \times{(\mathbf{r}-\mathbf{r'})} + (\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}}) \times (\mathbf{r}-\mathbf{r'}) \\
\end{align}
$$
We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Additionally, $\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}} = -3 \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^5}$, which when crossed with $\vb{r}-\vb{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
We can verify that $\nabla \times{(\mathbf{r}-\mathbf{r'})} = 0$, cancelling the first term. Additionally, $\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}} = -3 \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^5}$, which when crossed with $\mathbf{r}-\mathbf{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
## Section 2.4 - Electric and Magnetic Flux Densities
The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
The electric and magnetic flux density vectors are given by $\varepsilon_0 \mathbf{E}$ and $\mu_0 \mathbf{H}$.
Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
$$
\Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
\Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d S = Q_m
$$
Thus, applying divergence theorem,
$$
Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{E}} \dd V
Q_e = \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V
$$
$$
Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{H}} \dd V
Q_m = \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{H} d V
$$
Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
Since $Q_e = \int_V \rho_e d V$ and $Q_m = \int_V \rho_m d V$, we see that
$$
\begin{align}
\int_V \rho_e \dd V &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\rho_e &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\int_V \rho_e d V &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
\rho_e &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
\int_V \rho_m d V &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
\rho_m &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
\end{align}
$$

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@@ -2,22 +2,22 @@
## Section 3.1 - Work and Energy in Electrostatics and Magnetostatics
The force on charge $q$ is given by $\vb{F}(\vb{r}) = q_e \vb{E}(\vb{r})$ or $\vb{F}(\vb{r}) = q_m \vb{H}(\vb{r})$. If this charge is moved $\dd{\vb{l}} = \dd x \vu{x} + \dd y \vu{y} + \dd z \vu{z}$, the change in internal energy (work) this produces can be written as
The force on charge $q$ is given by $\mathbf{F}(\mathbf{r}) = q_e \mathbf{E}(\mathbf{r})$ or $\mathbf{F}(\mathbf{r}) = q_m \mathbf{H}(\mathbf{r})$. If this charge is moved $d{\mathbf{l}} = d x \hat{\mathbf{x}} + d y \hat{\mathbf{y}} + d z \hat{\mathbf{z}}$, the change in internal energy (work) this produces can be written as
$$
\dd{U}= - \vb{F} \vdot \dd{\vb{l}}
d{U}= - \mathbf{F} \cdot d{\mathbf{l}}
$$
Rewriting this, $\vb{F} = -\grad{U}$, with $U$ as potential energy. Now, we can denote this change in internal energy in terms of $q$ as follows:
Rewriting this, $\mathbf{F} = -\nabla{U}$, with $U$ as potential energy. Now, we can denote this change in internal energy in terms of $q$ as follows:
$$
\vb{E}(\vb{r}) = \frac{1}{q_e} \vb{F_e}(\vb{r}) = - \frac{1}{q_e} \grad{U_e(\vb{r})} = -\grad{V_e(\vb{r})}
\mathbf{E}(\mathbf{r}) = \frac{1}{q_e} \mathbf{F_e}(\mathbf{r}) = - \frac{1}{q_e} \nabla{U_e(\mathbf{r})} = -\nabla{V_e(\mathbf{r})}
$$
The units of electrostatic potential is Joule/Coulomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
$$
\vb{H}(\vb{r}) = \frac{1}{q_m} \vb{F_m}(\vb{r}) = - \frac{1}{q_m} \grad{U_m(\vb{r})} = -\grad{V_m(\vb{r})}
\mathbf{H}(\mathbf{r}) = \frac{1}{q_m} \mathbf{F_m}(\mathbf{r}) = - \frac{1}{q_m} \nabla{U_m(\mathbf{r})} = -\nabla{V_m(\mathbf{r})}
$$
The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Thus, the units of the magnetic field can be written as Amperes/meter.
@@ -25,45 +25,45 @@ The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Th
With this, we can calculate work. Moving a charge $q$ from $A$ to $B$, we see that
$$
\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e
\delta W = \int_A^B \mathbf{F} \cdot d{\mathbf{l}} = q_e \int_A^B \mathbf{E} \cdot d{\mathbf{l}} = -q_e \int_A^B \nabla{\mathbf{V}} \cdot d{\mathbf{l}} = -q_e \delta V_e
$$
Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus,
Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\mathbf{r'} = \mathbf{\infty}$ to $\mathbf{r'} = \mathbf{r}$. Thus,
$$
W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r}
W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\hat{\mathbf{r'}}}{r'^2} \cdot (\hat{\mathbf{r'}}) d{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r}
$$
$$
W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\hat{\mathbf{r'}}}{r'^2} \cdot (\hat{\mathbf{r'}}) d{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
$$
Letting the potential as $\vb{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
Letting the potential as $\mathbf{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}
V_e(\mathbf{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\mathbf{r}) = \frac{Q_m}{4 \pi \mu_0 r}
$$
Now, if we let the stationary charge $Q$ be located at $\vb{r'}$, we see that
Now, if we let the stationary charge $Q$ be located at $\mathbf{r'}$, we see that
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}}
V_e(\mathbf{r}) = \frac{Q_e}{4 \pi \varepsilon_0}{|\mathbf{r}-\mathbf{r'}|} \text{ and } V_m(\mathbf{r}) = \frac{Q_m}{4 \pi \mu_0}{|\mathbf{r}-\mathbf{r'}|}
$$
If we allow multiple charges, this becomes
$$
V_e(\vb{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}}
V_e(\mathbf{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_{ei}}{|\mathbf{r}-\mathbf{r_i}|}
$$
Taking this to its natural limit,
$$
V_e(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
V_e(\mathbf{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}}| d{V'}
$$
$$
V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
V_m(\mathbf{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}}| d{V'}
$$
## Section 3.2 - Energy of a Charge Distribution
@@ -71,43 +71,43 @@ $$
Given two point charges $Q_{e1}, Q_{e2}$ we know the work to bring them together is
$$
W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{|\mathbf{r_2} - \mathbf{r_1}|}
$$
Superposition applies here. The energy to create $N$ charges is
$$
W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
$$
For the sake of symmetry, sum overall charges and divide by 2.
