The Liénard-Wiechert Potentials and the Larmor radiation equation

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287

1 Introduction

These notes fill in some of the steps needed to calculate the fields and instantaneous power radiated by an accelerated charge. All results are fully relatvistic except when explicitly stated that the limit is for particle velocity much less than the speed of light.

2 Potentials of a point charge

The potentials in free space with the boundary conditions that the only sources of radiation are $\mbox{\boldmath$\space {J} $}$ and $\rho$ (i.e. no incoming waves from infinity) in the Lorentz gauge are

$$\mbox{\boldmath$ {A} $} \mbox{\boldmath$ {r} $} {\textstyle\frac{1}{c}} \int \delta \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r'} $} {\frac{\mbox{\boldmath$ {J} $}(\mbox{\boldmath$ {r'} $},t')}{\vert\mbox{\boldmath$ {r} $}- \mbox{\boldmath$ {r'} $}\vert}}$$ =

$$\Phi \mbox{\boldmath$ {r} $} \int \delta \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r'} $} {\frac{\rho(\mbox{\boldmath$ {r'} $},t')}{\vert\mbox{\boldmath$ {r} $}- \mbox{\boldmath$ {r'} $}\vert}}$$ = (1)

The charge and current densities of a point charge q at position $\mbox{\boldmath$\space {r} $}$₀(t) are

$$\rho \mbox{\boldmath$ {r} $} \delta^{3}_{} \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r} $}$$ =

$$\mbox{\boldmath$ {J} $} \mbox{\boldmath$ {r} $} {\frac{d\mbox{\boldmath$ {r} $}_0(t)}{dt}} \delta^{3}_{} \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r} $} \equiv \mbox{\boldmath$ {v} $} \delta^{3}_{} \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r} $}$$ = (2)

It is readily verified that these are the correct expressions. The charge density is zero everywhere but the position of the point charge and the integral of the charge density is q . Similarly, the current density is zero everywhere but the position of the point charge, and the time integral of the current over a surface element through which the charge passes is zero for times less than the time when the charge passes through the surface, and q for larger times.

Plugging in the charge and current densities, the $\mbox{\boldmath$\space {r'} $}$ integral can be done immediately to give

$$\mbox{\boldmath$ {A} $} \mbox{\boldmath$ {r} $} {\frac{q}{c}} \int \delta \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r_0(t')} $} {\frac{\mbox{\boldmath$ {v} $}(t')}{\vert\mbox{\boldmath$ {r} $}- \mbox{\boldmath$ {r} $}_0(t')\vert}}$$ =

$$\Phi \mbox{\boldmath$ {r} $} \int \delta \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r_0(t')} $} {\textstyle\frac{1}{\vert\mbox{\boldmath$ {r} $}- \mbox{\boldmath$ {r} $}_0(t')\vert}}$$ = (3)

The delta function integral is given in the usual way by changing variables to its argument in the region of the zeroes of its argument. For example if x₀ is the only zero of f (x) in the range of integration, then

$$\int \delta {\textstyle\frac{1}{\left \vert\left ( \frac{df(x)}{dx} \right )_{x=x_0} \right \vert}}$$ (4)

Applying this to the integrals above gives

$$\mbox{\boldmath$ {A} $} \mbox{\boldmath$ {r} $} {\frac{q \mbox{\boldmath$ {v} $}(t_{\rm ret})}{c \vert\mbox{\boldm... ...eft [\mbox{\boldmath$ {r} $}-\mbox{\boldmath$ {r} $}_0(t_{\rm ret}) \right ] }}$$ =

$$\Phi \mbox{\boldmath$ {r} $} {\frac{q}{\vert\mbox{\boldmath$ {r} $}-\mbox{\boldmath$ {r} $}_0(t... ...eft [\mbox{\boldmath$ {r} $}-\mbox{\boldmath$ {r} $}_0(t_{\rm ret}) \right ] }}$$ = (5)

where the retarded time is given by the solution of the equation

$$\mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r} $}$$ (6)

These are the Liénard-Wiechert potentials.

