Jefimenko's Equations [Electromagnetism I Studied #37]

Using the Lorenz gauge from the previous post, we’ll make $V$ and $A$ more general by taking ’time’ into account too. Both potentials then show the time delay explicitly, hehehe. The Coulomb gauge gives the same causal $E$ and $B$; its instantaneous-looking scalar potential is balanced by the vector potential.

The Lorenz-gauge wave equations for V and A, with the sources rho and J

is what we’ve got~~ For now, let’s say $\rho$ and $J$ are constant~~~ in time (easy does it, step by step):

With rho and J constant in time: the Laplacian of V is minus rho over epsilon0, and the Laplacian of A is minus mu0 J

We already solved this back in Electromagnetism 1, right?

The static solutions: V(r) and A(r) as integrals of rho(r′) and J(r′) over the distance eta

Taking time into account

But hey, we said we were going to take time into account, right? The $\rho$ and $J$ at time $t$ won’t determine $A$ and $V$ simultaneously and instantaneously. $E$ and $B$ come at the speed of light, so $V$ and $A$ come at the speed of light too, and they’re set by whatever has had time to arrive.

So $V$ and $A$ at time $t$ were made by the $\rho$ and $J$ of some teeeny bit earlier time $t_{ret}$ , yup.

Okay, now let me modify the equations.

The potentials with the sources taken at the retarded time: rho(r′, t_ret) and J(r′, t_ret) inside the integrals

The potential $V$ at position $r$ at time $t$ /// comes from the $\rho$ at position $r'$ at the retarded time $t_{\text{ret}}$.

Alriiight~ let me express that retarded time $t_{ret}$ in terms of $t$. $V$ (and $A$) at some time $t$ comes from the $\rho$ (and $J$) at a time $t'$ before $t$, and that time $t'$ is the time it takes to travel the distance $\eta$ at the speed of light $c$ (since electromagnetic fields come at the speed of light).

is what we can say. So that equation from before integrates $\rho$ and $J$ at the retarded time, and that’s why they make a point of calling it the retarded potential, apparently~

The retarded potentials V_ret(r,t) and A_ret(r,t): integrals of rho and J taken at the retarded time

An example: a very long straight wire

Example 10.2 in the author’s handwriting: a very long straight wire carries a current I(t) that is 0 up to t = 0 and a steady I0 after that; find the fields this produces

Sketch: the wire along z carrying the current I, a point at distance s, and the distance eta to a bit of wire at height z

Let’s solve a problem before moving on. It’s a teeny~~ bit of an unfamiliar concept for me, so… the concpt… er, I mean the concept is kinda bewildering.

“Find the electric and magnetic fields at point P, y’all~” it says, but don’t you need some charge before you can find anything -_-!!!!

Yup, the wire is neutral, so $V = 0$. Let’s focus on the vector potential. First let me write down the formula we derived for the retarded vector potential~

Rewriting it nicely for the problem~~ (integrating the current density $J$ over the wire’s cross-section leaves the line current $I(t)$)

The same formula for the wire: A_ret(s,t) as an integral over z, from minus to plus infinity, of J(t_ret) over eta

(We said it’s an infinitely long wire, so the integration range is just like that!~)

First, sta~~~rt!!! For the first $s/c$ there’ll be no change in the potential, and once more than $s/c$ seconds have passed, $A$ will gradu-gradu-gradually start to grow. This is how we can think about it~

Three sketches: as time goes on, a longer and longer stretch of the wire (red) is close enough for its influence to have reached the point

As time flows on~~…. the range that has an effect… hehe

Eta squared equals z squared plus s squared; at the edge of the stretch eta equals ct, so c squared t squared equals z squared plus s squared

Then we can do the integral like this

A(s,t) along z-hat as an integral of dz over the root of z squared plus s squared, between minus and plus the root of (ct) squared minus s squared

Doing the integral gives a logarithm: A along z-hat, proportional to ln of (ct plus the root of (ct) squared minus s squared) over s

This is the retarded vector potential, and using the relation $E = -\nabla V - \frac{\partial A}{\partial t}$ to find $E$ gives

The electric field’s z-component: minus mu0 I0 c over 2 pi times the root of (ct) squared minus s squared

Using the relation $B = \nabla \times A$ to find $B$ gives

The magnetic field: mu0 I0 over 2 pi s, times ct over the root of (ct) squared minus s squared, along phi-hat

That was just one example. Right~ up until then we were looking at potentials that take causality into account.