$$
W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
$$
Rearranging, we see the following:
$$
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i})
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\mathbf{r_i})
$$
We can rewrite this as a Riemann sum and convert to an integral.
$$
W_e = \frac{1}{2} \int_V p_e(\vb{r}) V_e(\vb{r}) \dd V ; \quad
W_m = \frac{1}{2} \int_V p_m(\vb{r}) V_m(\vb{r}) \dd V
W_e = \frac{1}{2} \int_V p_e(\mathbf{r}) V_e(\mathbf{r}) d V ; \quad
W_m = \frac{1}{2} \int_V p_m(\mathbf{r}) V_m(\mathbf{r}) d V
$$
We can also express this as
$$
W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\mathbf{r})\rho_e(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} d{V'} d{V}
$$
$$
W_m = \frac{1}{2} \frac{1}{4 \pi \mu_0} \int_V \int_{V'} \frac{\rho_m(\vb{r})\rho_m(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
W_m = \frac{1}{2} \frac{1}{4 \pi \mu_0} \int_V \int_{V'} \frac{\rho_m(\mathbf{r})\rho_m(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} d{V'} d{V}
$$
Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previously. If we were to determine the potential based on a different set of charges, the factor would be absent.
@@ -115,28 +115,28 @@ Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previo
We can now write an expression for energy of a charge density in terms of the field that it produces.
$$
W = \frac{\varepsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V
W = \frac{\varepsilon_0}{2} \int_V (\nabla \cdot \mathbf{E}(\mathbf{r})) V(\mathbf{r}) d V
$$
Simplifying, we see that
$$
W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad
W_m = \frac{\mu_0}{2} \int_{V} H^2(\vb{r}) \dd V
W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\mathbf{r}) d V ; \quad
W_m = \frac{\mu_0}{2} \int_{V} H^2(\mathbf{r}) d V
$$
## Section 3.3 - The Poisson and Laplace Equations
We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V_m(\vb{r})}$
We know that $\mathbf{E}(\mathbf{r}) = -\nabla \cdot V_e(\mathbf{r})$ and $\mathbf{H}(\mathbf{r}) = -\nabla \cdot V_m(\mathbf{r})$
Combined this, as well as the first of the Maxwell equations, we see that
$$
\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\varepsilon_0}
\nabla \cdot \mathbf{E} = -\nabla \cdot \nabla V_e = - \nabla^2{V_e} = \frac{\rho_e}{\varepsilon_0}
$$
$$
\div{\vb{H}} = -\div{\grad{V_m}} = - \laplacian{V_m} = \frac{\rho_m}{\mu_0}
\nabla \cdot \mathbf{H} = -\nabla \cdot \nabla V_m = - \nabla^2{V_m} = \frac{\rho_m}{\mu_0}
$$
The last inequality is called the Poisson Equation, or the inhomogeneous Laplace equation.
@@ -144,19 +144,19 @@ The last inequality is called the Poisson Equation, or the inhomogeneous Laplace
To solve this equation, we define a Green function as follows:
$$
\laplacian G(\vb{r}, \vb{r'}) = \delta(\vb{r} - \vb{r'})
\nabla^2 G(\mathbf{r}, \mathbf{r'}) = \delta(\mathbf{r} - \mathbf{r'})
$$
Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation.
$$
V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = - \int_V G(\mathbf{r}, \mathbf{r'}) \frac{\rho_e(\mathbf{r'})}{\varepsilon_0} d{V'}
$$
This is the specific solution. Let $\psi(\vb{r})$ be a solution to the homogenous equation. We can state the following:
This is the specific solution. Let $\psi(\mathbf{r})$ be a solution to the homogenous equation. We can state the following:
$$
V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = \psi(\mathbf{r}) - \int_V G(\mathbf{r}, \mathbf{r'}) \frac{\rho_e(\mathbf{r'})}{\varepsilon_0} d{V'}
$$
We will consider the potential of a point charge. THat is, the limit of potential is zero as distance approaches infinity.
@@ -164,36 +164,36 @@ We will consider the potential of a point charge. THat is, the limit of potentia
Recall the potential of a point charge:
$$
V_e(\vb{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}}
V_e(\mathbf{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|}
$$
We know that $- \laplacian{V(\vb{r})} = \div{\vb{E}(\vb{r})}$. Thus, recall the electric field of a point charge.
We know that $- \nabla^2{V(\mathbf{r})} = \nabla \cdot \mathbf{E}(\mathbf{r})$. Thus, recall the electric field of a point charge.
$$
\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\varepsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\varepsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3}
\mathbf{E}(\mathbf{r}) = -\nabla{V(\mathbf{r})} = \frac{Q_e}{\varepsilon_0} \nabla({\frac{-1}{4\pi|\mathbf{r}-\mathbf{r'}|}}) = \frac{Q_e}{\varepsilon_0} \frac{\mathbf{r} - \mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
$$
Taking the divergence, we find that
$$
- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
= \frac{Q_e}{\varepsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\varepsilon_0} \delta(\vb{r} - \vb{r'})
- \nabla^2{V(\mathbf{r})} = G(\mathbf{r}, \mathbf{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \nabla^2({\frac{-1}{4\pi|\mathbf{r}-\mathbf{r'}|}})
= \frac{Q_e}{\varepsilon_0} \nabla \cdot \frac{\mathbf{r} - \mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = \frac{Q_e}{\varepsilon_0} \delta(\mathbf{r} - \mathbf{r'})
$$
Thus, we see that
$$
\laplacian {\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}} = \delta(\vb{r} - \vb{r'}) \quad \Rightarrow \quad G(\vb{r}, \vb{r'}) = {\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}
\nabla^2 {\frac{-1}{4\pi}}{|\mathbf{r}-\mathbf{r'}|} = \delta(\mathbf{r} - \mathbf{r'}) \quad \Rightarrow \quad G(\mathbf{r}, \mathbf{r'}) = {\frac{-1}{4\pi}}{|\mathbf{r}-\mathbf{r'}|}
$$
Finally,
$$
V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = \int_{V'} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|} \frac{\rho_e}{\varepsilon_0} d{V'}
$$
$$
V_m(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_m}{\mu_0} \dd{V'}
V_m(\mathbf{r}) = \int_{V'} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|} \frac{\rho_m}{\mu_0} d{V'}
$$
## Section 3.4 - The Laplace and Poisson Equations with Boundary Conditions
@@ -215,52 +215,52 @@ We want a simple way to write the Green function.
