Accelerated charges

Let us calculate the electric and magnetic fields observed at position $x^i$ and time $t$ due to a charge $e$ whose retarded position and time are ${x^i}'$ and $t'$, respectively. From now on $(x^i, t)$ is termed the field point and $({x^i}', t')$ is termed the source point. It is assumed that we are given the retarded position of the charge as a function of its retarded time: i.e., ${x^i}'(t')$. The retarded velocity and acceleration of the charge are

$$u^i = \frac{d{x^i}'}{dt'},$$ (1594)

and

$$\dot{u}^i = \frac{d{u^i}}{dt'},$$ (1595)

respectively. The radius vector ${\bf r}$ is defined to extend from the retarded position of the charge to the field point, so that $r^i = x^i - {x^i}'$. (Note that this is the opposite convention to that adopted in Sects. 10.18 and 10.19). It follows that

$$\frac{d{\bf r}}{dt'}= -{\bf u}.$$ (1596)

The field and the source point variables are connected by the retardation condition

$$r(x^i, {x^i}') = \left[ (x^i-{x^i}') (x_i-{x_i}')\right]^{1/2} = c (t-t').$$ (1597)

The potentials generated by the charge are given by the Liénard-Wiechert formulae

$${\bf A}(x^i, t) = \frac{\mu_0 e}{4\pi} \frac{\bf u}{s},$$ (1598)

$$\phi(x^i, t) = \frac{e}{4\pi\epsilon_0} \frac{1}{s},$$ (1599)

where $s = r - {\bf r}\!\cdot\!{\bf u}/c$ is a function both of the field point and the source point variables. Recall that the Liénard-Wiechert potentials are valid for accelerating, as well as uniformly moving, charges.

The fields ${\bf E}$ and ${\bf B}$ are derived from the potentials in the usual manner:

$${\bf E} = - \nabla\phi - \frac{\partial {\bf A}}{\partial t},$$ (1600)

$${\bf B} = \nabla\times{\bf A}.$$ (1601)

However, the components of the gradient operator $\nabla$ are partial derivatives at constant time, $t$, and not at constant time, $t'$. Partial differentiation with respect to the $x^i$ compares the potentials at neighbouring points at the same time, but these potential signals originate from the charge at different retarded times. Similarly, the partial derivative with respect to $t$ implies constant $x^i$, and, hence, refers to the comparison of the potentials at a given field point over an interval of time during which the retarded coordinates of the source have changed. Since we only know the time variation of the particle's retarded position with respect to $t'$ we must transform $\partial/\partial t\vert _{x^i}$ and $\partial/\partial x^i\vert _t$ to expressions involving $\partial/\partial t'\vert _{x^i}$ and $\partial/ \partial x^i\vert _{t'}$.

Now, since ${x^i}'$ is assumed to be given as a function of $t'$, we have

$$r(x^i, {x^i}'(t') ) \equiv r(x^i, t') = c (t-t'),$$ (1602)

which is a functional relationship between $x^i$, $t$, and $t'$. Note that

$$\left(\frac{\partial r}{\partial t'}\right)_{x^i} = - \frac{{\bf r}\!\cdot \!{\bf u}}{r}.$$ (1603)

It follows that

$$\frac{\partial r}{\partial t} = c\left(1-\frac{\partial t'}{... ...{{\bf r}\!\cdot\!{\bf u}}{r} \frac{\partial t'}{\partial t},$$ (1604)

where all differentiation is at constant $x^i$. Thus,

$$\frac{\partial t'}{\partial t} = \frac{1}{1- {\bf r}\!\cdot\! {\bf u}/r c} = \frac{r}{s},$$ (1605)

giving

$$\frac{\partial}{\partial t} = \frac{r}{s} \frac{\partial}{\partial t'}.$$ (1606)

Similarly,

$$\nabla r = - c \nabla t' = \nabla'r + \frac{\partial r}{ \... ...c{{\bf r}}{r} - \frac{{\bf r}\!\cdot\!{\bf u}}{r} \nabla t',$$ (1607)

where $\nabla'$ denotes differentiation with respect to $x^i$ at constant $t'$. It follows that

$$\nabla t' = - \frac{{\bf r}}{s c},$$ (1608)

so that

$$\nabla = \nabla' - \frac{{\bf r}}{s c} \frac{\partial}{\partial t'}.$$ (1609)

