Electromagnetic radiation

Electromagnetic radiation is radiation of electromagnetic waves. It occurs when charged particles accelerate, which causes them to emit energy that is carried off to infinity through electromagnetic waves1.

Electromagnetic waves can be of any frequency. Since this leads to a massive (infinite, actually) range of possible waves, called the electromagnetic spectrum, they are conventionally categorized in seven "types", from most to least energetic:

  1. Gamma rays
  2. X-rays
  3. Ultraviolet light
  4. Visible light, or simply just light
  5. Infrared light
  6. Microwaves
  7. Radio waves

Sources

Consider a localized source near the origin. To determine the energy radiated from the source at some time t0t_{0}, we encase the source in a sphere of very large radius rr, large enough that the source looks like a point at the center of the sphere. The radiant power passing through the surface is given, according to Poynting's theorem, by integrating the Poynting vector over the surface

P(r,t)=∮SS⋅da=1μ0∮S(E×B)⋅daP(r,t)=\oint_{\mathcal{S}} \mathbf{S}\cdot d\mathbf{a}=\frac{1}{\mu_{0}}\oint_{\mathcal{S}}(\mathbf{E}\times \mathbf{B})\cdot d\mathbf{a}

since the electromagnetic energy density does not change in time. Since electromagnetic waves travel at the speed of light, thus causing the fields E\mathbf{E} and B\mathbf{B} to depend not on current time but on retarded time, the energy that we feel now at the surface was emitted by the source in the past, at the retarded time t0=t−r/ct_{0}=t-r/c. The power radiated is then

Prad(t0)=lim⁡r→∞P(r,t0+rc)P_\text{rad}(t_{0})=\lim_{ r \to \infty } P\left( r,t_{0}+ \frac{r}{c} \right)

where t0t_{0} is held constant. This is the power that is radiated from the source in a spherical fashion up to infinity. This transfer is irreversible: radiation is emitted and never comes back.

The area of the sphere goes like ∼r2\sim r^{2}, so for the power not to vanish at r→∞r\to \infty, the Poynting vector must decrease at most like ∼r−2\sim r^{-2}. Any faster and it dilutes to zero at infinity when spread over a spherical surface. Since S∝E×B\mathbf{S}\propto \mathbf{E}\times \mathbf{B}, this is like saying the the product of EE and BB must decrease at most like ∼r−2\sim r^{-2}.

Let's see what happens with static fields: Coulomb's law says that electrostatic fields go like ∼r−2\sim r^{-2} and the Biot-Savart law says the same for magnetostatic fields. Their product, then, is ∼r−4\sim r^{-4}; way too fast for what we need. Clearly then, it is impossible for a static configuration to radiate: it just cuts out too fast.

When we add in a time-dependent source, it's Jefimenko's equations that tell us the fields. These include our previous static terms, but also include terms that go like ∼r−1\sim r^{-1}. Multiply them together and you get something that satisfies our condition of not decreasing faster than ∼r−2\sim r^{-2}. It is these terms alone that are responsible for all electromagnetic radiation; everything else cuts out too fast and can be safely ignored.

A discussion on the physics of radiation from an arbitrary source can be found below. That said, starting from the general case is probably not a good idea; Electric dipole > Radiation and Magnetic dipole > Radiation are probably a better choice.

Radiation from an arbitrary source

In an ideal world, we'd be able to determine the exact radiation at any point from any source analytically. This is a bit of a dream scenario and in practice it's not at all possible to solve for most source distributions. That said, not all hope is lost. While we can't solve everything, we can do some truly general considerations.

To start, let's consider some generic charge distribution of any shape whatsoever, with our only prerequisite being that it's a finite volume near the origin.