It simplifies the notation (although the first two hide some of the dependencies) to write

$$\mbox{\boldmath$ {R} $} \mbox{\boldmath$ {r} $} \mbox{\boldmath$ {r} $}$$ =

$$\left [ f \right]$$ = f (t_ret)

$$\mbox{\boldmath$ {\beta} $} {\frac{\mbox{\boldmath$ {v} $}}{c}}$$ = (7)

so that the potentials can be written

$$\mbox{\boldmath$ {A} $} \mbox{\boldmath$ {r} $} \left [ \frac{q \mbox{\boldmath$ {\beta} $} }{R(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})} \right]$$

$$\Phi \mbox{\boldmath$ {r} $} \left [ \frac{q }{R(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})} \right]$$ (8)

3 The Fields

The fields can be calculated by taking the derivatives

$$\mbox{\boldmath$ {B} $} \mbox{\boldmath$ {\nabla} $} \mbox{\boldmath$ {A} $}$$ =

$$\mbox{\boldmath$ {E} $} \mbox{\boldmath$ {\nabla} $} \Phi {\textstyle\frac{1}{c}} {\frac{\partial\mbox{\boldmath$ {A} $}}{\partial t}}$$ = (9)

so the rest of this section simply describes taking these derivatives carefully using the chain rule.

Expanding the equation for the retarded time keeping $\mbox{\boldmath$\space {r} $}$ constant gives the differential as

$$\mbox{\boldmath$ {\beta} $} \cdot \mbox{\boldmath$ {\hat {R}} $}$$ (10)

or

$${\frac{\partial t_{\rm ret}}{\partial t}} \left [ \frac{1}{1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}}} \right]$$ (11)

Expanding keeping t,y,z constant gives

$$\left [ \frac{X dx}{c R} \right] \left [ \mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}} \right]d$$ (12)

or

$${\frac{\partial t_{\rm ret}}{\partial x}} \left [ \frac{1}{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})} \frac{X}{cR} \right]$$ (13)

where X,Y,Z are the components of $\mbox{\boldmath$\space {R} $}$ .

The chain rule gives

$${\frac{\partial}{\partial t}} \left [ F \right] \left [ \frac{1}{1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}}} \right] \left [ \dot F \right]$$ =

$$\mbox{\boldmath$ {\nabla} $} \left [ F \right] \left [ \mbox{\boldmath$ {\nabla} $} F \right] \left [ \frac{{\mbox{\boldmath$ {\hat {R}} $}}}{c (1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})} \right] \left [ \dot F \right]$$ = (14)

where dot is the time derivative with respect to the particle time, and the gradient inside the brackets is the spatial derivative with the particle time held constant.

Applying this to the time derivative of the vector potential

$${\frac{\partial\mbox{\boldmath$ {A} $}}{\partial t}} \left [ \frac{\dot {\mbox{\boldmath$ {\beta} $}}}{R(1-\mbox{\boldm... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right]$$ (15)

and to the gradient of the scalar potential

$$\mbox{\boldmath$ {\nabla} $} \Phi \left [ \frac{{\mbox{\boldmath$ {\hat {R}} $}} - \mbox{\boldmath$ ... ...oldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^2} \right ) \right]$$ (16)

the electric field is

 

$$\mbox{\boldmath$ {E} $} \left [ \frac{(1-\beta^2) ({\mbox{\boldmath$ {\hat {R}} $}} - \mbo... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right] {\frac{q}{c}} \left [ \frac{\dot {\mbox{\boldmath$ {\beta} $}}}{R(1-\mbox{\boldm... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right]$$ (17)

To get this into more standard form, write

$$\mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\beta} $} \dot{\mbox{\boldmath$ {\beta} $}} \mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\beta} $} \dot{\mbox{\boldmath$ {\beta} $}} \cdot \mbox{\boldmath$ {\hat {R}} $} \dot{\mbox{\boldmath$ {\beta} $}} \mbox{\boldmath$ {\beta} $} \cdot \mbox{\boldmath$ {\hat {R}} $}$$ (18)

which shows that the numerator of the second term can be rewritten to give

$$\mbox{\boldmath$ {E} $} \left [ \frac{(1-\beta^2) ({\mbox{\boldmath$ {\hat {R}} $}} - \mbo... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right] {\frac{q}{c}} \left [\frac{ {\mbox{\boldmath$ {\hat {R}} $}} \times [({\mbox{\bo... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right]$$ (19)

which agrees with Jackson Eq. 14.14.

