- #1
Bob44
- 15
- 1
The Helmholtz theorem states that any vector field can be uniquely decomposed into longitudinal (curl free) and transverse (divergence free) parts
$$\mathbf{E}=\mathbf{E}_\parallel+\mathbf{E}_\perp.\tag{1}$$
We take the Coulomb gauge
$$\nabla \cdot \mathbf{A}_C=0\tag{2}$$
so that
$$
\begin{eqnarray}
\mathbf{E}_\parallel &=& -\nabla \Phi_C,\tag{3}\\
\mathbf{E}_\perp &=& - \frac{\partial \mathbf{A}_C}{\partial t}.\tag{4}
\end{eqnarray}
$$
By Gauss's law we have
$$\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{E}_\parallel=\frac{\rho}{\varepsilon_0}.\tag{5}$$
By substituting eqn.##(3)## into eqn.##(5)## we obtain the Poisson equation
$$\nabla^2 \Phi_C=-\frac{\rho}{\varepsilon_0}\tag{6}$$
which has the general instantaneous solution
$$\Phi_C(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{7}$$
Therefore by eqn.##(3)## the longitudinal electric field ##\mathbf{E}_\parallel=-\nabla\Phi_C(\mathbf{r},t)## depends instantaneously on the charge density ##\rho(\mathbf{r}',t)##.
As the longitudinal electric field ##\mathbf{E}_\parallel## is measurable it seems like we have a problem with causality.
Now consider extended electrodynamics as described in this paper by Lee Hively and Andrew Loebl.
It can be derived from the classical electromagnetic Lagrangian with the addition of Feynman's gauge fixing term
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j^\mu A_\mu - \frac{1}{2}(\partial_\mu A^\mu)^2.\tag{8}$$
By using a Lagrangian equivalent to eqn.##(8)## Feynman was able to derive quantum electrodynamics using the path integral approach without the infinities due to overcounting physically identical states that occurs with the fully gauge-invariant standard electromagnetic Lagrangian.
The equations of motion of eqn.##(8)## are given by
$$\partial_\nu F^{\nu\mu}+\partial^\mu(\partial_\alpha A^\alpha)=j^\mu.\tag{9}$$
By using the identity
$$\partial_\nu F^{\nu\mu}=\Box A^\mu-\partial^\mu(\partial_\alpha A^\alpha)\tag{10}$$
we can substitute eqn.##(10)## into eqn.##(9)## in order to derive the wave equations
$$\Box A^\mu=j^\mu\tag{11}$$
without any need to explicitly fix a gauge condition as done in standard electrodynamics.
The equations of motion ##(9)## are equivalent to the following modified inhomogeneous Maxwell equations
$$
\begin{eqnarray}
\nabla \cdot \mathbf{E}&=&\frac{\rho}{\varepsilon_0}-\frac{\partial C}{\partial t},\tag{12}\\
\nabla \times \mathbf{B}&=&\mu_0 \mathbf{J} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}+\nabla C\tag{13}
\end{eqnarray}
$$
where the scalar ##C## is given by
$$C=\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}.\tag{14}$$
In order that the scalar field ##C## in eqn.##(14)## is physical we must obey a restricted gauge invariance such that ##\Box \Lambda=0## when ##\phi\rightarrow\phi-\partial\Lambda/\partial t## and ##\mathbf{A}\rightarrow\mathbf{A}+\nabla\Lambda##.
If we take the divergence of eqn.##(13)## and substitute in eqn.##(12)## we find the following wave equation for the scalar ##C## given by
$$\nabla^2 C - \frac{1}{c^2}\frac{\partial^2 C}{\partial t^2}=-\mu_0\left(\frac{\partial \rho}{\partial t}+\nabla \cdot \mathbf{J}\right).\tag{15}$$
As the righthand side is zero by the continuity relation then eqn.##(15)## reduces to a homogeneous equation.
In ##3##D vector form eqn.##(11)## gives the following wave equations for the scalar potential ##\phi## and vector potential ##\mathbf{A}##:
$$
\begin{eqnarray}
\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}&=&-\frac{\rho}{\varepsilon_0},\tag{16}\\
\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2}&=&-\mu_0\mathbf{J}\tag{17}
\end{eqnarray}
$$
with the general retarded solutions
$$
\begin{eqnarray}
\phi(\mathbf{r},t)&=&\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}d^3r',\tag{18}\\
\mathbf{A}(\mathbf{r},t)&=&\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}d^3r'\tag{19}
\end{eqnarray}
$$
where retarded time ##t_r=t-|\mathbf{r}-\mathbf{r}'|/c##.
Now the longitudinal electric field ##\mathbf{E}_\parallel## given by
$$\mathbf{E}_\parallel=-\nabla \phi-\frac{\partial \mathbf{A}_L}{\partial t}\tag{20}$$
depends on the retarded charge density ##\rho(\mathbf{r}',t_r)## and the retarded longitudinal current density ##\mathbf{J}_L(\mathbf{r}',t_r)## so that causality is maintained.
