Subfield: Electromagnetism

The problem concerns investigation of transforming the electron microscope with magnetic guiding of the electron beam (which is accelerated with the potential difference $U=$ $511 \mathrm{kV}$ ) into a proton microscope (in which the proton beam is accelerated with the potential difference $-U$ ). For this purpose, solve the following two problems:

Question: A. An electron after leaving a device, which accelerated it with the potential difference $U$, falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_{1}, L_{2}, \ldots, L_{n}$. The known currents in the coils are $i_{1}, i_{2}, \ldots, i_{n}$, respectively.

What should the currents $i_{1}$ ', $i_{2}$ ', $\ldots, i_{n}$ ' in the coils $L_{1}, L_{2}, \ldots, L_{n}$ be, in order to guide the proton (initially accelerated with the potential difference $-U$ ) along the same trajectory (and in the same direction) as that of the electron?

HINT: The problem can be solved by finding a condition under which the equation describing the trajectory is the same in both cases. It may be helpful to use the relation:

$$
\boldsymbol{p} \frac{d}{d t} \boldsymbol{p}=\frac{1}{2} \frac{d}{d t} \boldsymbol{p}^{2}=\frac{1}{2} \frac{d}{d t} p^{2}
$$

Solution:
PART A

At the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference $U=511 \mathrm{kV}$ equals to its rest energy $E_{0}$. Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws.

The relativistic equation of motion of a particle with the charge $e$ in the magnetic field B has the following form:

$$
\frac{d}{d t} \mathbf{p}=\mathbf{F}_{L}
$$

where $\mathbf{p}=m_{0} \gamma \mathbf{v}$ denotes the momentum of the particle (vector) and

$$
\mathbf{F}_{L}=e \mathbf{v} \times \mathbf{B}
$$

is the Lorentz force (its value is $e v B$ and its direction is determined with the right hand rule). $m_{0}$ denotes the (rest) mass of the particle and $v$ denotes the velocity of the particle. The quantity $\gamma$ is given by the formula:

$$
\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
$$

The Lorentz force $\mathbf{F}_{L}$ is perpendicular to the velocity $\mathbf{v}$ of the particle and to its momentum $\mathbf{p}=m_{0} \boldsymbol{\gamma}$. Hence,

$$
\mathbf{F}_{L} \cdot \mathbf{v}=\mathbf{F}_{L} \cdot \mathbf{p}=0
$$

Multiplying the equation of motion by $\mathbf{p}$ and making use of the hint given in the text of the problem, we get:

$$
\frac{1}{2} \frac{d}{d t} p^{2}=0
$$



It means that the value of the particle momentum (and the value of the velocity) is constant during the motion:

$$
p=m_{0} v \gamma=\text { const; } \quad v=\text { const. }
$$

The same result can be obtained without any formulae in the following way:

The Lorentz force $\mathbf{F}_{L}$ is perpendicular to the velocity $\mathbf{v}$ (and to the momentum $p$ as $\left.\mathbf{p}=m_{0} \mathcal{W}\right)$ and, as a consequence, to the trajectory of the particle. Therefore, there is no force that could change the component of the momentum tangent to the trajectory. Thus, this component, whose value is equal to the length of $\mathbf{p}$, should be constant: $\mathbf{p}=$ const. (The same refers to the component of the velocity tangent to the trajectory as $\mathbf{p}=m_{0} \mathcal{W}$ ).

Let $s$ denotes the path passed by the particle along the trajectory. From the definition of the velocity, we have:

$$
\frac{d s}{d t}=v
$$

Using this formula, we can rewrite the equation of motion as follows:

$$
\begin{gathered}
v \frac{d}{d s} \mathbf{p}=\frac{d s}{d t} \frac{d}{d s} \mathbf{p}=\frac{d}{d t} \mathbf{p}=\mathbf{F}_{L}, \\
\frac{d}{d s} \mathbf{p}=\frac{\mathbf{F}_{L}}{v} .
\end{gathered}
$$

Dividing this equation by $p$ and making use of the fact that $p=$ const, we obtain:

$$
v \frac{d}{d s} \frac{\mathbf{p}}{p}=\frac{\mathbf{F}_{L}}{v p}
$$

and hence

$$
\frac{d}{d s} \mathbf{t}=\frac{\mathbf{F}_{L}}{v p}
$$

where $\mathbf{t}=\mathbf{p} / p=\mathbf{v} / v$ is the versor (unit vector) tangent to the trajectory. The above equation is exactly the same for both electrons and protons if and only if the vector quantity:

$$
\frac{\mathbf{F}_{L}}{v p}
$$

is the same in both cases.

