Physics of fields — physics

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Spring constant

In physics, Hooke's law is an empirical law which states

${\vec {F}}=-k\,{\vec {x}}$

where

( $x$ ) is the displacement. The distance the spring is displaced.
( $F$ ) is the restoring force
The spring constant $k$ is a constant factor characteristic of the spring (i.e., its stiffness)
- The spring constant is negative because the restoring force points in the opposite direction of the displacement vector

The spring constant $k$ can be defined as the force per displacement:

$k={\frac {F}{x}}$

The larger $k$ is the stronger the force and the less motion it "permits". The displacement "x" is given by:

$x={\frac {F}{k}}$

Simple harmonic oscillator

A simple harmonic oscillator is an oscillator that is neither driven nor damped.

It consists of a mass $m$ , which when displaced from its equilibrium position at the point $x = 0$ , experiences a single restoring force F which pulls the mass back toward the equilibrium position and is proportional to the displacement x:

$\vec F = -k \vec x,$ where k is a positive constant.

Balance of forces (Newton's second law) for the system is $F = m a = m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = m\ddot{x} = -k x.$

Solving this differential equation, we find that the motion is described by the function

$x(t)=A\sin(\omega t+\varphi ),$

with a frequency that is independent of the amplitude:

$\omega = \sqrt{\frac{k}{m}}$

The motion is periodic, repeating itself in a sinusoidal fashion with constant amplitude A. The position at a given time t also depends on the phase $\varphi$ , which determines the starting point on the sine wave.

The period and frequency are determined by the size of the mass m and the force constant k, while the amplitude and phase are determined by the starting position and velocity.

Gravitational field

According to Newton's law of universal gravitation, the magnitude of the attractive force ( $F$ ) between two bodies each with a spherically symmetric density distribution is directly proportional to the product of their masses, $m 1$ and $m 2$ , and inversely proportional to the square of the distance, $r$ , directed along the line connecting their centres of mass:

${\vec {F}}={\frac {G_{c}}{4\pi r^{2}}}m_{1}m_{2}{\hat {\mathbf {r} }}$

Where

$\vec F$ = The force between the masses which is negative (meaning they attract each other) because $G c$ is negative
$4\pi r^2$ is the surface area of a sphere of radius $r$
G_c is the corrected Gravitational constant $G_{c}=-4\pi G$ $G_{c}=-4\pi G$ where
- $G$ is the gravitational constant. (A constant of proportionality).
$\hat{\mathbf{r}}$ = a unit vector pointing outward from the origin (the source mass)

The gravitational field is the force per mass on a test mass due to the mass at distance "r" that is the source of the gravitational field.

${\text{Grav field}}={\frac {\vec {F}}{m_{t}}}=acc={\frac {G_{c}}{4\pi r^{2}}}m_{s}{\hat {\mathbf {r} }}$

It has units of acceleration because $F=ma$ and is negative because $G c$ is negative

The bigger $G c$ is the stronger the force is.

Colloquially, the gravitational constant is also called "Big G", distinct from "small g" ( $g$ ), which is the local gravitational field of Earth (also referred to as free-fall acceleration). Where $M_\oplus$ is the mass of Earth and $r_{\oplus}$ is the radius of Earth, the two quantities are related by:

$g_{acc}={\frac {G_{c}}{4\pi r_{\oplus }^{2}}}M_{\oplus }{\hat {\mathbf {r} }}$

Gravitational displacement

If we define the displacement of the gravitational field to be

$\mathbf {\vec {D_{g}}} \equiv {\frac {\mathbf {\vec {g}} }{G_{c}}}={\frac {G_{c}}{G_{c}4\pi r^{2}}}m_{s}{\hat {\mathbf {r} }}={\frac {m_{s}}{4\pi r^{2}}}{\hat {\mathbf {r} }}$

Then it follows that

$\nabla \cdot \mathbf {D_{g}} =\nabla \cdot {\frac {m_{s}}{4\pi r^{2}}}{\hat {\mathbf {r} }}=\rho _{g}$

Where

$D g$ = The gravitational displacement which is positive because both g and $G c$ are negative
$\nabla$ = divergence
$\rho_g$ = mass density = mass/volume

why 4 pi?

It is tempting to imagine the mass in a single point at the center but then where does the $4\pi$ come from? It comes from the fact that $\rho$ gives the mass density at a point and for that single point all that matters (assuming a spherical distribution of matter) is the shell of matter containing that point.

Imagine a hollow spherical shell of density $\rho$ , surface area $4\pi r^2$ , arbitrarily small thickness dt (not zero), and therefore infinitesimal mass $m_{shell}=4\pi r^{2}\cdot dt\cdot \rho .$ Centering the sphere on the origin and calculating $\nabla \cdot \mathbf {D_{g}}$ along the x axis.

$\nabla \cdot \mathbf {D_{g}} ={\frac {\partial D_{gx}}{\partial x}}+{\frac {\partial D_{gy}}{\partial y}}+{\frac {\partial D_{gz}}{\partial z}}={\frac {\partial D_{gx}}{\partial x}}+0+0$

This reduces the problem to one dimension. By setting $dt=\partial x$ (which we are free to do since it is arbitrary) we see that at the surface of the sphere ${\partial D_{gx}}$ is just how much $D_{gx}$ changes as you go from one side of the shell to the other. By the shell theorem the value inside the shell is zero and the value of the other side is ${\frac {4\pi r^{2}\cdot dt\cdot \rho }{4\pi r^{2}}}$ . So our answer is ${\frac {dt\cdot \rho }{\partial x}}=\rho$ because $dt=\partial x$ .

To calculate the total mass we would then have to integrate $\nabla \cdot \mathbf {D_{g}}$ over the surface of the sphere (thus introducing a factor of $4\pi r^2$ ) to get our original $m_{shell}$ .

An integral over a volume tells how much mass is within the volume. Greens theorem says that the integral over the volume is equal to the number of field line crossing the surface of that volume. This is obviously true since every field line ends at a unit of mass. Mass is the source of the gravitational field lines.

$\int _{V}(\nabla \cdot \mathbf {F} )\,dV\;=\;\oint _{\partial V}\mathbf {F} \cdot d\mathbf {A}$

Electric field

The force between two separated electric charges with spherical symmetry (in the vacuum of classical electromagnetism) is given by Coulomb's law:

${\vec {F_{\text{C}}}}={\frac {\varepsilon _{0}^{-1}}{4\pi r^{2}}}q_{1}q_{2}{\hat {\mathbf {r} }}$

Where

$4\pi r^{2}.$ is the surface area of a sphere of radius $r$
q₁ is charge one
q₂ is charge two
r is the distance between their centres.
The constant $\varepsilon _{0}^{-1}$ is $\frac{1}{\varepsilon_0}$ ${\frac {1}{\varepsilon _{0}}}$ where
- $\varepsilon_0$ is the vacuum electric permittivity commonly denoted $ε 0$ (pronounced "epsilon nought" or "epsilon zero"), is the value of the absolute dielectric permittivity of classical vacuum. It may also be referred to as the permittivity of free space, the electric constant, or the distributed capacitance of the vacuum. It is a measure of how dense of an electric field is "permitted" to form in response to electric charges and relates the units for electric charge to mechanical quantities such as length and force.
$\hat{\mathbf{r}}$ is a unit vector from the origin (source charge)

The the electric field $E$ can be derived from Coulomb's law. By choosing one of the point charges to be the source, and the other to be the test charge, it follows from Coulomb's law that the magnitude of the electric field $E$ created by a single source point charge $q_{s}$ at a certain distance from it r in vacuum is given by

$\mathbf {\vec {E}} ={\frac {\vec {F_{\text{C}}}}{q_{t}}}={\frac {\varepsilon _{0}^{-1}}{4\pi r^{2}}}q_{s}{\hat {\mathbf {r} }}$

The bigger $\varepsilon _{0}^{-1}$ is the bigger the force.

Displacement field

The charge density displacement field "D" is defined as

$\mathbf {{\vec {D}}_{0}} \equiv {\frac {\vec {\mathbf {E} }}{\varepsilon _{0}^{-1}}}$

where

$\varepsilon_0$ is the vacuum permittivity (also called permittivity of free space),
E is the electric field,

The charge density displacement field satisfies Gauss's law:

$\nabla \cdot \mathbf {{\vec {D}}_{0}} =\rho$

Where

$\nabla \cdot$ is divergence. Divergence is nonzero where field lines begin and end. Electric field lines begin and end at charges.
$\rho$ is charge density

In many problems, it is more convenient to work with $D$ than with $E$

Displacement current

Differentiating this equation with respect to time defines the displacement current density ( $J D$ ):

$\mathbf {\vec {J}} _{\mathrm {\vec {D}} }={\frac {\partial \mathbf {\vec {D}} }{\partial t}}$

The term on the right hand side is present in free space (where there is no charge) but nevertheless it does have an associated magnetic field, just as a current does due to charge motion. Some authors apply the name displacement current to the term. Thus:

$I_{\mathrm {D} }=\iint _{S}\mathbf {\vec {J}} _{\mathrm {D} }\cdot \operatorname {d} \!\mathbf {S} =\iint _{S}{\frac {\partial \mathbf {\vec {D}} }{\partial t}}\cdot \operatorname {d} \!\mathbf {S} ={\frac {\partial }{\partial t}}\iint _{S}\mathbf {\vec {D}} \cdot \operatorname {d} \!\mathbf {S} ={\frac {\partial \Phi _{\mathrm {D} }}{\partial t}}\,.$

This is the displacement current as it was originally conceived by Maxwell. Maxwell made no special treatment of the vacuum, treating it as a material dielectric medium.

Magnetic field

The magnetic force between two wires of equal length L is given by:

$F={\frac {\mu _{0}}{2\pi rL}}LI_{1}LI_{2}$

where

$2\pi rL$ $2\pi rL$ is the surface area (lateral area) of an open cylinder (having neither top nor bottom)
- r = distance between wires (and the radius of the cylinder)
- L = length of both wires
$\mu_0$ is the vacuum permeability. The bigger $\mu_0$ is the bigger the force
$I$ ₁ = current in wire 1
$I$ ₂ = current in wire 2

So the magnetic field is defined as:

$B={\frac {F}{LI_{t}}}={\frac {\mu _{0}}{2\pi rL}}LI_{s}({\hat {\mathbf {I} }}\times {\hat {\mathbf {r} }})$

where

${\hat {\mathbf {I} }}$ = unit vector along current direction
$\hat{\mathbf{r}}$ = unit vector from wire toward the point of observation

Which gives

$F=LIB=QvB$

The magnetic force $F$ (see Lorentz force) acting on a point particle (like an electron) is given by

$\mathbf {F} =Q\left(\mathbf {v} \times \mathbf {B} \right)$

Where

$F$ is the force on the particle,
$Q$ is the electric charge of the point particle
$v$ is the velocity of the point particle
× denotes the cross product.
$B$ is the magnetic field the particle is moving in

Magnetic displacement field

The magnetic-charge density displacement field $H 0$ is defined:

$\mathbf {H_{0}} \equiv {\frac {1}{\mu _{0}}}\mathbf {B}$

where

$\mu_0$ is the vacuum permeability,

In a vacuum, $B$ and $H 0$ are proportional to each other. Inside a material they are different (see H and B inside and outside magnetic materials). The SI unit of the $H 0$ -field is the ampere per metre (A/m), and the CGS unit is the oersted (Oe).

