User:Alfred Centauri/sandbox

http://en.wikipedia.org/wiki/Help:Displaying_a_formula

TL

(Note: In the following, subscript notation is used for 1st and 2nd partial derivatives.)

For an ideal lossless two conductor transmission line (TL), the voltage across the conductors and the current through the conductors must satisfy the following equations:

$V_x+L{I}_t=0$

$I_x+C{V}_t=0$

where L and C are the inductance and capacitance per unit length of the TL.

These equations lead to the following wave equations for the voltage and current:

$V_{xx}-LC{V}_{tt}=0$

$I_{xx}-LC{I}_{tt}=0$

Consider such a TL of unit length that is unterminated at both ends. The current at the ends of the TL is constrained to be zero. With these boundary conditions, the solution to the current wave equation is:

$I(x,t)=\sum _{n=1}^{\mathrm{\infty }}{a}_n(t)\sin (n\pi x)$

where

$a_n(t)=2[(\cos (\frac{n\pi t}{\sqrt{LC}})\underset{0}{\overset{1}{\int }}I(x,0)\sin (n\pi \tau )d\tau )+(\sin (\frac{n\pi t}{\sqrt{LC}})\underset{0}{\overset{1}{\int }}{I}_t(x,0)\sin (n\pi \tau )d\tau )]$

For

$I(x,t)=0$

the only solution is the trivial solution:

$a_n(t)=0$

which can only be true when the initial conditions are:

$I(x,0)=I_t(x,0)=0$

Thus:

$V_x=0\Rightarrow V(x,t)=constant$

That is, for a finite length ideal lossless TL that is unterminated at both ends, there is no possible superposition of waves that give the solution V(x,t) = constant, I(x,t) = 0.

works for me. Pfalstad 02:23, 15 October 2005 (UTC)

Capacitor Weirdness

{I found this in the Displacement Current article}

Over the years, many scientists and engineers have questioned whether or not displacement current "causes" magnetic fields. The following simple derivation shows how displacement current and charges interact to form magnetic fields.

The capacitance of a parallel plate capacitor with plates of area A and plate separation of d may be expressed as:

$C={\epsilon }_R{ϵ}_O\frac{A}{d}$ (1)

Where ${ϵ}_R$ and ${ϵ}_0$ are dielectric and free space permittivity, respectively.

Another basic equation relates the uniform charge, Q stored in a capacitor, to the capacitance C, and the Voltage across the plates, V.

$Q=CV$ (2)

Divide both sides of equation (2) by the distance between the plates, d:

$\frac{Q}{d}=C\frac{V}{d}$ (3)

Assuming a uniform electric field directed in the +z direction, $E_z=\frac{V}{d}$, we may rewrite equation (3) as follows:

$\frac{Q}{d}=C{E}_z$ (4)

Substitute the value for C from equation 1 into equation (4):

$\frac{Q}{d}={\epsilon }_R{ϵ}_O\frac{A}{d}{E}_z$ (5)

Multiply both sides of equation 5 by d, and take the partial derivative of equation 5 with respect to time to yield:

$\frac{\partial Q}{\partial t}={\epsilon }_R{ϵ}_{O}A\frac{\partial E_z}{\partial t}$ (6)

Using the constituitive relation ${\epsilon }_R{ϵ}_O\overrightarrow{E}=\overrightarrow{D}$, equation (6) may be written as:

$\frac{\partial Q}{\partial t}=A\frac{\partial D_Z}{\partial t}$ (7)

The point form of Ampere’s law, as modified by Maxwell, is often written as:

$\mathrm{\nabla }\times \overrightarrow{H}=J+\frac{\partial \overrightarrow{D}}{\partial t}$ (8)

Applying Ampere's law to our capacitor, we have:

$\frac{\partial H_y}{\partial x}-\frac{\partial H_x}{\partial y}=J_z+A^{-1}\frac{\partial Q}{\partial t}$ (9)

{What follows is nonsense}

But $A^{-1}\frac{\partial Q}{\partial t}$ is another way of expressing J. In this case, since the electrical field is Transverse to the plane of the conductor, we define a new term,

