Discovering vector potential from moving field analysis

Question

First I would like to remind Lorentz transformation of length and time as a matrix:
$$ begin{pmatrix} 
ct'
x' 
y'
z' 
end{pmatrix} = begin{pmatrix}
gamma & -  frac{v}{c} gamma & 0 & 0 
- frac{v}{c} gamma & gamma & 0 & 0 
0 & 0 & 1 & 0 
0 & 0 & 0 & 1
end{pmatrix} cdot begin{pmatrix} ct  x  y  z end{pmatrix}  $$
The Transform matrix 4x4 should be the same for all four vectors in Special Relativity.
Look at the picture below:

We know only scalar potential in both cases. My goal is to find vector potentials and to prove that
$$ vec{E} = - vec{nabla} phi - frac{partial vec{A}}{partial t} $$
Let me use Lorentz transformation for some imaginery four electric vector:
$$ begin{pmatrix} 
lambda phi'
A_x' 
A_y'
A_z' 
end{pmatrix} = begin{pmatrix}
gamma & -  frac{v}{c} gamma & 0 & 0 
- frac{v}{c} gamma & gamma & 0 & 0 
0 & 0 & 1 & 0 
0 & 0 & 0 & 1
end{pmatrix} cdot begin{pmatrix} lambda phi  A_x  A_y  A_z end{pmatrix}  $$
$$ phi '= - frac{Q}{4 pi epsilon_0 r'}  $$
and
$$ phi = - frac{1}{4 pi epsilon_0 } cdot frac{Q}{r' - vec{r'} cdot frac{vec{v}}{c} } $$
$lambda$ is some constant for a while
It is looking even good until I begin manipulate the algebra to get $ A_x' $
What do You think about this approach?
Maybe Am I totally wrong?
Please help

Rob Tan · Accepted Answer

If I understood well you are trying to deduce an expression for $boldsymbol{A}$. I'll try to give it just by a retarded-time discussion without explicity passing from Lorentz transformations. You should consider a particle with charge $q_i$ moving along a trajectory $boldsymbol{r}_i(t)$ with a velocity $dot{boldsymbol{r}}_i(tau)doteqtext{d}boldsymbol{r}_i(tau)/text{d}tau$ where $tau$ will be the retarded time, defined at the end of the argumentation. Let's start from the equations of the Lorentz gauge
begin{gather*}
frac{1}{mu}nablacdotboldsymbol{A}
+varepsilonfrac{partialphi}{partial t}=0

nabla^2phi
-muvarepsilonfrac{partial^2phi}{partial t^2}
=
-frac{rho}{varepsilon}

nabla^2boldsymbol{A}
-muvarepsilonfrac{partial^2 boldsymbol{A}}{partial t^2}
=
-muboldsymbol{j}
end{gather*}
which have solutions
begin{gather*}
phi(boldsymbol{r},t)
=
frac{1}{4piepsilon}
intlimits_V
frac{
rholeft(boldsymbol{r}^prime,t
-displaystyle{
frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}
}right)
}{
|boldsymbol{r}-boldsymbol{r}^prime|
}
text{d}^3{boldsymbol{r}^prime}

boldsymbol{A}(boldsymbol{r},t)
=
frac{mu}{4pi}
intlimits_V
frac{
boldsymbol{j}left(boldsymbol{r}^prime,t
-displaystyle{
frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}
}right)
}{
|boldsymbol{r}-boldsymbol{r}^prime|
}
text{d}^3{boldsymbol{r}^prime}
end{gather*}
but you know what $rho,boldsymbol{j}$ are for a moving particle
begin{gather*}
rho_i(boldsymbol{r},t)
=
q_idelta(boldsymbol{r}-boldsymbol{r}_i(t))

boldsymbol{j}_i(boldsymbol{r},t)
=
q_iboldsymbol{u}_i(t)delta(boldsymbol{r}-boldsymbol{r}_i(t))

phi_i(boldsymbol{r},t)
=
frac{1}{4piepsilon}
intlimits_V
frac{q_ideltaleft(boldsymbol{r}^prime-boldsymbol{r}_ileft(t
-displaystyle{frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}}right)right)
}{
|boldsymbol{r}-boldsymbol{r}^prime|
}
text{d}^3{boldsymbol{r}^prime}

boldsymbol{A}_i(boldsymbol{r},t)
=
frac{mu}{4pi}
intlimits_V
frac{q_idot{boldsymbol{r}}_ileft(t-displaystyle{frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}}right)deltaleft(boldsymbol{r}^prime-boldsymbol{r}_ileft(t
-displaystyle{frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}}right)right)
}
{|boldsymbol{r}-boldsymbol{r}^prime|
}
text{d}^3{boldsymbol{r}^prime}
end{gather*}
But you don't like very much this integral so you operate a substitution
begin{gather*}
t^prime
doteq
t
-displaystyle{frac{|boldsymbol{r}-boldsymbol{r}^prime|}{c}}

