I'd like to understand the physical meaning of the Feynman's vector potential definition:
$$ A_{m}^{(b)}(x) = e_b \int \delta (xb_{\mu}xb^{\mu})db_m(b), \qquad m=0,1,2,3 $$
(component m of the vector potential of the particle b at the point x)
Here
$\delta$ is the Dirac delta function
$ xb_{\mu}xb^{\mu} $ is the 4-norm of $x - b$
So the $\delta$ function is non-zer $\iff x$ is in the boundary of the light cone of b.
Can you give me any clues?
Thanks!
On
Okay, I'm still not very sure about the fact that there's no distance function in the integrand, but if you assume it is somehow absorbed in the definition of $db_m(b)$, you can make some sort of sense out of this expression.
First you should recall that the classical electromagnetic potentials and their relations to the position and motion of a point particle. The electric potential generated by a charge density $\rho(x)$ is usually written as
$$ A_0(x) = V(x) = C_1 \int \frac{\rho(y)}{|x - y|} dy$$
and the magnetic vector potential generated by a current is usually written as (Biot-Savart law form)
$$ \vec{A}(x) = C_2 \int \frac{\vec{J}(y)}{|x-y|} dy$$
where $C_1,C_2$ are constants which depends on the unit chosen. Then in this case, in units where the speed of light $c = 1$, $(\rho,\vec{J})$ is a four-vector representing the movement of the particles. (Recall that an object with velocity $\vec{v}$ is traversing $\vec{v}$ spatial units in 1 time unit, so the four-vector corresponding to the tangent vector of its world line will be $(1,\vec{v})$.
Now, you also realize that the signal (electromagnetic potential) should propagate at the speed of light. So while the above classical notion is valid for small object velocities, they are only ultimately valid in the electrostatics/magnetostatics cases, in which case there is a preferred time direction (the one in which everything stays constant) and we don't have to worry about this propagation); in other words, in those cases the particles on the past light cone can be transported forward using the stationarity to particles on your instantaneous surface and the computations can be done there.
By direct analogy then, the proper way to compute the electromagnetic vector four potential is to set $j$ the four-current of your charge distribution (so that conservation of charge demands that the space-time divergence of $j$ vanishes, and we also write $\hat{j}$ for the unit time vector in the $j$ direction) is
$$ A_\nu(x) = k\int \frac{j_\nu(y)}{(x-y)_\mu \hat{j}^\mu} \delta((x-y)_\mu(x-y)^\mu) dy $$
where $x,y$ are space-time events. ($k$ is just some normalizing constant, usually set to $\frac{1}{4\pi\epsilon_0}$). Basically the idea is not to compute $A$'s dependence on $j$, but rather $j$'s contribution to $A$. The influence of $j$ originating from space-time event $y$ will propagate out along the light cone (the $\delta$ function term). In the inertial frame corresponding to the instantaneous velocity $\hat{j}(y)$, the propagation is uniform in directions (because instantaneous at $y$, the distribution is at rest), and decays as $1/r$ where $r$ is the spatial distance, in $\hat{j}(y)$'s rest frame. A direct computation tells you that $r = (x-y)_\mu \hat{j}^\mu$. So the integrand tells you the infinitesmial contribution by $j(y)$ at space-time event $x$ to the four-potential. Then you just add them all up. If you look carefully at the integrand, you'd notice that it is the convolution of the Green's function for the linear wave equation against $j$! So in fact from this expression you recover the differential form of the law for the four-potential
$$ \nabla_\mu\nabla^\mu A = 4\pi k j $$
which is just equation (25.22) in volume 2 of Feynman's lectures. If you just take the formula above, and restrict $j$ to be the tangent vector to the world line of a charged particle, you will recover the equation you wrote down, if you assume $dB_m(b)$ is given by $dy$ times the fraction that appears in the formula above.
The 4-vector density current is \begin{equation*} j^{\mu} := \rho \frac{\partial x^{\mu}}{\partial \tau} = \rho {\dot{x}}^{\mu} = \rho (c, \mathbf{\dot{x}}) = (\rho c, \mathbf{j}) \end{equation*} where $\tau$ is the proper time.\ We can ask ${\dot{x}}^{\mu}$ to be a 4-versor, or in other words to be of unit Minkowski norm. Usually we define the vector potential $\mathbf{A}$ in terms of $\mathbf{j}$, and the scalar potential $\phi$ in terms of the current density $\rho$ \begin{equation*} \mathbf{A}(\mathbf{x}) = C_1 \int_{space} \frac{\mathbf{j}(\mathbf{y})}{r}d\mathbf{y}, \qquad \phi(\mathbf{x}) = C_2 \int_{space} \frac{\rho(\mathbf{y})}{r}d\mathbf{y} \end{equation*} where $r$ is the spatial distance between $\mathbf{x}$ and $\mathbf{y}$, where $C_1$ and $C_2$ are costants. \ Let's put all together in the 4-vector $A$ \begin{equation*} A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} dy \end{equation*} But we want $A$ to be non-zero at the point $\mathbf{x}$ only when the light comes from $\mathbf{y}$ to $\mathbf{x}$, or when the event $x$ os on the boundary of the light cone centered in $y$. To impose this is sufficient to multiply the integrand with $\delta((x-y)^{\nu} (x-y)_{\nu})$ \begin{equation*} A_{\mu}(x) = k \int_{spacetime} \frac{j_{\mu}(y)}{r} \delta((x-y)^{\nu} (x-y)_{\nu}) dy \end{equation*} We can simplify the expression of $A$, posing \begin{equation*} \int \delta(f(\alpha,\beta))g(\alpha,\beta) d\beta := \sum_i \frac{g(\alpha,\beta^i)}{| \frac{\partial f}{\partial \beta}(\alpha, \beta^i) |} \end{equation*} where $\alpha$ is fixed, and the sum is extended to all the $\beta^i$ for which $f(\alpha,\beta^i) = 0$. Doing like this we have, speaking about a single particle $b$,
\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = e_b \int \delta((x-b)^{\nu} (x-b){\nu}) \dot{b}{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{| (x-b)^{\nu} \dot{b}_{\nu} |} \end{equation*}
If we only consider the events $x$ that $(x-b)^{\nu} (x-b)_{\nu} = 0$, we have \begin{equation*} r = \lvert \mathbf{x-b} \rvert = (x-b)^0 \end{equation*} In the rest frame of the current flux $\dot{b} = (1,0,0,0)$. So we find \begin{equation*} r = (x-b)^0 = (x-b)^{\nu}{\dot{b}}_{\nu} \end{equation*} and
\begin{equation*} \int \delta((x-b)^{\nu} (x-b){\nu})j{\mu}(\beta) d\beta = \frac{e_b}{2} \sum_i \frac{j_{\mu}(\beta^i)}{r} \end{equation*}
Finally we notice $j_{\mu} d\beta = e_b (db_{\mu} / d\beta) d\beta = e_b db_{\mu}$, and so we define the 4-vector potential as follows The 4 dimensional vector potential of particle $b$ at the point $x$, with charge $e_b$ is
\begin{equation}\label{A} A_{\mu}^{(b)}(x) := e_b \int \delta((x-b)^{\nu} (x-b){\nu}) db{\mu} \end{equation}
where we integrate over the whole world line of $b$.