# Why do we need to add $\frac{r'}{c}\frac{d}{dt}\left(\frac{\bf e_{r'}}{r'^2}\right)$ to retarded Coulomb field in retarded electric field's formula?

The retarded electric field is given by: $$\mathbf E= \frac{q}{4\pi\epsilon_0}\left[\frac{e_{r'}}{r'^2} + \frac{r'}{c}\frac{d}{dt} \left(\frac{\mathbf e_{r'}}{r'^2}\right) + \frac{1}{c^2}\frac{d^2}{dt^2} e_{r'}\right]$$

Regarding the second term, Feynman remarks:

[...] The other terms tell us that the laws of electricity do not say that all the fields are the same as the static ones, but just retarded (which is what people sometimes like to say). To the “retarded Coulomb field” we must add the other two terms. The second term says that there is a “correction” to the retarded Coulomb field which is the rate of change of the retarded Coulomb field multiplied by $r′/c$, the retardation delay. In a way of speaking, this term tends to compensate for the retardation in the first term. The first two terms correspond to computing the “retarded Coulomb field” and then extrapolating it toward the future by the amount $r′/c$, that is, right up to the time $t\;!$ The extrapolation is linear, as if we were to assume that the “retarded Coulomb field” would continue to change at the rate computed for the charge at the point (2′). If the field is changing slowly, the effect of the retardation is almost completely removed by the correction term, and the two terms together give us an electric field that is the “instantaneous Coulomb field”—that is, the Coulomb field of the charge at the point (2)—to a very good approximation.

I'm having problem in comprehending what Feynman wants to say. There is no doubt that the first term is retarded Coulomb field; but what about the second term?

Feynman tells, it is a correction, a compensation to the retardation of the first term.

My questions are:

$\bullet$ Why do we need a correction, a compensation to the retardation of the first term?

$\bullet$ How does the second term make the correction? How does it compensate the retardation?

# EDIT:

Okay, I got the point that the second term, being a derivative, when multiplied by the retardation time, gives the net change in electric field at that point during the retardation time. so, when I add it to the first term, I get the field at that point at time $t\;_,$ isn't it?

Thus, the two term adds to give the electric field at $t\;.$

But, since cause precedes effect, wouldn't the electric field at time $t$ be caused by the charge at $t-r/c\;?$ But the two term seems to give the condition at $t$, not $t-r/c\;.$ Doesn't this violate causality?

I am really missing something and messing the retarded field; but wouldn't the electric field at time $t$ be caused by the charge configuration at time $t-r/c\;?$ But the second term in the definition of the electric field adds the change in electric field in $r/c$ to the first term to give the configuration of the charge at time $t$ and not $t-r/c$ - this violates causality as the field at $t$ would be caused by the charge at $t-r/c\;.$

So, can anyone please explain where I'm mistaking?

• Doesn't your quote already answer your question? "If the field is changing slowly, the effect of the retardation is almost completely removed by the correction term" - if the field is changing slowly, it should not differ "too much" from the field of a stationary charge, and the correction term ensures that. – ACuriousMind Jan 5 '16 at 1:21

the second term, being a derivative, when multiplied by the retardation time, gives the net change in electric field at that point during the retardation time. so, when I add it to the first term, I get the field at that point at time $t\;_,$ isn't it?

No. If the retarded Coulomb changed linearly in time, then the first two terms added together would give you the Coulomb now. Instead they merely together give a linear time extrapolation. Like using $x(0)+tv(0),$ in general it isn't giving you the current value. Only when it actually changed linearly in time.

But, since cause precedes effect, wouldn't the electric field at time $t$ be caused by the charge at $t-r/c\;?$

If you did that to the potential then all would be fine.

But the two term seems to give the condition at $t$, not $t-r/c\;.$ Doesn't this violate causality?

I'm not sure anyone knows what you mean. You are evaluating the field at time $t$ and it is based on the prior times; isn't that exactly what the the expression is saying? Isn't everything in the expression evaluated at the retarded time and the result is the current field?

wouldn't the electric field at time $t$ be caused by the charge configuration at time $t-r/c\;?$

It is. And you either base in on the position, velocity and acceleration at the retarded time, or you base it on those three terms of Feynman which are based on other things, but things also evaluated at the retarded time.

this violates causality as the field at $t$ would be caused by the charge at $t-r/c\;.$

Firstly, the field is caused by the charge at the retarded time. That's what you want. And Again, you are just making a mathematical error on your part. The derivative is not a magic thing that can see what you average rate of change over some future tine interval is going to be. It's like an instantaneous velocity versus an average velocity. The instantaneous velocity is just about the now and doesn't care if tomorrow you start moving at half the speed of light. But if you talked about your average velocity over the next coming week, that would care if you start moving at half light-speed starting tomorrow. But derivatives are instantaneous changes, not average changes.

The derivatives in Feynman's equation are all instantaneous things. They don't know what happens later. In particular they don't know if someone is going to grab that charge tomorrow and stick it on a train going at half light speed. They are just about what the charge is doing at that one instant in retarded time. And that's where and when the derivative is taken.

Let's start by noting that Coulomb is wrong. When charges far away accelerate at time t-r'/c then you feel a totally different field than the Coulomb field. And sure, when something is moving at a constant velocity then in another frame they are at rest, and when at rest. Then you expect Coulomb. So since you expect Coulomb when at rest: go ahead and include a term for that. And you expect a different frame to get a Coulomb force when it moves at constant velocity, so include a term for that so that the first two terms give that force. And finally those terms still aren't enough when it accelerates so add another term to get it to actually match experiments.

Matching experiments is the key, and its what we want to do ultimately.

• +1; thanks for responding. All the terms in the right is computed at $t-r'/c$ - the retarded terms. But ultimately it boils down to the question: why do we need the correction? Why do we need the extrapolation? Coulomb's field - the first term gives the field at $r'$ at time $t-r'/c_;$ why do we then need to extrapolate by adding the second term? The field needed time to reach $r'\;_;$ the time taken is $r'/c$. So, the time at which the field reached $r'$ is $t_;$ this means the field that was present at $t-r/c$ reached $r'$ at $t.$ [...] – user36790 Jan 6 '16 at 16:09
• ... So, the Coulomb term itself gives the field at $r'$ at $t\;.$ Then why do we need to extrapolate? – user36790 Jan 6 '16 at 16:12
• @user36790 I don't understand the question. Coulomb is only accurate in statics. If you used it outside of statics you'd get totally wrong answers. Totally totally wrong. Look at that third term, it doesn't have all those 1/r factors so at large distance that can dominate when there is acceleration at time t-r'/c. Basically you have to argue why Coulomb is good when thibgs accelerate, and you won't win. – Timaeus Jan 6 '16 at 16:15
• @user36790 I added two more paragraphs. – Timaeus Jan 6 '16 at 16:20
• Could you help me in this question? I would be grateful if you help. – user36790 Jan 13 '16 at 14:54