nyuuzyou/wikis huggingface dataset:
Macroscopic Electric Field in Body

== Theorem ==
Let $B$ be a body of matter.
Let $P$ be a point inside $B$ whose position vector is $\mathbf r$.
The macroscopic electric field at $P$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V' + \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all surfaces} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
where:
:$\d V'$ is an infinitesimal volume element
:$\d S'$ is an infinitesimal area element
:$\mathbf r'$ is the position vector of $\d V'$ or $\d S'$ as appropriate
:$\map \rho {\mathbf r'}$ is the macroscopic charge density of the macroscopic electric field at $\mathbf r'$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
From Field Generated by Macroscopic Charge Density, the electric field at $P$ generated by the macroscopic charge density within $B$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$
From Field Generated by Surface Charge Density, the electric field at $P$ generated by the surface charge density over $B$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all surfaces} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
We have that the Electric Field satisfies Principle of Superposition.
Hence the total macroscopic electric field at $P$ is given by the sum of these:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V' + \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all surfaces} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field: $(1.10)$
Category:Macroscopic Electric Fields
Macroscopic Electric Field in Conductor in Static Equilibrium is Zero

== Theorem ==
Let $B$ be a body made out of an electrically conducting substance.
Let $B$ be in static equilibrium.
Then the macroscopic electric field within $B$ is everywhere zero.
== Proof ==
By definition of static equilibrium there are no forces upon $B$.
That includes forces caused by an electric field.
From Distribution of Macroscopic Electric Charge within Conductor, the macroscopic electric field is constant throughout $B$.
An electric field arises through differences in electric charges.
As we have seen, there are no such differences.
Hence the result.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Electrical Conductors
Category:Macroscopic Electric Fields
Category:Static Equilibrium
Magic Constant of Magic Cube

== Theorem ==
The magic constant of a magic cube of order $n$ is given by:
:$C_n = \dfrac {n \paren {n^3 + 1} } 2$
== Proof ==
Let $M_n$ denote a magic cube of order $n$.
By Sum of Terms of Magic Cube, the total of all the entries in $M_n$ is given by:
:$T_n = \dfrac {n^3 \paren {n^3 + 1}} 2$
There are $n^2$ rows in $M_n$, each one with the same magic constant.
Thus the magic constant $C_n$ of the magic cube $M_n$ is given by:
=== Sequence ===
Category:Magic Cubes
Magic Constant of Magic Cube/Sequence

== Sequence of Magic Constants of Magic Cubes ==
The sequence of the magic constants of magic cubes of order $n$ begins:
:$1, (9,) \, 42, 130, 315, 651, 1204, 2052, 3285, 5005, 7326, 10 \, 374, 14 \, 287, 19 \, 215, 25 \, 320, 32 \, 776, \ldots$
However, note that while $9 = \dfrac {2 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.
Category:Magic Cubes
Magic Constant of Magic Square

== Theorem ==
The magic constant of a magic square of order $n$ is given by:
:$S_n = \dfrac {n \paren {n^2 + 1} } 2$
== Proof ==
Let $M_n$ denote a magic square of order $n$.
By Sum of Terms of Magic Square, the total of all the entries in a magic square of order $n$ is given by:
:$T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$
There are $n$ rows in $M_n$, each one with the same magic constant.
Thus the magic constant $S_n$ of the magic square $M_n$ is given by:
=== Sequence ===
Category:Magic Squares
Magic Constant of Magic Square/Sequence

== Sequence of Magic Constants of Magic Squares ==
The sequence of the magic constants of magic squares of order $n$ begins:
:$1, (5,) \, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, \ldots$
However, note that while $5 = \dfrac {2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist.
Category:Magic Squares
Magic Constant of Order 3 Magic Hexagon

== Theorem ==
The magic constant of the order 3 magic hexagon is $38$.
It is noted that the central cell is $5$, the same as that of the order 3 magic square.
== Proof ==
== Sources ==
* 
* : $38$
* : $19$
* : $38$
* 
Category:Magic Hexagons
Category:38
Magic Constant of Order 3 Magic Square/Proof 1