Let us assume all charge is contained in a sphere with radius $R$ centered at the origin. Then, for points $r$ far from the origin, the Green function can be written as
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4\pi \sqrt{r^2 - 2\vb{r} \vdot \vb{r'} + r'^2}}
= \frac{1}{4\pi r}(1 - 2 \vu{r} \vdot \vu{r'} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}
\frac{1}{4 \pi|\mathbf{r}-\mathbf{r'}|} = \frac{1}{4\pi \sqrt{r^2 - 2\mathbf{r} \cdot \mathbf{r'} + r'^2}}
= \frac{1}{4\pi r}(1 - 2 \hat{\mathbf{r}} \cdot \hat{\mathbf{r'}} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}
$$
This inverse square root term $(1 - 2 \vu{r} \vdot \vu{r'} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}$ can be expanded as a power series in $\frac{r'}{r}$.
This inverse square root term $(1 - 2 \hat{\mathbf{r}} \cdot \hat{\mathbf{r'}} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}$ can be expanded as a power series in $\frac{r'}{r}$.
The first two terms of this power series are simple enough.
$$
G(\vb{r}, \vb{r'}) = \frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} \approx \frac{1}{4 \pi r} ( 1 + \frac{\vu{r} \vdot \vb{r'}}{r}); \quad \text{ for} r > r'
G(\mathbf{r}, \mathbf{r'}) = \frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} \approx \frac{1}{4 \pi r} ( 1 + \frac{\hat{\mathbf{r}} \cdot \mathbf{r'}}{r}); \quad \text{ for} r > r'
$$
Applying this to the equation for voltage, we see that
$$
V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx
\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'}
= \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \varepsilon_0 r^2}
V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\mathbf{r}, \mathbf{r'}) p_e(\mathbf{r'}) d{V'} \approx
\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\hat{\mathbf{r}} \cdot \mathbf{r'}}{r}) p_e(\mathbf{r'}) d{V'}
= \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\hat{\mathbf{r}} \cdot \mathbf{p}}{4 \pi \varepsilon_0 r^2}
$$
**Definition**. The first and second terms of this equation are the *monopole* and *dipole* terms respectively.
**Definition**. We define $\vb{p}$ as the *electric dipole moment*, and in the magnetic version, $\vb{m}$ as the *magnetic dipole moment* as follows:
**Definition**. We define $\mathbf{p}$ as the *electric dipole moment*, and in the magnetic version, $\mathbf{m}$ as the *magnetic dipole moment* as follows:
$$
\vb{p} = \int_{V'} \vb{r'} \rho_e(\vb{r'}) \dd{V'}
\mathbf{p} = \int_{V'} \mathbf{r'} \rho_e(\mathbf{r'}) d{V'}
$$
Notably, the moments only depend on the charge density, not the point at which the field is being examined. That is, this integral only needs to be computed once.
To compute higher-order terms, let $\epsilon = 2\frac{r'}{r}\vu{r}\vdot\vu{r'}-(\frac{r'}{r})^2$. Now we can expand $(1-\epsilon)^{-\frac{1}{2}}$.
To compute higher-order terms, let $\varepsilon = 2\frac{r'}{r}\hat{\mathbf{r}}\cdot\hat{\mathbf{r'}}-(\frac{r'}{r})^2$. Now we can expand $(1-\varepsilon)^{-\frac{1}{2}}$.
$$
(1-\epsilon)^{-\frac{1}{2}} = 1 + \frac{1}{2}\epsilon + \frac{3}{8}\epsilon^2 + \frac{5}{16}\epsilon^3 + \ldots
(1-\varepsilon)^{-\frac{1}{2}} = 1 + \frac{1}{2}\varepsilon + \frac{3}{8}\varepsilon^2 + \frac{5}{16}\varepsilon^3 + \ldots
$$
However, we want an expansion in terms of $t = \frac{r'}{r}$. To do this, we write the expansion as
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4 \pi r} \sum_{n=0}^{\infty} (\frac{r'}{r}) P_n(\vu{r} \vdot \vu{r'})
\frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} = \frac{1}{4 \pi r} \sum_{n=0}^{\infty} (\frac{r'}{r}) P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})
$$
Here, $P_n(\vu{r} \vdot \vu{r'})$ is a polynomial. Because $\abs{\vb{r} - \vb{r'}}$ is symmetric, we can say that if $r' > r$ instead, simply switch the two. Thus the equation becomes
Here, $P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})$ is a polynomial. Because $|\mathbf{r}-\mathbf{r'}|$ is symmetric, we can say that if $r' > r$ instead, simply switch the two. Thus the equation becomes
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4 \pi} \sum_{n=0}^{\infty} (\frac{r'^n_{<}}{r^{n+1}_{>}}) P_n(\vu{r} \vdot \vu{r'})
\frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} = \frac{1}{4 \pi} \sum_{n=0}^{\infty} (\frac{r'^n_{<}}{r^{n+1}_{>}}) P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})
$$
Where $r_>$ is the greater of $r, r'$, and $r_<$ the lesser.