50%

Potentials

The retarded scalar potential is

V(r,t)=14πε0∫Vρ(r′,t−r/c)rdτ′V(\mathbf{r},t)=\frac{1}{4\pi \varepsilon_{0}}\int_{\mathcal{V}} \frac{\rho(\mathbf{r}',t- \mathfrak{r}/c)}{\mathfrak{r}}d\tau'

where r=r2+r′2−2r⋅r′\mathfrak{r}=\sqrt{ r^{2}+r'^{2}-2\mathbf{r}\cdot \mathbf{r}' }. As in other situations, like Electric dipole > Radiation and Magnetic dipole > Radiation, we are going to consider the source as very small compared to the distances that we are considering

r′≪r(Small source approximation)r'\ll r\tag{Small source approximation}

Since r′r' is a variable of integration, this means that r′≪rr'\ll r at all points during the integration, i.e. the entire source is small. In this approximation we get

r=r1−2r^⋅r′r+r′2r2\mathfrak{r}=r\sqrt{ 1- 2\frac{\hat{\mathbf{r}}\cdot \mathbf{r}'}{r}+ \frac{r'^{2}}{r^{2}}}

Careful with the change from r\mathbf{r} to r^\hat{\mathbf{r}} in the cross-term! r′2/r2r'^{2}/r^{2} is second-order and therefore negligible in the small source approximation. The rest of the root can use the binomial expansion to first order (1−x≃1−x/2\sqrt{ 1-x }\simeq1-x/2):

r≃r(1−r^⋅r′r)\mathfrak{r}\simeq r\left( 1- \frac{\hat{\mathbf{r}}\cdot\mathbf{r}'}{r} \right)

and so our distribution is actually dependent on

ρ(r′,t−rc)≃ρ(r′,t−rc+r^⋅r′c)\rho\left( \mathbf{r}',t- \frac{\mathfrak{r}}{c} \right)\simeq \rho\left( \mathbf{r}',t- \frac{r}{c}+ \frac{\hat{\mathbf{r}}\cdot\mathbf{r}'}{c} \right)

Expanding ρ\rho in a Taylor series in tt about the retarded time at the origin

t0≡t−rct_{0}\equiv t- \frac{r}{c}

yields

ρ(r′,t−rc)≃ρ(r′,t0)+ρ˙(r′,t0)(r^⋅r′c)+12ρ¨(r^⋅r′c)2+13!ρ... (r^⋅r′c)3+…\rho\left( \mathbf{r}',t- \frac{\mathfrak{r}}{c} \right)\simeq \rho(\mathbf{r}',t_{0})+\dot{\rho}(\mathbf{r}',t_{0})\left( \frac{\hat{\mathbf{r}}\cdot\mathbf{r}'}{c} \right)+ \frac{1}{2}\ddot{\rho}\left( \frac{\hat{\mathbf{r}}\cdot \mathbf{r}'}{c} \right)^{2}+ \frac{1}{3!}\dddot{\rho}\, \left( \frac{\hat{\mathbf{r}}\cdot \mathbf{r}'}{c} \right)^{3}+\ldots

using dot notation for time derivatives. We do this because we wish to truncate to first order, but to do so must accept a second approximation:

r′≪c∣ρ¨/ρ˙∣,c∣ρ...,ρ˙∣1/2,c∣ ρ.... ,ρ˙∣1/3…(Approximation 2)r'\ll \frac{c}{\lvert \ddot{\rho}/\dot{\rho} \rvert }, \frac{c}{\lvert \dddot{\rho},\dot{\rho }\rvert^{1/2} }, \frac{c}{\lvert \ \ddddot{\rho}\ ,\dot{\rho} \rvert^{1/3} }\ldots\tag{Approximation 2}

and other further terms. Now, interpreting this is not an easy task, as we do not have an a priori idea of what ρ\rho is, besides some charge distribution. To gather some insight, recall that in simple oscillating systems, like electric and magnetic dipoles, these all end up being c/ωc/\omega, with ω\omega the angular frequency of oscillation, which is directly proportional to the wavelength λ\lambda. In those cases, we call it the large wavelength approximation, as it expects us to keep the wavelength larger than the size of the dipole. We can generalize this idea to mean "keep the size of whatever distribution small compared to the wavelength that's being emitted", with "size" being a vague, qualitative term more so than a precise quantity. But in general, we can't be so precise. In practice, this approximation, alongside the small source approximation, really just means "only keep the first-order terms in r′r', that is, (r′)±1(r')^{\pm1}".