By Gauss's law for magnetism the divergence of H is zero:

${\begin{aligned}\nabla \cdot \mathbf {H} \,\,\,&=0\;\rho _{m}\\\end{aligned}}$

where

$\nabla \cdot$ is divergence. Divergence is nonzero where field lines begin and end. Magnetic field lines go around in complete circles. They have no beginning nor end.

This means that the magnetic monopole density is zero because (free) magnetic monopoles dont exist.

Energy in fields

The potential energy stored in a spring is given by

${\frac {k}{2}}x^{2}={\frac {1}{2}}F\cdot x$

The total energy per unit volume stored by the gravitational field is

${\frac {E_{nergy}}{L^{3}}}={\frac {G_{c}}{2}}|\mathbf {D_{g}} |^{2}={\frac {1}{2}}(D_{g}\cdot g)$

which is negative because G and g are negative. All physical systems move spontaneously and effortlessly toward lower potential energy. Therefore masses attract thereby creating larger gravitational fields with more negative energy.

The total energy per unit volume stored by the electromagnetic field (in a vacuum) is the electric energy density plus the magnetic energy density:

${\frac {E_{nergy}}{L^{3}}}=u_{\text{EM}}={\frac {\varepsilon _{0}^{-1}}{2}}|\mathbf {D} |^{2}+{\frac {\mu _{0}}{2}}|\mathbf {H} |^{2}={\frac {1}{2}}(\mathbf {E} \cdot \mathbf {D} +\mathbf {B} \cdot \mathbf {H} )$

where

$ε$ is the permittivity of the medium in which the field exists,
$\mu$ its magnetic permeability,
$E$ is the electric field vector
$B$ is the magnetic field vector

Both terms are equal in a propagating wave.

Because electromagnetic field energy is positive like charges repel and opposite charges attract.

Units and scaling factors

physical quantities

Scaling factor = S
G	FL²/MM	1/S²
k	FL²/QQ	1/S²
μ₀	FT²/QQ	1/S²
Distance	L	S
Time	T	S
Mass	M	S³
Charge	Q	S³
Density	M/L³	1
velocity v	L/T	1
Acc	L/T²	1/S
Force	M*Acc	S²
Energy	FL	S³
Power	E/T	S²
Grav f	F/M	1/S
E	F/Q	1/S
D	E/k	S
B	F/IL	1/S
H	B/μ₀	1/S³
I	Q/T	S²
V	E/Q	1
C	Q/V	S³
L	VT²/Q	1/S

$F=Ma={\frac {ML}{T^{2}}}$

$k={\frac {F_{s}}{L}}$

${\frac {k}{2}}x^{2}={\frac {1}{2}}F\cdot x$

$G_{c}=F{\frac {L^{2}}{M^{2}}}=Ma{\frac {L^{2}}{M^{2}}}={\frac {aL^{2}}{M}}$

${\text{Grav field}}={\frac {F}{M}}={\frac {Ma}{M}}=a={\frac {L}{T^{2}}}$

${\frac {E}{L^{3}}}={\frac {1}{G_{c}}}a\cdot a={\frac {M}{aL^{2}}}a\cdot a={\frac {Ma}{L^{2}}}={\frac {MaL}{L^{3}}}={\frac {E}{L^{3}}}$

$\varepsilon _{0}^{-1}={\frac {FL^{2}}{Q^{2}}}$

$|\mathbf {E} |={\frac {F}{Q}}$

$\mathbf {D} \equiv {\frac {\mathbf {E} }{\varepsilon _{0}^{-1}}}={\frac {F}{Q}}{\frac {Q^{2}}{FL^{2}}}={\frac {Q}{L^{2}}}$

$\nabla \cdot \mathbf {D} ={\frac {\partial D_{x}}{\partial x}}+{\frac {\partial D_{y}}{\partial y}}+{\frac {\partial D_{z}}{\partial z}}={\frac {D}{L}}={\frac {Q}{L^{3}}}$

$\mathbf {J} _{\mathrm {D} }={\frac {\partial \mathbf {D} }{\partial t}}={\frac {Q}{L^{2}}}{\frac {1}{T}}={\frac {I}{L^{2}}}$

${\frac {E}{L^{3}}}=(\mathbf {E} \cdot \mathbf {D} )={\frac {F}{Q}}{\frac {Q}{L^{2}}}={\frac {F}{L^{2}}}={\frac {FL}{L^{3}}}={\frac {E}{L^{3}}}$

${\frac {E}{L^{3}}}=\varepsilon _{0}^{-1}|\mathbf {D} |^{2}={\frac {FL^{2}}{Q^{2}}}{\frac {Q}{L^{2}}}{\frac {Q}{L^{2}}}={\frac {F}{L^{2}}}={\frac {FL}{L^{3}}}={\frac {E}{L^{3}}}$

$\mu _{0}={\frac {FL^{2}}{L^{2}I^{2}}}={\frac {F}{I^{2}}}={\frac {ML}{T^{2}}}{\frac {T^{2}}{Q^{2}}}={\frac {ML}{Q^{2}}}$

$B={\frac {F}{LI}}={\frac {FT}{LQ}}={\frac {MLT}{T^{2}LQ}}={\frac {M}{TQ}}$

$\mathbf {H} \equiv {\frac {1}{\mu _{0}}}\mathbf {B} ={\frac {I^{2}}{F}}{\frac {F}{LI}}={\frac {I}{L}}={\frac {Q}{LT}}$

${\frac {E}{L^{3}}}=(\mathbf {B} \cdot \mathbf {H} )={\frac {F}{LI}}{\frac {I}{L}}={\frac {F}{L^{2}}}={\frac {FL}{L^{3}}}={\frac {E}{L^{3}}}$

${\frac {E}{L^{3}}}=\mu _{0}|\mathbf {H} |^{2}={\frac {F}{I^{2}}}{\frac {I}{L}}{\frac {I}{L}}={\frac {F}{L^{2}}}={\frac {FL}{L^{3}}}={\frac {E}{L^{3}}}$

The kinetic energy of a moving mass is $E_{k}=mv^{2}$ so its hardly surprising that energy has units of $M L^2 T^{-2}$ yet this remains true even for energy in electric and magnetic fields even though there is no mass associated with them (except $E=mc^2$ )

Speed of light

The wave velocity in a system of identical masses "m" and identical springs of spring constant "k" of any number of dimenstions "n" where each mass is connected to its 2n nearest neighbors only and where the wavelength is large compared to the distance "a" between masses is

$a{\sqrt {\frac {k}{m}}}$

In classical physics, light is described as a type of electromagnetic wave. The classical behaviour of the electromagnetic field is described by Maxwell's equations, which predict that the speed $c$ with which electromagnetic waves (such as light) propagate in vacuum is related to the distributed capacitance and inductance of vacuum, otherwise respectively known as the electric constant $ε 0$ and the magnetic constant $μ 0$ , by the equation

$c={\sqrt {\frac {\varepsilon _{0}^{-1}}{\mu _{0}}}}.$

Or

$c^{2}\mu =\varepsilon _{0}^{-1}$

Maxwells equations

Polarization and magnetization

An external electric field that is applied to a dielectric material, causes a displacement of bound charged elements, its molecules gain electric dipole moment and the dielectric is said to be polarized. Polarization density (or electric polarization, or simply polarization) is the vector field that expresses the volumetric density of permanent or induced electric dipole moments in a dielectric material

A bound charge is a charge that is associated with an atom or molecule within a material. It is called "bound" because it is not free to move within the material like free charges. Positive charged elements are displaced in the direction of the field, and negative charged elements are displaced opposite to the direction of the field. The molecules may remain neutral in charge, yet an electric dipole moment forms.

Polarization can be compared to magnetization, which is the measure of the corresponding response of a material to a magnetic field in magnetism.

The origin of the magnetic moments responsible for magnetization can be either microscopic electric currents resulting from the motion of electrons in atoms, or the spin of the electrons or the nuclei. Net magnetization results from the response of a material to an external magnetic field.

The definitions of electric and magnetic displacements with polarization and magnetization are:

$\mathbf {D} \ =\ {\frac {1}{\varepsilon _{0}^{-1}}}\mathbf {E} +{\color {red}\mathbf {P} }\quad \ =\ \varepsilon _{0}(1+\chi _{\text{e}})\mathbf {E} \ =\ \varepsilon _{\text{r}}\varepsilon _{0}\mathbf {E} \ =\ \varepsilon \mathbf {E} ={\frac {1}{\varepsilon ^{-1}}}\mathbf {E}$

${\begin{aligned}\mathbf {H} &={\frac {1}{\mu _{0}}}\mathbf {B} -{\color {red}\mathbf {M} }\quad ={\frac {1}{\mu _{0}}}\mathbf {B} -{\frac {1}{\mu _{0}}}{\frac {\chi _{m}}{(1+\chi _{m})}}\mathbf {B} ={\frac {1}{\mu _{0}}}\left(1-{\frac {\chi _{m}}{1+\chi _{m}}}\right)\mathbf {B} ={\frac {1}{\mu _{0}}}\left({\frac {1}{1+\chi _{m}}}\right)\mathbf {B} ={\frac {1}{\mu _{0}}}\mu _{r}\mathbf {B} ={\frac {\mathbf {B} }{\mu }}\end{aligned}}$

where

$ε 0$ is the permittivity of free space
ε_r = ε / ε₀ is the relative permittivity
- $ε$ is the complex frequency-dependent permittivity of the material
$\chi_\text{e}$ is the electric susceptibility;
$μ 0$ the permeability of free space.
$μ$ is the permeability
$μ r$ is the relative permeability
$χ m$ is Volume magnetic susceptibility
$P$ is the polarization field
$M$ is the magnetization field

which are defined in terms of microscopic bound charges and bound currents respectively.

The macroscopic bound charge density $ρ b$ and bound current density $J b$ in terms of polarization $P$ and magnetization $M$ are then defined as

${\begin{aligned}\rho _{\text{b}}&=-\nabla \cdot \mathbf {P} ,\\\mathbf {J} _{\text{b}}&=\nabla \times \mathbf {M} +{\frac {\partial \mathbf {P} }{\partial t}}.\end{aligned}}$

If we define the total, bound, and free charge and current density by

${\begin{aligned}\rho &=\rho _{\text{b}}+\rho _{\text{f}},\\\mathbf {J} &=\mathbf {J} _{\text{b}}+\mathbf {J} _{\text{f}},\end{aligned}}$

and use the defining relations above to eliminate $D$ , and $H$ , the "macroscopic" Maxwell's equations reproduce the "microscopic" equations.

Displacement with polarization

The electric charge density displacement field "D" is defined as

$\mathbf {D} \equiv {\frac {\mathbf {E} }{\varepsilon _{0}^{-1}}}+\mathbf {P} ,$

where

$\varepsilon_0$ is the vacuum permittivity (also called permittivity of free space),
E is the electric field, and
P is the (macroscopic) density of the permanent and induced electric dipole moments in the material, called the polarization density.

Let a volume $d V$ be isolated inside the dielectric. Due to polarization the positive bound charge $\mathrm d q_b^+$ will be displaced a distance $\mathbf{d}$ relative to the negative bound charge $\mathrm d q_b^-$ , giving rise to a dipole moment $\mathrm d \mathbf p = \mathrm d q_b\mathbf d$ .