== Theorem ==
== Proof ==
Let $M_3$ denote the order $3$ magic square
By Sum of Terms of Magic Square, the total of all the entries in $M_3$ is given by:
:$T_3 = \dfrac {3^2 \left({3^2 + 1}\right)} 2 = \dfrac {9 \times 10} 2 = 45$
As there are $3$ rows of $M_3$, the magic constant of $M_3$ is given by:
:$S_3 = \dfrac {45} 3 = 15$
Category:Magic Constant of Order 3 Magic Square
Magic Constant of Order 3 Magic Square/Proof 2

== Theorem ==
== Proof ==
Let $M_n$ denote the magic square of order $n$.
By Magic Constant of Magic Square, the magic constant of $M_n$ is given by:
:$S_n = \dfrac {n \left({n^2 + 1}\right)} 2$
Setting $n = 3$:
:$S_3 = \dfrac {3 \times 10} 2 = 15$
Category:Magic Constant of Order 3 Magic Square
Magic Constant of Order 4 Magic Square/Proof 1

== Theorem ==
== Proof ==
Let $M_4$ denote an order $4$ magic square
By Sum of Terms of Magic Square, the total of all the entries in $M_4$ is given by:
:$T_4 = \dfrac {4^2 \left({4^2 + 1}\right)} 2 = \dfrac {16 \times 17} 2 = 136$
As there are $4$ rows of $M_4$, the magic constant of $M_4$ is given by:
:$S_4 = \dfrac {136} 4 = 34$
Category:Magic Constant of Order 4 Magic Square
Magic Constant of Order 4 Magic Square/Proof 2

== Theorem ==
== Proof ==
Let $M_n$ denote the magic square of order $n$.
By Magic Constant of Magic Square, the magic constant of $M_n$ is given by:
:$S_n = \dfrac {n \left({n^2 + 1}\right)} 2$
Setting $n = 4$:
:$S_4 = \dfrac {4 \times 17} 2 = 34$
Category:Magic Constant of Order 4 Magic Square
Magic Constant of Order 5 Magic Square/Proof 1

== Theorem ==
== Proof ==
Let $M_5$ denote an order $5$ magic square
By Sum of Terms of Magic Square, the total of all the entries in $M_5$ is given by:
:$T_5 = \dfrac {5^2 \paren {5^2 + 1} } 2 = \dfrac {25 \times 26} 2 = 325$
As there are $5$ rows of $M_5$, the magic constant of $M_5$ is given by:
:$S_5 = \dfrac {325} 5 = 65$
Category:Magic Constant of Order 5 Magic Square
Magic Constant of Order 5 Magic Square/Proof 2

== Theorem ==
== Proof ==
Let $M_n$ denote the magic square of order $n$.
By Magic Constant of Magic Square, the magic constant of $M_n$ is given by:
:$S_n = \dfrac {n \left({n^2 + 1}\right)} 2$
Setting $n = 5$:
:$S_5 = \dfrac {5 \times 26} 2 = 65$
Category:Magic Constant of Order 5 Magic Square
Magic Constant of Smallest Prime Magic Square

== Theorem ==
The magic constant of the smallest prime magic square is $111$.
== Proof ==
The smallest prime magic square (including $1$) is:
As can be seen by inspection, the sums of the elements in the rows, columns and diagonals are $111$:
== Sources ==
* : $111$
* : $111$
Category:Prime Magic Squares
Category:111
Magic Constant of Smallest Prime Magic Square with Consecutive Primes

== Theorem ==
The magic constant of the smallest prime magic square whose elements are consecutive odd primes is $4 \, 440 \, 084 \, 513$.
== Proof ==
The smallest prime magic square whose elements are consecutive odd primes is:
As can be seen by inspection, the sums of the elements in the rows, columns and Diagonal of Array is $4 \, 440 \, 084 \, 513$:
Category:Prime Magic Squares
Category:4,440,084,513
Magic Constant of Smallest Prime Magic Square with Consecutive Primes from 3

== Theorem ==
The magic constant of the smallest prime magic square whose elements are consecutive odd primes from $3$ upwards is $4514$.
== Proof ==
The smallest prime magic square whose elements are the first consecutive odd primes is:
The fact that this is the smallest is proven here.
The sum of the first $144$ prime numbers can either be calculated or looked up: it is $54 \, 169$.
We see that $2$ is not included in this prime magic square, but instead $1$ is used.
So the total of all the elements of this prime magic square is $54 \, 169 - 2 + 1 = 54 \, 168$.
There are $12$ rows, and $12$ columns, each with the same magic constant.
This magic constant must therefore be $\dfrac {54 \, 168} {12} = 4514$.
As can be seen by inspection, the sums of the elements in the rows, columns and diagonals is $4514$.
== Sources ==
* : $144$
* : $144$
::but beware: he makes a mistake and reports it as $4515$.
Category:Prime Magic Squares
Category:4514
Magic Cube/Examples/Order 3