We now put the expansion of ρ\rho, to first order only, alongside

1r≃1r(1+r^⋅r′r)=1r(1+r⋅r′r2)\frac{1}{\mathfrak{r}}\simeq \frac{1}{r}\left( 1+ \frac{\hat{\mathbf{r}}\cdot\mathbf{r}'}{r} \right)=\frac{1}{r}\left( 1+ \frac{\mathbf{r}\cdot \mathbf{r}'}{r^{2}} \right)

in the formula for the retarded potential:

V(r,t)≃14πε0∫V1r(1+r⋅r′r2)[ρ(r′,t0)+ρ˙(r′,t0)(r^⋅r′c)]dτ′=14πε0r∫V[ρ(r′,t0)+ρ(r′,t0)(r⋅r′r2)++ρ˙(r′,t0)(r^⋅r′c)+r⋅r′r2ρ˙(r,t0)(r^⋅rc)]=…\begin{align} V(\mathbf{r},t)&\simeq\frac{1}{4\pi \varepsilon_{0}}\int_{\mathcal{V}} \frac{1}{r}\left( 1+ \frac{\mathbf{r}\cdot \mathbf{r}'}{r^{2}} \right)\left[ \rho(\mathbf{r}',t_{0})+ \dot{\rho}(\mathbf{r}',t_{0})\left( \frac{\hat{\mathbf{r}}\cdot \mathbf{r}'}{c} \right) \right]d\tau' \\ &=\frac{1}{4\pi \varepsilon_{0}r}\int_{\mathcal{V}}\Big[ \rho(\mathbf{r}',t_{0})+\rho(\mathbf{r}',t_{0})\left( \frac{\mathbf{r}\cdot \mathbf{r}'}{r^{2}} \right)+ \\ &\qquad \qquad \quad+\dot{\rho}(\mathbf{r}',t_{0})\left( \frac{\hat{\mathbf{r}}\cdot \mathbf{r}'}{c} \right)+ \frac{\mathbf{r}\cdot \mathbf{r}'}{r^{2}}\dot{\rho}(\mathbf{r},t_{0})\left( \frac{\hat{\mathbf{r}}\cdot \mathbf{r}}{c} \right)\Big] \\ &=\ldots \end{align}

The last term in the integral drops out because it is second order in r′\mathbf{r}' (since we have (r^⋅r′)(r^⋅r′)(\hat{\mathbf{r}}\cdot \mathbf{r}')(\hat{\mathbf{r}}\cdot \mathbf{r}')). We are left with

V(r,t)≃14πε0r[∫Vρ(r′,t0)dτ′+r^r⋅∫Vr′ρ(r′,t0)dτ′+r^c⋅ddt∫Vr′ρ(r′,t0)dτ′]V(\mathbf{r},t)\simeq \frac{1}{4\pi \varepsilon_{0}r}\left[ \int_{\mathcal{V}}\rho(\mathbf{r}',t_{0})d\tau'+ \frac{\hat{\mathbf{r}}}{r}\cdot \int_{\mathcal{V}}\mathbf{r}'\rho(\mathbf{r}',t_{0})d\tau'+ \frac{\hat{\mathbf{r}}}{c}\cdot \frac{d}{dt} \int_{\mathcal{V}}\mathbf{r}'\rho(\mathbf{r}',t_{0})d\tau' \right]

The first integral is just the total charge QQ at time t0t_{0}, and since charge is a conserved quantity, it remains the same at all times. The other two integrals are the electric dipole moment p(t0)\mathbf{p}(t_{0}) at time t0t_{0}, so

V(r,t)≃14πε0[Qr+r^⋅p(t0)r2+r^⋅p˙(t0)rc]\boxed{V(\mathbf{r},t)\simeq \frac{1}{4\pi \varepsilon_{0}}\left[ \frac{Q}{r}+ \frac{\hat{\mathbf{r}}\cdot \mathbf{p}(t_{0})}{r^{2}}+ \frac{\hat{\mathbf{r}}\cdot \dot{\mathbf{p}}(t_{0})}{rc} \right]}

What we got here is basically an extended multipole expansion, truncated at the first order. The first term is the monopole term, the second is the dipole term, and the third is a new term that does not appear in the multipole expansion. That is because the multipole expansion is considered in static scenarios where p˙=0\dot{\mathbf{p}}=0.

