Is the Liénard-Wiechert electric field conservative?

Question

I know that an accelerated charge should emit an e.m. field and loose energy. Therefore, the Liénard-Wiechert (L.W.) electric field of an accelerated charge should be non-conservative.

But I checked first what happens when the charge is not accelerated, i.e. moves with a constant velocity. I expected to find a conservative field as in the case when the charge is at rest. A charge moving with constant velocity doesn't radiate. But it seems that this is not what happened.

Given the scalar potential $\phi$ and vector potential $\vec A$, the electric field is

$$ \ (1) \ \vec E = - \nabla \phi - \frac {∂ \vec A}{∂t},$$

where

$$ (2) \ \phi (r, t) = \frac {1}{4 \pi \epsilon _0} \left( \frac {q}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r},$$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \ \vec A = \frac {\mu _0 c}{4 \pi} \left( \frac {q \ \vec {\beta} _s}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r} = \frac {\vec \beta _s (t_r)}{c} \phi (r, t).$$

see the article.

I assume that for constant velocity of the charge, $t_r = t$. A field that obeys

$$ \ (4) \ \vec F(\vec r) = \nabla V(r)$$

is conservative, i.e.

$$ \ (5) \ \int_{\vec {r_1}}^{\vec {r_2}} \vec F \ d \vec {\ell} = V(\vec {r_2}) - V(\vec {r_1}).$$

So, I expected that for the constant velocity the formula (1) will turn into (4), i.e. that I would get that $\vec A$ does not depend on time. But this doesn't happen. Why? A charge in movement with constant velocity shouldn't radiate, its electric field should be conservative.

Do I make a confusion, do I make a mistake?

Answer 1

An Electric Field is only conservative if it is static. The propagation of E with a L-W field contradicts this, so it is not conservative.

Answer 2

For a constant velocity of the charge, does $t_r=t$?

No. Consider a charge moving in the positive x direction at speed $v > 0$, and assume for simplicity that at $t=0$, the particle is at $x=y=z=0$ Then at $t=0$ for a point at $x=d > 0$ the retarded time $t_r$ is at a time on the past light cone.

For the charge $x(t)=vt$, so we want a retarded time $t_r$ such that light from the charge at $x(t_r)=vt_r$ just now got to $x=d\neq 0$. It is now $t=0$, so in the time interval between $t=t_r$ and $t=0$ light travelled from $x=x(t_r)=vt_r$ to $x=d$, so it travelled a distance $d-vt_r$ in a duration $-t_r$. So $d-vt_r=-ct_r$ and we can solve for $t_r$ to get:

$$ t_r = \frac{-d}{c-v}. $$

So the place is $x=d$, the time is $t=0$, the velocity is a constant $\vec{v}=(v,0,0)$, but $ t_r = \frac{-d}{c-v} \neq 0 = t $. So $t\neq t_r$ even for constant velocity. In fact $t_r$ is always strictly less than $t$ unless the charge moves at lightspeed or above, space wraps around, or the charge happens to be right on top of you.

There is a related issue, namely ...

When can you ignore retarded time?

You can ignore retarded time when the thing you want to compute: position, velocity, acceleration, etc. is the same at the retarded time as at the current time. This can hold in static situations. You can find the charge density then or now, but if it is the same, you can just look at it now. Rightfully you should look at it then. It's like looking at an old bank statement if you know no money has gone in or out of the account. It's not the proper way to know the current balance, but it gives the right answer in that situation.