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Please correct me if I am wrong in my statements

While it is said that electromagnetic waves are formed by the oscillation (acceleration) of charges which forms 'kinks' in their electric field which is being transmitted as some distorted wave like fields at the speed of light.

This changing electric field causes a magnetic field which also propagates perpendicular to the electric field. And it is also said that these electromagnetic waves carry energy which is radiated from the oscillating charge which can be quantized by an equation.

This might be a redundant question, but I really want to know

Why don't static charges radiate energy the same as accelerating charges?

Also I can understand the potential energy due to a static charge due to its electric field: Let's take a source charge which is positive and currently at rest. If another positive charge is brought closer and closer to the source charge it's potential energy increases ( Here I can understand the energy due to the source charge, because as the distance between the two charges decreases the repulsive force among the charges increase, which also results in the increase in their potential energies)

But in the case of electromagnetic energy (or) radiation I can't understand how is the energy stored, formed or carried away by the changing electric fields of the accelerating charge. In the static charge case it was evident that the electrostatic potential energy is governed by the Coulomb's law. But I genuinely don't understand how is electromagnetic energy formed or carried away by the fields which also drains out the energy of the oscillating charge, but not in the case of static charge(The electric fields don't drain out the energy of the static charge).

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3 Answers 3

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  1. The essential difference between the electric field $\mathbf{E}(t,\mathbf{x})$ generated by an accelerated charge and the electric field $\mathbf{E}(\mathbf{x})$ generated by a charge at rest is their (leading) depence on the distance $R$ between the position of the charge and the point where the electric field is measured. In the first case, the electric field falls off as $|\mathbf{E}|\sim 1/R+\mathcal{O}(1/R^2)$. This is the behaviour of your "kink" and is a characteristic feature of electromagnetic "radiation". On the other hand, in the second case of a pure Coulomb field you have $|\mathbf{E}|\sim 1/R^2$.

  2. In both cases the density of the electromagnetic field is given by $\eta \sim\mathbf{E}^2+\mathbf{B}^2$, where the prefactor depends on your favourite unit system (Heaviside, Gauss, SI). Likewise the energy flow density is given by the Poynting vector $\mathbf{S} \sim \mathbf{E} \times \mathbf{B}$. $\eta$ and $\mathbf{S}$ fulfill the continuity equation $\dot{\eta}+\mathbf{\nabla} \cdot \mathbf{S} =-\mathbf{j} \cdot \mathbf{E}$, where $\mathbf{j}$ denotes the current density.

  3. As the name says, the electrostatic potential makes only sense in the static case, where $\mathbf{\nabla} \times \mathbf{E} = - \dot{\mathbf{B}}=0$.

  4. Details can be found in all good books on electrodynamics like Landau-Lifshitz vol.2, Jackson, etc., but also in the Feynman Lectures on Physics.

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The concept of 'energy', as you try to apply it here, comes from classical mechanics and is as such not strictly applicable to electromagnetic waves (neither for quantum mechanics). If you want to describe the cause for the emission of light in classical terms, you could probably say that it is related to a kind of friction force on the moving (accelerated) charges, but this probably doesn't get you very far in practice. However, on the basis of Maxwell's equation, the difference in the nature of the static and electromagnetic (induced) fields is at least formally fairly obvious, as you can write for instance for the electric field (using Gaussian cgs-units)

$$\vec{E} = -\nabla{\phi} -\frac{1}{c}\cdot\frac{\partial{\vec{A}}}{\partial{t}}$$

where $\phi=q/r'$ is the electrostatic potential of a charge $q$ at distance $r'$ and $\vec{A}=\vec{v'}/c\cdot \phi$ is the vector potential of that charge having velocity $\vec{v'}$ (the primes here shall indicate that the position and velocity are taken at the retarded time i.e. when the signal was emitted by the charge).

The first term is simply the static Coulomb field, and the second term is the induced field responsible for the emission of electromagnetic radiation. As you can see, the latter is only different from zero if the vector potential $\vec{A}$ changes with time (either because the distance $r'$ changes or the charge $q$ changes). If you do a detailed calculation of the time derivative of the vector potential, you find indeed that this contains a term depending on the acceleration of the charge which decreases $\propto 1/r'$. This is the radiation term.