We now move on to the retarded vector potential

A(r,t)=μ04π∫VJ(r′,t−r/c)rdτ′\mathbf{A}(\mathbf{r},t)=\frac{\mu_{0}}{4\pi}\int_{\mathcal{V}} \frac{\mathbf{J}(\mathbf{r}',t- \mathfrak{r}/c)}{\mathfrak{r}}d\tau'

To first order in r′r', we can just replace r\mathfrak{r} with rr (i.e. a zeroth order approximation, you'll see why in a moment), so

A(r,t)≃μ04πr∫VJ(r′,t0)dτ′\mathbf{A}(\mathbf{r},t)\simeq \frac{\mu_{0}}{4\pi r}\int_{\mathcal{V}}\mathbf{J}\left( \mathbf{r}', t_{0} \right)d\tau'

Now, it can be found that

∫VJdτ′=p˙\int_{\mathcal{V}}\mathbf{J}d\tau'=\dot{\mathbf{p}}

by evaluating ∫V∇⋅(xJ)dτ\int_{\mathcal{V}}\nabla\cdot(x\mathbf{J})d\tau. So

A(r,t)≃μ04πrp˙(t0)\boxed{\mathbf{A}(\mathbf{r},t)\simeq \frac{\mu_{0}}{4\pi r}\dot{\mathbf{p}}(t_{0})}

This is why we could accept a zeroth order approximation r≃r\mathfrak{r}\simeq r, since p\mathbf{p} is already first order in r′r'. A first order approximation of r\mathfrak{r} would've ended up becoming a second (or higher) order one.

Fields

As for the fields, we only really care about the radiation zone, so

r−2 terms and below are negligible(Far field approximation)r^{-2}\text{ terms and below are negligible}\tag{Far field approximation}

In the electric field, we need to take the Gradient of VV. Because of the far field approximation, the first term drops out (∇r−1∝r−2\nabla r^{-1}\propto r^{-2}). The second term is already second order on its own so it also drops out. The third term requires us to take the gradient of t0t_{0}:

∇t0=−1c∇r=−1cr^\nabla t_{0}=- \frac{1}{c}\nabla r=- \frac{1}{c}\hat{\mathbf{r}}

and hence

V(r,t)≃∇[14πϵ0r^⋅p˙(t0)rc]≃14πε0r^⋅p¨(t0)c∇t0=−r^⋅p¨(t0)4πε0c2r^V(\mathbf{r},t)\simeq \nabla\left[ \frac{1}{4\pi \epsilon_{0}} \frac{\hat{\mathbf{r}}\cdot \dot{\mathbf{p}}(t_{0})}{rc} \right]\simeq \frac{1}{4\pi \varepsilon_{0}}\frac{\hat{\mathbf{r}}\cdot \ddot{\mathbf{p}}(t_{0})}{c}\nabla t_{0}=- \frac{\hat{\mathbf{r}}\cdot \ddot{\mathbf{p}}(t_{0})}{4\pi \varepsilon_{0}c^{2}}\hat{\mathbf{r}}

The time derivative of A\mathbf{A} is

∂A∂t=μ04πrp¨(t0)\frac{ \partial \mathbf{A} }{ \partial t } = \frac{\mu_{0}}{4\pi r} \ddot{\mathbf{p}}(t_{0})

So

E(r,t)≃μ04πr[(r^⋅r¨)r^−p¨]=μ04πrr^×(r^×p¨)\boxed{\mathbf{E}(\mathbf{r},t)\simeq \frac{\mu_{0}}{4\pi r}[(\hat{\mathbf{r}}\cdot \ddot{\mathbf{r}})\hat{\mathbf{r}}-\ddot{\mathbf{p}}]=\frac{\mu_{0}}{4\pi r}\hat{\mathbf{r}}\times(\hat{\mathbf{r}}\times \ddot{\mathbf{p}})}

