I Something about retarded potentials for oscillating electric dipole

AI Thread Summary
The discussion focuses on the potential vector for an oscillating electric dipole, expressed as A = (μ₀I₀d/4π)(cos(ω(t-r/c))/r), where various parameters define the system. The dipole moment is represented as p = (kdI₀/ω)sin(ωt), allowing the potential vector to be rewritten in terms of the time derivative of the dipole moment, A = (μ₀/4π)(ṗ(t-r/c)/r). The charge-current distribution is expanded to first order, yielding the electrostatic Coulomb field of a stationary charge and additional terms that contribute to the potential vector. Ultimately, the final equation for the potential vector is A(t,x) = (μ₀/4π)(ṗ(t-r/c)/r). This analysis provides insights into the behavior of oscillating dipoles in electromagnetic fields.
Salmone
Messages
101
Reaction score
13
In a problem of an oscillating electric dipole, under appropriate conditions, one can find, for the potential vector calculated at the point ##\vec{r}##, the expression ##\vec{A}=\hat{k}\frac{\mu_0I_0d}{4\pi}\frac{cos(\omega(t-r/c))}{r}## where: ##\hat{k}## is the direction of the ##z-axis## where the dipole is oscillating, ##I_0## is the current (##I(t)=I_0cos(\omega t)##), ##d## is the distance between the charges of the dipole and ##r## is the distance between the origin of the system and the point where I want to calculate the potential vector. Let ##\vec{p}=\hat{k}qd=\frac{\hat{k}dI_0}{\omega}sin(\omega t)## be the dipole moment, it is possible to rewrite the potential vector as ##\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{\dot p(t-r/c)}}{r}## where ##\vec{\dot p}## is the derivative with respect to time.
 
Physics news on Phys.org
You can start with the source of a point charge,
$$\rho(t,\vec{x})=q \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(t,\vec{x})= q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
where ##\vec{y}(t)## is the trajectory of the charge.

Now we assume that
$$\vec{y}(t)=\vec{d} \sin(\omega t).$$
For ##r=|\vec{x}|\gg |\vec{d}|## we can expand the charge-current distribution up to first order in ##\vec{d}##,
$$\rho(t,\vec{x})=q \delta^{(3)}(\vec{x}) - q \vec{y}(t) \cdot \vec{\nabla} \delta^{(3)}(\vec{x}) + \mathcal{O}(\vec{d}^2), \quad \vec{j}(t,\vec{x})=q \dot{\vec{y}}(t) \delta^{(3)}(\vec{x}) + \mathcal{O}(\vec{d}^2).$$
From the first term of ##\rho## (of order ##\mathcal{O}(d^0)##) you get the electrostatic Coulomb field of a charge at rest in the origin (which you can easily verify using the retarded potential too).

For the terms of order ##\mathcal{O}(d)## you get for ##\vec{A}## (in SI units)
$$\vec{A}(t,\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(t-|\vec{x}-\vec{x}'|/c,\vec{x}') \frac{1}{|\vec{x}-\vec{x}'|}=\frac{\mu_0 q \omega \vec{d}}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \cos[\omega (t-|\vec{x}-\vec{x}'|/c] \frac{\delta^{(3)}(\vec{x}')}{|\vec{x}-\vec{x}'|} = \frac{\mu_0 q \omega \vec{d}}{4 \pi r} \cos[\omega (t-r/c)].$$
With ##I_0=q \omega## that's the solution you are looking for.

The final equation, of course, must read
$$\vec{A}(t),\vec{x})=\frac{\mu_0}{4 \pi} \frac{\dot{\vec{P}}(t-r/c)}{r}.$$
 
I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...
Back
Top