According to my textbook, the correct formula for the electric field produced by a moving charge is $$\vec E=-\frac{q}{4\pi\epsilon_0}\left[\frac{\hat e_{r'}}{r^{'2}}+\frac{r'}{c}\frac{d}{dt}\left(\frac{\hat e_{r'}}{r^{'2}}\right)+\dfrac{1}{c^2}\dfrac{d^2\hat e_{r'}}{dt^2}\right]$$ where $r'$ is the retarded position and $\hat e_{r'}$ is the unit retarded position vector.

Now since the first two terms fall as $\dfrac{1}{r^{'2}}$ and the third term $\dfrac{1}{c^2}\dfrac{d^2\hat e_{r'}}{dt^2}$ by $\dfrac{1}{r'}$, only the third term is significant at large distances from the charge, and I now quote:

This term permits a charged particle to influence another charged particle at a great distance through the 1/r electric field. This is referred to as electromagnetic radiation and light is a familiar example of this phenomenon. It is obvious from the formula that only accelerating charges produce radiation.

Now, my questions:

  1. Why is the "at large distances" constraint needed? If the distance is small then the first two terms will also influence the radiation right?
  2. If the charge is in uniform motion, then the part of the third term $\dfrac{1}{c^2}\dfrac{d^2\hat e_{r'}}{dt^2}$ which falls as $\dfrac{1}{r'}$ is 0 but the first two terms are not. So, at large distances, even though very small, can non-accelerated charges produce radiation?

EDIT: Here's the link to the textbook: Physics I: Oscillations and Waves notes by S. Bharadwaj and S.P. Khastgir, https://drive.google.com/file/d/1CTGX7fXzeb_rHn5oqDto-3REWJKoR83S/view?usp=sharing. I am referencing pg 61.

  • 1
    $\begingroup$ You refer twice to the "third term", so it is not quite clear which term you are talking about. Besides there's no term in you equation that behaves as $1/r$. Also, please cite the textbook that you use. $\endgroup$
    – Roger V.
    Commented Feb 5, 2023 at 12:40
  • $\begingroup$ @RogerVadim I calculated $\dfrac{d^2\hat e_{r'}}{dt^2}$ as equal to $$\dfrac{1}{r'}(\vec{\ddot r'}-\ddot r'\hat e_{r'})-\dfrac{1}{r^{'2}}(\vec{\dot r'}-\dot r\hat e_{r'}')2\dot r'$$ so I think it does depend on $\dfrac{1}{r'}$. I have made other necessary corrections. $\endgroup$ Commented Feb 5, 2023 at 12:45
  • $\begingroup$ Some comments removed. If you want to answer a question, please write an answer instead. Comments are for improving and clarifying the post being commented on. $\endgroup$
    – ACuriousMind
    Commented Feb 5, 2023 at 16:07
  • $\begingroup$ In a general séance in order to emit radiation, the energy should be given to a radiation, when charge does not decelerate, no energy is lost to the radiation. $\endgroup$ Commented Feb 6, 2023 at 9:32
  • $\begingroup$ Hi insipidintegrator. Welcome to Phys.SE. Linking to private clouds, dropbox, etc, is for various reasons not acceptable on SE, cf. this meta post. $\endgroup$
    – Qmechanic
    Commented Feb 6, 2023 at 9:49

3 Answers 3


The answer to your first question is that the nature of the $1/r$ term is different and it is the only one that should be treated as radiation.

The relevance of the $1/r$ decay is that we live in three dimensions. In three dimensions, a field that falls as $1/r^2$ will not transmit energy. Think about a point charge in space producing some EM field. The flux of energy of an EM field is described by its Poynting vector, which is generally proportional to $|\mathbf{E}|^2$. If the field falls off as $1/r^2$, the amplitude of the Poynting vector will fall off as $1/r^4$. Integrating this over a sphere containing the charge, with surface area proportional to $r^2$ (this is where the fact that our space is three-dimensional comes in), we see that the amount of energy crossing the sphere is proportional to $1/r^2$ and thus falls off quickly as the distance from the charge increases. For a far-away sphere, the amount of energy is zero and there is no energy being radiated away into the infinite. Thus, even though the $1/r^2$ terms are relevant for short-distance dynamics, they are fundamentally irrelevant for long distances, because they do not radiate energy away.