$\mathbf P={\mathrm d q_b \over \mathrm d V}\mathbf d$

Since the charge $\mathrm d q_b$ bounded in the volume $d V$ is equal to $\rho_b \mathrm d V$ the equation for $P$ becomes:

$\mathbf {P} =\rho _{b}\mathbf {d}$

where $\rho_b$ is the density of the bound charge in the volume under consideration. It is clear from the definition above that the dipoles are overall neutral and thus $\rho_b$ is balanced by an equal density of opposite charges within the volume. Charges that are not balanced are part of the free charge.

The charge density displacement field satisfies Gauss's law in a dielectric:

$\nabla\cdot\mathbf{D} = \rho -\rho_\text{b} = \rho_\text{f}$

In this equation, $\rho_\text{f}$ is the number of free charges per unit volume. These charges are the ones that have made the volume non-neutral, and they are sometimes referred to as the space charge. This equation says, in effect, that the flux lines of D must begin and end on the free charges. In contrast $\rho_\text{b}$ , which is called the bound charge, is an effective density of the charges that are part of a dipole. In the example of an insulating dielectric between metal capacitor plates, the only free charges are on the metal plates and dielectric contains only dipoles. The net, unbalanced bound charge at the metal/dielectric interface balances the charge on the metal plate. If the dielectric is replaced by a doped semiconductor or an ionised gas, etc, then electrons move relative to the ions, and if the system is finite they both contribute to $\rho_\text{f}$ at the edges.

By the divergence theorem, Gauss's law for the field P can be stated in differential form as:

$-\rho _{b}=\nabla \cdot \mathbf {P} ,$

where $\nabla \cdot P$ is the divergence of the field P through a given surface containing the bound charge density $\rho_b$ .

In general, $P$ varies as a function of $E$ depending on the medium. In many problems, it is more convenient to work with $D$ and the free charges than with $E$ and the total charge.

Therefore, a polarized medium, by way of Green's theorem can be split into four components.

The bound volumetric charge density: $\rho_b = -\nabla \cdot \mathbf{P}$
The bound surface charge density: $\sigma_b = \mathbf{\hat{n}}_\text{out} \cdot \mathbf{P}$
The free volumetric charge density: $\rho_f = \nabla \cdot \mathbf{D}$
The free surface charge density: $\sigma_f = \mathbf{\hat{n}}_\text{out} \cdot \mathbf{D}$

Displacement current with polarization

Differentiating this equation with respect to time defines the displacement current density ( $J D$ ), which therefore has two components in a dielectric:

$\mathbf {J} _{\mathrm {D} }={\frac {\partial \mathbf {D} }{\partial t}}+{\frac {\partial \mathbf {P} }{\partial t}}\,.$

The second term on the right hand side, called polarization current density, comes from the change in polarization of the individual molecules of the dielectric material. Polarization results when, under the influence of an applied electric field, the charges in molecules have moved from a position of exact cancellation. The positive and negative charges in molecules separate, causing an increase in the state of polarization $P$ . A changing state of polarization corresponds to charge movement and so is equivalent to a current, hence the term "polarization current".

The first term on the right hand side is present in material media and in free (empty) space but it does have an associated magnetic field, just as a current does due to charge motion. Some authors apply the name displacement current to the first term by itself.

Thus,

$I_{\mathrm {D} }=\iint _{S}\mathbf {J} _{\mathrm {D} }\cdot \operatorname {d} \!\mathbf {S} =\iint _{S}{\frac {\partial \mathbf {D} }{\partial t}}\cdot \operatorname {d} \!\mathbf {S} ={\frac {\partial }{\partial t}}\iint _{S}\mathbf {D} \cdot \operatorname {d} \!\mathbf {S} ={\frac {\partial \Phi _{\mathrm {D} }}{\partial t}}\,.$

This polarization is the displacement current as it was originally conceived by Maxwell. Maxwell made no special treatment of the vacuum, treating it as a material medium. For Maxwell, the effect of $P$ was simply to change the relative permittivity $ε r$ in the relation $D = ε 0 ε r E$ .

Magnetic displacement with magnetization

The magnetic-charge density displacement field $H 0$ is defined:

$\mathbf{H}\equiv\frac{1}{\mu_0}\mathbf{B}-\mathbf{M}$

where

$\mu_0$ is the vacuum permeability,
$M$ is the magnetization vector.

In a vacuum, $B$ and $H 0$ are proportional to each other. Inside a material they are different (see H and B inside and outside magnetic materials). The SI unit of the $H 0$ -field is the ampere per metre (A/m), and the CGS unit is the oersted (Oe).

Magnetic displacement current with magnetization

${\begin{aligned}c\left(\nabla \times \mathbf {D} \right)&=-{\frac {1}{c}}{\frac {\partial \mathbf {H} }{\partial t}}\end{aligned}}$

Vector calculus

Vector operators

Standard (2,0) tensor convention used in mathematics, physics, and programming libraries alike:

T^{ij}={\begin{bmatrix}T^{11}\;e_{11}&T^{12}\;e_{12}&T^{13}\;e_{13}\\[3pt]T^{21}\;e_{21}&T^{22}\;e_{22}&T^{23}\;e_{23}\\[3pt]T^{31}\;e_{31}&T^{32}\;e_{32}&T^{33}\;e_{33}\end{bmatrix}}

where:

the first index i → row (vertical position)
the second index j → column (horizontal position)

Noncontracting outer product of a vector and a covector written with standard basis resulting in a (1,1) tensor:

{\begin{aligned}v^{1}\otimes c_{2}={\begin{bmatrix}v^{x}e_{x}\\v^{y}e_{y}\\v^{z}e_{z}\end{bmatrix}}\otimes {\begin{bmatrix}c_{x}e^{x}\\c_{y}e^{y}\\c_{z}e^{z}\end{bmatrix}}&={\begin{bmatrix}v^{x}e_{x}\!\otimes \!c_{x}e^{x}&v^{x}e_{x}\!\otimes \!c_{y}e^{y}&v^{x}e_{x}\!\otimes \!c_{z}e^{z}\\v^{y}e_{y}\!\otimes \!c_{x}e^{x}&v^{y}e_{y}\!\otimes \!c_{y}e^{y}&v^{y}e_{y}\!\otimes \!c_{z}e^{z}\\v^{z}e_{z}\!\otimes \!c_{x}e^{x}&v^{z}e_{z}\!\otimes \!c_{y}e^{y}&v^{z}e_{z}\!\otimes \!c_{z}e^{z}\end{bmatrix}}\\\\&={\begin{bmatrix}v^{x}c_{x}\,e_{x}\!\otimes \!e^{x}&v^{x}c_{y}\,e_{x}\!\otimes \!e^{y}&v^{x}c_{z}\,e_{x}\!\otimes \!e^{z}\\v^{y}c_{x}\,e_{y}\!\otimes \!e^{x}&v^{y}c_{y}\,e_{y}\!\otimes \!e^{y}&v^{y}c_{z}\,e_{y}\!\otimes \!e^{z}\\v^{z}c_{x}\,e_{z}\!\otimes \!e^{x}&v^{z}c_{y}\,e_{z}\!\otimes \!e^{y}&v^{z}c_{z}\,e_{z}\!\otimes \!e^{z}\end{bmatrix}}\\\\&={\begin{bmatrix}v^{x}c_{x}\,e_{x}^{x}&v^{x}c_{y}\,e_{x}^{y}&v^{x}c_{z}\,e_{x}^{z}\\v^{y}c_{x}\,e_{y}^{x}&v^{y}c_{y}\,e_{y}^{y}&v^{y}c_{z}\,e_{y}^{z}\\v^{z}c_{x}\,e_{z}^{x}&v^{z}c_{y}\,e_{z}^{y}&v^{z}c_{z}\,e_{z}^{z}\end{bmatrix}}\end{aligned}}

Each component (like

v^{x}c_{x}\,e_{x}^{x}

) is a scalar coefficient multiplied times a dyad.

Tensor contraction is denoted by labeling one covariant index and one contravariant index with the same letter, summation over that index being implied by the summation convention. The resulting contracted tensor inherits the remaining indices.

Euclidean (0,2) metric tensor (in Cartesian coordinates):

g_{ij}={\begin{bmatrix}1\;e^{11}&&0\;e^{12}\\[3pt]0\;e^{21}&&1\;e^{22}\end{bmatrix}}

Lowering indices by tensor contraction with the metric tensor to convert a (1,0) vector to a (0,1) covector (V flat):

{\begin{aligned}V^{\flat }=g_{ij}\;V^{i}e_{i}&={\begin{bmatrix}1\;e^{11}&&0\;e^{12}\\[3pt]0\;e^{21}&&1\;e^{22}\end{bmatrix}}{\begin{bmatrix}V^{1}\;e_{1}\\V^{2}\;e_{2}\end{bmatrix}}\\\\&={\begin{bmatrix}1\;e^{11}*V^{1}\;e_{1}\\1\;e^{22}*V^{2}\;e_{2}\end{bmatrix}}\\\\&={\begin{bmatrix}C_{1}\;e^{1}\\C_{2}\;e^{2}\end{bmatrix}}=C_{j}e^{j}\end{aligned}}

(because

1\;e^{11}*V2\;e_{2}\quad {\text{and}}\quad 1\;e^{22}*V1\;e_{1}\quad {\text{both}}\quad =0

)

Raising indices by tensor contraction with the inverse metric tensor (

g^{\mu \nu }g_{\nu \rho }=\delta _{\;\rho }^{\mu }

, see Kronecker delta) to convert a (0,1) covector to a (1,0) vector (V sharp):

(V^{\flat })^{\sharp }=g^{ij}\;C_{j}e^{j}=V^{i}e_{i}

Using Inner product to compute the length of a vector (trivial in Cartesian coordinates in Euclidean space):

{\begin{aligned}Length(V)^{2}=V\cdot V&=V^{j}e_{j}\;\;g_{ij}\;V^{i}e_{i}\\&=V^{j}e_{j}\;V_{j}e^{j}\\&={\begin{bmatrix}V1\;e_{1}\\V2\;e_{2}\end{bmatrix}}{\begin{bmatrix}V1\;e^{1}\\V2\;e^{2}\end{bmatrix}}\\&=(V1*V1)e^{1}e_{1}+(V2*V2)e^{2}e_{2}\\&=a^{2}+b^{2}=c^{2}\end{aligned}}

Every tensor is the sum of its symmetric part and antisymmetric part:

U_{ij\dots }=U_{\color {red}(ij)}+U_{\color {red}[ij]}

where:

$U_{\color {red}(ij)}={\frac {1}{2}}(U_{\color {red}ij}+U_{\color {red}ji})$ (symmetric part)
$U_{\color {red}[ij]}={\frac {1}{2}}(U_{\color {red}ij}-U_{\color {red}ji})$ (antisymmetric part)

A tensor A that is antisymmetric on indices $i$ and $j$ has the property that the contraction with a tensor B that is symmetric on indices $i$ and $j$ is identically 0.