$J_T=A^{-1}\frac{\partial Q}{\partial t}$ (10)

The term “J” has traditionally been taken to mean “longitudinal conduction current.” To differentiate it from $J_T$, let us rename J as $J_L$ standing for longitudinal current flow. With these re-definitions, we can now rewrite equation (9) in a form that is consistent with the spirit of Ampere:

$\mathrm{\nabla }\times H=J_L+J_T$ (11)

Conclusion Equations 8, 9 and 11 may be used interchangeably, but only if one understands that the source of the magnetic fields is the motion of charges, $\frac{\partial Q}{\partial t}$ .

Theoretical basis for impedance

For a system such as an electric circuit that is described mathematically by a linear system of non-homogeneous differential equations, the solutions to these equations are in general of a different form than the driving functions. For example, consider a simple circuit consisting of a voltage source and a capacitor. The current through the capacitor proportional to the derivative of the the voltage across the capacitor.

$i_C(t)=C\frac{d{v}_C(t)}{dt}$

In general, a function and its derivative are not of the same form. For example, let:

$v_C(t)=1+2t(V)$

The capacitor current is then given by:

$i_C(t)=2C(A)$

The ratio of the voltage and current associated with the capacitor is then:

$\frac{v_C(t)}{i_C(t)}=\frac{1+2t}{2C}$

This ratio clearly changes with time.

However, there is a special function, the complex exponential, that is of the same form as its derivative:

$\frac{d}{dt}{e}^{s_{0}t}=s_0{e}^{s_{0}t}$

Let the source voltage be given by:

$v_S(t)=e^{s_{0}t}$

where:

$s_0={\sigma }_0+j{\omega }_0$

The capacitor voltage is then given by:

$v_C(t)=e^{s_{0}t}(V)$

and the the capacitor current is:

$i_C(t)=s_{0}C{e}^{s_{0}t}(A)$

The ratio of this voltage and current is:

$\frac{v_C(t)}{i_C(t)}=\frac{1}{s_{0}C}$

which is clearly constant with time.

This ratio defines the impedance of the capacitor at the complex frequency s_0:

$Z_C(s_0)=\frac{1}{s_{0}C}$

Transconductance doubling

$i_O=2(I_B+I_S)\sinh \left(\frac{v_I-v_O}{V_T}\right)$

$v_O=i_O{R}_L=2(I_B+I_S)R_L\sinh \left(\frac{v_I-v_O}{V_T}\right)$

$\frac{d{i}_O}{d{v}_I}=\frac{2(I_B+I_S)}{V_T}\cosh \left(\frac{v_I-v_O}{V_T}\right)(1-\frac{d{v}_O}{d{v}_I})$

$\frac{v_I-v_O}{V_T}=\sinh {}^{-1}\left(\frac{v_O}{2(I_B+I_S)R_L}\right)$

$\cosh \left(\frac{v_I-v_O}{V_T}\right)=\cosh [\sinh {}^{-1}\left(\frac{v_O}{2(I_B+I_S)R_L}\right)]=\sqrt{{\left(\frac{v_O}{2(I_B+I_S)R_L}\right)}^2+1}$

$\frac{d{i}_O}{d{v}_I}=\frac{\sqrt{{\left(\frac{v_O}{R_L}\right)}^2+4{(I_B+I_S)}^2}}{V_T}(1-\frac{d{v}_O}{d{v}_I})$

$\frac{d{i}_O}{d{v}_I}=\frac{\frac{\sqrt{{\left(\frac{v_O}{R_L}\right)}^2+4{(I_B+I_S)}^2}}{V_T}}{1+\frac{\sqrt{{\left(\frac{v_O}{R_L}\right)}^2+4{(I_B+I_S)}^2}}{V_T}{R}_L}=\frac{1}{1+\frac{V_T}{\sqrt{v_O^2+4{(I_B+I_S)}^2{R}_L^2}}}\frac{1}{R_L}$