30
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@@ -12,11 +12,11 @@ We note that there is no electric fiend inside a conductor, as charges internal
Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symmetric charges, this charge may be calculated.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the voltage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \mathbf{E} \cdot d{\mathbf{l}}$. Since $\mathbf{E} = 0$ inside the conductor, the voltage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $d{\mathbf{l}}$. Now, $\mathbf{E} \cdot d{\mathbf{l}} = \mathbf{E}_s \cdot d{\mathbf{l}}_s + \mathbf{E}_p \cdot d{\mathbf{l}}_p = d{V_s} + d{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \nabla \cdot \mathbf{E} d{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
## Section 4.3 - Exercises involving conductors at fixed potentials
@@ -24,23 +24,23 @@ Consider a square with left and right potentials $V(0, y) = V(l, y) = V_1$ and $
In spherical polar coordinates, we see that with azimuthal symmetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S} = 0]$. With this, consider a surface $\dd{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.
**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S} = 0]$. With this, consider a surface $d{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.
Now, define $\Phi(\vb{r})$ as the difference between any two potential solutions to the Laplace equation at point $\vb{r}$. Since potential must be a constant,
Now, define $\Phi(\mathbf{r})$ as the difference between any two potential solutions to the Laplace equation at point $\mathbf{r}$. Since potential must be a constant,
$$
\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}}] \vdot \vu{n} \dd{S}
= \sum_{i=1}^N \Phi_i \int_{\dd{\Omega_i}} \grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S}
\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}}] \cdot \hat{\mathbf{n}} d{S}
= \sum_{i=1}^N \Phi_i \int_{d{\Omega_i}} \nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S}
$$
Thus, if potential is specified, $\Phi_i$ vanishes for that conductor. If the total charge is instead specified, the gradient vanishes because there is no difference in charge between any two points.
## Section 4.4 - Electric Field, Polarization Field, and Flux Density in the Presence of Conductors
**Definition**. A *bound charge* is any charge in a conductor that is bound to an atom and not free to be redistributed at the surface. We say that bound charges are the source of the polarization field $\vb{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
**Definition**. A *bound charge* is any charge in a conductor that is bound to an atom and not free to be redistributed at the surface. We say that bound charges are the source of the polarization field $\mathbf{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
$$
\div \vb{P} = - \rho_{eb}
\nabla \cdot \mathbf{P} = - \rho_{eb}
$$
This field is zero outside of a material, and if non-zero inside a material, will drop to zero at the surface discontinuously. If there is a component perpendicular to the surface, the discontinuity will generate curl. If there is a component parallel to the surface, it will generate divergence.
@@ -48,19 +48,19 @@ This field is zero outside of a material, and if non-zero inside a material, wil
**Definition**. Charges not bound are called *free*, with density denoted as $\rho_{ef}$. Combined with $\rho_{eb}$, they form the basis of the electric field. THat is,
$$
\varepsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
\varepsilon_0 \nabla \cdot \mathbf{E} = \rho_{ef} + \rho_{eb}
$$
**Definition**. The electric flux density field $\vb{D}$ is defined as
**Definition**. The electric flux density field $\mathbf{D}$ is defined as
$$
\vb{D} = \varepsilon_0 \vb{E} + \vb{P}
\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}
$$
Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
Both $\mathbf{D}$ and $\mathbf{P}$ have units of Coulombs/m^2. Additionally, we see that
$$
\div \vb{D} = \div (\varepsilon_0 \vb{E} + \vb{P}) = \div \varepsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
\nabla \cdot \mathbf{D} = \nabla \cdot (\varepsilon_0 \mathbf{E} + \mathbf{P}) = \nabla \cdot \varepsilon_0 \mathbf{E} + \nabla \cdot \mathbf{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
$$
## Section 4.5 - Induced Electric Charges, their Potentials and Fields
@@ -79,4 +79,4 @@ This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsi
## Section 4.7 - Forces on Charged Conductors in Electric Fields
We know that $\vb{F} = \int \vb{E}_{ext}(\vb{r}) \rho_e(\vb{r}) dV$, where $\vb{E}_{ext}(\vb{r})$ is the external electric field and $\rho_e(\vb{r})$ is the charge density of the object.
We know that $\mathbf{F} = \int \mathbf{E}_{ext}(\mathbf{r}) \rho_e(\mathbf{r}) dV$, where $\mathbf{E}_{ext}(\mathbf{r})$ is the external electric field and $\rho_e(\mathbf{r})$ is the charge density of the object.

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@@ -4,69 +4,67 @@
We want to work in a steady-state system. Thus, we restrict ourselves to currents that do not change in time.
With math, we see that $\div \vb{J}(\vb{r}) = -\frac{\partial \rho(\vb{r})}{\partial t}$. Since we are only considering a steady-state system, $\div \vb{J}_e = \div \vb{J}_m = 0$.
With math, we see that $\nabla \cdot \mathbf{J}(\mathbf{r}) = -\frac{\partial \rho(\mathbf{r})}{\partial t}$. Since we are only considering a steady-state system, $\nabla \cdot \mathbf{J}_e = \nabla \cdot \mathbf{J}_m = 0$.
**Definition**. The *conductance* of a material is $G = \frac{1}{R}$, where $R$ is the resistance of a material.
For a wire of uniform cross-sectional area, we see that $G = \sigma \frac{A}{L}$, where $A$ is the cross-sectional area, $L$ is the length of the wire, and $\sigma$ is the conductivity of a wire. Inverted, we see that $R$ = $\rho \frac{L}{A}$, where $\rho = \frac{1}{\sigma}$ is the resistivity of the wire.
**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\vb{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \vb{E}$
**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\mathbf{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \mathbf{E}$
## Section 5.2 - Currents and Curling Fields
We know that $\vb{J}_e = \curl{\vb{H}}$ and $\vb{J}_m = -\curl{\vb{E}}$. That is, current densities cause the opposing field to curl.
We know that $\mathbf{J}_e = \nabla \times{\mathbf{H}}$ and $\mathbf{J}_m = -\nabla \times{\mathbf{E}}$. That is, current densities cause the opposing field to curl.
For a wire with current $I_e$, we see that applying Stoke's theorem to the first equation,
$ \int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_{\partial S} = \vb{H} \vdot \dd{\vb{l}}$. Apply the identity $\curl{\vb{H}} = \vb{J}_e$ to the left side to see that $\int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_S \vb{J}_e \vdot \vu{n} \dd{S} = (I_e)_S$, or the current passing through the cross-sectional area. By the original equation, we see that $(I_e)_S = \vb{H} \vdot \dd{\vb{l}}$.