Scalar field:

f(x, y, z)

Vector:

\mathbf {\vec {v}} =v^{x}e_{x}+v^{y}e_{y}+v^{z}e_{z}

Infinitesimal displacement vector:

d\mathbf {\vec {v}} =dx\;e_{x}+dy\;e_{y}+dz\;e_{z}

Vector field:

\mathbf {\vec {F}} (x,y,z)={\begin{bmatrix}f^{x}e_{x}\\f^{y}e_{y}\\f^{z}e_{z}\end{bmatrix}}=f^{x}(x,y,z)\,e_{x}+f^{y}(x,y,z)\,e_{y}+f^{z}(x,y,z)\,e_{z}\quad =\quad f^{i}e_{i}

(Coordinate-free version)

Del

{\begin{aligned}&\nabla ^{\flat }={\begin{bmatrix}e^{x}{\dfrac {\partial }{\partial x}}\\e^{y}{\dfrac {\partial }{\partial y}}\\e^{z}{\dfrac {\partial }{\partial z}}\end{bmatrix}}=e^{i}\partial _{i}\quad =\quad e^{x}{\dfrac {\partial }{\partial x}}+e^{y}{\dfrac {\partial }{\partial y}}+e^{z}{\dfrac {\partial }{\partial z}}\end{aligned}}

Gradient:

{\text{grad}}(f)=\nabla ^{\flat }f(x,y,z)\quad =\quad \mathbf {e^{x}} {\frac {\partial f}{\partial x}}+\mathbf {e^{y}} {\frac {\partial f}{\partial y}}+\mathbf {e^{z}} {\frac {\partial f}{\partial z}}

In this article, the gradient operator is written as

\nabla ^{\flat }

to emphasize that it is a covector (1-form), not a vector. This is not a redefinition but a restoration of its true form: the differential of a scalar field f is inherently covectorial,

df=\nabla ^{\flat }f.

The familiar “vector gradient”

\nabla f

used in most texts is obtained only after applying the metric-dependent index-raising operation,

\nabla f=(df)^{\sharp }.

In other words, I am not adding a ‘flat’ to the symbol — I am removing an implicit ‘sharp’ that was assumed by historical convention.

The Directional derivative (

D_{v}f{\text{ or }}df(v)

) gives the rate of change of f at a given point when moving an infinitesimal distance in the direction and magnitude specified by the vector v. See Covariant derivative.

(\nabla ^{\flat }f)(v)=\nabla f\cdot v=e^{x}{\dfrac {\partial f}{\partial x}}\;v^{x}e_{x}+e^{y}{\dfrac {\partial f}{\partial y}}\;v^{y}e_{y}+e^{z}{\dfrac {\partial f}{\partial z}}\;v^{z}e_{z}\quad =\quad {\dfrac {\partial f}{\partial x}}\;v^{x}+{\dfrac {\partial f}{\partial y}}\;v^{y}+{\dfrac {\partial f}{\partial z}}\;v^{z}

Total differential:

df=\nabla ^{\flat }f\;d\mathbf {\vec {v}} \quad =\quad {\frac {\partial f}{\partial x}}dx+{\frac {\partial f}{\partial y}}dy+{\frac {\partial f}{\partial z}}dz

The (0,2) Hessian is the gradient of the gradient of a scalar. (The Hessian matrix is a symmetric tensor by the symmetry of second derivatives):

\nabla ^{\flat }(\nabla ^{\flat }f)={\begin{bmatrix}{\color {red}{\dfrac {\partial ^{2}f}{\partial x^{2}}}}&{\dfrac {\partial ^{2}f}{\partial x\,\partial y}}&{\dfrac {\partial ^{2}f}{\partial x\,\partial z}}\\[8pt]{\dfrac {\partial ^{2}f}{\partial y\,\partial x}}&{\color {red}{\dfrac {\partial ^{2}f}{\partial y^{2}}}}&{\dfrac {\partial ^{2}f}{\partial y\,\partial z}}\\[8pt]{\dfrac {\partial ^{2}f}{\partial z\,\partial x}}&{\dfrac {\partial ^{2}f}{\partial z\,\partial y}}&{\color {red}{\dfrac {\partial ^{2}f}{\partial z^{2}}}}\end{bmatrix}}

The Laplacian is the trace of the Hessian:

\nabla _{\flat }^{2}f=\nabla \cdot (\nabla f)={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}

The (1,1) Jacobian is the gradient of a vector field:

J^{i}{}_{j}\;=\;\nabla ^{\flat }F\;=\;{\frac {\partial f^{i}}{\partial x^{j}}}=J(F)={\begin{bmatrix}{\color {red}\partial _{x}f^{x}}&\partial _{x}f^{y}&\partial _{x}f^{z}\\\partial _{y}f^{x}&{\color {red}\partial _{y}f^{y}}&\partial _{y}f^{z}\\\partial _{z}f^{x}&\partial _{z}f^{y}&{\color {red}\partial _{z}f^{z}}\end{bmatrix}}

= Symmetric Jacobian + Anti-symmetric Jacobian

Symmetric Jacobian (order doesnt matter) is defined so that sym(sym(J))=sym(J):

sym(J)={\begin{bmatrix}\partial _{x}f^{x}&{\tfrac {1}{2}}(\partial _{x}f^{y}+\partial _{y}f^{x})&{\tfrac {1}{2}}(\partial _{x}f^{z}+\partial _{z}f^{x})\\[4pt]{\tfrac {1}{2}}(\partial _{y}f^{x}+\partial _{x}f^{y})&\partial _{y}f^{y}&{\tfrac {1}{2}}(\partial _{y}f^{z}+\partial _{z}f^{y})\\[4pt]{\tfrac {1}{2}}(\partial _{z}f^{x}+\partial _{x}f^{z})&{\tfrac {1}{2}}(\partial _{z}f^{y}+\partial _{y}f^{z})&\partial _{z}f^{z}\end{bmatrix}}

Anti-symmetric Jacobian (order matters) is likewise defined so that alt(alt(J))=alt(J):

alt(J)={\begin{bmatrix}0&{\tfrac {1}{2}}(\partial _{x}f^{y}-\partial _{y}f^{x})&{\tfrac {1}{2}}(\partial _{x}f^{z}-\partial _{z}f^{x})\\[4pt]{\tfrac {1}{2}}(\partial _{y}f^{x}-\partial _{x}f^{y})&0&{\tfrac {1}{2}}(\partial _{y}f^{z}-\partial _{z}f^{y})\\[4pt]{\tfrac {1}{2}}(\partial _{z}f^{x}-\partial _{x}f^{z})&{\tfrac {1}{2}}(\partial _{z}f^{y}-\partial _{y}f^{z})&0\end{bmatrix}}

Divergence is the trace of the Jacobian:

{\begin{aligned}{\text{div}}(\mathbf {\vec {F}} )=\nabla ^{\flat }\cdot {\mathbf {\vec {F}} (x,y,z)}&=\mathbf {e^{x}} {\frac {\partial f^{x}}{\partial x}}\mathbf {e_{x}} +\mathbf {e^{y}} {\frac {\partial f^{y}}{\partial y}}\mathbf {e_{y}} +\mathbf {e^{z}} {\frac {\partial f^{z}}{\partial z}}\mathbf {e_{z}} \\\\&={\frac {\partial f^{x}}{\partial x}}+{\frac {\partial f^{y}}{\partial y}}+{\frac {\partial f^{z}}{\partial z}}\end{aligned}}

Wedge is used to measure rotation of a field. Rotation from e_x to e_y is the negative of rotation from e_y to e_x hence the use of wedge product. Wedge is anti-symmetric but there is no factor of 1/2 because physics doesnt care about idempotency:

{\begin{aligned}\nabla ^{\flat }\wedge \mathbf {\vec {F^{\flat }}} (x,y,z)_{ij}&=(e^{i}\partial _{i})\wedge (f_{j}e^{j})\quad =\quad (\partial _{i}f_{j})\;e^{i}\wedge e^{j}\\\\&=(e^{i}\partial _{i})\otimes (f_{j}e^{j})-(f_{j}e^{j})\otimes (e^{i}\partial _{i})\quad =\quad \partial _{i}f_{j}\;e^{i}e^{j}-\partial _{i}f_{j}\;e^{j}e^{i}\\\\&{\overset {n=3}{=}}(\partial _{x}f_{y}-\partial _{y}f_{x})\;e^{x}\otimes e^{y}\;+\;(\partial _{y}f_{z}-\partial _{z}f_{y})\;e^{y}\otimes e^{z}\;+\;(\partial _{z}f_{x}-\partial _{x}f_{z})\;e^{z}\otimes e^{x}\\\\&{\overset {n=3}{=}}{\begin{bmatrix}0&\partial _{x}f_{y}-\partial _{y}f_{x}&\partial _{x}f_{z}-\partial _{z}f_{x}\\[4pt]-(\partial _{x}f_{y}-\partial _{y}f_{x})&0&\partial _{y}f_{z}-\partial _{z}f_{y}\\[4pt]-(\partial _{x}f_{z}-\partial _{z}f_{x})&-(\partial _{y}f_{z}-\partial _{z}f_{y})&0\end{bmatrix}}\\\\&{\overset {n=3}{=}}{\begin{vmatrix}\mathbf {e} ^{y}\wedge \mathbf {e} ^{z}&\mathbf {e} ^{z}\wedge \mathbf {e} ^{x}&\mathbf {e} ^{x}\wedge \mathbf {e} ^{y}\\[6pt]{\dfrac {\partial }{\partial x}}&{\dfrac {\partial }{\partial y}}&{\dfrac {\partial }{\partial z}}\\[6pt]f_{x}&f_{y}&f_{z}\end{vmatrix}}\\\\&{\overset {n=3}{=}}\left({\frac {\partial f_{z}}{\partial y}}-{\frac {\partial f_{y}}{\partial z}}\right)\mathbf {e} ^{y}\wedge \mathbf {e} ^{z}\quad +\quad \left({\frac {\partial f_{x}}{\partial z}}-{\frac {\partial f_{z}}{\partial x}}\right)\mathbf {e} ^{z}\wedge \mathbf {e} ^{x}\quad +\quad \left({\frac {\partial f_{y}}{\partial x}}-{\frac {\partial f_{x}}{\partial y}}\right)\mathbf {e} ^{x}\wedge \mathbf {e} ^{y}\end{aligned}}

Stokes' theorem:

{\begin{aligned}&\iint _{\Sigma }\left(\left({\frac {\partial f_{z}}{\partial y}}-{\frac {\partial f_{y}}{\partial z}}\right)\,\mathrm {d} y\wedge \mathrm {d} z+\left({\frac {\partial f_{x}}{\partial z}}-{\frac {\partial f_{z}}{\partial x}}\right)\,\mathrm {d} z\wedge \mathrm {d} x+\left({\frac {\partial f_{y}}{\partial x}}-{\frac {\partial f_{x}}{\partial y}}\right)\,\mathrm {d} x\wedge \mathrm {d} y\right)\\\\&=\oint _{\partial \Sigma }{\Bigl (}f_{x}\,\mathrm {d} x+f_{y}\,\mathrm {d} y+f_{z}\,\mathrm {d} z{\Bigr )}\end{aligned}}

The sum of all the infinitesimal current loops within the 2-D region (Σ) equals the outer current loop (dΣ) alone. This is true for all vector fields.

The Curl only works in 3 dimensions. The curl of the magnetic vector field is the electric current (which generates the magnetic field):

{\begin{aligned}{\text{curl}}(\mathbf {\vec {F}} )=\nabla \times F&{\overset {n=3}{=}}{\begin{vmatrix}\mathbf {e} _{x}&\mathbf {e} _{y}&\mathbf {e} _{z}\\[6pt]{\dfrac {\partial }{\partial x}}&{\dfrac {\partial }{\partial y}}&{\dfrac {\partial }{\partial z}}\\[6pt]f^{x}&f^{y}&f^{z}\end{vmatrix}}\\\\&{\overset {n=3}{=}}\left({\frac {\partial f^{z}}{\partial y}}-{\frac {\partial f^{y}}{\partial z}}\right)\mathbf {e} _{x}\quad +\quad \left({\frac {\partial f^{x}}{\partial z}}-{\frac {\partial f^{z}}{\partial x}}\right)\mathbf {e} _{y}\quad +\quad \left({\frac {\partial f^{y}}{\partial x}}-{\frac {\partial f^{x}}{\partial y}}\right)\mathbf {e} _{z}\end{aligned}}

In the xy plane curl measures counterclockwise rotation (

e_{x}\wedge e_{y}

) corresponding to positive z (out of the page).

