Bremmstrahlung

Bremmstrahlung or braking radiation is the electromagnetic radiation emitted by a point charge due to deceleration (or acceleration) in the direction of its velocity. The angular radiant power distribution is

dPdΩ=μ0q2a216π2csin⁡2θ(1−βcos⁡θ)5\frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi ^{2}c} \frac{\sin ^{2}\theta}{(1-\beta \cos \theta)^{5}}

and the total radiant power emitted is

P=μ0q2a2γ66πcP=\frac{\mu_{0}q^{2}a^{2}\gamma^{6}}{6\pi c}

where γ=1/1−v2/c2\gamma=1/\sqrt{ 1-v^{2}/c^{2} } is the usual relativistic coefficient.

Derivation

Consider a point charge moving in the same direction it is accelerating in, such that v\mathbf{v} and a\mathbf{a} are collinear at some retarded time trt_{r}. We care about the angular distribution of the radiant power emitted, so we use the Larmor formula in the Liénard generalization, specifically its solid angle

dPdΩ=q216π2ε0∣r^×(u×a)∣2(r^⋅u)5\frac{dP}{d\Omega}=\frac{q^{2}}{16\pi ^{2}\varepsilon_{0}} \frac{\lvert \hat{\boldsymbol{\mathfrak{r}}}\times(\mathbf{u}\times \mathbf{a}) \rvert ^{2}}{(\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{u})^{5}}

Since u=cr^−v\mathbf{u}=c \hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v} and v\mathbf{v} is collinear with a\mathbf{a}, we get u×a=c(r^×a)−v×a\mathbf{u}\times \mathbf{a}=c(\hat{\boldsymbol{\mathfrak{r}}}\times \mathbf{a})-\cancel{ \mathbf{v}\times \mathbf{a} }. Also r^⋅u=c−r^⋅v\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{u}=c-\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{v}. Thus

dPdΩ=q216π2ε0∣r^×(r^×a)∣2(c−r^⋅v)5\frac{dP}{d\Omega}=\frac{q^{2}}{16\pi ^{2}\varepsilon_{0}} \frac{\lvert \hat{\boldsymbol{\mathfrak{r}}}\times(\hat{\boldsymbol{\mathfrak{r}}}\times \mathbf{a}) \rvert ^{2}}{(c-\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{v})^{5}}

The triple product expands to

r^×(r^×a)=(r^⋅a)r^−a⇒∣r^×(r^×a)∣2=a2−(r^⋅a)2\hat{\boldsymbol{\mathfrak{r}}}\times(\hat{\boldsymbol{\mathfrak{r}}}\times \mathbf{a})=(\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{a})\hat{\boldsymbol{\mathfrak{r}}}-\mathbf{a}\quad\Rightarrow \quad \lvert \hat{\boldsymbol{\mathfrak{r}}}\times(\hat{\boldsymbol{\mathfrak{r}}}\times \mathbf{a}) \rvert^{2}=a^{2}-(\hat{\boldsymbol{\mathfrak{r}}}\cdot \mathbf{a})^{2}

If we set the frame of reference such that v\mathbf{v} is on the xx axis, then

dPdΩ=μ0q2a216π2csin⁡2θ(1−βcos⁡θ)5\frac{dP}{d\Omega}=\frac{\mu_{0}q^{2}a^{2}}{16\pi ^{2}c} \frac{\sin ^{2}\theta}{(1-\beta \cos \theta)^{5}}

where β≡v/c\beta\equiv v/c is the usual relativistic coefficient. Now, when β≃0\beta \simeq 0, the whole denominator below the sine goes away and we get the typical torus-shaped radiation emission. However, when β\beta is large, the quintic term at the denominator distorts the torus so that looks something like this:

80%

The squished blue shapes are the directions in which radiation is strongest. While there is still no radiation in the v\mathbf{v} direction, most of it is being emitted forwards. Notice how, due to the square of aa, the radiation distribution does not care about whether the charge is accelerating or decelerating. Either way, it gets emitted in the direction of the velocity. This kind of radiation is known as bremmstrahlung, or braking radiation since it is typically encountered when charges brake due to the presence of external fields.

The total power emitted is

P=∫SdPdΩdΩ=μ0q2a216π2c∫Ssin⁡2θ(1−βcos⁡θ)5sin⁡θdθdϕP=\int_{\mathcal{S}} \frac{dP}{d\Omega}d\Omega=\frac{\mu_{0}q^{2}a^{2}}{16\pi ^{2}c}\int_{\mathcal{S}} \frac{\sin ^{2}\theta}{(1-\beta \cos \theta)^{5}}\sin \theta d\theta d\phi

over a sphere S\mathcal{S} that contains the charge. The ϕ\phi integral is trivial since there is no ϕ\phi dependency and makes 2π2\pi. The θ\theta is solved with the substitution x≡cos⁡θx\equiv \cos \theta:

P=μ0q2a28πc∫−111−x2(1−βx)5dxP=\frac{\mu_{0}q^{2}a^{2}}{8\pi c}\int_{-1}^{1} \frac{1-x^{2}}{(1-\beta x)^{5}}dx

We then integrate by parts on numerator and denominator, which results in 43(1−β2)−3\frac{4}{3}(1-\beta ^{2})^{-3}, and so

P=μ0q2a2γ66πcP=\frac{\mu_{0}q^{2}a^{2}\gamma^{6}}{6\pi c}

In detectors

For an electron, the Particle process of bremmstrahlung due to some Atomic nucleus \ceZAX\ce{^{A}_{Z}X} is

\cee−+ZAX−>e−+γ +ZAX\ce{e^{-} + _{Z}^{A}X -> e^{-} + \gamma\ + _{Z}^{A}X}

It consists of the emission of a Photon with no change on the electron or the nucleus. The effects are largely kinematic and cause the deflection of the electron. This is useful in detector physics, where bremmstrahlung is the primary cause of energy loss for light particles traversing matter (also see Stopping power).