$ \int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_{\partial S} = \mathbf{H} \cdot d{\mathbf{l}}$. Apply the identity $\nabla \times{\mathbf{H}} = \mathbf{J}_e$ to the left side to see that $\int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_S \mathbf{J}_e \cdot \hat{\mathbf{n}} d{S} = (I_e)_S$, or the current passing through the cross-sectional area. By the original equation, we see that $(I_e)_S = \mathbf{H} \cdot d{\mathbf{l}}$.
If we assume cylindrical coordinates and that $\vb{H}(vb{r}) = H_\varphi(s) \vu{\varphi}$, then $\vb{H} \vdot \dd{\vb{l}} = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$, so then $(I_e)_S = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$. Thus, for $s > a$ (where $a$ is the radius of the wire), $2\pi s H_\varphi = I_e$, and for $s < a$, $2\pi s H_\varphi = I_e \frac{s^2}{a^2}$.
If we assume cylindrical coordinates and that $\mathbf{H}(vb{r}) = H_\varphi(s) \hat{\mathbf{\varphi}}$, then $\mathbf{H} \cdot d{\mathbf{l}} = \int_0^{2\pi} H_\varphi(S) s d{\varphi}$, so then $(I_e)_S = \int_0^{2\pi} H_\varphi(S) s d{\varphi}$. Thus, for $s > a$ (where $a$ is the radius of the wire), $2\pi s H_\varphi = I_e$, and for $s < a$, $2\pi s H_\varphi = I_e \frac{s^2}{a^2}$.
---
By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carrying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
By Helmholtz Theorem, we know that $\mathbf{H}(\mathbf{r}) = \nabla \times{\mathbf{A}(\mathbf{r})}$. For a current-carrying wire, $\mathbf{A}(\mathbf{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
$$
\vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
\mathbf{H}(\mathbf{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\mathbf{r}-\mathbf{r'}) \times d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|^3}
$$
---
Consider a current loop instead, on the $x-y$ plane and current $I$. Then, $r = z \vu{z}$ and $\dd{\vb{l'}} = R \vu{\varphi'} \dd{\phi'}$, and the magnetic field collapses to $\vb{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \vu{z}$
Consider a current loop instead, on the $x-y$ plane and current $I$. Then, $r = z \hat{\mathbf{z}}$ and $d{\mathbf{l'}} = R \hat{\mathbf{\varphi'}} d\phi'$, and the magnetic field collapses to $\mathbf{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \hat{\mathbf{z}}$
---
Consider some infinite bar magnet with height $h$ and width $w$. Then, the top and bottom surfaces will have a magnetic charge with density $\vb{J}_m^+ = M_0 \vb{b} \delta(z - h)$ and $\vb{J}_m^- = -M_0 \vb{v} \delta(z)$ respectively. By definition, $I_m = M_0 w v$.
Consider some infinite bar magnet with height $h$ and width $w$. Then, the top and bottom surfaces will have a magnetic charge with density $\mathbf{J}_m^+ = M_0 \mathbf{b} \delta(z - h)$ and $\mathbf{J}_m^- = -M_0 \mathbf{v} \delta(z)$ respectively. By definition, $I_m = M_0 w v$.
Now, consider a loop around only the top of the conductor. Then,
$$
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = I_m = M_o w v
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = I_m = M_o w v
$$
By definition,
$$
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = -\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S}
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = -\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S}
$$
Applying Stokes theorem,
$$
\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S} = M_0 w v
\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S} = M_0 w v
$$
## Section 5.3 - Forces on Moving Charges and Current
Consider an electric charge moving with velocity $\vb{v}$ in a magnetic parallel plate capacitor with charge densities $\plusminus \sigma_m$. That is, $\mu_0 \vb{H} = \sigma_m \vu{z}$. Then, we can apply theorems to see the resulting force.
Consider an electric charge moving with velocity $\mathbf{v}$ in a magnetic parallel plate capacitor with charge densities $\pm \sigma_m$. That is, $\mu_0 \mathbf{H} = \sigma_m \hat{\mathbf{z}}$. Then, we can apply theorems to see the resulting force.
**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
**Theorem**. *Lorentz Force Law* states that $\mathbf{F} = q_e \mathbf{v} \times \mu_0 \mathbf{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\mathbf{F} = q_e (\mathbf{E} + \mathbf{v} \times \mu_0 \mathbf{H})$.
**Theorem**. *Ampere's Force Law* states that generalizing the previous theorem, we can see that
$$
\dd{\vb{F}} = I_e \dd{\vb{L}} \cross \u_0 \vb{H}(\vb{r})
$$
$$d\mathbf{F} = I_e d{\mathbf{L}} \times \mu_0 \mathbf{H}(\mathbf{r})$$
## Section 5.4 - Multipole Expansion of a Vector Potential

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@@ -4,13 +4,13 @@
Each point $P \in \mathbb{R}^3$ can be written as $(x, y, z)$. However, we will utilize the convention below:
$$\vb{r} = x \vu{x} + y \vu{y} + z \vu{z}$$
$$\mathbf{r} = x \hat{\mathbf{x}} + y \hat{\mathbf{y}} + z \hat{\mathbf{z}}$$
However, when working with ambiguous unit basis, it may be better to use the following:
$$\vb{r} = r_1 \vu{e_1} + r_2 \vu{e_2} + r_3 \vu{e_3}$$
$$\mathbf{r} = r_1 \hat{\mathbf{e_1}} + r_2 \hat{\mathbf{e_2}} + r_3 \hat{\mathbf{e_3}}$$
Here, $r_1 = x, r_2 = y, r_3 = z$ and $\vu{e_1} = \vu{x_1}, \vdots$. That is, $\vb{r} = \sum r_i \vu{e_i}$.
Here, $r_1 = x, r_2 = y, r_3 = z$ and $\hat{\mathbf{e_1}} = \hat{\mathbf{x_1}}, \ldots$. That is, $\mathbf{r} = \sum r_i \hat{\mathbf{e_i}}$.