== Example of Magic Cube ==
Order $3$ magic cube:
:$\begin{array}{|c|c|c|}
\hline 2 & 13 & 27 \\
\hline 22 & 9 & 11 \\
\hline 18 & 20 & 4 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 16 & 21 & 5 \\
\hline 3 & 14 & 25 \\
\hline 23 & 7 & 12 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 24 & 8 & 16 \\
\hline 17 & 19 & 6 \\
\hline 1 & 15 & 26 \\
\hline
\end{array}$
Its magic constant is $42$.
== Proof ==
From Magic Constant of Magic Cube, we have that:
:$M_3 = \dfrac {3 \paren {3^3 + 1} } 2$
which (after arithmetic) gives $42$.
== Sources ==
* : $42$
Category:Examples of Magic Cubes
Magic Hexagon of Order 3 is Unique

== Theorem ==
Apart from the trivial order $1$ magic hexagon, there exists only one magic hexagon: the order $3$ magic hexagon:
== Proof ==
We first prove that only order $3$ magic hexagons exist apart from the order $1$ magic hexagon, by calculating the magic constant for each order.
If we ignore the center tile, the rest of the hexagon can be divided into $3$ parallelogram-like parts:
For example, the above can be divided into $\set {3, 17, 18, 19, 7, 1}$, $\set {11, 6, 8, 9, 14, 15}$ and $\set {2, 4, 13, 16, 12, 10}$.
Each part has $n \paren {n - 1}$ tiles.
Thus the order $n$ hexagon has $3 n \paren {n - 1} + 1$ tiles.
Since we require each row to have the same sum, and there are $2 n - 1$ rows:
:The sum of numbers in the hexagon $= \paren {2 n - 1} \times$ magic constant
By Sum of Arithmetic Sequence, the sum of numbers in the tiles is:
:$\dfrac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2}} 2$
Hence the magic constant is:
Since the magic constant is a sum of integers, it must be an integer.
Then:
:$32 \paren {\dfrac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1}}} = \dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1}$ is also an integer.
Since:
:$\dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1} = 72 n^3 - 108 n^2 + 90 n - 27 + \dfrac 5 {2 n - 1}$
We must require $\dfrac 5 {2 n - 1}$ to be an integer.
This happens only if $n = 1$ or $n = 3$.
Thus there are no magic hexagons that are not of order $1$ or $3$.
It remains to show that the order $3$ magic hexagon is unique.
== Sources ==
* : $19$
* : $19$
Category:Magic Hexagons
Magic Square/Examples/Order 3

== Example of Order $3$ Magic Square ==
Order $3$ magic square:
:$\begin {array} {|c|c|c|} \hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end {array}$
== Also known as ==
== Also see ==
* Smallest Magic Square is of Order 3
* Magic Constant of Order 3 Magic Square
== Historical Note ==
== Sources ==
* : $9$
* : $9$
* 
* 
* 
Category:Magic Squares
Magic Square/Examples/Order 3/Historical Note

== Historical Note on Order 3 Magic Square ==
The order 3 magic square was known to the ancient Chinese, where it features in that cultural mythology.
It is the earliest known example of a magic square.
:
It is connected with the mystical traditions of the Far East, and, with its obvious $4$ corner elements and $4$ side elements, evokes correspondences with various $4$-nesses in nature: the $4$ cardinal compass points, the $4$ seasons, and so on.
In Chinese legend going back to at least the $5$th century BCE, it was a gift of a turtle from the River Lo to the Emperor Yu the Great, who supposedly first controlled the Lo and the Yellow rivers.
== Also known as ==
== Sources ==
* 
* : $9$
* : Puzzles from China: The First Magic Square: $59$
* : $9$
* 
* 
Category:Historical Notes
Magic Square/Examples/Order 4

== Examples of Order $4$ Magic Squares ==
There are many order $4$ magic squares.
==== Dürer's Order $4$ Magic Square ====
This is one of the more famous ones, due to :
==== Moessner's Order $4$ Magic Square ====
This one, created by , has extra interesting properties:
== Also see ==
* Magic Constant of Order 4 Magic Square
* Number of Magic Squares of Order 4
Category:Magic Squares
Magic Square/Examples/Order 4/Alfred Moessner

