# Is Griffiths' formula for electric dipole radiation correct?

• Silviu
In summary, Griffiths considers two charged balls connected by a wire with charge going back and forth between them. He calculates the vector potential using this equation: $$A(r,t)=\frac{\mu_0}{4 \pi}\int_{-d/2}^{d/2}\frac{-q_0\omega sin[\omega(t-r'/c)]\hat{z}}{r'}dz$$ However, if he follows it properly through the book, this equation is derived from Biot-Savart law (and a proper choice of gauge). However, when introducing Biot-Savart, Griffiths emphasizes that the current density (or linear current in this case
Silviu
Hello! I am reading Griffiths derivation for the electric dipole radiation (actually my question would fit for the magnetic dipole radiation too). He considers 2 charged balls connected by a wire with charge going back and forth between them. Now, when he calculates the vector potential he uses this formula: $$A(r,t)=\frac{\mu_0}{4 \pi}\int_{-d/2}^{d/2}\frac{-q_0\omega sin[\omega(t-r'/c)]\hat{z}}{r'}dz$$
However, if I followed it properly through the book, this equation is derived from Biot-Savart law (and a proper choice of gauge). However, when introducing Biot-Savart, Griffiths emphasizes that the current density (or linear current in this case) must be infinite in extent i.e. started infinitely long in the past and uniform. However the current is not infinite in extent, nor uniform. I see that Griffiths takes into account the fact that the potential is retarded, but I am just a bit confused about using this formula derived from Biot-Savart. Is it that obvious that a formula based on a uniform, infinite in extent current is correct just by adding that ##t-r/c## term?

Delta2
Silviu said:
However, if I followed it properly through the book, this equation is derived from Biot-Savart law (and a proper choice of gauge). However, when introducing Biot-Savart, Griffiths emphasizes that the current density (or linear current in this case) must be infinite in extent i.e. started infinitely long in the past and uniform.
That formula is not Biot Savart, although it looks similar. It is the retarded potential which can be derived from Maxwell’s equations in the Lorenz gauge by using Green’s functions.

Delta2
Dale said:
That formula is not Biot Savart, although it looks similar. It is the retarded potential which can be derived from Maxwell’s equations in the Lorenz gauge by using Green’s functions.
Thank you for your reply. I know it is not Biot-Savart. What I was saying is that Griffiths uses Biot-Savart to derive it. His approach is to Biot-Savart to prove that ##\nabla \times B = \mu_0 J##. Then, using ##B=\nabla \times A ## and the gauge in which ##\nabla A = 0## he reaches the formula I sated above. My confusion is that in his approach the starting point is Biot-Savart. However, as he specifies, Biot Savart is for the case of a uniform current, of infinite extent. So why does this retarded potential formula holds, even if its starting point wouldn't hold in the case of a non-uniform current.

I can’t say anything about his specific derivation, but it is not necessary to start with Biot Savart. There are other derivations.

Silviu said:
Thank you for your reply. I know it is not Biot-Savart. What I was saying is that Griffiths uses Biot-Savart to derive it. His approach is to Biot-Savart to prove that ##\nabla \times B = \mu_0 J##. Then, using ##B=\nabla \times A ## and the gauge in which ##\nabla A = 0## he reaches the formula I sated above. My confusion is that in his approach the starting point is Biot-Savart. However, as he specifies, Biot Savart is for the case of a uniform current, of infinite extent. So why does this retarded potential formula holds, even if its starting point wouldn't hold in the case of a non-uniform current.
Ignoring , for the moment, how he arrives at ##\nabla \times B = \mu_0 J## (1), this equation is almost the full Maxwell- Ampere equation (which hold in any case of current density), it just misses a term ##\epsilon_0\mu_0\frac{\partial E}{\partial t}## (2) in the right hand side. For small enough frequencies and amplitudes of the electric field (for example smaller than ##10^{8}## cause the term ##\epsilon_0\mu_0## is of the order of ##10^{-16}##) we can take (1) as a good approximation of the full Maxwell-Ampere law so it holds approximately.