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  • $\begingroup$ Seems like I need to study more Vector calculus to understand Maxwell's equations in a more profound way. Also could you explain more about this retarded time. And could you link the article or book which has the complete derivation by Maxwell for the Electric field which takes into account both the static field and the induced field, so I could try to understand how he deduced the equation to find the energy of an electromagnetic wave. $\endgroup$ Commented Dec 27, 2023 at 18:29
  • $\begingroup$ Also is the energy emitted by an electromagnetic wave a continual transformation of the kinetic energy of the vibrating charge, eventually leading to the loss of its kinetic energy over time, bringing it to an equilibrium (static charge) state?. If that is true, could you explain the mechanism of this kinetic energy of the oscillating charge to the electromagnetic energy which is radiated from it, because I can't think of how this kinetic energy gets transmitted as radiated energy which in turn depletes its own kinetic energy. $\endgroup$ Commented Dec 27, 2023 at 18:43
  • $\begingroup$ @JeffyJames The equation for the electric field I wrote above simply combines the static field (as per Gauss' law) and the induced field (Faraday's law). The form in terms of the scalar and vector potential is fairly easy to derive (see page 4 in www2.ph.ed.ac.uk/~mevans/em/lec10.pdf or, if you want more explanatory details, chapter 18-6 in feynmanlectures.caltech.edu/II_18.html . The retardation of the field is not important for the resultant radiation field as such. It only means that you may have to take the propagation time of light into account to get the details right $\endgroup$
    – Thomas
    Commented Dec 28, 2023 at 16:26
  • $\begingroup$ @JeffyJames As I said already, it is questionable to try to understand the interaction with electromagnetic fields by means of concepts of classical mechanics. But if one insists of assigning energy and momentum to them, then one also has to take the conservation laws for the latter into account. And as an accelerated charges radiates primarily in the forward direction, then you would have to conclude that it is slowed down as a result (analogously to the recoil effect in classical mechanics). A continuous emission of radiation would therefore result in a continuous slowdown. $\endgroup$
    – Thomas
    Commented Dec 28, 2023 at 17:06
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it is said that electromagnetic waves are formed by the oscillation (acceleration) of charges which forms 'kinks' in their electric field which is being transmitted as some distorted wave like fields at the speed of light.

There are even three ways that charges radiate:
#1 A single electric charge emits photons during acceleration.
#2 The electron also emits photons when it is deflected laterally by a magnetic field.
#3 In addition, the electron emits photons during the relaxation from an excited state to an atomic nucleus.

This changing electric field causes a magnetic field which also propagates perpendicular to the electric field.

TL;DR
From the synchronous acceleration (in an electrical potential) of surface electrons on a conductor rod, we have recognised that an EM wave is created, which moves with c and consists of an oscillating electric and an oscillating magnetic field component, and they are perpendicular to each other.

Following the first paragraph, the EM wave, formed from many accelerated electrons, consists of photons.
Following the second paragraph, these photons have an electric and a magnetic field component.

Complementary and not entirely unimportant, the electron is both an electric field source and the source of a magnetic dipole field. Both static fields are constant in their strength (see NIST).
To put it more clearly, the electron is the source of two fields and the field structure of the emitted photons is based precisely on this duplicity of the electron.
TL;DR

it is also said that these electromagnetic waves carry energy which is radiated from the oscillating charge which can be quantized by an equation.

You can look at an electromagnetic wave unquantised. For radio technology, it is enough for a smallest amount of the EM wave to hit a distant conductor to extract energy from it. In detail, the polarised Em wave simply accelerates the surface electrons on the receiver conductor in a reversal process.

However, if you remove your receiver very, very far from the source of any EM radiation, then you will ultimately only be able to receive individual photons. This is the renewed proof that EM radiation can always be treated quantised.


Why don't static charges radiate energy the same as accelerating charges? …

It would be better if you asked an extra question with a link to your first question. This corresponds more to PSE's policy.

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  • $\begingroup$ Is the energy emitted by an electromagnetic wave a continual transformation of the kinetic energy of the vibrating charge, eventually leading to the loss of its kinetic energy over time, bringing it to an equilibrium (static charge) state?. If that is true, could you explain the mechanism of this kinetic energy of the oscillating charge to the electromagnetic energy which is radiated from it, because I can't think of how this kinetic energy gets transmitted as radiated energy which in turn depletes its own kinetic energy. $\endgroup$ Commented Dec 28, 2023 at 16:08
  • $\begingroup$ Jeffy, a charge accelerates when attracted by an electrical potential. In the case of the EM wave, it is accelerated forward and backward by a wave generator. In the process, the electrons emit photons; and resist the electrical potential. The generator must therefore pump energy into the movement of the electrons, which emit them as (desired) power dissipation as a polarised stream of photons. $\endgroup$ Commented Dec 28, 2023 at 17:11

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