The curl of the vector potential is calculated the same way,

$$\mbox{\boldmath$ {\nabla} $} \mbox{\boldmath$ {A} $} \left [ \frac{{\mbox{\boldmath$ {\hat {R}} $}} \times \mbox{\boldm... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^2} \right] {\frac{q}{c}} \left [\frac{ {\mbox{\boldmath$ {\hat {R}} $}} \times [({\mbox{\bo... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right]$$ =

$$\left [ \frac{(1-\beta^2){\mbox{\boldmath$ {\hat {R}} $}} \times \... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right] {\frac{q}{c}} \left [\frac{ {\mbox{\boldmath$ {\hat {R}} $}} \times [({\mbox{\bo... ...-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^3} \right]$$ = (20)

and comparing with Eq. 17 the magnetic induction is

$$\mbox{\boldmath$ {B} $} \mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {E} $}$$ (21)

which agrees with Jackson Eq. 14.13.

4 Larmor formula

To find the radiation from the moving particle as a function of its time, look far from the particle for fields that go like r ^{- 1}. These are given by the last term in the field expressions above. The energy radiated into solid angle d$\Omega$ in time interval dt is given by the Poynting vector at large distances, or

$$\lim_{R\rightarrow \infty}^{} {\frac{c R^2}{4\pi}} \mbox{\boldmath$ {\hat {R}} $} \cdot \mbox{\boldmath$ {E} $} \mbox{\boldmath$ {B} $} \Omega$$ dE =

$${\frac{q^2}{4\pi c}} {\frac{ \vert{\mbox{\boldmath$ {\hat {R}} $}} \times [({\mbox{\bol... ...t^2}{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^6}} \Omega$$ = (22)

Generally we are not interested in the energy detected by an observer in time interval dt far from our charge. Rather we are interested in the amount of energy the particle loses to radiation in a particular direction during a change in the particles time. That is, we want dE/dt_particle not dE/dt , and since the particle is evaluated at the retarded time, the power radiated into solid angle d$\Omega$ by the particle is then

$${\frac{dP}{d\Omega}} {\frac{q^2}{4\pi c}} {\frac{ \vert{\mbox{\boldmath$ {\hat {R}} $}} \times [({\mbox{\bol... ...t^2}{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^5}}$$ = (23)

where the extra factor 1 - $\mbox{\boldmath$\space {\beta} $}$ $\cdot$ $\mbox{\boldmath$\space {\hat {R}} $}$ is dt/dt_ret calculated above. This is Jackson Eq. 14.38.

The cross product can be expanded to give

$$\mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\beta} $} \dot{\mbox{\boldmath$ {\beta} $}} \dot{\beta} \beta^{2}_{} \dot{\mbox{\boldmath$ {\beta} $}} \cdot \mbox{\boldmath$ {\beta} $} \mbox{\boldmath$ {\hat {R}} $} \cdot \mbox{\boldmath$ {\beta} $} \dot{\beta} \mbox{\boldmath$ {\beta} $} \cdot \dot{\mbox{\boldmath$ {\beta} $}} \dot{\mbox{\boldmath$ {\beta} $}} \cdot \mbox{\boldmath$ {\hat {R}} $} \dot{\mbox{\boldmath$ {\beta} $}} \cdot \mbox{\boldmath$ {\hat {R}} $} \mbox{\boldmath$ {\hat {R}} $} \cdot \mbox{\boldmath$ {\beta} $} \dot{\mbox{\boldmath$ {\beta} $}}$$ (24)

Taking the z axis along $\beta$ , the $\phi$ integrations are trivial and the integrals needed to find the total power radiated are of the form

$$\int_{-1}^{1} {\textstyle\frac{1}{(1-\beta x)^5}} {\frac{2(1+\beta^2)}{(1-\beta^2)^4}}$$ =

$$\int_{-1}^{1} {\frac{x}{(1-\beta x)^5}} {\frac{2\beta(5+\beta^2)}{3 (1-\beta^2)^4}}$$ =

$$\int_{-1}^{1} {\frac{x^2}{(1-\beta x)^5}} {\frac{2(1+5\beta^2)}{ 3(1-\beta^2)^4}}$$ = (25)

with the results

$${\frac{q^2}{4\pi c}} \int \Omega {\frac{ (\dot \beta)^2(1+\beta^2)-(\dot {\mbox{\boldmath$ {\beta} ... ...^2 }{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^5}} {\frac{q^2}{c (1-\beta^2)^4}} \left [ (\dot \beta)^2(1+\beta^2) -(\dot {\mbox{\boldmath$ {\beta} $} }\cdot \mbox{\boldmath$ {\beta} $})^2 \right]( \beta^{2}_{}$$ =