Dale
The trick is that you have an indealized charge-current distribution which makes the exact retarded solution very simple. The general case is much more involved. I've just developed a nice treatment of the Hertzian dipole for my E&M lecture. The idea is to start with a harmonically oscillating charged particle, and it's most convenient to use the complexified treatment, i.e., one makes all the fields complex valued and understands as the physical values its real part. Then we consider a point charge harmonically oscillating along the 3-axis with angular frequency ##\omega##, i.e.,
$$\vec{y}(t)=d \vec{e}_3 \exp(-\mathrm{i} \omega t).$$
Then the charge-current distribution is given by
$$\rho(t,\vec{r})=q \delta^{(3)}[\vec{r}-\vec{y}(t)], \quad \vec{j}(t,\vec{r}) = q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)].$$
It's crucial to note that this obeys the continuity equation (charge conservation):
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
If you'd use this to calculate the scalar and vector potentials of the electromagnetic field, you'd get the Lienard-Wiechert potentials, but that's a pretty cumbersome calculation.

Instead the idea is to assume that ##d \ll \lambda##, where ##\lambda=2 \pi c/\omega## is the wavelength of the radiation expected to be produced by the harmonically oscillating charge. Then, if you look at the fields not too close to the origin of the coordinate system, it's justified to expand the charge-current distribution in powers of ##\vec{y}## and ##\dot{\vec{y}}##. It's important to make this consistent such that the continuity equation holds at each order. The first non-trivial order is the expansion up to order ##d##, i.e.,
$$\begin{split} \rho&=\rho_0 + \rho_1=q \delta^{(3)}(\vec{r}) - q \vec{y}(t) \cdot \vec{\nabla} \delta^{(3)}(\vec{r}),\\ \vec{j} &= \vec{j}_1=q \dot{\vec{y}}(t) \delta^{(3)}(\vec{r}). \end{split}$$
Indeed, using this approximation for charge and current densities still obeys the continuity equation exactly, which is crucial to get physically meaningful results for the em. field.

The first term ##\rho_0## is just the static (time-averaged) charge distribution of a point charge ##q## sitting at the origin. Integrating this with the retarded potential leads to the corresponding Coulomb potential in Lorenz gauge, i.e.,
$$\Phi_0(t,\vec{r}) = \frac{q}{4 \pi \epsilon_0 r}.$$
The next order is indeed electric-dipole radiation (i.e., the ##\ell=1## piece in the spherical multipole expansion), i.e., the lowest multipole order giving rise to the radiation of electromagnetic waves. It's very easy to see in this Cartesian approach that there is no "monopole radiation" in electromagnetics.

The rest of the calculation is most simply done with starting to calculate the vector potential. The retarded Green's function of the D'Alembert operator is
$$D_{\text{ret}}(t,\vec{r})=\frac{1}{4 \pi r} \delta(t-r/c).$$
Thus the vector potential is
$$\vec{A}_1(t,\vec{r}) = \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\mu_0 \vec{j}_1(t-r'/c,\vec{r'}}{4 \pi |\vec{r}-\vec{r}'|} = -\frac{\mathrm{i} \mu_0 \omega q \vec{d}}{4 \pi r} \exp(\mathrm{i} k r - \mathrm{i} \omega t).$$
From this it's a piece of cake to calculate
$$\vec{B}_1=\vec{\nabla} \times \vec{A}_1.$$
Instead of also calculating ##\Phi_1##, it's easier to use the Maxwell-Ampere equation to directly obtain ##\vec{E}_1##, making use of the fact that the time dependence of all fields is simply a factor ##\exp(-\mathrm{i} \omega t)##:
$$\vec{E}_1=\frac{\mathrm{i} c^2}{\omega} \vec{\nabla} \times \vec{B}_1.$$

Last edited:
Silviu said:
Is it that obvious that a formula based on a uniform, infinite in extent current is correct just by adding that ##t-r/c## term?
No.