Potential fields within the Special Theory of Relativity

Energy of a free particle

In classical mechanics, the energy of a free particle is just the kinetic energy denoted by $T$:

$T=\frac{1}{2}m|\overrightarrow{v}{|}^2=\frac{|\overrightarrow{p}{|}^2}{2m}$

where m is the mass, $\overrightarrow{v}$ is the velocity, and $\overrightarrow{p}$ is the classical momentum of the particle:

$\overrightarrow{p}=m\overrightarrow{v}$

In relativistic mechanics, the total energy of a free particle is given by:

$E=\gamma m{c}^2=\sqrt{(|\overrightarrow{p}|c{)}^2+(m{c}^2{)}^2}$

where $c$ is the speed of light, $\gamma$ is the Lorentz factor and the relativistic momentum $\overrightarrow{p}$ is given by:

$\overrightarrow{p}=\gamma m\overrightarrow{v}$

and the kinetic energy is given by:

$T=E-E_0=\gamma m{c}^2-m{c}^2=m{c}^2(\gamma -1)$

where $E_0$ is the rest energy.

Energy of a particle in a potential field

In classical mechanics, the total energy of a particle in a potential is the sum of the kinetic energy, $T$ and the potential energy, $V$.

$E=T+V$

so that the classical kinetic energy is given by:

$T=E-V$

In relativistic mechanics, a potential energy cannot be simply added to or subtracted from the kinetic energy to get the total energy since, within the Special Theory of Relativity (STR), energy is not a scalar quantity. Instead, energy is a component of a four-vector. For example, the energy-momentum four-vector describing the kinetic energy-momentum of a free particle is:

$\mathbf{𝐩}=(-\frac{E}{c},\overrightarrow{p})$

where $E$ is the relativistic energy of a free particle: $E=E_0+T$.

Potential energy-momentum

As the kinetic energy of a particle is a component of a four-vector, it follows that the potential energy of a particle must also be a component of a four-vector - a potential energy-momentum four-vector. The total energy must also be a component of a four-vector - the total energy-momentum four-vector which must be the sum of the kinetic and potential energy-momentum four-vectors:

$\mathbf{𝐏}=\mathbf{𝐩}+\mathit{\varphi }(\mathbf{𝐱})=m\mathbf{𝐔}+\mathit{\varphi }(\mathbf{𝐱})$

where $\mathbf{𝐏}$ is the total energy-momentum four-vector (total four-momentum), $\mathbf{𝐩}$ is the kinetic energy-momentum four-vector (kinetic four-momentum), $\mathit{\varphi }$ is the potential energy-momentum four-vector (potential four-momentum), $\mathbf{𝐱}=(-ct,\overrightarrow{x})$ is the position four-vector and $\mathbf{𝐔}=(-\gamma c,\gamma \overrightarrow{v})$ is the four-velocity:

The Lagrangian mechanics

Classical Lagrangian

In the classical (non-relativistic) Lagrangian mechanics, the action ${𝒮}$ of the particle in a potential is defined by:

${𝒮}[\overrightarrow{x}(t),\overrightarrow{v}(t)]\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\underset{t_1}{\overset{t_2}{\int }}L(\overrightarrow{x},\overrightarrow{v},t)dt$

where $L(\overrightarrow{x},\overrightarrow{v},t)$ is the Lagrangian of the system. According to Hamilton's principle, the action of a system is stationary:

$\frac{\delta {𝒮}}{\delta \overrightarrow{x}(t)}=0$

By the calculus of variations, the action is found to be stationary when the Euler-Lagrange equation holds :

$\frac{d}{dt}\left(\frac{\partial L}{\partial \overrightarrow{v}}\right)-\frac{\partial L}{\partial \overrightarrow{x}}=0$

The Lagrangian, as the time rate of change of the action, is an energy and so must be a function of the kinetic and potential energy of the particle. For the particle in a potential system, the Lagrangian is defined to be:

$L=\frac{1}{2}m|\overrightarrow{v}{|}^2-V(\overrightarrow{x})$

Inserting the Lagrangian into the Euler-Lagrange equation yields the familar Newtonian equation of motion:

