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

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


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

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

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?

  • $\begingroup$ your (1) defines a conservative field (or rather, is one of the defining properties, like the integral one you wrote after). If you meant to ask if the energy is conserved in electromagnetic interactions in the general case (like in the Lienard-Wiechert potential) the answer is clearly yes, but this is not equivalent to the electromagnetic field being conservative $\endgroup$
    – glS
    Jan 30 '15 at 18:16

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.

  • $\begingroup$ Why? What is the problem with a time-varying field? $\endgroup$
    – Sofia
    Jan 30 '15 at 16:03
  • $\begingroup$ A time varying electric field gives rise to a magnetic field, and the resulting magnetic flux means a non-conservative field (see Faraday's law). $\endgroup$
    – Paul
    Jan 30 '15 at 16:38
  • $\begingroup$ my problem is that the vector $\vec A$ looks like a gradient. And the rotor from a gradient is zero. Then, by Stokes' theorem, if $\int_{\Sigma} \nabla \ \text x \ \vec A \ d \Sigma = 0$ where $\Sigma$ is a surface, the integral of $A$ along any closed line on the surface $\Sigma$ is zero, s.t. (1) is satisfied. $\endgroup$
    – Sofia
    Jan 30 '15 at 17:27
  • $\begingroup$ What you say is correct; however, the definition of the electric field as a gradient only works in a static field $\endgroup$
    – Paul
    Jan 30 '15 at 18:13
  • 1
    $\begingroup$ No, it's just that any changing magnetic field caused an electric field to be non-conservative. $\endgroup$
    – Paul
    Jan 30 '15 at 19:11

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.

  • $\begingroup$ Thank you for considering my question. But it's only tomorrow that I'll be able to look. I had a tour-de-force in explaining something to someone from my distant homeland. I consumed most of my sleep hours. It's not everyday that such things happen. So, with God's help, let's talk tomorrow. $\endgroup$
    – Sofia
    Feb 2 '15 at 2:13
  • $\begingroup$ Well, my mission of explanation ended with success. What a nostalgic event! How the world is small! To meet someone from my homeland in the Phys.SE. Well, I don't know what is the hour in your country, but I rush to sleep. Good night ! $\endgroup$
    – Sofia
    Feb 2 '15 at 2:18

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