Hodge star:

In 3 dimensions:

{\star }(\mathbf {u} \wedge \mathbf {v} )=\mathbf {u} \times \mathbf {v} \qquad {\star }(\mathbf {u} \times \mathbf {v} )=\mathbf {u} \wedge \mathbf {v}

{\begin{aligned}\\{\star }\,dx&=dy\wedge dz\\{\star }\,dy&=dz\wedge dx\\{\star }\,dz&=dx\wedge dy\end{aligned}}\qquad {\begin{aligned}{\star }\,1&=e_{x}\wedge e_{y}\wedge e_{z}\\{\star }\,e_{x}&=e_{y}\wedge e_{z}\\{\star }\,e_{y}&=e_{z}\wedge e_{x}\\{\star }\,e_{z}&=e_{x}\wedge e_{y}\end{aligned}}\qquad {\begin{aligned}\star (e_{x}\wedge e_{y}\wedge e_{z})&=1\\\star (e_{y}\wedge e_{z})&=e_{x}\\\star (e_{z}\wedge e_{x})&=e_{y}\\\star (e_{x}\wedge e_{y})&=e_{z}\end{aligned}}

In 4 dimensions

with metric signature (++++): $\alpha \wedge (\star \alpha )=|\alpha |^{2}\,dt\wedge dx\wedge dy\wedge dz$
with metric signature (-+++): $\alpha \wedge (\star \alpha )=-|\alpha |^{2}\,dt\wedge dx\wedge dy\wedge dz$

Levi-Civita symbol is either +, -, or zero:

e_{i}\wedge e_{j}\wedge e_{k}=\varepsilon _{ijk}\,e_{1}\wedge e_{2}\wedge e_{3}

Tensor version

Preliminary math

Let the orthonormal basis in four-dimensional Euclidean spacetime be

${e_{0},e_{1},e_{2},e_{3}}$

where:

$e_{\mu }\cdot e_{\nu }=\delta _{\mu \nu }$ $e_{\mu }\cdot e_{\nu }=\delta _{\mu \nu }$ because they are orthogonal and normalized unit vectors
- $\delta _{\mu \nu }={\begin{cases}0&{\text{if }}\mu \neq \nu \\1&{\text{if }}\mu =\nu \end{cases}}$

And orthogonal dual basis in cotangent space:

${e^{0},e^{1},e^{2},e^{3}}$

where

$e^{i}=g^{ij}\,e_{j}$ by definition of dual basis

The metric tensor is the identity tensor (so raising/lowering indices does nothing to component values):

$g_{\mu \nu }=\delta _{\mu \nu }$

the metric signature is (+,+,+,+).

$e_{0}^{2}=e_{1}^{2}=e_{2}^{2}=e_{3}^{2}=1$

With coordinates

$x^{\mu }e_{\mu }\quad =\quad x^{0}e_{0}+x^{1}e_{1}+x^{2}e_{2}+x^{3}e_{3}$

Where:

$x^{0}\equiv ic_{0}t$ = imaginary distance

With four-velocity

${\begin{aligned}U^{\mu }e_{\mu }={\frac {dx^{\mu }}{d\tau }}e_{\mu }\quad &=\quad {\frac {ic_{0}\;d(t)}{d\tau }}e_{0}+{\frac {dx^{1}}{d\tau }}{\frac {dt}{dt}}e_{1}+{\frac {dx^{2}}{d\tau }}{\frac {dt}{dt}}e_{2}+{\frac {dx^{3}}{d\tau }}{\frac {dt}{dt}}e_{3}\quad \\\\&=\quad ic_{0}\gamma \;e_{0}+{\frac {dx^{1}}{dt}}\gamma \;e_{1}+{\frac {dx^{2}}{dt}}\gamma \;e_{2}+{\frac {dx^{3}}{dt}}\gamma \;e_{3}\end{aligned}}$

where

${\frac {dt}{d\tau }}=\gamma$

The four-current is charge density times the four velocity

$J^{\mu }e_{\mu }=\rho _{0}\,U^{\mu }e_{\mu }=\rho _{0}\,{\dfrac {dx^{\mu }}{d\tau }}e_{\mu }\quad =\quad J^{0}e_{0}+J^{1}e_{1}+J^{2}e_{2}+J^{3}e_{3}$

Where

$J^{0}=\rho \;ic_{0}\gamma \;e_{0}$ = charge with imaginary velocity (through time)

The spacetime gradient acts equally in all four directions:

$e^{\mu }\partial _{\mu }\quad =\quad {\begin{bmatrix}{\begin{array}{l}e^{0}\,\partial _{(ict)}\\e^{1}\,\partial _{x}\\e^{2}\,\partial _{y}\\e^{3}\,\partial _{z}\end{array}}\end{bmatrix}}={\begin{bmatrix}{\begin{array}{r}{\frac {1}{ic_{0}}}e^{0}{\frac {\partial }{\partial t}}\\e^{1}{\frac {\partial }{\partial x}}\\e^{2}{\frac {\partial }{\partial y}}\\e^{3}{\frac {\partial }{\partial z}}\end{array}}\end{bmatrix}}$

Note: It may feel strange to treat a derivative as a (co)vector, since differentiation is an operation, not a quantity. However, the spacetime gradient $\partial =e^{\mu }\partial _{\mu }$ behaves as if it were a (co)vector because it acts linearly and independently along each coordinate direction.

The antisymmetric displacement tensor

$e_{\mu \nu }F^{\mu \nu }={\begin{bmatrix}{\begin{array}{lllllll}{\color {red}e_{00}0}&&-e_{01}\,D_{tx}\,ic_{0}&&-e_{02}\,D_{ty}\,ic_{0}&&-e_{03}\,D_{tz}\,ic_{0}\\[6pt]{\color {red}e_{10}\,D_{xt}\,ic_{0}}&&{\phantom {-}}e_{11}\,0&&{\phantom {-}}e_{12}\,H_{xy}&&-e_{13}\,H_{xz}\\[6pt]{\color {red}e_{20}\,D_{yt}\,ic_{0}}&&-e_{21}\,H_{yx}&&{\phantom {-}}e_{22}\,0&&{\phantom {-}}e_{23}\,H_{yz}\\[6pt]{\color {red}e_{30}\,D_{zt}\,ic_{0}}&&{\phantom {-}}e_{31}\,H_{zx}&&-e_{32}\,H_{zy}&&{\phantom {-}}e_{33}\,0\end{array}}\end{bmatrix}}$

The spacetime gradient of the displacement tensor is

$e^{\mu }\partial _{\mu }F^{\mu \nu }e_{\mu \nu }={\begin{bmatrix}{\begin{array}{lllllll}{\color {red}{\frac {1}{ic_{0}}}e^{0}\partial _{t}({\phantom {-}}e_{00}0)}&+&{\color {red}e^{1}\partial _{x}({\phantom {-}}e_{10}D_{xt}ic_{0})}&+&{\color {red}e^{2}\partial _{y}({\phantom {-}}e_{20}D_{yt}ic_{0})}&+&{\color {red}e^{3}\partial _{z}({\phantom {-}}e_{30}D_{zt}ic_{0})}\\{\frac {1}{ic_{0}}}e^{0}\partial _{t}(-e_{01}D_{tx}ic_{0})&+&e^{1}\partial _{x}({\phantom {-}}e_{11}0)&+&e^{2}\partial _{y}(-e_{21}H_{yx})&+&e^{3}\partial _{z}({\phantom {-}}e_{31}H_{zx})\\{\frac {1}{ic_{0}}}e^{0}\partial _{t}(-e_{02}D_{ty}ic_{0})&+&e^{1}\partial _{x}({\phantom {-}}e_{12}H_{xy})&+&e^{2}\partial _{y}({\phantom {-}}e_{22}0)&+&e^{3}\partial _{z}(-e_{32}H_{zy})\\{\frac {1}{ic_{0}}}e^{0}\partial _{t}(-e_{03}D_{tz}ic_{0})&+&e^{1}\partial _{x}(-e_{13}H_{xz})&+&e^{2}\partial _{y}({\phantom {-}}e_{23}H_{yz})&+&e^{3}\partial _{z}({\phantom {-}}e_{33}0)\end{array}}\end{bmatrix}}={\begin{bmatrix}{\begin{array}{r}ic_{0}\rho \;e_{0}\\j_{x}\;e_{1}\\j_{y}\;e_{2}\\j_{z}\;e_{3}\end{array}}\end{bmatrix}}=J^{\nu }e_{\nu }$

Which yields Gauss and Ampère–Maxwell

${\begin{aligned}{\color {red}ic_{0}\left(\nabla \cdot \mathbf {D} \right)}&{\color {red}=ic_{0}\rho }\\-{\frac {\partial \mathbf {D} }{\partial t}}+\left(\nabla \times \mathbf {H} \right)&=\mathbf {J} \end{aligned}}$

The hodge dual of the displacement tensor is

$e_{\mu \nu }(*F)^{\mu \nu }\;=\;e_{\mu \nu }{\tfrac {1}{2}}\,\varepsilon ^{\mu \nu \alpha \beta }\,g_{\alpha \gamma }\,g_{\beta \delta }\,F^{\gamma \delta }={\begin{bmatrix}{\begin{array}{lllllll}{\color {red}e_{00}\,0}&&-e_{01}H_{tx}&&-e_{02}H_{ty}&&-e_{03}H_{tz}\\[6pt]{\color {red}e_{10}H_{xt}}&&{\phantom {-}}e_{11}\,0&&-e_{12}\,D_{xy}\,ic_{0}&&{\phantom {-}}e_{13}\,D_{xz}\,ic_{0}\\[6pt]{\color {red}e_{20}H_{yt}}&&{\phantom {-}}e_{21}\,D_{yx}\,ic_{0}&&{\phantom {-}}e_{22}\,0&&-e_{23}\,D_{yz}\,ic_{0}\\[6pt]{\color {red}e_{30}H_{zt}}&&-e_{31}\,D_{zx}\,ic_{0}&&{\phantom {-}}e_{32}\,D_{zy}\,ic_{0}&&{\phantom {-}}e_{33}\,0\end{array}}\end{bmatrix}}$