Problem solved?
$$\mathbf{E}=\mathbf{E}_\parallel+\mathbf{E}_\perp.\tag{1}$$
We take the Coulomb gauge
$$\nabla \cdot \mathbf{A}_C=0\tag{2}$$
so that
$$
\begin{eqnarray}
\mathbf{E}_\parallel &=& -\nabla \Phi_C,\tag{3}\\
\mathbf{E}_\perp &=& - \frac{\partial \mathbf{A}_C}{\partial t}.\tag{4}
\end{eqnarray}
$$
By Gauss's law we have
$$\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{E}_\parallel=\frac{\rho}{\varepsilon_0}.\tag{5}$$
By substituting eqn.##(3)## into eqn.##(5)## we obtain the Poisson equation
$$\nabla^2 \Phi_C=-\frac{\rho}{\varepsilon_0}\tag{6}$$
which has the general instantaneous solution
$$\Phi_C(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{7}$$
Therefore by eqn.##(3)## the longitudinal electric field ##\mathbf{E}_\parallel=-\nabla\Phi_C(\mathbf{r},t)## depends instantaneously on the charge density ##\rho(\mathbf{r}',t)##.
As the longitudinal electric field ##\mathbf{E}_\parallel## is measurable it seems like we have a problem with causality.
Now consider extended electrodynamics as described in this paper by Lee Hively and Andrew Loebl.
It can be derived from the classical electromagnetic Lagrangian with the addition of Feynman's gauge fixing term
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j^\mu A_\mu - \frac{1}{2}(\partial_\mu A^\mu)^2.\tag{8}$$
By using a Lagrangian equivalent to eqn.##(8)## Feynman was able to derive quantum electrodynamics using the path integral approach without the infinities due to overcounting physically identical states that occurs with the fully gauge-invariant standard electromagnetic Lagrangian.
The equations of motion of eqn.##(8)## are given by
$$\partial_\nu F^{\nu\mu}+\partial^\mu(\partial_\alpha A^\alpha)=j^\mu.\tag{9}$$
By using the identity
$$\partial_\nu F^{\nu\mu}=\Box A^\mu-\partial^\mu(\partial_\alpha A^\alpha)\tag{10}$$
we can substitute eqn.##(10)## into eqn.##(9)## in order to derive the wave equations
$$\Box A^\mu=j^\mu\tag{11}$$
without any need to explicitly fix a gauge condition as done in standard electrodynamics.
The equations of motion ##(9)## are equivalent to the following modified inhomogeneous Maxwell equations
$$
\begin{eqnarray}
\nabla \cdot \mathbf{E}&=&\frac{\rho}{\varepsilon_0}-\frac{\partial C}{\partial t},\tag{12}\\
\nabla \times \mathbf{B}&=&\mu_0 \mathbf{J} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}+\nabla C\tag{13}
\end{eqnarray}
$$
where the scalar ##C## is given by
$$C=\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}.\tag{14}$$
In order that the scalar field ##C## in eqn.##(14)## is physical we must obey a restricted gauge invariance such that ##\Box \Lambda=0## when ##\phi\rightarrow\phi-\partial\Lambda/\partial t## and ##\mathbf{A}\rightarrow\mathbf{A}+\nabla\Lambda##.
If we take the divergence of eqn.##(13)## and substitute in eqn.##(12)## we find the following wave equation for the scalar ##C## given by
$$\nabla^2 C - \frac{1}{c^2}\frac{\partial^2 C}{\partial t^2}=-\mu_0\left(\frac{\partial \rho}{\partial t}+\nabla \cdot \mathbf{J}\right).\tag{15}$$
As the righthand side is zero by the continuity relation then eqn.##(15)## reduces to a homogeneous equation.
In ##3##D vector form eqn.##(11)## gives the following wave equations for the scalar potential ##\phi## and vector potential ##\mathbf{A}##:
$$
\begin{eqnarray}
\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}&=&-\frac{\rho}{\varepsilon_0},\tag{16}\\
\nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2}&=&-\mu_0\mathbf{J}\tag{17}
\end{eqnarray}
$$
with the general retarded solutions
$$
\begin{eqnarray}
\phi(\mathbf{r},t)&=&\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}d^3r',\tag{18}\\
\mathbf{A}(\mathbf{r},t)&=&\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}d^3r'\tag{19}
\end{eqnarray}
$$
where retarded time ##t_r=t-|\mathbf{r}-\mathbf{r}'|/c##.
Now the longitudinal electric field ##\mathbf{E}_\parallel## given by
$$\mathbf{E}_\parallel=-\nabla \phi-\frac{\partial \mathbf{A}_L}{\partial t}\tag{20}$$
depends on the retarded charge density ##\rho(\mathbf{r}',t_r)## and the retarded longitudinal current density ##\mathbf{J}_L(\mathbf{r}',t_r)## so that causality is maintained.
Problem solved?