== Example of Order $4$ Magic Square ==
This example of an order $4$ magic square is due to :
:$\begin {array} {|c|c|c|c|} \hline 12 & 13 & 1 & 8 \\ \hline 6 & 3 & 15 & 10 \\ \hline 7 & 2 & 14 & 11 \\ \hline 9 & 16 & 4 & 5 \\ \hline \end {array}$
== Also see ==
* Properties of Moessner's Order $4$ Magic Square
== Historical Note ==
== Sources ==
* 
* : $16$
* : $16$
Category:Magic Squares
Magic Square/Examples/Order 4/Dürer

== Example of Order $4$ Magic Square ==
This example of an order $4$ magic square is due to :
:$\begin {array} {|c|c|c|c|} \hline 16 & 3 & 2 & 13 \\ \hline 5 & 10 & 11 & 8 \\ \hline 9 & 6 & 7 & 12 \\ \hline 4 & 15 & 14 & 1 \\ \hline \end {array}$
== Historical Note ==
== Sources ==
* 
* : $16$
* : Rational Amusements for Winter Evenings: $153$
* : $16$
* 
* 
* 
Category:Magic Squares
Magic Square/Examples/Order 4/Dürer/Historical Note

== Historical Note on Dürer's Magic Square ==
Dürer's magic square appears in 's engraving Melencolia I:
:
The date of the engraving, $1514$, appears in the bottom row of the magic square.
== Sources ==
* : $16$
* : $16$
* 
* 
* 
Category:Historical Notes
Magic Square/Examples/Order 5

== Examples of Order $5$ Magic Squares ==
:$\begin {array} {|c|c|c|c|c|} \hline 23 & 6 & 19 & 2 & 15 \\ \hline 10 & 18 & 1 & 14 & 22 \\ \hline 17 & 5 & 13 & 21 & 9 \\ \hline 4 & 12 & 25 & 8 & 16 \\ \hline 11 & 23 & 7 & 20 & 3 \\ \hline \end {array}$
== Also see ==
* Magic Constant of Order 5 Magic Square
Category:Magic Squares
Magic Square of Order 3 is Unique

== Theorem ==
Up to rotations and reflections, the magic square of order $3$ is unique:
== Proof ==
Let $M_3$ denote the magic square of order $3$.
Each row, column and diagonal of $M_3$ must be a different set of $3$ elements of $\N_9$, where $\N_9$ denotes the set $\set {1, 2, 3, 4, 5, 6, 7, 8, 9}$.
The sets of $3$ elements of $\N_9$ adding to $15$ can be stated:
:$\set {1, 5, 9}, \set {1, 6, 8}$
:$\set {2, 4, 9}, \set {2, 5, 8}, \set {2, 6, 7}$
:$\set {3, 4, 8}, \set {3, 5, 7}$
:$\set {4, 5, 6}$
The number of rows, columns and diagonals of $M_3$ passing through each cell of $M_3$ depends upon where in $M_3$ that cell is positioned.
:The corner cells are each on one row, one column and one diagonal
:The edge cells are each on one row and one column
:The center cells is on one row, one column and both diagonals.
Thus the center cell has to contain an integer which is in at least $4$ of the $3$-element subsets of $\N^*_9$ above.
There is only one such element, which is $5$.
Thus $5$ goes in the center square of $M_3$.
By a similar analysis, it is seen that:
:the corner cells contain $2, 4, 6, 8$
:the edge cells contain $1, 3, 7, 9$.
, let $2$ be placed in the upper left corner of $M_3$.
There is no $3$-element subset of $\N^*_9$ adding to $15$ containing both $1$ and $2$.
It follows that $1$ cannot be placed on either the upper edge or left hand edge.
Thus $1$ must go into the right hand edge or the bottom edge.
, let $1$ be placed on the right hand edge.
Up to rotations and reflections, the placements of these three elements is unique.
Hence $M_3$ so far looks like this:
:$\begin {array} {|c|c|c|}
\hline 2 & & \\
\hline & 5 & 1 \\
\hline & & \\
\hline
\end {array}$
The population of the remaining cells is forced:
:$\begin {array} {|c|c|c|}
\hline 2 & & \\
\hline 9 & 5 & 1 \\
\hline & & 8 \\
\hline
\end {array}$
:$\begin {array} {|c|c|c|}
\hline 2 & & 6 \\
\hline 9 & 5 & 1 \\
\hline 4 & & 8 \\
\hline
\end {array}$
:$\begin {array} {|c|c|c|}
\hline 2 & 7 & 6 \\
\hline 9 & 5 & 1 \\
\hline 4 & 3 & 8 \\
\hline
\end {array}$
== Sources ==
* : $9$
* : Puzzles from China: The First Magic Square: $59$
* : $9$
Category:Magic Squares
Magma/Examples/Real Numbers

