STEM Problem Detail

Problem Statement

Prompt

Provide step by step solution of Problem 2.10. only given in the image .

Solution Rendering...

## Solution 2.10

**a)** For nonrelativistic motion of an accelerating electron with velocity $$v = \beta c$$, the electric component of the radiation field is (see Jackson, Chapter 14)
\begin{equation}
\mathbf{E} = \frac{-e}{c} \left[ \frac{\hat{\mathbf{n}} \times (\hat{\mathbf{n}} \times \boldsymbol{\dot{\beta}})}{R} \right]_{\text{ret}}, \tag{11.105}
\end{equation}
at a distance $$R$$ in the direction $$\hat{\mathbf{n}}$$ from the electron. The expression in brackets is evaluated at the retarded time $$t' = t - R/c$$.

The magnetic radiation field is given by $$\mathbf{B} = \hat{\mathbf{n}} \times \mathbf{E}$$, and the energy flux by the Poynting vector:
\begin{equation}
S = \frac{c}{4\pi} (\mathbf{E} \times \mathbf{B}) = \frac{c}{4\pi} |\mathbf{E}|^2 \hat{\mathbf{n}}. \tag{11.106}
\end{equation}

The angular dependence in $$|\mathbf{E}|$$ comes from the cross-product $$\hat{\mathbf{n}} \times \boldsymbol{\dot{\beta}}$$. Let $$\theta$$ be the angle between the line connecting the electron and the nucleus, and the line connecting the electron and the observation point (at time $$t'$$). Then $$\hat{\mathbf{n}} \times \boldsymbol{\dot{\beta}}$$ is proportional to $$\sin \theta$$, and from (11.106) the radiated power varies as $$\sin^2 \theta$$.

**b)** To find the polarization of the radiation, we observe that the vector $$\hat{\mathbf{n}} \times \boldsymbol{\dot{\beta}}$$ defines the normal to the plane containing the electron, the nucleus and the observer. From equation (11.105) we see that the electric field is perpendicular to this normal --- i.e., it is in the plane and also that it is perpendicular to the line between the charge and the observer. This is sufficient to define its polarization. Alternatively, we could write the (unnormalized) polarization vector as $$\boldsymbol{\epsilon} = -\hat{\mathbf{n}} \times (\hat{\mathbf{n}} \times \hat{\mathbf{z}})$$, where $$\hat{\mathbf{z}}$$ is a unit vector pointing from the electron to the proton.

**c)** For $$v \ll c$$ we can ignore the complication of having to evaluate the fields at retarded time. So
\begin{equation}
|\mathbf{E}| \approx \frac{e}{c} \frac{|\hat{\mathbf{n}} \times (\hat{\mathbf{n}} \times \boldsymbol{\dot{\beta}})|}{R} = \frac{e}{cR} |\boldsymbol{\dot{\beta}}| \sin \theta, \tag{11.107}
\end{equation}
and therefore the Poynting vector is
\begin{equation}
S \approx \frac{e^2}{4\pi c R^2} \dot{\beta}^2 \sin^2 \theta \, \hat{\mathbf{n}}. \tag{11.108}
\end{equation}

We can integrate this over a sphere of radius $$R$$ centered on the charge to get the total power radiated:
\begin{align}
\mathcal{P} &= \int d\Omega \, R^2 S \cdot \mathbf{n} = \int d\Omega \frac{e^2 a^2}{4\pi c^3} \sin^2 \theta \nonumber \\
&= \frac{e^2 a^2}{4\pi c^3} \int_{0}^{2\pi} d\phi \int_{-1}^{1} d(\cos \theta) (1 - \cos^2 \theta) = \frac{2e^2 a^2}{3c^3} \tag{11.109}
\end{align}
(where $$a/c = \dot{\beta}$$), which is the well-known Larmor formula.

By assumption we are neglecting the force on the electron due to radiation reaction, and so the acceleration of the electron is due simply to the electrostatic force between it and the nucleus, $$F = Ze^2/r^2$$. Hence our expression for the energy radiated per unit time when the electron is at a distance $$r$$ from the nucleus is
\begin{equation}
\mathcal{P}(r) = \frac{2e^2}{3c^3} \left( \frac{Ze^2}{mr^2} \right)^2 = \frac{2Z^2 e^6}{3m^2 c^3 r^4}, \tag{11.110}
\end{equation}
where $$m$$ is the mass of the electron.

**d)** To evaluate the total power loss, we have to use the simplifying assumption that the total energy radiated is small enough that we may ignore its effect on the motion of the electron. (In fact this is no more than restating that we are ignoring radiation reaction.) Hence we can use conservation of energy,
\begin{equation}
-\frac{Ze^2}{r_0} = \frac{1}{2}mv^2 - \frac{Ze^2}{r}, \tag{11.111}
\end{equation}
to find the velocity squared:
\begin{equation}
v^2 = \frac{2Ze^2}{m} \left( \frac{1}{r} - \frac{1}{r_0} \right). \tag{11.112}
\end{equation}

The total energy radiated is
\begin{equation}
\mathcal{E} = \int dt \, \mathcal{P} = \int dr' \left| \frac{dt}{dr'} \right| \mathcal{P}(r'), \tag{11.113}
\end{equation}
where $$dt/dr = 1/v$$, which is less than zero because the particle is falling inwards. Substituting for the speed of the particle from equation (11.112) gives us
\begin{equation}
\mathcal{E} = \sqrt{\frac{m}{2Ze^2}} \int_{r}^{r_0} \frac{dr'}{(1/r' - 1/r_0)^{1/2}} \frac{2Z^2 e^6}{3m^2 c^3 r'^4}. \tag{11.114}
\end{equation}

We have to evaluate the integral
\begin{equation}
I = \int_{r}^{r_0} \frac{dr'}{(1/r' - 1/r_0)^{1/2}} \frac{1}{r'^4}. \tag{11.115}
\end{equation}

If we make the substitution $$1/r' - 1/r_0 = w^2$$, $$dr'/r'^2 = -2w \, dw$$ and define $$w_0^2 \equiv 1/r - 1/r_0$$, then this becomes
\begin{align}
I &= \int_{w_0}^{0} \frac{-2w \, dw}{w} \left( w^2 + \frac{1}{r_0} \right)^2 \nonumber \\
&= 2 \left[ \frac{1}{r_0^2} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{1/2} + \frac{2}{3r_0} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{3/2} + \frac{1}{5} \left( \frac{1}{r} - \frac{1}{r_0} \right)^{5/2} \right]. \tag{11.116}
\end{align}

Thus the total radiated power is
\begin{equation}
\mathcal{E} = \left( \frac{2Z}{m} \right)^{3/2} \frac{e^5}{45c^3 r_0^{5/2}} \left( \frac{r_0}{r} - 1 \right)^{1/2} \left[ 8 + \frac{4r_0}{r} + \frac{3r_0^2}{r^2} \right]. \tag{11.117}
\end{equation}

Loading solution...

Problem #40