$\frac{d}{dt}\left(\frac{\partial L}{\partial \overrightarrow{v}}\right)=\frac{d}{dt}m\overrightarrow{v}=\frac{d\overrightarrow{p}}{dt}$

$\frac{\partial L}{\partial \overrightarrow{x}}=-\mathrm{\nabla }V(\overrightarrow{x})$

$\Rightarrow \frac{d\overrightarrow{p}}{dt}=-\mathrm{\nabla }V(\overrightarrow{x})$

Lorentz invariant Lagrangian

The classical momentum $\overrightarrow{p}$ and the classical Lagrangian $L$ are related by:

$\overrightarrow{p}=\frac{\partial L}{\partial \overrightarrow{v}}$

Relating the total energy-momentum four-vector of a particle to its relativistic (Lorentz invariant) Lagrangian in the same manner gives:

$P_{\mu }=p_{\mu }+{\varphi }_{\mu }=m{U}_{\mu }+{\varphi }_{\mu }=\frac{\partial L}{\partial U^{\mu }},\mu =0,1,2,3$

Integrating with respect to the four-velocity gives:

$L=\frac{1}{2}m{U}_{\mu }{U}^{\mu }+{\varphi }_{\mu }{U}^{\mu }+C$

where C is the constant of integration with respect to $\mathbf{𝐔}$.

Integrating $\mathbf{𝐏}$ with respect to the particle's four-velocity yields a Lorentz invariant Lagrangian that contains a kinetic term and a potential term. The constant of integration could be a function of $\mathbf{𝐱}$ only but, as will be shown later, this would destroy the Lorentz invariance of the Lagrangian. Thus, $C$ must be a constant with respect to both $\mathbf{𝐔}$ and $\mathbf{𝐱}$.

In the classical limit ($c\to \mathrm{\infty }$), the classical Lagrangian should be recovered from the Lorentz invariant Lagrangian:

$\underset{c\to \mathrm{\infty }}{\lim }L=-\frac{1}{2}m{c}^2+\frac{1}{2}m|\overrightarrow{v}{|}^2-V(\overrightarrow{x})+C$

where $V=c{\varphi }^0$ (before taking the limit!) in the reference frame where ${\varphi }^i=0$.

The classical Lagrangian is recovered if:

$C=\frac{1}{2}m{c}^2$

$\Rightarrow L=\frac{1}{2}m{c}^2+\frac{1}{2}m{U}_{\mu }{U}^{\mu }+{\varphi }_{\mu }{U}^{\mu }$

The Lorentz covariant Euler-Lagrange equation

The relativistic (Lorentz covariant) Euler-Lagrange equation is:

$\frac{d}{d\tau }\left(\frac{\partial L}{\partial U^{\mu }}\right)-\frac{\partial L}{\partial x^{\mu }}=0$

where $\tau$ is the proper time:

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

Insert the Lorentz invariant Lagrangian into the Lorentz covariant Euler Lagrange equation:

$\frac{d}{d\tau }\left(\frac{\partial L}{\partial U^{\mu }}\right)=\frac{d}{d\tau }(m{U}_{\mu }+{\varphi }_{\mu })=m\frac{d{U}_{\mu }}{d\tau }+\frac{\partial {\varphi }_{\mu }}{\partial x^{\nu }}\frac{\partial x^{\nu }}{\partial \tau }=\frac{d{p}_{\mu }}{d\tau }+{\varphi }_{\mu }{,}_{\nu }{U}^{\nu }$

$\frac{\partial L}{\partial x^{\mu }}=\frac{\partial {\varphi }_{\nu }}{\partial x^{\mu }}{U}^{\nu }={\varphi }_{\nu }{,}_{\mu }{U}^{\nu }$

Putting this together and rearranging yields the equation of motion for the particle:

$\frac{d{p}_{\mu }}{d\tau }=({\varphi }_{\nu }{,}_{\mu }-{\varphi }_{\mu }{,}_{\nu })U^{\nu }$

