Liénard-Wiechert potentials

The Liénard-Wiechert potentials are the retarded potentials of a point charge $q$ moving on a specific trajectory:

$\mathbf{v}$ is the velocity of the charge at the retarded time and $\boldsymbol{\mathfrak{r}}$ is the distance between the retarded position (the position at the retarded time) to the field point $\mathbf{r}$.

These potentials lead to the general electric and magnetic field of a moving point charge:

where $\mathbf{u}\equiv c \hat{\mathfrak{r}}-\mathbf{v}$. The first term of the electric field is known as the velocity field (or generalized Coulomb field) and the second as the acceleration field (or radiation field).

In case of a point charge moving with constant velocity, these reduce to

where $\mathbf{R}\equiv \mathbf{r}-\mathbf{v}t$ is the vector from the present location $\mathbf{v}t$ of the charge to $\mathbf{r}$, and $\theta$ is the angle between $\mathbf{R}$ and $\mathbf{v}$.

Derivation#

The proof begins from the general form of retarded potentials. The electric potential is

where, critically, $\rho$ is dependent on the retarded time $t_{r}$. This form is almost identical to the simultaneous case, just with $t_{r}$ added in, which begs the suggestion that the solution for a point charge $q$ might be the same

with the caveat that $\mathfrak{r}$ is now the distance from the retarded position instead of the current one. But that would be wrong, for the $t_{r}$ dependency in $\rho$ fundamentally changes what $\int \rho(\mathbf{r}',t_{r})d\tau'$ is. Whereas without $t_{r}$, $\rho(\mathbf{r}')$ refers to the charge distribution at one instant, with $t_{r}$ it is the charge distribution spread not just in space but also throughout time, as the retardation leads to a distorted charge depending on how far you are from the origin.

It pays to describe this phenomenon with a more easily imagined object: a train. Say you're looking at a train that's coming right towards you¹. Though you will not notice it in the real world because it's so tiny, you will see the train as longer than if were standing to the side, or behind it. This is a fundamentally geometric effect due to the finite speed of light. The light from the back of the train was emitted earlier than the one we receive at the same time from the front, since the train is moving towards you. If the train were still, the difference in distance covered by light from the front and back would be exactly the length of the train, but since it's moving, the light from the front has to cover less distance because it's moving towards. By the time the light from the back reaches, the train moved forward and the light you receive simultaneously from the front is after the motion, since the train covered space that the light would've had to cover itself. This means that you see the current front of the train, but the past back.

The train looks longer by an amount determined by how much space the train covers while the light from the back attempts to reach you. If the light from the back needs to travel a distance $L'$ at speed $c$ for it to reach you, the train moves by a distance $L'-L$ at speed $v$. Since these two take the same amount of time we can state

The train looks longer by a factor $(1-v/c)^{-1}\geq1$. If the train were going away, it would look shorter by a factor $(1+v/c)^{-1}\leq 1$.

If instead you are at angle $\theta$ from the train, and it is far enough or short enough that the light beams from the front and back can be considered parallel, the past light from the back must instead travel $L'\cos \theta$ (and the train still moves the same distance $L'-L$), which yields

Same as before, but we notice that only the part of velocity coming towards you (more formally, parallel to your line of sight) is significant, as shown by the cosine. The perpendicular component does not matter: if you are watching the train from the side ($\theta \simeq90°$), the distortion factor falls back to $1$ and you see no distortion. We can therefore generalize from $v\cos \theta$ to $\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{v}$, where $\hat{\boldsymbol{\mathfrak{r}}}$ is the unit vector from the train to the observer. The volume of the train $\tau$, then, will be shrunk to a new volume $\tau'$ by a factor

Back to electrodynamics, we can just change the object's volume $\tau$ to the total charge $q$ within the volume. Thus, our integral $(1)$ then solves almost in the original way, but the charge $q$ is now modified by the same factor as seen in $(2)$:

where $\mathbf{v}$ is now the velocity of the charge. It might be surprising that this correction is significant for a point charge also, since it has no volume, but the correction factor in $(2)$ actually makes no reference to the size of the train/charge/whatever, it only cares about the velocity relative to the observer. It is fair to say that anything that's moving must abide to this phenomenon².