boldsymbol{d}_i(t)
doteq
boldsymbol{r}-boldsymbol{r}_i(t)
end{gather*}
such that the condition posed by the $delta$ becomes $boldsymbol{r}^prime=boldsymbol{r}-boldsymbol{d}_i(t^prime)$. So now you have
begin{gather*}
phi_i(boldsymbol{r},t)
=
frac{1}{4piepsilon}
intlimits_V
frac{q_ideltaleft(boldsymbol{r}^prime-boldsymbol{r}_i(t^prime)right)}{|boldsymbol{d}_i(t^prime)|}
text{d}^3{boldsymbol{r}^prime}
equiv
frac{1}{4piepsilon}
intlimits_{t^{primeprime}}
intlimits_V
frac{q_ideltaleft(boldsymbol{r}^prime-boldsymbol{r}_i(t^{primeprime})right)}{|boldsymbol{d}_i(t^{primeprime})|}
delta(t^{primeprime}-t^prime)
text{d}^3{boldsymbol{r}^prime}text{d}{t^{primeprime}}

boldsymbol{A}_i(boldsymbol{r},t)
=
frac{mu}{4pi}
intlimits_V
frac{q_idot{boldsymbol{r}}_i(t^prime)deltaleft(boldsymbol{r}^prime-boldsymbol{r}_i(t^prime)right)}{|boldsymbol{d}_i(t^prime)|}
text{d}^3{boldsymbol{r}^prime}
equiv
frac{mu}{4pi}
intlimits_{t^{primeprime}}intlimits_V
frac{q_idot{boldsymbol{r}}_i(t^{primeprime})deltaleft(boldsymbol{r}^prime-boldsymbol{r}_i(t^{primeprime})right)}{|boldsymbol{d}_i(t^{primeprime})|}
delta(t^{primeprime}-t^prime)
text{d}^3{boldsymbol{r}^prime}text{d}{t^{primeprime}}

phi_i(boldsymbol{r},t)
=
frac{1}{4piepsilon}
intlimits_{t^{primeprime}}
frac{q_ideltaleft(t^{primeprime}-t+displaystyle{frac{|boldsymbol{r}-boldsymbol{r}_i(t^{primeprime})|}{c}}right)}{|boldsymbol{d}_i(t^{primeprime})|}
text{d}{t^{primeprime}}

boldsymbol{A}_i(boldsymbol{r},t)
=
frac{mu}{4pi}
intlimits_{t^{primeprime}}
frac{q_idot{boldsymbol{r}}_i(t^{primeprime})deltaleft(t^{primeprime}-t+displaystyle{frac{|boldsymbol{r}-boldsymbol{r}_i(t^{primeprime})|}{c}}right)}{|boldsymbol{d}_i(t^{primeprime})|}
text{d}{t^{primeprime}}
end{gather*}
You see? Now the integral is on the time domain! We are near the end. Now a new variable substitution
begin{equation*}
t^{primeprimeprime}
=
t^{primeprime}
-t
+frac{|boldsymbol{d}_i(t^{primeprime})|}{c}
Longrightarrow
text{d}{t^{primeprimeprime}}
=
text{d}{t^{primeprime}}
+frac{1}{c}frac{text{d}|boldsymbol{d}_i(t^{primeprime})|}{text{d}t^{primeprime}}text{d}{t^{primeprime}}
end{equation*}
but defining
begin{equation*}
boldsymbol{n}_i(t^{primeprime})
doteq
frac{boldsymbol{d}_i(t^{primeprime})}{|boldsymbol{d}_i(t^{primeprime})|}
end{equation*}
You will see that
begin{equation*}
{text{d}|boldsymbol{d}_i(t^{primeprime})|}{text{d}t^{primeprime}}=-boldsymbol{n}_i(t^{primeprime})cdotdot{boldsymbol{r}}_i(t^{primeprime})
end{equation*}
and define
begin{equation*}
kappa_i(t^{primeprime})
doteq
1-frac{1}{c}boldsymbol{n}_i(t^{primeprime})cdotdot{boldsymbol{r}}_i(t^{primeprime})
end{equation*}
such that
begin{gather*}
phi_i(boldsymbol{r},t)
=
frac{1}{4piepsilon}
intlimits_{t^{primeprimeprime}}
frac{q_ideltaleft(t^{primeprimeprime}right)}{|boldsymbol{d}_i(t^{primeprime})|kappa_i(t^{primeprime})}
text{d}{t^{primeprimeprime}}

boldsymbol{A}_i(boldsymbol{r},t)
=
frac{mu}{4pi}
intlimits_{t^{primeprimeprime}}
frac{q_idot{boldsymbol{r}}_i(t^{primeprime})deltaleft(t^{primeprimeprime}right)}{|boldsymbol{d}_i(t^{primeprime})|kappa_i(t^{primeprime})}
text{d}{t^{primeprimeprime}}
end{gather*}
and finally the last definition
begin{gather*}
tau
+frac{|boldsymbol{r}-boldsymbol{r}_i(tau)|}{c}
doteq
t

phi_i(boldsymbol{r},t)
=
frac{1}{4piepsilon}
frac{q_i}{|boldsymbol{r}-boldsymbol{r}_i(tau)|kappa_i(tau)}

boldsymbol{A}_i(boldsymbol{r},t)
=
frac{mu}{4pi}
frac{q_idot{boldsymbol{r}}_i(tau)}{|boldsymbol{r}-boldsymbol{r}_i(tau)|kappa_i(tau)}
end{gather*}
That are the potential of Liénard-Wiechart. The expression of the electric field that you cited is just a consequence of Maxwell equations and so electromagnetic field can be obtained by the potentials, being careful with the gradient and the time derivative. Hope this helps

Discovering vector potential from moving field analysis

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