In 3 plus 1 form, this equation is written as:

$\frac{d{p}_0}{dt}=\frac{d\gamma mc}{dt}=\overrightarrow{v}\cdot (-\mathrm{\nabla }{\varphi }_0+\frac{1}{c}\frac{\partial \overrightarrow{\varphi }}{\partial t})$

$\frac{d\overrightarrow{p}}{dt}=\frac{d\gamma m\overrightarrow{v}}{dt}=c(\mathrm{\nabla }{\varphi }_0-\frac{1}{c}\frac{\partial \overrightarrow{\varphi }}{\partial t})+\overrightarrow{v}\times (\mathrm{\nabla }\times \overrightarrow{\varphi })$

The spatial fields

To simply the form of these equations, make the following identifications with the field terms above in parenthesis:

$\overrightarrow{\mathfrak{𝔈}}\equiv \mathrm{\nabla }{\varphi }_0-\frac{1}{c}\frac{\partial \overrightarrow{\varphi }}{\partial t}=-\mathrm{\nabla }{\varphi }^0-\frac{1}{c}\frac{\partial \overrightarrow{\varphi }}{\partial t}$

$\overrightarrow{\mathfrak{𝔅}}\equiv \mathrm{\nabla }\times \overrightarrow{\varphi }$

Substituting yields:

$\frac{d{p}_0}{dt}=\frac{d\gamma mc}{dt}=-\overrightarrow{\mathfrak{𝔈}}\cdot \overrightarrow{v}$

$\frac{d\overrightarrow{p}}{dt}=\frac{d\gamma m\overrightarrow{v}}{dt}=c\overrightarrow{\mathfrak{𝔈}}+\overrightarrow{v}\times \overrightarrow{\mathfrak{𝔅}}$

The unit of the $\mathfrak{𝔈}$ and $\mathfrak{𝔅}$ fields is momentum per meter or alternately, force per speed.

The tensor field

Taking another look at the four-vector equation:

$\frac{d{p}_{\mu }}{d\tau }=({\varphi }_{\nu }{,}_{\mu }-{\varphi }_{\mu }{,}_{\nu })U^{\nu }={\mathfrak{𝔉}}_{\mu \nu }{U}^{\nu }$

The term in parenthesis is just the exterior derivative of the potential energy-momentum four-vector $\mathit{\varphi }$ defining ${\mathfrak{𝔉}}_{\mu \nu }$, an anti-symmetric tensor field of rank 2. This equation can be written in component free form:

$\frac{d\mathbf{𝐩}}{d\tau }=\mathbf{𝐝}\mathit{\varphi }\mathbf{(}\mathbf{𝐔}\mathbf{)}=\mathfrak{𝔉}\mathbf{(}\mathbf{𝐔}\mathbf{)}$

The unit of the $\mathfrak{𝔉}$ field is momentum per meter or alternately, force per speed.

The exterior derivative of the tensor field

One of the properties of the exterior derivative is the following identity:

$\mathbf{𝐝}\mathfrak{𝔉}={\mathbf{𝐝}}^{\mathbf{𝟐}}\mathit{\varphi }\equiv 0$

In component form, this is:

${\mathfrak{𝔉}}_{\alpha \beta ,\gamma }+{\mathfrak{𝔉}}_{\beta \gamma ,\alpha }+{\mathfrak{𝔉}}_{\gamma \alpha ,\beta }\equiv 0$

In 3 plus 1 form, this is:

$\mathrm{\nabla }\cdot \overrightarrow{\mathfrak{𝔅}}\equiv 0$

$\frac{1}{c}\frac{\partial }{\partial t}\overrightarrow{\mathfrak{𝔅}}+\mathrm{\nabla }\times \overrightarrow{\mathfrak{𝔈}}\equiv 0$

The divergence of the tensor field

The divergence of the field $\mathfrak{𝔉}$ gives the source of the field:

$\mathrm{\nabla }\cdot \mathfrak{𝔉}=\mathfrak{𝔤},{\mathfrak{𝔉}}^{\alpha \beta }{,}_{\beta }={\mathfrak{𝔤}}^{\alpha }$.