Denoting corresponding quantities for protons with the same symbols as for the electrons, but with primes, one gets that the condition, under which both electrons and protons can move along the same trajectory, is equivalent to the equality:

$$
\frac{\mathbf{F}_{L}}{v p}=\frac{\mathbf{F}_{L}^{\prime}}{v^{\prime} p^{\prime}}
$$

However, the Lorentz force is proportional to the value of the velocity of the particle, and the directions of any two vectors of the following three: $\mathbf{t}$ (or $\mathbf{v}$ ), $\mathbf{F}_{\mathrm{L}}, \mathbf{B}$ determine the direction of the third of them (right hand rule). Therefore, the above condition can be written in the following form:

$$
\frac{e \mathbf{B}}{p}=\frac{e^{\prime} \mathbf{B}^{\prime}}{p^{\prime}}
$$



Hence,

$$
\mathbf{B}^{\prime}=\frac{e}{e^{\prime}} \frac{p^{\prime}}{p} \mathbf{B}=\frac{p^{\prime}}{p} \mathbf{B}
$$

This means that at any point the direction of the field $\mathbf{B}$ should be conserved, its orientation should be changed into the opposite one, and the value of the field should be multiplied by the same factor $p^{\prime} / p$. The magnetic field $\mathbf{B}$ is a vector sum of the magnetic fields of the coils that are arbitrarily distributed in the space. Therefore, each of this fields should be scaled with the same factor $-p^{\prime} / p$. However, the magnetic field of any coil is proportional to the current flowing in it. This means that the required scaling of the fields can only be achieved by the scaling of all the currents with the same factor $-p^{\prime} / p$ :

$$
i_{n}^{\prime}=-\frac{p^{\prime}}{p} i_{n}
$$

Now we shall determine the ratio $p^{\prime} / p$. The kinetic energies of the particles in both cases are the same; they are equal to $E_{k}=e|U|=511 \mathrm{keV}$. The general relativistic relation between the total energy $E$ of the particle with the rest energy $E_{0}$ and its momentum $p$ has the following form:

$$
E^{2}=E_{0}^{2}+p^{2} c^{2}
$$

where $\mathrm{c}$ denotes the velocity of light. energies:

The total energy of considered particles is equal to the sum of their rest and kinetic

$$
E=E_{0}+E_{k}
$$

Using these formulae and knowing that in our case $E_{k}=e|U|=E_{e}$, we determine the momenta of the electrons $(p)$ and the protons $(p ')$. We get:

a) electrons:

$$
\begin{gathered}
\left(E_{e}+E_{e}\right)^{2}=E_{e}^{2}+p^{2} c^{2}, \\
p=\frac{E_{e}}{c} \sqrt{3} .
\end{gathered}
$$

b) protons

$$
\begin{gathered}
\left(E_{p}+E_{e}\right)^{2}=E_{p}^{2}+p^{\prime 2} c^{2}, \\
p^{\prime}=\frac{E_{e}}{c} \sqrt{\left(\frac{E_{p}}{E_{e}}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}} .
\end{gathered}
$$

Hence,

$$
\frac{p^{\prime}}{p}=\frac{1}{\sqrt{3}} \sqrt{\left(\frac{E_{p}}{E}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}} \approx 35.0
$$

and

$$
i_{n}^{\prime}=-35.0 i_{n} \text {. }
$$



It is worthwhile to notice that our protons are 'almost classical', because their kinetic energy $E_{k}\left(=E_{e}\right)$ is small compared to the proton rest energy $E_{p}$. Thus, one can expect that the momentum of the proton can be determined, with a good accuracy, from the classical considerations. We have:

$$
\begin{gathered}
E_{e}=E_{k}=\frac{p^{\prime 2}}{2 m_{p}}=\frac{p^{\prime 2} c^{2}}{2 m_{p} c^{2}}=\frac{p^{\prime 2} c^{2}}{2 E_{p}}, \\
p^{\prime}=\frac{1}{c} \sqrt{2 E_{e} E_{p}} .
\end{gathered}
$$

On the other hand, the momentum of the proton determined from the relativistic formulae can be written in a simpler form since $E_{\mathrm{p}} / E_{\mathrm{e}} \gg 1$. We get:

$$
p^{\prime}=\frac{E_{e}}{c} \sqrt{\left(\frac{E_{p}}{E_{e}}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}}=\frac{E_{e}}{c} \sqrt{2 \frac{E_{p}}{E_{e}}+1} \approx \frac{E_{e}}{c} \sqrt{2 \frac{E_{p}}{E_{e}}}=\frac{1}{c} \sqrt{2 E_{e} E_{p}} .
$$

In accordance with our expectations, we have obtained the same result as above.

Final answer: $i_{n}^{\prime}=-35.0 i_{n}$