The retarded time for a point charge is determined implicitly by the equation

where $\mathbf{r}'$ is the trajectory of the charge and $\mathbf{r}'(t_{r})$ is the location of the charge at time $t_{r}$, the retarded position so to speak. The left side is the distance the signal must travel, whereas $t-t_{r}$ on the right is the time it takes to do so. We can call $\boldsymbol{\mathfrak{r}}=\mathbf{r}-\mathbf{r}'(t_{r})$ the distance from the retarded position to the point $\mathbf{r}$. This is important because it allows us to claim that there is only ever one point $\mathbf{r}'(t_{r})$ in communication with $\mathbf{r}$ at any given time. In fact, if there were two different retarded times $t_{r_{1}}$ and $t_{r_{2}}$ (and associated positions) such that $\mathfrak{r}_{1}=c(t-t_{r_{1}})$ and $\mathfrak{r}_{2}=c(t-t_{r_{2}})$, the then we'd have $\mathfrak{r}_{1}-\mathfrak{r}_{2}=c(t_{r_{2}}-t_{t_{1}})$. But this implies that the average speed of the particle in the direction of $\mathbf{r}$ must be $c$. Even ignoring components in other directions, which could very well set the speed to be greater than $c$ — an impossibility on its own — no charged particle can ever travel at exactly $c$, so this is really impossible³. Therefore, there can only every be one retarded time/position for any point $\mathbf{r}$. This is important because it guarantees that it is unique everywhere and it is never a superposition of two retarded potentials from the same particle:

Similarly, since the current density is $\rho \mathbf{v}$ and $\mathbf{v}$ comes out of the volume integral, in practice you get the same result: the vector potential is

Fields#

Now that we have the potentials, it's possible to calculate the electric field and magnetic field of the point charge⁴. The potential expression of the fields is

Differentiation is greatly complicated by the fact that more than one variable is dependent on retarded time, $\boldsymbol{\mathfrak{r}}=\mathbf{r}-\mathbf{w}(t_{r})$ and $\mathbf{v}=\dot{\mathbf{w}}(t_{r})$, with $t_{r}\equiv t_{r}(\mathbf{r},t)$ itself being dependent on $\mathbf{r}$ and $t$, so the calculation is rather arduous. It is, however, possible.

Let's start from the Gradient of $V$:

We need the gradient at the end. Normally, without retarded time, we'd have $\nabla \mathfrak{r}=\hat{\boldsymbol{\mathfrak{r}}}$. However, with retarded time, we have $\mathfrak{r}=c(t-t_{r})$, which means $\nabla \mathfrak{r}=-c\nabla t_{r}$. The second piece of gradient is

We'll need to calculate all of these one by one, no way around it. Term one:

where $\mathbf{a}\equiv \dot{\mathbf{v}}$ is the acceleration of the point charge. Term two:

The first part is

The second part is essentially the same process as term one:

just instead of getting $\dot{\mathbf{v}}=\mathbf{a}$ we get $\dot{\mathbf{w}}=\mathbf{v}$. Combined we get:

Inside of term three:

Inside of term four:

where we used the same logic as term three, just with $\mathbf{v}=\dot{\mathbf{w}}$ instead of $\mathbf{a}=\dot{\mathbf{v}}$. Put all of this back in $(3)$ and use $\mathbf{A}\times(\mathbf{B}\times \mathbf{C})=\mathbf{B}(\mathbf{A}\cdot \mathbf{C})-\mathbf{C}(\mathbf{A}\cdot \mathbf{B})$ on the triple products to get

Put this and $\nabla \mathfrak{r}=-c\nabla t_{r}$ in the $\nabla V$ and you get

We now need $\nabla t_{r}$. To do so, we'll use a trick to reverse engineer its value in terms. Recall that $\nabla \mathfrak{r}=-c\nabla t_{r}$, but $\mathfrak{r}=\lvert \boldsymbol{\mathfrak{r}} \rvert=\sqrt{ \boldsymbol{\mathfrak{r}}\cdot \boldsymbol{\mathfrak{r}} }$. We can use this to our advantage:

Notice how $(\boldsymbol{\mathfrak{r}}\cdot \nabla)\boldsymbol{\mathfrak{r}}$ is something like term two, so it must be