Vectors are an $\mathbb{R}$-module, with addition defined component-wise. Additionally, the scalar (dot) product and vector (cross) product are defined as usual.
@@ -24,27 +24,27 @@ Regarding time, classically, time is universal.
**Definition**. The *mass* $m$ of an object characterizes its translational inertia, and is measured in kilograms (kg)
**Definition**. The *force* $\vb{F}$ exerted on an object is a push or pull on said object and is measured in Newtons (N). Note that force is a vector.
**Definition**. The *force* $\mathbf{F}$ exerted on an object is a push or pull on said object and is measured in Newtons (N). Note that force is a vector.
## Section 1.4 - Newton's First and Second Laws; Inertial Frames
**Definition**. A *point mass* or *particle* is a convenient fiction, in which an object with mass has no size. It may move in space but has no internal degrees of freedom. Additionally, it may not have any rotational or vibrational kinetic energy.
**Theorem**. Newton's First Law. In the absence of external forces, a particle moves with constant velocity $\vb{v}$.
**Theorem**. Newton's First Law. In the absence of external forces, a particle moves with constant velocity $\mathbf{v}$.
**Theorem**. Newton's Second Law. Given any particle with mass $m$, the net force $\vb{F}$ on the particle is always equal to the particle's mass times its acceleration. That is,
**Theorem**. Newton's Second Law. Given any particle with mass $m$, the net force $\mathbf{F}$ on the particle is always equal to the particle's mass times its acceleration. That is,
$$\vb{F} = m\vb{a} = m \ddot{\vb{r}}$$
$$\mathbf{F} = m\mathbf{a} = m dot{\mathbf{r}}$$
This can also be rewritten in terms of momentum. We know that momentum $\vb{p}$ can be written as $\vb{p} = m\vb{v} = m \dot{\vb{r}}$. Then,
This can also be rewritten in terms of momentum. We know that momentum $\mathbf{p}$ can be written as $\mathbf{p} = m\mathbf{v} = m \dot{\mathbf{r}}$. Then,
$$\vb{F} = m\vb{a} = m \dot{\vb{p}} = m \ddot{\vb{r}}$$
$$\mathbf{F} = m\mathbf{a} = m \dot{\mathbf{p}} = m dot{\mathbf{r}}$$
If we have a constant force $\vb{F} = F_0 \vu{x}$, we can write $\ddot{\vu{x}}(t) = \frac{F_0}{m}$. Then,
If we have a constant force $\mathbf{F} = F_0 \hat{\mathbf{x}}$, we can write $dot{\hat{\mathbf{x}}}(t) = \frac{F_0}{m}$. Then,
$$\dot{\vb{x}}(t) = \int \ddot{\vb{x}} dt = v_0 + \frac{F_0}{m}t$$
$$\dot{\mathbf{x}}(t) = \int dot{\mathbf{x}} dt = v_0 + \frac{F_0}{m}t$$
$$\vb{x}(t) = \int \dot{\vb{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
$$\mathbf{x}(t) = \int \dot{\mathbf{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
**Definition**. An *inertial frame* is a reference frame relative to some fixed frame if they are moving with constant velocity in regards to each other. Otherwise, the frames are *non-inertial*.
@@ -54,17 +54,17 @@ $$\vb{x}(t) = \int \dot{\vb{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
**Definition**. Force pairs that operate in the same line as each other (eg. gravitational attraction) are called *central forces*.
Recall that the change in momentum of any particle can be defined as $\dot{\vb{p}}_1 = \vb{F}_1 = \vb{F}_{12} + \vb{F}_{1}^{ext}$. Then, $\dot{\vb{p}}_2 = \vb{F}_{21} + \vb{F}_2^{ext}$. As $\vb{P} = \vb{p}_1 + \vb{p}_2$, we can see that
Recall that the change in momentum of any particle can be defined as $\dot{\mathbf{p}}_1 = \mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{1}^{ext}$. Then, $\dot{\mathbf{p}}_2 = \mathbf{F}_{21} + \mathbf{F}_2^{ext}$. As $\mathbf{P} = \mathbf{p}_1 + \mathbf{p}_2$, we can see that
$$\dot{\vb{P}} = \vb{p}_1 + \vb{p}_2 = \vb{F}_{12} +\vb{F}_{21} + \vb{F}_{1}^{ext} + \vb{F}_{2}^{ext} = \vb{F}_{1}^{ext} + \vb{F}_{2}^{ext} = \vb{F}^{ext}$$
$$\dot{\mathbf{P}} = \mathbf{p}_1 + \mathbf{p}_2 = \mathbf{F}_{12} +\mathbf{F}_{21} + \mathbf{F}_{1}^{ext} + \mathbf{F}_{2}^{ext} = \mathbf{F}_{1}^{ext} + \mathbf{F}_{2}^{ext} = \mathbf{F}^{ext}$$
From this, we can see that if $\vb{F}^{ext} = 0$, then $\vb{P}$ is a constant.
From this, we can see that if $\mathbf{F}^{ext} = 0$, then $\mathbf{P}$ is a constant.
This argument can be generalized to multi-particle systems. Consider particle $\alpha$. Then, $\dot{\vb{p}}_\alpha = \vb{F}_\alpha = \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} + \vb{F}_\alpha^{ext}$. We can then see that
This argument can be generalized to multi-particle systems. Consider particle $\alpha$. Then, $\dot{\mathbf{p}}_\alpha = \mathbf{F}_\alpha = \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} + \mathbf{F}_\alpha^{ext}$. We can then see that
$$\dot{\vb{P}} = \sum_\alpha \vb{p}_\alpha = \sum_\alpha \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} + \sum_\alpha \vb{F}_\alpha^{ext}$$
$$\dot{\mathbf{P}} = \sum_\alpha \mathbf{p}_\alpha = \sum_\alpha \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} + \sum_\alpha \mathbf{F}_\alpha^{ext}$$
With $\sum_\alpha \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} = \sum_\alpha \sum_{\beta > \alpha} \vb{F}_{\alpha\beta}( + \vb{F}_{\beta\alpha})$, we can see that $\dot{\vb{P}} = \sum_\alpha \vb{F}_\alpha^{ext}$.