Griffiths starts with the static potentials $$V(\vec r) = \frac 1 {4 \pi \varepsilon_0} \int {\frac {\rho ({\vec r}^\prime)} {|\vec r - {\vec r}^\prime|} \, d {\vec \tau}^\prime} \\ \vec A (\vec r) = \frac {\mu_0} {4 \pi} \int {\frac {\vec J ({\vec r}^\prime)} {|\vec r - {\vec r}^\prime|} \, d {\vec \tau}^\prime}$$ He generalizes them by evaluating ##\rho## and ##\vec J## at the retarded time for each source point: $$t_r = t - \frac {|\vec r - {\vec r}^\prime|} c$$ to get $$V({\vec r}, t) = \frac 1 {4 \pi \varepsilon_0} \int {\frac {\rho ({\vec r}^\prime, t_r)} {|\vec r - {\vec r}^\prime|} \, d {\vec \tau}^\prime} \\ \vec A ({\vec r}, t) = \frac {\mu_0} {4 \pi} \int {\frac {\vec J ({\vec r}^\prime, t_r)} {|\vec r - {\vec r}^\prime|} \, d {\vec \tau}^\prime}$$ Then he says:
Griffiths said:
Well, that all sounds reasonable—and surprisingly simple. But are we sure it's right? I didn't actually derive these formulas for ##V## and ##\vec A##; all I did was invoke a heuristic argument ("electromagnetic news travels at the speed of light") to make them seem plausible. To prove them, I must show that they satisfy the inhomogeneous wave equation (10.16) and meet the Lorentz condition (10.12). In case you think I'm being fussy, let me warn you that if you apply the same argument to the fields [##\vec E## and ##\vec B##] you'll get entirely the wrong answer [...]
He then shows that the retarded potentials satisfy (10.16) and leaves it as an exercise to show that they satisfy (10.12).

(Actually he does only the scalar potential explicitly, and says that "essentially the same argument would serve for the vector potential".)

[This is from the 3rd edition. The 4th edition may be different.]

vanhees71 and Delta2

## 1. What is electric dipole radiation?

Electric dipole radiation is the emission of electromagnetic waves from an oscillating electric dipole. An electric dipole is a pair of equal and opposite charges separated by a small distance, which can be created by the separation of positive and negative charges within a molecule or atom.

## 2. How is electric dipole radiation produced?

Electric dipole radiation is produced when an electric dipole oscillates or vibrates. This oscillation can be caused by an external electromagnetic field or by the intrinsic vibrations of the dipole itself. As the dipole oscillates, it creates a changing electric field that in turn produces electromagnetic waves.

## 3. What are the properties of electric dipole radiation?

Electric dipole radiation has several key properties, including directionality, polarization, and energy. The direction of the radiation is perpendicular to the axis of oscillation of the dipole. The radiation is also polarized, meaning that the electric field oscillates in a specific direction. Additionally, the energy of the radiation is proportional to the frequency of the oscillation and the strength of the dipole moment.

## 4. What are the applications of electric dipole radiation?

Electric dipole radiation has many practical applications, including in communication technologies such as antennas and satellite transmissions. It is also used in medical imaging techniques like MRI, where the oscillation of the electric dipole moments in the body's atoms can be detected and used to create images. Additionally, electric dipole radiation is used in spectroscopy to study the energy levels and structure of molecules.

## 5. How is electric dipole radiation related to other forms of electromagnetic radiation?

Electric dipole radiation is a type of electromagnetic radiation, along with other forms such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. These forms of radiation all have different frequencies and wavelengths, but they are all produced by the oscillation of electric and magnetic fields. Electric dipole radiation specifically is in the lower frequency range of the electromagnetic spectrum, with wavelengths ranging from a few centimeters to hundreds of meters.

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