The spacetime gradient of the dual displacement tensor is

$e^{\mu }\partial _{\mu }*F^{\mu \nu }e_{\mu \nu }={\begin{bmatrix}{\begin{array}{lllllll}{\color {red}{\frac {1}{ic_{0}}}\,e^{0}\partial _{t}({\phantom {-}}e_{00}0)}&+&{\color {red}e^{1}\partial _{x}({\phantom {-}}e_{10}H_{xt})}&+&{\color {red}e^{2}\partial _{y}({\phantom {-}}e_{20}H_{yt})}&+&{\color {red}e^{3}\partial _{z}({\phantom {-}}e_{30}H_{zt})}\\{\frac {1}{ic_{0}}}\,e^{0}\partial _{t}(-e_{01}H_{tx})&+&e^{1}\partial _{x}({\phantom {-}}e_{11}0)&+&e^{2}\partial _{y}(\,{\phantom {-}}e_{21}D_{yx}ic_{0})&+&e^{3}\partial _{z}(\,-e_{31}D_{zx}ic_{0})\\{\frac {1}{ic_{0}}}\,e^{0}\partial _{t}(-e_{02}H_{ty})&+&e^{1}\partial _{x}(\,-e_{12}D_{xy}ic_{0})&+&e^{2}\partial _{y}({\phantom {-}}e_{22}0)&+&e^{3}\partial _{z}(\,{\phantom {-}}e_{32}D_{zy}ic_{0})\\{\frac {1}{ic_{0}}}\,e^{0}\partial _{t}(-e_{03}H_{tz})&+&e^{1}\partial _{x}(\,{\phantom {-}}e_{13}D_{xz}ic_{0})&+&e^{2}\partial _{y}(\,-e_{23}D_{yz}ic_{0})&+&e^{3}\partial _{z}({\phantom {-}}e_{33}0)\end{array}}\end{bmatrix}}=0$

Which yields Gauss–magnetic and Faraday

${\begin{aligned}{\color {red}\nabla \cdot \mathbf {H} }&{\color {red}=0\;\rho _{m}}\\-{\frac {1}{ic_{0}}}{\frac {\partial \mathbf {H} }{\partial t}}+ic_{0}\left(\nabla \times \mathbf {D} \right)&=0\end{aligned}}$

The complex displacement tensor is

${\mathcal {F}}^{\mu \nu }=F^{\mu \nu }+*F^{\mu \nu }$

All of Maxwells equations are therefore reduced to

$\partial _{\mu }{\mathcal {F}}^{\mu \nu }=J^{\nu }$

Units (SI) — explicit and consistent

Coordinates: $[x^{\mu }]=\mathrm {m}$ $[x^{\mu }]=\mathrm {m}$
- (since $x^{0}=ict$ makes time a length).
Basis vectors encode direction only: abstract/unitless; units live in components.
Derivatives: $[\partial _{\mu }]=\mathrm {m} ^{-1}$
Derivatives: $[\partial _{0}]=(1/ic_{0})\partial _{t}=s/ms=\mathrm {m} ^{-1}$
Fields:
- $[\mathbf {D} ]=\mathrm {Asm^{-2}}$
- $[\mathbf {H} ]=\mathrm {Am^{-1}}$
$F_{0i}=c_{0}D=Am^{-1}$
$F_{ij}=H=\mathrm {Am^{-1}}$
Charge/current:
- $[\rho ]=\mathrm {Cm^{-3}}$
- $[J^{k}]=\mathrm {Am^{-2}}$
- $[J^{0}]=[ic_{0}\rho ]=\mathrm {Am^{-2}}$ (so all four components of $J^{\mu }$ share the same units).
Constants:
- $[\mu _{0}]=\mathrm {Hm^{-1}} =\mathrm {VsA^{-1}m^{-1}}$ ,
- $\mu _{0}\varepsilon _{0}c_{0}^{2}=1$ as usual.

Schrödinger equation

Starting from Maxwell’s equations in vacuum, define the complex Riemann–Silberstein vector

$\mathbf {F} =\mathbf {E} +i\,c\mathbf {B} .$

Maxwell’s curl equations then combine into a single, Schrödinger-like form:

$i\,{\frac {\partial \mathbf {F} }{\partial t}}=c\,(\nabla \times \mathbf {F} ),$

where the curl operator acts as a Hamiltonian generating time evolution. Squaring this equation yields the standard wave equation

${\frac {\partial ^{2}\mathbf {F} }{\partial t^{2}}}=c^{2}\nabla ^{2}\mathbf {F} ,$

analogous to the Klein–Gordon equation for massless particles.

$\ \left(\ {\frac {1}{\ c^{2}}}{\frac {\ \partial ^{2}}{\ \partial t^{2}\ }}-\nabla ^{2}+{\frac {\ m^{2}c^{2}\ }{\hbar ^{2}}}\ \right)\ \psi (\ t,\mathbf {x} \ )=0\$

The Schrödinger equation can be linearized versions of second-order wave equations, just like the Riemann–Silberstein equation for electromagnetism.

For a massless particle (photon), the wave equation is

${\frac {\partial ^{2}\psi }{\partial t^{2}}}=c^{2}\nabla ^{2}\psi ,$

and it can be factorized as

$\left({\frac {\partial }{\partial t}}-c\nabla \right)\left({\frac {\partial }{\partial t}}+c\nabla \right)\psi =0.$

Each factor corresponds to a one-directional propagation equation:

${\frac {\partial \psi }{\partial t}}=\pm c\,({\hat {\mathbf {n} }}\cdot \nabla )\psi .$

Schrödinger did the same kind of move but for massive waves.

Adding a rest-mass term and taking the low-velocity limit gives the familiar Schrödinger equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi .$

So Schrödinger’s equation looks like a diffusion equation with an imaginary diffusion coefficient:

${\frac {\partial \psi }{\partial t}}=i{\frac {\hbar }{2m}}\nabla ^{2}\psi .$

That’s what gives it the oscillatory (wave-like) character.

The diffusion equation can be trivially derived from the continuity equation, which states that a change in density in any part of the system is due to inflow and outflow of material into and out of that part of the system. Effectively, no material is created or destroyed: $\frac{\partial\phi}{\partial t}+\nabla\cdot\mathbf{j}=0,$ where j is the flux of the diffusing material. The diffusion equation can be obtained easily from this when combined with the phenomenological Fick's first law, which states that the flux of the diffusing material in any part of the system is proportional to the local density gradient: $\mathbf{j}=-D(\phi,\mathbf{r})\,\nabla\phi(\mathbf{r},t).$

Electromagnetic waves

Start with a simple traveling wave:

${\frac {\partial \psi }{\partial t}}=\pm c\,{\frac {\partial \psi }{\partial x}}.$

This says:

The field changes in time exactly as fast as it moves through space — at speed c.

Differentiate again to get the wave equation:

${\frac {\partial ^{2}\psi }{\partial t^{2}}}=c^{2}{\frac {\partial ^{2}\psi }{\partial x^{2}}}.$

Its solutions are just functions of $x\pm ct$ :

$\psi (x,t)=f(x-ct)+g(x+ct).$

Right-moving and left-moving waves.

The vector case: curl replaces ∂/∂x

When we move from a 1D scalar wave to a 3D vector wave, the simple spatial derivative becomes a curl:

${\frac {\partial \mathbf {F} }{\partial t}}=-i\,c\,(\nabla \times \mathbf {F} ).$

That’s exactly the Riemann–Silberstein equation, just without the i in front of time derivative if we pick the right-hand circular polarization convention.

Taking another time derivative yields:

${\frac {\partial ^{2}\mathbf {F} }{\partial t^{2}}}=-c^{2}\,(\nabla \times (\nabla \times \mathbf {F} )).$

Use the vector identity (and remember $\nabla \cdot \mathbf {F} =0$ ):

$\nabla \times (\nabla \times \mathbf {F} )=-\nabla ^{2}\mathbf {F} .$

So you get the vector wave equation:

${{\frac {\partial ^{2}\mathbf {F} }{\partial t^{2}}}=c^{2}\nabla ^{2}\mathbf {F} },$

which is the exact analog of the scalar wave equation.

${\frac {\partial ^{2}u}{\partial t^{2}}}=c^{2}\left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}\right)$

where

$c$ is a fixed non-negative real coefficient representing the propagation speed of the wave
$u$ is a scalar field representing the displacement or, more generally, the conserved quantity (e.g. pressure or density)
$x,y,$ and $z$ are the three spatial coordinates and $t$ being the time coordinate.

In a region with no charges ( $ρ = 0$ ) and no currents ( $J = 0$ ), such as in vacuum, Maxwell's equations reduce to: ${\begin{aligned}\nabla \cdot \mathbf {E} &=0,&\nabla \times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}=0,\\\nabla \cdot \mathbf {B} &=0,&\nabla \times \mathbf {B} -\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}=0.\end{aligned}}$

Taking the curl $(\nabla\times)$ of the curl equations, and using the curl of the curl identity

$\nabla \times \left(\nabla \times \mathbf {A} \right)\ =\ \nabla (\nabla {\cdot }\mathbf {A} )\,-\,\nabla ^{2\!}\mathbf {A}$

we obtain ${\begin{aligned}\mu _{0}\varepsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}-\nabla ^{2}\mathbf {E} =0,\\\mu _{0}\varepsilon _{0}{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}-\nabla ^{2}\mathbf {B} =0.\end{aligned}}$

The quantity $\mu _{0}\varepsilon _{0}$ has the dimension (T/L)². Defining $c=(\mu _{0}\varepsilon _{0})^{-1/2}$ , the equations above have the form of the standard wave equations ${\begin{aligned}{\frac {1}{c^{2}}}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}-\nabla ^{2}\mathbf {E} =0,\\{\frac {1}{c^{2}}}{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}-\nabla ^{2}\mathbf {B} =0.\end{aligned}}$

Note about pseudovectors

Unlike the dot product, the cross product depends on a choice of an orientation of the space (this is why the space must be oriented). The cross product is invariant under a rotation of the basis but is changed into its opposite by an odd permutation of the basis vectors. Therefore, the cross product is a pseudovector.

Since position $r$ , linear momentum $p$ and force $F$ are all true vectors, both the angular momentum $L$ and the moment of a force $M$ are pseudovectors or axial vectors.

The wedge product generalizes the cross product to higher dimensions while avoiding pseudovectors entirely. But does so at the cost of complicating the equations. See Wikipedia:Mathematical_descriptions_of_the_electromagnetic_field#Geometric_algebra_formulations

A vector changes direction under space reflection, while a pseudovector does not. The relationship between electric current and magnetic fields is illustrated by the hand rule, indicating that they cannot both be vectors or pseudovectors. The magnetic field is identified as a pseudovector because it does not change sign under spatial transformations, unlike the electric field, which behaves as a true vector. The electromagnetic field tensor reveals that the electric field is the time-space component, while the magnetic field is the space-space component, confirming their distinct behaviors. Ultimately, under Lorentz transformations, both fields require the complete EM field tensor for accurate representation.

Reference: https://www.physicsforums.com/threads/what-distinguishes-a-vector-from-a-pseudovector-in-electromagnetism.632321/

Natural units

Quantity	Planck	Stoney	Atomic	Particle and atomic physics	Strong	Schrödinger
Defining constants	$c$ , $G$ , $\hbar$ , $k_\text{B}$ , $k_{\text{e}}$	$c$ , $G$ , $e$ , $k_{\text{e}}$	$e$ , $m_\text{e}$ , $\hbar$ , $k_{\text{e}}$	$c$ , $m_\text{e}$ , $\hbar$ , $\varepsilon_0$	$c$ , $m_{\text{p}}$ , $\hbar$	$\hbar$ , $G$ , $e$ , $k_{\text{e}}$
Speed of light $c$	$1$	$1$	${1}/{\alpha }$	$1$	$1$	${1}/{\alpha }$
Reduced Planck constant $\hbar$	$1$	${1}/{\alpha }$	$1$	$1$	$1$	$1$
Elementary charge $e$	$\sqrt{\alpha}$	$1$	$1$	$\sqrt{4\pi\alpha}$	—	$1$
Vacuum permittivity $\varepsilon_0$	${1}/{4\pi }$	${1}/{4\pi }$	${1}/{4\pi }$	$1$	—	${1}/{4\pi }$
Gravitational constant $G$	$1$	$1$	${\eta _{\mathrm {e} }}/{\alpha }$	$\eta _{\mathrm {e} }$	$\eta _{\mathrm {p} }$	$1$

where:

$α$ is the fine-structure constant ( $α = e2 \ 4πε0ħc$ ≈ 0.007297)
$ηe = Gme2 \ ħc$ ≈ 1.7518 × 10^-45

$ηp = Gmp2 \ ħc$ ≈ 5.9061 × 10^-39

A dash (—) indicates where the system is not sufficient to express the quantity.