With $\sum_\alpha \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} = \sum_\alpha \sum_{\beta > \alpha} \mathbf{F}_{\alpha\beta}( + \mathbf{F}_{\beta\alpha})$, we can see that $\dot{\mathbf{P}} = \sum_\alpha \mathbf{F}_\alpha^{ext}$.
---

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@@ -2,13 +2,13 @@
## Section 2.1 - Air Resistance
**Definition**. The *drag*, or resistive force on an object due to the atmosphere, is denoted as $\vb{f}$. Note that this is **not** the force density, but the overall force. In most cases, this force directly opposes the direction of motion. If not, the other component is known as *lift*, however this is mostly negligible.
**Definition**. The *drag*, or resistive force on an object due to the atmosphere, is denoted as $\mathbf{f}$. Note that this is **not** the force density, but the overall force. In most cases, this force directly opposes the direction of motion. If not, the other component is known as *lift*, however this is mostly negligible.
We define air resistance as $\vb{f} = -f(v) \vu{v}$. We consider two types in this text: linear, where $f(v) = f_{lin} = bv$, and quadratic, where $f(v) = f_{quad} = cv^2$. Note that often times we consider both, and state that $f(v) = f_{lin} + f_{quad} = bv + cv^2$.
We define air resistance as $\mathbf{f} = -f(v) \hat{\mathbf{v}}$. We consider two types in this text: linear, where $f(v) = f_{lin} = bv$, and quadratic, where $f(v) = f_{quad} = cv^2$. Note that often times we consider both, and state that $f(v) = f_{lin} + f_{quad} = bv + cv^2$.
The linear term comes from viscous drag and is generally proportional to the viscosity of the medium, while quadratic drag tends to arise from the particle needing to accelerate the mass of air which it is constantly colliding against.
In some cases, we can calculate these coefficients. With $D$ as the diameter of a spherical object, and $\beta$ and $\gamma$ as properties of the medium, we can state that $b = \beta D$ and $c = \gamma D^2$. In air at STP, $\beta = 1.6 \times 10^{-4} \text{N} \vdot \text{s}/\text{m}^2$, and $\gamma = 0.25 \text{N} \vdot \text{s}^2/\text{m}^4$.
In some cases, we can calculate these coefficients. With $D$ as the diameter of a spherical object, and $\beta$ and $\gamma$ as properties of the medium, we can state that $b = \beta D$ and $c = \gamma D^2$. In air at STP, $\beta = 1.6 \times 10^{-4} \text{N} \cdot \text{s}/\text{m}^2$, and $\gamma = 0.25 \text{N} \cdot \text{s}^2/\text{m}^4$.
Oftentimes, one factor is far more impactful than the other, meaning that the smaller of the two may be neglected. To do so, compute the following ratio:
@@ -18,9 +18,9 @@ Note that the result is expected to be of the same order of magnitude as the *Re
## Section 2.2 - Linear Air Resistance
First, consider the case in which the drag force is negligible. Then, we see that with $\vb{F}_g = \vb{w} = mg$ and $\vb{F}_{drag} = \vb{f} = -b\vb{v}$, then Newton's second law tells us that
First, consider the case in which the drag force is negligible. Then, we see that with $\mathbf{F}_g = \mathbf{w} = mg$ and $\mathbf{F}_{drag} = \mathbf{f} = -b\mathbf{v}$, then Newton's second law tells us that
$$\vb{F} = m \dot{\vb{v}} = m\vb{g} - b\vb{b}$$
$$\mathbf{F} = m \dot{\mathbf{v}} = m\mathbf{g} - b\mathbf{b}$$
This separates into two equations:
@@ -58,7 +58,7 @@ We can find the horizontal range by setting $y(R) = 0$. This results in an equat
## Section 2.4 - Quadratic Air Resistance
In the case of quadratic air resistance, we see that $\vb{f} = -c v^2 \vu{v}$. Then, we can see that in the horizontal case, $m\frac{dv}{v_x^2} = -c dt$. Thus,
In the case of quadratic air resistance, we see that $\mathbf{f} = -c v^2 \hat{\mathbf{v}}$. Then, we can see that in the horizontal case, $m\frac{dv}{v_x^2} = -c dt$. Thus,
$$v_x(t) = \frac{v_{x0}}{1+cv_{x0}t/m} = \frac{v_{x0}}{1+\tau/t}$$

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@@ -24,11 +24,11 @@
$$PV = nRT$$
In SI units, $R = 8.31 \frac{\text{J}}{\text{mol} \vdot \text{K}}$.
In SI units, $R = 8.31 \frac{\text{J}}{\text{mol} \cdot \text{K}}$.
**Definition**. Recall that one *mole* of a substance is $6.022 \cross 10^{23}$ units of said substance. This constant, $N_A$, is *Avogadro's Number*.
**Definition**. Recall that one *mole* of a substance is $6.022 \times 10^{23}$ units of said substance. This constant, $N_A$, is *Avogadro's Number*.
Using Avogadro's Number, we can rewrite the Ideal Gas law in terms of molecules, with $N = \text{number of molecules of a gas}$ and $n \vdot N_A = N$. Thus,
Using Avogadro's Number, we can rewrite the Ideal Gas law in terms of molecules, with $N = \text{number of molecules of a gas}$ and $n \cdot N_A = N$. Thus,
$$PV = NkT$$
@@ -72,7 +72,7 @@ $$\frac{1}{2} m \overline{v^2} = \frac{1}{2} m (\overline{v_x^2} + \overline{v_y
This is the average translational kinetic energy for an ideal gas.
**Definition**. A useful unit to measure energy on this scale is the **electron-volt** (eV), which is the kinetic energy gained by an electron that has been accelerated through a voltage difference of one volt. Note that $1 \text{eV} = 1.6 \cross 10^{-19} \text{J}$.