== Examples of Magmas ==
Let $\R$ be the set of real numbers.
$\struct{\R, +}$, $\struct{\R, -}$ and $\struct{\R, \times}$ are all magmas.
== Sources ==
* : $\S 4.4$. Gruppoids, semigroups and groups: Example $76$
Category:Examples of Magmas
Magma Subset Product with Self

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$.
Then $\struct {T, \circ}$ is a magma $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.
== Proof ==
By definition:
:$T \circ T = \set {x = a \circ b: a, b \in T}$
=== Necessary Condition ===
Let $\struct {T, \circ}$ be a magma.
Then $T$ is closed.
That is:
:$\forall x, y \in T: x \circ y \in T$
Thus:
:$x \circ y \in T \circ T \implies x \circ y \in T$
=== Sufficient Condition ===
Let $T \circ T \subseteq T$.
Then:
:$x \circ y \in T \circ T \implies x \circ y \in T$
That is, $T$ is closed.
Therefore $\struct {T, \circ}$ is a magma by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
Category:Subset Products
Category:Magmas
Magma is Submagma of Itself

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a submagma of itself.
== Proof ==
By definition, a magma is an algebraic structure $\struct {S, \circ}$ where $\circ$ is closed.
That is:
:$\forall x, y \in S: x \circ y \in S$
From Set is Subset of Itself, $S \subseteq S$.
By definition, $\struct {T, \circ}$ is a submagma of $S$ if:
:$\forall x, y \in T: x \circ y \in T$
It follows from the above that $\struct {S, \circ}$ is, by definition, a submagma of itself.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.3$
Category:Magmas
Category:Submagmas
Magnetic Force on Moving Charged Particle

== Theorem ==
Let $p$ be a particle with electric charge $q$ moving with velocity $\mathbf v$ through a magnetic field with magnetic field strength $\mathbf B$.
The force $\mathbf F$ exerted on $p$ by the magnetic field is given by:
:$\mathbf F = q \mathbf v \times \mathbf B$
where $\times$ denotes the vector cross product.
== Proof ==
== Also see ==
* Electric Force on Charged Particle
== Sources ==
* 
* 
Category:Magnetic Fields
Category:Electric Charge
Category:Force
Category:Electromagnetism
Magnitude/Examples/Arbitrary Example 1

== Example of Magnitude ==
Let $\mathbf v$ be the position vector in space defined as:
:$\mathbf v = x \mathbf i + y \mathbf j + z \mathbf k$
The magnitude of $\mathbf v$ is given by:
:$\size {x \mathbf i + y \mathbf j + z \mathbf k} = \sqrt {x^2 + y^2 + z^2}$
== Sources ==
* 
* 
Category:Examples of Magnitude
Magnitude/Examples/Arbitrary Example 2

== Example of Magnitude ==
Let $\mathbf v$ be the position vector in the plane defined as:
:$\mathbf v = x \mathbf i + y \mathbf j$
The magnitude of $\mathbf v$ is given by:
:$\size {x \mathbf i + y \mathbf j} = \sqrt {x^2 + y^2}$
== Sources ==
* 
* 
Category:Examples of Magnitude
Magnitude and Direction of Equilibrant

== Theorem ==
Let $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ be a set of $n$ forces acting on a particle $B$ at a point $P$ in space.
The equilibrant $\mathbf E$ of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ is:
:$\mathbf E = -\ds \sum_{k \mathop = 1}^n \mathbf F_k$
That is, the magnitude and direction of $\mathbf E$ is such as to balance out the effect of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$.
== Proof ==
From Newton's First Law of Motion, the total force on $B$ must equal zero in order for $B$ to remain stationary.
That is, $\mathbf E$ must be such that:
:$\mathbf E + \ds \sum_{k \mathop = 1}^n \mathbf F_k = \bszero$
That is:
:$\mathbf E = -\ds \sum_{k \mathop = 1}^n \mathbf F_k$
Category:Force
Category:Equilibrants
Magnitude of Coriolis Acceleration