The Fine-structure constant, $α$ , in terms of other fundamental physical constants:

\alpha ={\frac {1}{2\varepsilon _{0}}}{\frac {e^{2}}{hc}}={\frac {\mu _{0}}{2}}{\frac {e^{2}c}{h}}={\frac {k_{\text{e}}e^{2}}{\hbar c}}={\frac {c\mu _{0}}{2R_{\text{K}}}}={\frac {e^{2}}{2}}{\frac {Z_{0}}{h}}

System Quantity	Planck			Stoney	Schrödinger	Atomic			"Natural"		Quantum chromodynamics
System Quantity	original	with L–H	with Gauss	Stoney	Schrödinger	Hartree	Rydberg	new	with L–H	with Gauss	original	with L–H	with Gauss
Speed of light $c \,$	$1\,$	$1\,$	$1\,$	$1\,$	$\frac{1}{\alpha} \$	$\frac{1}{\alpha} \$	$\frac{2}{\alpha} \$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$
Reduced Planck constant $\hbar = \frac{h}{2 \pi}$	$1\,$	$1\,$	$1\,$	$\frac{1}{\alpha} \$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$
Vacuum permittivity $\varepsilon_0 \,$	$\frac{1}{4 \pi \alpha}$	$1\,$	$\frac{1}{4\pi}$	$\frac{1}{4\pi}$	$\frac{1}{4\pi}$	$\frac{1}{4\pi}$	$\frac{1}{4\pi}$	$\frac{1}{4 \pi \alpha}$	$1\,$	$\frac{1}{4\pi}$	$\frac{1}{4 \pi \alpha}$	$1\,$	$\frac{1}{4\pi}$
Coulomb constant $k_{e}={\frac {1}{4\pi \epsilon _{0}}}\,$	$\alpha$	$\frac{1}{4\pi}$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$\alpha$	$\frac{1}{4\pi}$	$1\,$	$\alpha$	$\frac{1}{4\pi}$	$1\,$
Gravitational constant $G \,$	$1\,$	$\frac{1}{4\pi}$	$1\,$	$1\,$	$1\,$	${\frac {\alpha _{G}}{\alpha }}\,$	${\frac {8\alpha _{G}}{\alpha }}\,$	$\alpha_G$	${\frac {\alpha _{G}}{{m_{\text{e}}}^{2}}}\approx 6.71\times 10^{-57}{\text{eV}}^{-2}$	${\frac {\alpha _{G}}{{m_{\text{e}}}^{2}}}\approx 6.71\times 10^{-57}{\text{eV}}^{-2}$	$\mu ^{2}\alpha _{G}$	$\mu ^{2}\alpha _{G}$	$\mu ^{2}\alpha _{G}$
Boltzmann constant $k_\text{B} \,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$	$1\,$
Elementary charge $e \,$	$1\,$	$\sqrt{4\pi\alpha} \,$	$\sqrt{\alpha} \,$	$1\,$	$1\,$	$1\,$	$\sqrt{2} \,$	$1\,$	$\sqrt{4\pi\alpha}$	$\sqrt{\alpha}$	$1\,$	$\sqrt{4\pi\alpha} \,$	$\sqrt{\alpha} \,$
Electron rest mass $m_\text{e} \,$	${\sqrt {\alpha _{G}}}\,$	${\sqrt {4\pi \alpha _{G}}}\,$	${\sqrt {\alpha _{G}}}\,$	${\sqrt {\frac {\alpha _{G}}{\alpha }}}\,$	${\sqrt {\frac {\alpha _{G}}{\alpha }}}\,$	$1\,$	$\frac{1}{2} \,$	$1\,$	$m_{e}\approx 511{\text{ keV}}$	$m_{e}\approx 511{\text{ keV}}$	$\frac{1}{\mu}$	$\frac{1}{\mu}$	$\frac{1}{\mu}$
Proton rest mass $m_\text{p} \,$	$\mu {\sqrt {\alpha _{G}}}\,$	$\mu {\sqrt {4\pi \alpha _{G}}}\,$	$\mu {\sqrt {\alpha _{G}}}\,$	$\mu {\sqrt {\frac {\alpha _{G}}{\alpha }}}\,$	$\mu {\sqrt {\frac {\alpha _{G}}{\alpha }}}\,$	$\mu \,$	$\frac{\mu}{2} \,$	$\mu \,$	$\mu m_{\text{e}}\approx 938{\text{ MeV}}$	$\mu m_{\text{e}}\approx 938{\text{ MeV}}$	$1\,$	$1\,$	$1\,$
Vacuum permeability $\mu_0 = \frac{1}{\epsilon_0 c^2} \,$	$4 \pi \alpha$	$1\,$	$4\pi$	$4\pi$	$4 \pi \alpha^2$	$4 \pi \alpha^2$	$\pi \alpha^2$	$4 \pi \alpha$	$1\,$	$4\pi$	$4 \pi \alpha$	$1\,$	$4\pi$
Impedance of free space $Z_0 = \frac{1}{\epsilon_0 c} = \mu_0 c \,$	$4 \pi \alpha$	$1\,$	$4\pi$	$4\pi$	$4 \pi \alpha$	$4 \pi \alpha$	$2\pi\alpha$	$4 \pi \alpha$	$1\,$	$4\pi$	$4 \pi \alpha$	$1\,$	$4\pi$
Bohr radius $a_0=\frac{4\pi \epsilon_0 \hbar^2}{m_{\text{e}} e^2} = \frac{\hbar}{m_{\text{e}} c \alpha}$	${\frac {1}{\alpha {\sqrt {\alpha _{G}}}}}$	${\frac {1}{\alpha {\sqrt {4\pi \alpha _{G}}}}}$	${\frac {1}{\alpha {\sqrt {\alpha _{G}}}}}$	${\frac {1}{\sqrt {\alpha ^{3}\alpha _{G}}}}$	${\sqrt {\frac {\alpha }{\alpha _{G}}}}\,$	$1\,$	$1\,$	${\frac {1}{\alpha }}$	${\frac {1}{m_{e}\alpha }}\approx 2.68\times 10^{-4}{\text{eV}}^{-1}$	${\frac {1}{m_{e}\alpha }}\approx 2.68\times 10^{-4}{\text{eV}}^{-1}$	${\frac {\mu }{\alpha }}$	${\frac {\mu }{\alpha }}$	${\frac {\mu }{\alpha }}$
Bohr magneton $\mu_B=\frac{e \hbar}{2 m_e}$	${\frac {1}{2{\sqrt {\alpha _{G}}}}}$	${\frac {\alpha }{4\alpha _{G}}}$	${\frac {\alpha }{4\alpha _{G}}}$	${\frac {1}{\sqrt {4\alpha \alpha _{G}}}}$	${\sqrt {\frac {\alpha }{4\alpha _{G}}}}\,$	$\frac{1}{2}$	$\sqrt{2}$	$\frac{1}{2}$	${\frac {\sqrt {4\pi \alpha }}{2m_{e}}}\approx 2.96\times 10^{-7}{\text{eV}}^{-1}$	${\frac {\sqrt {\alpha }}{2m_{e}}}\approx 8.36\times 10^{-8}{\text{eV}}^{-1}$	${\frac {\mu }{2}}$	${\frac {\mu {\sqrt {4\pi \alpha }}}{2}}$	${\frac {\mu {\sqrt {\alpha }}}{2}}$
Josephson constant $K_\text{J} =\frac{e}{\pi \hbar} \,$	$\frac{1}{\pi} \,$	$\sqrt{\frac{4\alpha}{\pi}} \,$	$\frac{\sqrt{\alpha}}{\pi} \,$	$\frac{\alpha}{\pi} \,$	$\frac{1}{\pi} \,$	$\frac{1}{\pi} \,$	$\frac{\sqrt{2}}{\pi} \,$	$\frac{1}{\pi} \,$	$\sqrt{\frac{4\alpha}{\pi}} \,$	$\frac{\sqrt{\alpha}}{\pi} \,$	$\frac{1}{\pi} \,$	$\sqrt{\frac{4\alpha}{\pi}} \,$	$\frac{\sqrt{\alpha}}{\pi} \,$
von Klitzing constant $R_\text{K} =\frac{2 \pi \hbar}{e^2} \,$	$2 \pi \,$	$\frac{1}{2\alpha}$	$\frac{2\pi}{\alpha} \,$	$\frac{2\pi}{\alpha} \,$	$2 \pi \,$	$2 \pi \,$	$\pi\,$	$2 \pi \,$	$\frac{1}{2\alpha}$	$\frac{2 \pi}{\alpha}$	$2 \pi \,$	$\frac{1}{2\alpha}$	$\frac{2\pi}{\alpha} \,$
Rydberg constant $R_\infty={\frac {m_{\text{e}}e^{4}}{8\epsilon_0^2h^3c}}=\frac{\alpha^2 m_\text{e} c}{4 \pi \hbar}$	${\frac {\sqrt {\alpha ^{4}\alpha _{G}}}{4\pi }}$	${\sqrt {\frac {\alpha ^{4}\alpha _{G}}{4\pi }}}$	${\frac {\sqrt {\alpha ^{4}\alpha _{G}}}{4\pi }}$	${\frac {\sqrt {\alpha ^{5}\alpha _{G}}}{4\pi }}$	${\frac {\sqrt {\alpha \alpha _{G}}}{4\pi }}$	${\frac {\alpha }{4\pi }}$	${\frac {\alpha }{4\pi }}$	${\frac {\alpha ^{2}}{4\pi }}$	${\frac {\alpha ^{2}m_{\text{e}}}{4\pi }}\approx 2.17{\text{eV}}$	${\frac {\alpha ^{2}m_{\text{e}}}{4\pi }}\approx 2.17{\text{eV}}$	${\frac {\alpha ^{2}}{4\pi \mu }}$	${\frac {\alpha ^{2}}{4\pi \mu }}$	${\frac {\alpha ^{2}}{4\pi \mu }}$
Stefan–Boltzmann constant $\sigma ={\frac {\pi ^{2}k_{\text{B}}^{4}}{60\hbar ^{3}c^{2}}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}\alpha ^{3}}{60}}\,$	${\frac {\pi ^{2}\alpha ^{2}}{60}}\,$	${\frac {\pi ^{2}\alpha ^{2}}{60}}\,$	${\frac {\pi ^{2}\alpha ^{2}}{240}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$	${\frac {\pi ^{2}}{60}}\,$

where:

$α$ is the fine-structure constant, k_ee^2<\sup>ħc = $(e q Planck) 2 \approx 0.0072973525693$

$α G$ is the gravitational coupling constant, Gm_e^2<\sup>ħc = $(m e m Planck) 2 \approx 1.7518 \times 10 -45$