**Definition**. A useful unit to measure energy on this scale is the **electron-volt** (eV), which is the kinetic energy gained by an electron that has been accelerated through a voltage difference of one volt. Note that $1 \text{eV} = 1.6 \times 10^{-19} \text{J}$.
Note that the average speed in this model can be obtained as follows:
@@ -90,7 +90,7 @@ We are familiar with energy in the form of $\frac{1}{2}ab_{x, y, z}^2$, where $a
If an object contains $N$ molecules, each with $f$ degrees of freedom, the total (average) thermal energy is
$$U_\text{thermal} = N \vdot f \vdot \frac{1}{2}{k}{T} $$
$$U_\text{thermal} = N \cdot f \cdot \frac{1}{2}{k}{T} $$
In monoatomic molecules, each molecule has $3$ degrees of freedom, corresponding to the translational position. In diatomic gasses, there are $2$ additional rotational degrees of freedom.
@@ -120,7 +120,7 @@ Note that with heat engines, we often see $Q - W$, where $W$ instead represents
**Theorem**. The First Law of Thermodynamics. $\Delta U = Q + W$, along with the Law of Conservation of Energy. In other words, you can't win the game.
**Definition**. The SI unit of energy is the *Joule*, where $1J = 1 \text{kg} \vdot \text{m}^2 / \text{s}^2$.
**Definition**. The SI unit of energy is the *Joule*, where $1J = 1 \text{kg} \cdot \text{m}^2 / \text{s}^2$.
**Definition**. The imperial unit of energy is the *calorie*, or the amount of energy to heat one gram of water by $1 \degree \text{C}$. The exact conversion factor is defined as $1 \text{J} = 4.184 \text{cal}$
@@ -140,11 +140,11 @@ We know that the force exerted by the piston is equal to the pressure times the
However, this assumes constant pressure. For a non-constant pressure, $P(V)$, we know that $W = \int F dx = - \int P(V) dV$.
**Definition**. In *isothermal compression*, we see that $\delta T = 0$. Then, $T$ is a constant, so $nRT$ is constant meaning $PV$ is constant. Then, as $P(V) = (P_0V_0)/(V)$, we can see that $W = \int_V_1^V_2 P(V) dV = nRT \ln(V_i/V_f) = PV \ln(P_f/P_i)$
**Definition**. In *isothermal compression*, we see that $\delta T = 0$. Then, $T$ is a constant, so $nRT$ is constant meaning $PV$ is constant. Then, as $P(V) = (P_0V_0)/(V)$, we can see that $W = \int_{V_1}^{V_2} P(V) dV = nRT \ln(V_i/V_f) = PV \ln(P_f/P_i)$
**Definition**. The line formed by isothermal compression on a $PV$-diagram is called an *isotherm*.
**Definition**. In *adiabatic compression*, we assume that $Q = 0$. Then, we see that $PV^\gamma$ is constant, with $\gamma$ being the adiabatic constant. Note that as $U = \frac{f}{2}NkT = W$, we can see that $V_2 T_2^{f/2} = V_1 T_1^{f/2}$, and \$gamma = (f+2)(f)$. Additionally, we see that $W$ = $\frac{1}{1-\gamma}\frac{P_2 V_2 - P_1 V_1}$
**Definition**. In *adiabatic compression*, we assume that $Q = 0$. Then, we see that $PV^\gamma$ is constant, with $\gamma$ being the adiabatic constant. Note that as $U = \frac{f}{2}NkT = W$, we can see that $V_2 T_2^{f/2} = V_1 T_1^{f/2}$, and \$gamma = (f+2)(f)$. Additionally, we see that $W$ = $\frac{P_2 V_2 - P_1 V_1}{1-\gamma}$
## Section 1.6 - Heat Capacity
@@ -154,7 +154,7 @@ However, this assumes constant pressure. For a non-constant pressure, $P(V)$, we
In a system, we can note that
$$$C = \frac{Q}{\Delta T} = \frac{\Delta U - W}{\Delta T}$$
$$C = \frac{Q}{\Delta T} = \frac{\Delta U - W}{\Delta T}$$
Consider the case in which volume is constant. Under these circumstances, we can calculate the *heat capacity at constant volume*, and as volume does not change, $W = 0$, so

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@@ -22,7 +22,7 @@ Assuming the external magnetic field points up, we note that an up-dipole contai
## Section 2.2 - The Einstein Model of a Solid
Consider a collection of microscopic systems that can each store any discrete amount of quantized energy. If each system is a harmonic oscillator, we know from [Quantum Mechanics](../../todo.md) that the potential energy is $\frac{1}{2}k_s x^2$, where $k_s$ is the spring constant. Then, the size of energy units is $hf$, where $h$ is Planck's constant ($h = 6.63 \times 10^{-34} \text{J} \vdot \text{s}$) and $f$ is the natural frequency of the oscillator ($f = \frac{1}{2\pi} \sqrt{k_s / m}$).
Consider a collection of microscopic systems that can each store any discrete amount of quantized energy. If each system is a harmonic oscillator, we know from [Quantum Mechanics](../../todo.md) that the potential energy is $\frac{1}{2}k_s x^2$, where $k_s$ is the spring constant. Then, the size of energy units is $hf$, where $h$ is Planck's constant ($h = 6.63 \times 10^{-34} \text{J} \cdot \text{s}$) and $f$ is the natural frequency of the oscillator ($f = \frac{1}{2\pi} \sqrt{k_s / m}$).
**Definition**. For a three-dimensional solid, each particle can oscillate in three dimensions. Thus, if there are $N$ oscillators, there are $N/3$ particles. A solid modeled as such is known as an *Einstein solid*.

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@@ -24,5 +24,9 @@ plugins:
- git-authors
extra_javascript:
- javascripts/mathjax.js
- https://unpkg.com/mathjax@3/es5/tex-mml-chtml.js
- javascripts/katex.js
- https://unpkg.com/katex@0/dist/katex.min.js
- https://unpkg.com/katex@0/dist/contrib/auto-render.min.js
extra_css:
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