The retarded potentials V_ret(r,t) and A_ret(r,t) again

The electric field

Alright then, let’s look at the electric field!!!! As we saw in the example abov-e, $E = -\nabla V - \frac{\partial A}{\partial t}$ : we’ll go with this relation. First let me work out grad $V$.

The gradient of V: it goes inside the integral, and the product rule splits it into grad rho(r′, t_ret) times 1/eta plus rho(r′, t_ret) times grad(1/eta)

Now the problem is $\nabla \rho(r', t_{ret})$ , this guy.

Why does this turn out like this….?

I’ll use the chain rule: simple, but hard to think up.

The chain rule in the author’s colours: the derivative of rho goes through t_ret and eta and comes out as rho-dot times minus 1/c times grad eta

Haa… how do people even think of stuff like this? lolol Amazing, really lol Alright, let’s move on. Oh, and let me also jot down something from Chapter 1 here.

I’ll apply this too.

The gradient of V with those results plugged in

Waitt a sec!!! If you take del once more here, it comes back to the result with the d’Alembertian, but I’ll skip that!

And now it’s $\frac{\partial A}{\partial t}$ , this guy’s turn:

Its value: mu0 over 4 pi times the integral of J-dot(r′, t_ret) over eta

At last we can write down $E(r,t)$.

E = −grad V − ∂A/∂t with both pieces put in

The pieces combined into one integral for E

This is the Reallll ultra-rigorous expression for $E$, taking time into account and the change of $\rho$ too, and riding the momentum, let’s compute $B$ too.

The magnetic field

To compute this, if we flip to page 23 for a sec,

A product rule from page 23: the curl of A/g equals g times the curl of A plus A cross grad g, all over g squared

there’s this. Slamming this formula in,

The curl of the retarded A worked out with that rule, taking g = eta

What we’ve got to examine this time is the curl of $J$: we’ll look at just the $x$-component and then generalize. Like this.

Both partial derivatives by the chain rule through t_ret, in the author’s colours

So the x-component of the curl of J comes out in terms of J-dot and grad eta, that is, eta-hat

The equation right above — if you do it yourself, it amazingly turns out like that. Try verifying it at least once!! hehe

Putting it all together, Jefimenko’s equation for B: mu0 over 4 pi times the integral of (J over eta squared plus J-dot over c eta) cross eta-hat

Aaaah~~ dang, that’s complicated, y’know.

This, they say, is what Jefimenko published in 1966 (oh, that’s not so long ago): Jefimenko’s equations~~

I seriously want to switch majors T_T T_T T_T

lolololololol

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the long-wire example, the current $I_0$ belongs in front of every integral, and inside the roots $ct^2$ means $(ct)^2$: $\vec{A}(s,t) = \frac{\mu_0 I_0}{2\pi}\ln\frac{ct + \sqrt{(ct)^2 - s^2}}{s}\,\hat{z}$, so $\vec{E} = -\frac{\mu_0 I_0 c}{2\pi\sqrt{(ct)^2 - s^2}}\,\hat{z}$, a vector along $-\hat{z}$.
The gradient is $\nabla(1/\eta) = -\hat{\eta}/\eta^2$ (with the hat), and the $\dot{\rho}$ term needs one more $1/\eta$ in $\nabla V$ and in $\vec{E}$. With the sources at $\vec{r}'$ and $\hat{\eta}$ on the first term, Jefimenko’s equation for E is $\vec{E} = \frac{1}{4\pi\varepsilon_0}\int\left[\frac{\rho(\vec{r}',t_{ret})}{\eta^2}\hat{\eta} + \frac{\dot{\rho}(\vec{r}',t_{ret})}{c\eta}\hat{\eta} - \frac{\dot{\vec{J}}(\vec{r}',t_{ret})}{c^2\eta}\right]d\tau'$.
In the curl of $\vec{A}$, the second term is $-\vec{J}\times\nabla(1/\eta) = +\vec{J}\times\hat{\eta}/\eta^2$; the second chain-rule row is $\partial J_y/\partial z$; and $\nabla\times\vec{J} = +\frac{1}{c}\dot{\vec{J}}\times\hat{\eta}$ (plus sign). With these, the final $\vec{B}$, Jefimenko’s equation for B, is right.

Taking time into account

An example: a very long straight wire

The electric field

The magnetic field

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