$µ$ is the proton-to-electron mass ratio, m_pm_e ≈ $1836.15267247$

Proposed system

		L	T	M
o	4^128
	4^112			3.57 * 10^28 SM
	4^96		8.14 my	8.3 * 10^18 SM
	4^80		16.62 hours	1.93 billion SM
a	4^64	4.176 k		0.4569 SM
	4^48	972 nm		2.115 * 10^20 kg
	4^32	0.226 fm		4.925 * 10^10 kg
	4^16			11.47 kg
i	4^0			2.67 μg
	4^-16
	4^-32			0.085 m_u
	4^-48			0.019 eV
	4^-64			4.4 peV
	4^-80

	0	$C_{0}V_{0}$
4^0	1	$C_{1}V_{1}$
	2	$C_{2}V_{1}$
	3	$C_{3}V_{1}$

4^1	4	$C_{1}V_{4}$
	8	$C_{2}V_{4}$
	12	$C_{3}V_{4}$

4^2	16	$C_{1}V_{16}$
4^4	256	$C_{1}V_{256}$
4^8	65536
4^16	4,294,967,296
4^32
4^64
4^128

255 =

(C_{3}V_{4}C_{3})V_{16}\;(C_{3}V_{4}C_{3})V_{1}

256^2 =

C_{1}V_{256}\;C_{2}V_{power}

$\alpha ={\frac {e^{2}}{4\pi \epsilon _{0}\hbar c}}={\frac {1}{2\varepsilon _{0}}}{\frac {e^{2}}{hc}}={\frac {\mu _{0}}{2}}{\frac {e^{2}c}{h}}={\frac {k_{\text{e}}e^{2}}{\hbar c}}={\frac {c\mu _{0}}{2R_{\text{K}}}}={\frac {e^{2}}{2}}{\frac {Z_{0}}{h}}$

e = 1 electron charge
c $={\sqrt {\frac {\varepsilon _{0}^{-1}}{\mu _{0}}}}$ = 1 unit of velocity
$\varepsilon _{0}^{-1}=1\,{\frac {F_{e}\cdot 4\pi L^{2}}{Q^{2}}}$

\hbar ={\frac {1}{\alpha }}=137.035999177

$G_{c}=1{\frac {F_{g}\cdot 4\pi L^{2}}{M^{2}}}$
$k_B$ = 1 or 2?

Numerical results:
$L=1.226838563896828\times 10^{-35}\ {\text{m}}$

L\cdot 2^{128}=4.176\,{\text{km}}

$T=4.092292955204457\times 10^{-44}\ {\text{s}}$

T\cdot 2^{128}=13.9\,{\text{microseconds}}

$M=2.6293187157825153\times 10^{-9}\ {\text{kg}}$

M\cdot 2^{128}=0.45\,{\text{solar masses}}

Mass of electron = ${\frac {9.1093837015\times 10^{-31}}{2.6293187157825153\times 10^{-9}}}=3.4649\times 10^{-22}$

Mass of proton = ${\frac {1.67262192369\times 10^{-27}}{2.6293187157825153\times 10^{-9}}}=6.3626\times 10^{-19}$

Derived units: velocity, acceleration, force, energy, momentum, pressure, volts

Fields: Gravitational, Gravitomagnetic, Electric, Magnetic

Per = -n, per volume, per area, per distance, per time, per cycle, per degree of freedom

Squared, cubed, 4th power

Inverse or 1/

Negative

Angle as fundamental unit

The Planck constant has the same dimensions as action and as angular momentum (both with unit J·s = kg·m²·s⁻¹). The Planck constant is fixed at $h$ = 6.62607015 × 10⁻³⁴ J⋅Hz⁻¹ as part of the definition of the SI units.

Alternatively, if the radian were considered a base unit, then $h$ would have the dimension of action (unit J·s), while $\hbar$ would have the dimension of angular momentum (unit J·s·rad⁻¹), instead.

This value is used to define the SI unit of mass, the kilogram: "the kilogram [...] is defined by taking the fixed numerical value of $h$ to be 6.62607015 × 10^-34 when expressed in the unit J⋅s, which is equal to kg⋅m²⋅s⁻¹, where the metre and the second are defined in terms of speed of light $c$ and duration of hyperfine transition of the ground state of an unperturbed caesium-133 atom $Δ ν Cs$ ." Technologies of mass metrology such as the Kibble balance measure the kilogram by fixing the Planck constant.

From Radian:

Plane angle may be defined as $θ = s\r$ , where $θ$ is the magnitude in radians of the subtended angle, $s$ is circular arc length, and $r$ is radius. One radian corresponds to the angle for which $s = r$ , hence $1 radian = 1 m\m = 1$ . However, $rad$ is only to be used to express angles, not to express ratios of lengths in general. A similar calculation using the area of a circular sector $θ = 2A\r2$ gives 1 radian as 1 m²\m² = 1. The key fact is that the radian is a dimensionless unit equal to 1. In SI 2019, the SI radian is defined accordingly as 1 rad = 1. It is a long-established practice in mathematics and across all areas of science to make use of $rad = 1$ .

Giacomo Prando writes "the current state of affairs leads inevitably to ghostly appearances and disappearances of the radian in the dimensional analysis of physical equations". For example, an object hanging by a string from a pulley will rise or drop by $y = rθ$ centimetres, where $r$ is the magnitude of the radius of the pulley in centimetres and $θ$ is the magnitude of the angle through which the pulley turns in radians. When multiplying $r$ by $θ$ , the unit radian does not appear in the product, nor does the unit centimetre—because both factors are magnitudes (numbers). Similarly in the formula for the angular velocity of a rolling wheel, $ω = v\r$ , radians appear in the units of $ω$ but not on the right hand side. Anthony French calls this phenomenon "a perennial problem in the teaching of mechanics". Oberhofer says that the typical advice of ignoring radians during dimensional analysis and adding or removing radians in units according to convention and contextual knowledge is "pedagogically unsatisfying".

In 1993 the American Association of Physics Teachers Metric Committee specified that the radian should explicitly appear in quantities only when different numerical values would be obtained when other angle measures were used, such as in the quantities of angle measure (rad), angular speed (rad\s), angular acceleration (rad\s²), and torsional stiffness (N⋅m\rad), and not in the quantities of torque (N⋅m) and angular momentum (kg⋅m²/s).

At least a dozen scientists between 1936 and 2022 have made proposals to treat the radian as a base unit of measurement for a base quantity (and dimension) of "plane angle". Quincey's review of proposals outlines two classes of proposal. The first option changes the unit of a radius to meters per radian, but this is incompatible with dimensional analysis for the area of a circle, $π r 2$ . The other option is to introduce a dimensional constant. According to Quincey this approach is "logically rigorous" compared to SI, but requires "the modification of many familiar mathematical and physical equations". A dimensional constant for angle is "rather strange" and the difficulty of modifying equations to add the dimensional constant is likely to preclude widespread use.

In particular, Quincey identifies Torrens' proposal to introduce a constant $η$ equal to 1 inverse radian (1 rad⁻¹) in a fashion similar to the introduction of the constant ε₀. With this change the formula for the angle subtended at the center of a circle, $s = rθ$ , is modified to become $s = ηrθ$ , and the Taylor series for the sine of an angle $θ$ becomes:

$\operatorname {Sin} \theta =\sin \ x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =\eta \theta -{\frac {(\eta \theta )^{3}}{3!}}+{\frac {(\eta \theta )^{5}}{5!}}-{\frac {(\eta \theta )^{7}}{7!}}+\cdots ,$

where

$x=\eta \theta =\theta /{\text{rad}}$ is the angle in radians.

The capitalized function $Sin$ is the "complete" function that takes an argument with a dimension of angle and is independent of the units expressed, while $sin$ is the traditional function on pure numbers which assumes its argument is a dimensionless number in radians. The capitalised symbol $\operatorname {Sin}$ can be denoted $\sin$ if it is clear that the complete form is meant.

Current SI can be considered relative to this framework as a natural unit system where the equation $η = 1$ is assumed to hold, or similarly, 1 rad = 1. This radian convention allows the omission of $η$ in mathematical formulas.

Defining radian as a base unit may be useful for software, where the disadvantage of longer equations is minimal. For example, the Boost units library defines angle units with a plane_angle dimension, and Mathematica's unit system similarly considers angles to have an angle dimension.

Metric system

From Wikipedia:Template:SI prefixes (infobox)

SI prefixes
Prefix		Base 10	Decimal	Adoption
Name	Symbol	Base 10	Decimal	Adoption
quetta	Q	10³⁰	1000000000000000000000000000000	2022
ronna	R	10²⁷	1000000000000000000000000000	2022
yotta	Y	10²⁴	1000000000000000000000000	1991
zetta	Z	10²¹	1000000000000000000000	1991
exa	E	10¹⁸	1000000000000000000	1975
peta	P	10¹⁵	1000000000000000	1975
tera	T	10¹²	1000000000000	1960
giga	G	10⁹	1000000000	1960
mega	M	10⁶	1000000	1873
kilo	k	10³	1000	1795
hecto	h	10²	100
deca	da	10¹	10
—	—	10⁰	1	—
deci	d	10⁻¹	0.1	1795
centi	c	10⁻²	0.01
milli	m	10⁻³	0.001
micro	μ	10⁻⁶	0.000001	1873
nano	n	10⁻⁹	0.000000001	1960
pico	p	10⁻¹²	0.000000000001	1960
femto	f	10⁻¹⁵	0.000000000000001	1964
atto	a	10⁻¹⁸	0.000000000000000001	1964
zepto	z	10⁻²¹	0.000000000000000000001	1991
yocto	y	10⁻²⁴	0.000000000000000000000001	1991
ronto	r	10⁻²⁷	0.000000000000000000000000001	2022
quecto	q	10⁻³⁰	0.000000000000000000000000000001	2022

References

This page uses content (with sometimes significant modification) from Wikipedia: Hooke's law, Maxwell's equations, Gravitational constant, Coulomb's law, Vacuum permittivity, Electric displacement field, Magnetic field, Displacement current, Energy density, Speed of light, Schrödinger equation, Natural units, Electromagnetic tensor, Mathematical descriptions of the electromagnetic field, Antisymmetric tensor, Directional derivative, Exterior derivative, Diffusion equation, Polarization density, magnetization, Harmonic oscillator

$\color {red}\left({\frac {{\frac {Q}{L^{3}}}L}{L}}\right)$	$\nabla \cdot \mathbf {D}$	=	$\rho$	$\color {red}\left({\frac {Q}{L^{3}}}\right)$
$\color {red}\left({\frac {Q}{L^{3}}}v+{\frac {{\frac {Q}{L^{3}}}L}{T}}\right)$	$\mathbf {J} +{\frac {\partial \mathbf {D} }{\partial t}}$	=	$\nabla \times \mathbf {H}$	$\color {red}\left({\frac {{\frac {Q_{v}}{L^{3}}}L}{L}}\right)$
$\color {red}\left({\frac {Q_{v}}{L^{3}}}\right)$	$0\;\rho _{m}$	=	$\nabla \cdot \mathbf {H}$	$\color {red}\left({\frac {{\frac {Q_{v}}{L^{3}}}L}{L}}\right)$
$\color {red}{\frac {Q}{L^{3}}}vv=v^{2}\left({\frac {{\frac {Q}{L^{3}}}L}{L}}\right)$	$c_{0}^{2}\left(\nabla \times \mathbf {D} \right)$	=	$-{\frac {\partial \mathbf {H} }{\partial t}}$	$\color {red}\left({\frac {{\frac {Q_{v}}{L^{3}}}L}{T}}\right)={\frac {Q_{v}}{L^{3}}}v=J_{m}$