The Curl of A\mathbf{A} is

∇×A=μ04πr[∇×p˙(t0)]=μ04πr[(∇t0)×p¨(t0)]=−μ04πcr[r^×p¨(t0)]\nabla\times \mathbf{A}=\frac{\mu_{0}}{4\pi r}[\nabla \times \dot{\mathbf{p}}(t_{0})]=\frac{\mu_{0}}{4\pi r}[(\nabla t_{0})\times \ddot{\mathbf{p}}(t_{0})]=- \frac{\mu_{0}}{4\pi cr}[\hat{\mathbf{r}}\times \ddot{\mathbf{p}}(t_{0})]

and so

B(r,t)≃−μ04πcrr^×p¨=−1cr^×E\boxed{\mathbf{B}(\mathbf{r},t)\simeq - \frac{\mu_{0}}{4\pi cr}\hat{\mathbf{r}}\times \ddot{\mathbf{p}}=- \frac{1}{c}\hat{\mathbf{r}}\times \mathbf{E}}

The fields are in phase, mutually perpendicular and transverse. The ratio of their amplitudes is exactly the speed of light: E0/B0=cE_{0}/B_{0}=c. This matches the behavior that an electromagnetic wave in the vacuum is supposed to have.

Electromagnetic waves

Analyzing the electromagnetic waves is more complicated than in a simple oscillating case, as we don't have a clear angular frequency ω\omega to use. Either way, we can make some general statements. We'll write the fields in spherical coordinates, with the zz axis in the direction of p¨(t0)\ddot{\mathbf{p}}(t_{0}):

E(r,θ,t)≃μ0p¨(t0)4πrsin⁡θ θ^,B(r,θ,t)≃μ0p¨(t0)4πcrsin⁡θ ϕ^\mathbf{E}(r,\theta,t)\simeq \frac{\mu_{0}\ddot{p}(t_{0})}{4\pi r}\sin \theta\ \hat{\boldsymbol{\theta}},\qquad \mathbf{B}(r,\theta,t)\simeq \frac{\mu_{0}\ddot{p}(t_{0})}{4\pi cr} \sin \theta\ \hat{\boldsymbol{\phi}}

The Poynting vector is

S(r,t)≃μ016π2cr2[p¨(t0)]2sin⁡2θ r^\mathbf{S}(\mathbf{r},t)\simeq \frac{\mu_{0}}{16\pi ^{2}cr^{2}}[\ddot{p}(t_{0})]^{2}\sin ^{2}\theta\ \hat{\mathbf{r}}

The radiant power passing through a large sphere of radius rr is

P(r,t)=∮SS(r,t)⋅da=μ06πc[p¨(t−rc)]2P(r,t)=\oint_{\mathcal{S}}\mathbf{S}(\mathbf{r},t)\cdot d\mathbf{a}=\frac{\mu_{0}}{6\pi c}\left[ \ddot{p}\left( t- \frac{r}{c} \right) \right]^{2}

and the total radiated power is

Prad(t0)=lim⁡r→∞P(r,t)≃μ06πc[p¨(t0)]2P_\text{rad}(t_{0})=\lim_{ r \to \infty } P(r,t)\simeq \frac{\mu_{0}}{6\pi c}[\ddot{p}(t_{0})]^{2}

We can see that if p(t)=p0cos⁡(ωt)p(t)=p_{0}\cos(\omega t), we get p¨(t)=−ω2p0cos⁡(ωt)\ddot{p}(t)=-\omega ^{2}p_{0}\cos(\omega t). Substitute this in the formulas above to get the electric dipole radiation again. More generally, set p(t)=qr(t)\mathbf{p}(t)=q\mathbf{r}(t), where qq is an electric charge and r(t)\mathbf{r}(t) is the distance of qq from the origin. Then p¨(t)=qa(t)\ddot{\mathbf{p}}(t)=q\mathbf{a}(t), where a(t)\mathbf{a}(t) is the acceleration of the charge, so the power is

Prad(t0)=μ0q2a26πcP_\text{rad}(t_{0})=\frac{\mu_{0}q^{2}a^{2}}{6\pi c}

This is known as the Larmor formula, which gives the radiant power of an accelerating point charge. Notice how it is dependent on the square of the acceleration.

Footnotes

  1. Radiation as expressed here makes sense only in the context a localized source, that is, a source of finite extent, since the idea of "energy carried off to infinity" no longer makes sense when the source itself extends to infinity. Such shapes of course do not exist in the real world, but are common in simplified formulations of phenomena, such as infinite wires, infinite planes, etc. ↩