If however, the electric field falls off as $1/r$, then the Poynting vector is proportional to $1/r^2$ and the energy crossing a sphere at a distance $r$ is independent of $r$. There is energy leaving the charge and going into infinity. This is the only way that a charge can produce radiation since radiation carries energy away, and thus a radiating charge must have some acceleration. The energy comes from whatever mechanism is accelerating the charge. This also answers your second question: only the $1/r$ term can produce radiation, so the charge must be accelerated to radiate.

Another way of answering your second question is to think about relativity. Suppose the charge is moving with constant velocity, and consider an inertial frame in which it is static. Then the charge produces no magnetic field and thus cannot emit any energy. So there is no radiation being emitted from it in this reference frame. But this implies that there is no energy being radiated away in any reference frame since the question of whether there exists radiation is a physical one and not frame dependent.

  • $\begingroup$ Poynting vector is in direction of propogation and field's distance is perpendicular to it, both are different. $\endgroup$ Commented Feb 5, 2023 at 13:13

In your formula, the only term that remains important at any arbitrarily large distance from the source is the third term, because the other terms fall off more quickly with distance and at large distances they are negligible. It turns out this acceleration-dependent term describes emission of EM waves that also describe light and all kinds of EM radiation. That's why this part of the field is called the radiation part.

The other two terms can produce field that varies in time, e.g. the field of a uniformly moving charged particle passing by. But this time-dependent electric field is Coulomb-like (falls off quickly with distance) and is not called radiation.

Not every time-dependent EM field is called radiation. For another example, consider a circuit with a resistor and a capacitor that is being charged (and the circuit has very low self-inductance): electric field everywhere changes in time, especially inside the capacitor. But in and near the capacitor, most of this time-dependent field is due to the first two terms and not due to the third term.

The third term is proportional to acceleration of the charge and it becomes important in AC circuits if we have an inductor in the circuit. When current changes, there are accelerating charges and in inductor they produce significant induced electric field. This also produces some radiation that carries energy away from the device. Often that is undesirable so the AC circuit boards are designed to make the emissions as low as possible (quality ground plane, optimized geometry of traces on the board). In other cases, the emission of radiation is the goal, and design is adapted to make it reliable and effective (emitting antenna).


Yes, from given expression for electric field by a moving charge it's possible, let's examine it in detail. The first term in expression is electric field, second term is magnetic field and perpendicular to electric field, and third term is loss of electric field in direction of motion and called as radiation of electric field. On solving given equation we have, $$\vec E=E_0\left(\left(1-\dfrac{v^2}{c^2}\right)\hat{e_r}'+\dfrac{v}{c}\hat{\phi}\right)$$ where,$\quad \text{I}^{st}\ \text{term}=E_0\hat{e_r}'=-\dfrac{q}{4\pi\epsilon r'^2}\hat{e_r}'\ $ and,$\quad \quad \quad \text{II}^{nd}\ \text{term}=E_0\dfrac{v}{c}\hat{\phi}$

Now suppose if charge is moving into the page then magnetic field is anti-clockwise and direction of third term is oot of the page. It is clear that square of charge speed is acceleration of charge and that depends upon radius of loop. At short distance, it's contribution is effective, at large distance contribution from second term is more as it's implication of Biot-Savart's law. And for very far distance, only first term contribute.

Now explanation for yes. If speed of charge equals to $c$, then first and third term cancel each other and only second term remains. So that contributes for electric field at distance but not along the durection of motion.

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    $\begingroup$ This is wrong. Charges in uniform motion do not radiate. The field pattern of a non-accelerating charge simply “moves” along with the charge. The lack of radiation is obvious when one analyzes the situation in the rest frame of the charge. $\endgroup$
    – Ghoster
    Commented Feb 6, 2023 at 23:09
  • $\begingroup$ @Ghoster At speed equal to $c$, I just discussed about possibility of electric field at distance. Why don't you ever question about validity of Gauss law or how static charge produce electric field, there is no divergence without flow. $\endgroup$ Commented Feb 7, 2023 at 2:58
  • 3
    $\begingroup$ We know of no massless charged particles, and charged particles with mass cannot move at speed $c$. I don’t question Gauss’ Law because, unlike you, I accept mainstream physics. This site is exclusively for mainstream physics, as you can read in this link. $\endgroup$
    – Ghoster
    Commented Feb 7, 2023 at 6:01

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