== Theorem ==
Let $B$ be a body in a rotating frame of reference $\FF$.
The magnitude of the Coriolis acceleration $\mathbf a$ on $B$ caused by the Coriolis force is given by:
:$\size {\mathbf a} = 2 v \omega$ where:
:$v$ is the magnitude of the velocity of $B$ relative to $\FF$
:$\omega$ is the magnitude of the angular velocity of $\FF$ relative to an inertial frame.
== Proof ==
== Sources ==
* 
* 
Category:Coriolis Forces
Magnitude of Coriolis Force

== Theorem ==
Let $B$ be a body in a rotating frame of reference $\FF$.
The magnitude of the Coriolis force $\mathbf F$ on $B$ is given by:
:$\size {\mathbf F} = 2 m v \omega$ where:
:$m$ is the mass of $B$
:$v$ is the magnitude of the velocity of $B$ relative to $\FF$
:$\omega$ is the magnitude of the angular velocity of $\FF$ relative to an inertial frame.
== Proof ==
== Sources ==
* 
* 
Category:Coriolis Forces
Magnitude of Electric Field caused by Point Charge

== Theorem ==
Let $q$ be a point charge.
Let $\map {\mathbf E} {\mathbf r}$ be the electric field strength due to $q$ at a point $P$ whose position vector is $\mathbf r$.
The magnitude of the electric field strength due to $q$ at $P$ is given by:
Then:
:$\size {\map {\mathbf E} {\mathbf r} } = \dfrac {\size q} {4 \pi \epsilon_0 r^2}$
where:
:$r$ is the distance $P$ is from $q$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
The following diagram shows the points of equal magnitude of $\map {\mathbf E} {\mathbf r}$ as dotted circles:
:
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field
Category:Electric Fields
Category:Point Charges
Magnitude of Lorentz-Fitzgerald Contraction

== Theorem ==
Let $B$ be a body moving in an inertial frame of reference $\FF_1$.
Let an observer $O$ be in a different inertial frame of reference $\FF_2$.
Let $v$ be the magnitude of the relative velocity of $\FF_1$ $\FF_2$.
Let $l$ be the length of $B$ in the direction of the relative velocity of $\FF_1$ $\FF_2$.
Let $l'$ be the length of $B$ in the direction of the relative velocity of $\FF_1$ $\FF_2$ as measured by $O$.
Then:
:$l' = l \sqrt {1 − \dfrac {v^2} {c^2} }$
where $c$ denotes the speed of light.
== Proof ==
== Sources ==
* 
* 
Category:Lorentz-Fitzgerald Contraction
Category:Relativistic Mechanics
Magnitude of Projection of Complex Number on Another

== Theorem ==
Let $z_1$ and $z_2$ denote complex numbers in vector form.
Let $\map {\pr_1} {z_1, z_2}$ denote the projection of $z_1$ on $z_2$.
Then:
:$\cmod {\map {\pr_1} {z_1, z_2} } = \dfrac {\cmod {z_1 \circ z_2} } {\cmod {z_2} }$
where:
:$z_1 \circ z_2$ denotes complex dot product
:$\cmod {z_2}$ denotes complex modulus.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product: $3.$
Category:Complex Dot Product
Category:Complex Modulus
Magnitude of Relative Velocity

== Theorem ==
Let $A$ and $B$ be bodies in space, both moving in the same direction.
Let $\mathbf v_A$ and $\mathbf v_B$ denote the velocities of $A$ and $B$.
Let $\mathbf v_{AB}$ denote the velocity of $A$ relative to $B$.
Then:
:$v_{AB} = \dfrac {\size {v_A - v_B} } {1 - \dfrac {v_A v_B} {c^2} }$
where:
:$v_{AB}$, $v_A$ and $v_B$ are the magnitudes of $\mathbf v_{AB}$, $\mathbf v_A$ and $\mathbf v_B$ respectively
:$c$ denotes the speed of light.
== Proof ==
== Sources ==
* 
* 
Category:Relative Velocity
Category:Relativistic Mechanics