Equation (1600) yields

$$\frac{4\pi \epsilon_0}{e} {\bf E} = \frac{\nabla s}{s^2} - \frac{\partial} {\partial t} \frac{\bf u}{s c^2},$$ (1610)

or

$$\frac{4\pi \epsilon_0}{e} {\bf E} = \frac{\nabla' s}{s^2... ...+ \frac{r {\bf u}}{s^3 c^2} \frac{\partial s}{\partial t'}.$$ (1611)

However,

$$\nabla' s = \frac{{\bf r}}{r} - \frac{{\bf u}}{c},$$ (1612)

and

$$\frac{\partial s}{\partial t'} = \frac{\partial r}{\partial ... ...}{r} - \frac{ {\bf r}\!\cdot\!\dot{\bf u}}{c} + \frac{u^2}{c}.$$ (1613)

Thus,

$$\frac{4\pi \epsilon_0}{e} {\bf E} = \frac{1}{s^2 r} \l... ...t\!\dot{\bf u}}{c}\right) - \frac{r}{s^2 c^2} \dot{\bf u},$$ (1614)

which reduces to

$$\frac{4\pi \epsilon_0}{e} {\bf E} = \frac{1}{s^3} \left(... ...- \frac{r {\bf u}}{c}\right)\times\dot{\bf u}\right] \right).$$ (1615)

Similarly,

$$\frac{4\pi}{\mu_0 e} {\bf B} = \nabla\times\frac{\bf u}{s... ... - \frac{{\bf u}}{s^2} \frac{\partial s}{\partial t'} \right),$$ (1616)

or

$$\frac{4\pi}{\mu_0 e} {\bf B} = -\frac{{\bf r}\times{\bf u... ...{\bf r}\!\cdot\!\dot{\bf u}}{c} - \frac{u^2}{c}\right)\right],$$ (1617)

which reduces to

$$\frac{4\pi}{\mu_0 e} {\bf B} = \frac{{\bf u}\times{\bf r}... ... \frac{r {\bf u}}{c}\right)\times\dot{\bf u}\right] \right).$$ (1618)

A comparison of Eqs. (1615) and (1618) yields

$${\bf B} = \frac{{\bf r}\times{\bf E}}{r c}.$$ (1619)

Thus, the magnetic field is always perpendicular to ${\bf E}$ and the retarded radius vector ${\bf r}$. Note that all terms appearing in the above formulae are retarded.

The electric field is composed of two separate parts. The first term in Eq. (1615) varies as $1/r^2$ for large distances from the charge. We can think of ${\bf r}_u = {\bf r} - r {\bf u}/c$ as the virtual present radius vector: i.e., the radius vector directed from the position the charge would occupy at time $t$ if it had continued with uniform velocity from its retarded position to the field point. In terms of ${\bf r}_u$, the $1/r^2$ field is simply

$${\bf E}_{\rm induction} = \frac{e}{4\pi \epsilon_0} \frac{1-u^2/c^2}{s^3} {\bf r}_u.$$ (1620)

We can rewrite the expression (1538) for the electric field generated by a uniformly moving charge in the form

$${\bf E} = \frac{e}{4\pi \epsilon_0} \frac{1-u^2/c^2}{r_0^{ 3}(1-u^2/c^2 + u_r^{ 2}/c^2)^{3/2}} {\bf r}_0,$$ (1621)

where ${\bf r}_0$ is the radius vector directed from the present position of the charge at time $t$ to the field point, and $u_r={\bf u}\!\cdot\!{\bf r}_0/r_0$. For the case of uniform motion, the relationship between the retarded radius vector ${\bf r}$ and the actual radius vector ${\bf r}_0$ is simply

$${\bf r}_0 = {\bf r} - \frac{r}{c} {\bf u}.$$ (1622)

It is straightforward to demonstrate that

$$s = r_0 \sqrt{1-u^2/c^2 + u_r^{ 2}/c^2}$$ (1623)

in this case. Thus, the electric field generated by a uniformly moving charge can be written

$${\bf E} = \frac{e}{4\pi \epsilon_0} \frac{1-u^2/c^2}{s^3} {\bf r}_0.$$ (1624)

Since ${\bf r}_u = {\bf r}_0$ for the case of a uniformly moving charge, it is clear that Eq. (1620) is equivalent to the electric field generated by a uniformly moving charge located at the position the charge would occupy if it had continued with uniform velocity from its retarded position.

The second term in Eq. (1615),

$${\bf E}_{\rm radiation} = \frac{e}{4\pi \epsilon_0 c^2} \frac{{\bf r}\times ({\bf r}_u\times\dot{\bf u})}{s^3},$$ (1625)

is of order $1/r$, and, therefore, represents a radiation field. Similar considerations hold for the two terms of Eq. (1618).