$\mathfrak{𝔤}$ is a four-vector and is the source of $\mathfrak{𝔉}$. The unit of $\mathfrak{𝔤}$ is momentum per square meter, a current density four-vector or a four-current.

Written in 3 plus 1 form, this is:

$\mathrm{\nabla }\cdot (-\mathrm{\nabla }{\varphi }^0-\frac{1}{c}\frac{\partial }{\partial t}\overrightarrow{\varphi })=\mathrm{\nabla }\cdot \overrightarrow{\mathfrak{𝔈}}={\mathfrak{𝔤}}^0$

$\frac{1}{c^2}\frac{\partial }{\partial t}c(\mathrm{\nabla }{\varphi }^0+\frac{1}{c}\frac{\partial \overrightarrow{\varphi }}{\partial t})+\mathrm{\nabla }\times (\mathrm{\nabla }\times \overrightarrow{\varphi })=-\frac{1}{c}\frac{\partial \overrightarrow{\mathfrak{𝔈}}}{\partial t}+\mathrm{\nabla }\times \overrightarrow{\mathfrak{𝔅}}=\overrightarrow{\mathfrak{𝔤}}$

The conserved four-current

The source of $\mathfrak{𝔉}$ is $\mathfrak{𝔤}$, the current density four-vector - a four-current. As $\mathfrak{𝔉}$ is antisymmetric, the divergence of $\mathfrak{𝔤}$ is identically zero:

${\mathfrak{𝔤}}^{\mu }{,}_{\mu }=\frac{1}{c}\frac{\partial g^0}{dt}+\mathrm{\nabla }\cdot \overrightarrow{g}\equiv 0$

Thus, $\mathfrak{𝔤}$ satisfies a continuity equation and thus, is a conserved current. The quantity $\rho$ that is conserved is:

${\rho }_M=\frac{g^0}{c}$

where ${\rho }_M$ is a moment (mass X displacement) volume density.

A current density is the product of a volume density and a velocity:

${\mathfrak{𝔤}}^{\mu }=U^{\mu }{\rho }_{M_0}=\frac{U^{\mu }}{\gamma }\gamma {\rho }_{M_0}=\frac{U^{\mu }}{\gamma }{\rho }_M$

where ${\rho }_M\equiv \gamma {\rho }_{M_0}$ and ${\rho }_{M_0}$ is the invariant (Lorentz scalar) moment density:

$(c{\rho }_{M_0}{)}^2={\mathfrak{𝔤}}_{\mu }{\mathfrak{𝔤}}^{\mu }$

By interpreting $\mathbf{𝐠}$ as a four-current, $\mathbf{𝐠}$ is constrained to be a time-like four-vector, i.e., a tachyonic current density is presumed to be unphysical.

Energy-momentum waves

In the special case where $\mathfrak{𝔤}$ is the zero four-vector, the field equations become:

${\mathfrak{𝔉}}^{\alpha \beta }{,}_{\beta }=0=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{\varphi })-◻\mathit{\varphi }$

where $◻$ is the D'Alembert operator

In 3 plus 1 form, this is written as:

$\mathrm{\nabla }\cdot \overrightarrow{\mathfrak{𝔈}}=0$

$\mathrm{\nabla }\times \overrightarrow{\mathfrak{𝔅}}=\frac{1}{c}\frac{\partial \overrightarrow{\mathfrak{𝔈}}}{\partial t}$

While $\mathfrak{𝔤}$ contains a term that depends on the divergence of the four-potential ($\mathrm{\nabla }\cdot \mathit{\varphi }$), the equations of motion for the particle do not so that the divergence of $\mathit{\varphi }$ is a degree of freedom. The field equations for $\mathfrak{𝔤}$ = 0 become:

$\mathrm{\nabla }\cdot \mathit{\varphi }=0\Rightarrow ◻\mathit{\varphi }=0$

This is a wave equation for the four-potential $\mathit{\varphi }$. In 3 plus 1 form, this wave equation becomes:

${\mathrm{\nabla }}^2{\varphi }_0=\frac{1}{c^2}\frac{{\partial }^2{\varphi }_0}{\partial t^2}$

${\mathrm{\nabla }}^2\overrightarrow{\varphi }=\frac{1}{c^2}\frac{{\partial }^2\overrightarrow{\varphi }}{\partial t^2}$

Thus, waves propagating with the speed $c$ can exist in the four-potential field as energy-momentum waves.