Meanwhile $\nabla\times \boldsymbol{\mathfrak{r}}$ is the inside of term four, which we've already established in $(4)$, so

Thus

and so

Plug this into the gradient of $V$ and we finally get

A similar calculation leads to

For brevity, we define the vector $\mathbf{u}\equiv c \hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v}$, and combining these two yields the electric field:

We are not done yet however, as we still lack the magnetic field. The curl of $\mathbf{A}$ is

$\nabla\times \mathbf{v}$ is the inside of term three from before, $\nabla\times \mathbf{v}=-\mathbf{a}\times \nabla t_{r}$. Meanwhile, we also know the gradient $\nabla V$. Together:

Notice how similar the square brackets here are to the ones in the electric field formula. In fact, take the electric field's square brackets and expand the triple product as

which is, evidently, almost identical; the only difference the usage of $\mathbf{u}$ instead of $\mathbf{v}$. This would be a problem, but recall that $\mathbf{u}=c \hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v}$. When you take the Vector product of this and $\boldsymbol{\mathfrak{r}}$, then the term $c \hat{\boldsymbol{\mathfrak{r}}}$ disappears since the vector product of two parallel vectors is zero, so $\boldsymbol{\mathfrak{r}}\times \mathbf{u}=\boldsymbol{\mathfrak{r}}\times(-\mathbf{v})$. We can use this fact to change all the $\mathbf{v}$ vectors inside of $\nabla\times \mathbf{A}$ into $-\mathbf{u}$ vectors, since they're all going through this vector product anyway. But this makes the square brackets truly identical, and so

The magnetic field of a point charge is entirely determined by its electric field and is always perpendicular to both it and vector to the retarded position.

Let's analyze what we've gotten after all this work. The electric field is now made up of two terms. The second term vanishes is only nonzero if the particle accelerates. If it doesn't, we are left with

Furthermore, if the particle also does not accelerate, then $\mathbf{u}=c \hat{\boldsymbol{\mathfrak{r}}}$, so

but this is just the traditional field of a static point charge! Clearly, our field is here is a true generalization of the classical formula, to which it reduces in the static case. In fact, the first term, the one that appears when $\mathbf{a}=0$, is called the velocity field (or sometimes generalized Coulomb field). The second term is instead called the acceleration field (or sometimes radiation field). Analogous terminology is used for the magnetic field. Notably, the velocity field goes like $\sim \mathfrak{r}^{-2}$, but the acceleration field only goes like $\sim \mathfrak{r}^{-1}$, meaning it falls off much slower, but is only present for accelerating charges. This has important consequences, as it is this behavior that causes the acceleration field to generate electromagnetic radiation, hence its alternative name.

Due to its importance, we should calculate the velocity field in a more practical scenario. Imagine a point charge traveling at constant velocity (which implies a straight trajectory $\mathbf{w}=\mathbf{v}t$). We have $\mathbf{a}=0$ and the electric field reduces to the aforementioned

We want to get rid of $\mathbf{u}$, so

It can be found that

Defining $\mathbf{R}\equiv \mathbf{r}-\mathbf{v}t$, the vector from the present location $\mathbf{v}t$ of the charge to $\mathbf{r}$, this can be written as

where $\theta$ is the angle between $\mathbf{R}$ and $\mathbf{v}$. Thus

There's a few of things to notice here. The fields points over $\hat{\mathbf{R}}$, which is vector from the present position of the charge. This is curious because the potential here is the retarded one, not the present one. Furthermore, the presence of a $\sin ^{2}\theta$ term means that the intensity of the field depends on the orientation with respect to the charge. It is not spherically uniform any longer, being strongest at 90° angles and suppressed when on the line of motion of the particle.

The field also depends on the ratio between particle speed and light speed, with faster moving charges being penalized by a factor $(1-v^{2}/c^{2})$ on the axis of motion. At 90° angles, on the other hand, the field is boosted by a factor $1/\sqrt{ 1-v^{2}/c^{2} }$.

As for the magnetic field, we have

and so

The magnetic field inherits the orientation dependence from $\mathbf{E}$, being strongest when at a 90° angle from the particle.

When $v^{2}\ll c^{2}$, these approximately reduce to

which are the usual point charge electric field and a "point charge Biot-Savart law".