$${\frac{q^2}{4\pi c}} \int \Omega {\frac{ -{\mbox{\boldmath$ {\hat {R}} $}} \cdot\mbox{\boldmath$ {\... ...}} }{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^5}} {\frac{q^2}{c (1-\beta^2)^4}} \left [ -\frac{2}{3}(\dot \beta)^2 \beta^2 (5+\beta^2) +\frac{2}{3... ...ath$ {\beta} $} \cdot \dot {\mbox{\boldmath$ {\beta} $}})^2 (5+\beta^2) \right]$$ =

$${\frac{q^2}{4\pi c}} \int \Omega {\frac{ -(\dot {\mbox{\boldmath$ {\beta} $}} \cdot {\mbox{\boldmat... ...^2 }{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^5}} {\frac{q^2}{c (1-\beta^2)^4}} \left [ -2 (\mbox{\boldmath$ {\beta} $} \cdot \dot {\mbox{\boldmath$ {\beta} $}})^2 - \frac{1}{3}\dot \beta^2(1-\beta^2) \right]$$ =

$${\frac{q^2}{4\pi c}} \int \Omega {\frac{ -({\mbox{\boldmath$ {\hat {R}} $}} \cdot [\mbox{\boldmath$... ...^2 }{(1-\mbox{\boldmath$ {\beta} $} \cdot {\mbox{\boldmath$ {\hat {R}} $}})^5}} {\frac{q^2}{c (1-\beta^2)^4}} \left [ -\frac{1}{3}(1-\beta^2) [\beta^2 (\dot \beta)^2-(\mbox{\boldmath$ {\beta} $} \cdot \dot {\mbox{\boldmath$ {\beta} $}})^2] \right]\,$$ = (26)

Adding the results together gives

$${\frac{2 q^2}{3 c (1-\beta^2)^4}} \left [ \dot \beta^2 (1-\beta^2)^2 +(\mbox{\boldmath$ {\beta} $} \cdot \dot {\mbox{\boldmath$ {\beta} $}})^2 (1-\beta^2) \right]$$ P =

$${\frac{2 q^2}{3c (1-\beta^2)^3}} \left [ \dot \beta^2 (1-\beta^2) +(\mbox{\boldmath$ {\beta} $} \cdot \dot {\mbox{\boldmath$ {\beta} $}})^2 \right]$$ =

$${\frac{2 q^2}{3c (1-\beta^2)^3}} \left [ \dot \beta^2 -\vert\mbox{\boldmath$ {\beta} $} \times \dot {\mbox{\boldmath$ {\beta} $}}\vert^2 \right]$$ = (27)

and the last expression agrees with Jackson Eq. 14.26.

5 Nonrelativistic Limit

Taking the nonrelativistic limit where $\beta$ $\ll$ 1 , the angular distribution and the total power radiated are the Larmor results

$${\frac{dP}{d\Omega}} {\frac{q^2}{4\pi c^3}} \mbox{\boldmath$ {\hat {R}} $} \dot{\mbox{\boldmath$ {v} $}}$$ =

$${\frac{2q^2}{3c^3}} \dot{\mbox{\boldmath$ {v} $}}$$ P = (28)

It is considerably easier to derive these results from the potentials by taking their nonrelativistic limit first and dropping the velocity dependent terms in the derivatives. Taking the curl will give the magnetic induction. Since derivatives of R will give terms that go like 1/r ² etc., they will not contribute to the radiation field. Derivatives of R with respect to t_ret will give terms higher order in the velocity, and these can be dropped in the nonrelativistic limit. The only term that survives is the derivative of the velocity in the numerator of $\mbox{\boldmath$\space {A} $}$ with respect to t_ret and the gradient of ct_ret gives - $\mbox{\boldmath$\space {\hat {R}} $}$. The radiation field for nonrelativistic motion is therefore

$$\mbox{\boldmath$ {B} $} \mbox{\boldmath$ {r} $} \rightarrow \infty \rightarrow {\frac{q ({\mbox{\boldmath$ {\hat {R}} $}} \times \dot {\mbox{\boldmath$ {v} $}})}{c^2 R}}$$ (29)

With dt_ret/dt = 1 nonrelativistically, the radiated power per solid angle is

$${\frac{dP}{d\Omega}} {\frac{c R^2}{4\pi}} \mbox{\boldmath$ {B} $} \mbox{\boldmath$ {r} $} \rightarrow \infty {\frac{q^2}{4\pi c^3}} \mbox{\boldmath$ {\hat {R}} $} \dot{\mbox{\boldmath$ {v} $}}$$ (30)

and integrating over solid angle gives the usual Larmor formula.