Point sources

Let the invariant four-current be equal to $-c{\rho }_{M_0}$:

${\mathfrak{𝔤}}_{\mu }{\mathfrak{𝔤}}^{\mu }=-c{\rho }_{M_0}$

Then, there is a Lorentz frame where:

$\mathfrak{𝔤}=(-c{\rho }_{M_0},\overrightarrow{0})$

In this frame:

${\mathfrak{𝔤}}^0=c{\rho }_{M_0}=\frac{1}{c}\frac{\partial }{\partial t}(\mathrm{\nabla }\cdot \mathit{\varphi })-◻{\varphi }_I=-{\mathrm{\nabla }}^2{\varphi }_I$

$\overrightarrow{\mathfrak{𝔤}}=\overrightarrow{v}{\rho }_M=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{\varphi })$

The equation for ${\mathfrak{𝔤}}^0$ is Poisson's equation. A solution is for ${\rho }_M$ to be zero everywhere except at the spatial origin where ${\rho }_M$ is infinite:

${\rho }_M=\frac{M(t)}{m^3}\delta (r)\Rightarrow {\varphi }_I=\frac{M(t)}{4\pi r}$

This is the solution for a point source at rest with a (coordinate) time varying moment charge of M(t).

The charge of a point source is constant

Except at the spatial origin, the four-current for the point source solution is:

${\mathfrak{𝔤}}^0=0$

$\overrightarrow{\mathfrak{𝔤}}=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{\varphi })=\mathrm{\nabla }\left(\frac{1}{c}\frac{\partial }{\partial t}\frac{M(t)}{4\pi r}\right)$

But, unless $\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{\varphi })$ vanishes, this is a space-like four-current which conflicts with the requirement that the four-current be time-like. Thus, in this case:

$\mathrm{\nabla }\cdot \mathit{\varphi }=0\Rightarrow \frac{\partial {\varphi }_I}{\partial t}=0\Rightarrow {\varphi }_I=\frac{M}{4\pi r}$

where $M$ is a constant. The requirement that the four-current is time-like leads to the requirement that ${\varphi }^0={\varphi }_I$ is constant in time in this frame. But, this must be true in any other Lorentz frame:

$\mathit{\varphi }=(\gamma {\varphi }_I,\gamma {\varphi }_I\frac{\overrightarrow{v}}{c})\Rightarrow \mathrm{\nabla }\cdot \mathit{\varphi }=0\Rightarrow \frac{\partial {\varphi }^0}{\partial t}=0$

Thus, the requirement that the four-current is time-like implies that, for point source, $\mathrm{\nabla }\cdot \mathit{\varphi }=0$ and that the charge of a point source is constant. More importantly, the fact that any $\mathit{\varphi }$ can be represented as a superposition of point sources leads to the generalization of these results. Thus, the following results are true in general:

$\mathrm{\nabla }\cdot \mathit{\varphi }=0$

$\frac{\partial {\varphi }^0}{\partial t}=0$

Point source as a particle

The point source in the previous section is located at the spatial origin. In another Lorentz frame where the point source has a relative velocity $\overrightarrow{u}$, the position of the point source is:

$\mathbf{𝐱}=(\gamma ct,\gamma \overrightarrow{u}t)$

and force four-current $\mathfrak{𝔤}$ in this frame is

$\mathfrak{𝔤}=(\gamma c\rho ,\gamma \overrightarrow{u}\rho )$

The point action charge at rest at the spatial origin in one frame is seen as moving point charge in another constituting a force current.