Mathematical Model of the Motion of a Charged Particle in Matter in the Presence of a Magnetic Field
Abstract Generated abstract
This paper formulates equations for the motion of relativistic charged particles, especially protons, in matter under an external magnetic field, motivated by calculations of combined radiation shielding. Starting from the Lorentz force equation, it adds an effective resistance term representing ionization energy loss along the trajectory and derives the corresponding three-dimensional relativistic equations in Cartesian coordinates. The equations are algebraically transformed to solve explicitly for the highest derivatives, yielding a form suitable for analog computer modeling, with the nonrelativistic limit given as a special case. A cylindrical-coordinate version and relations between particle kinetic energy and velocity are also presented for use in computational modeling.
Full Text
Mathematical Model of the Process of Motion of a Charged Particle in Matter in the Presence of a Magnetic Field
V. M. Razin
(Presented by the scientific seminar of the Department of Computer Engineering)
In calculations of active biological protection, it is useful to obtain the equations of motion of relativistic protons in matter in the presence of a magnetic field, since combined methods of shielding may prove optimal in cases where the weight of protective screens must be bounded from above.
As is known [1], the equation of motion of a charged particle in a magnetic field in the absence of matter, in vector form, has the form:
\[ m\frac{d\vec v}{dt}=\frac{e}{c}[\vec v\vec B], \tag{1} \]
where \(m\) is the mass of the particle; \(e\) is the charge; \(c\) is the speed of light in vacuum; \(\vec v\) is the velocity vector; \(\vec B\) is the magnetic induction vector.
Since we shall be interested in particles whose velocities are comparable with the speed of light, equation (1) should be written in the form of the relation
\[ \frac{d\vec p}{dt}=\frac{e}{c}[\vec v\vec B], \tag{2} \]
where \(\vec p=m\vec v\) is the momentum of the particle, whose mass \(m\) depends on the energy, i.e., on the velocity of the particle.
When a particle passes through matter, various interactions of the particle with the medium will occur, as a result of which the energy of the particle will decrease along the trajectory of motion. In the energy range of protons of interest to us, the main losses are energy losses due to ionization of atoms of the shielding medium. Energy losses due to ionization will be equivalent to the action of a certain force acting on the particle (proton) in the direction opposite to the direction of the velocity vector. The magnitude of this force will be equal to the specific ionization energy per element of trajectory length \(l\), i.e., the force is equal to \(\dfrac{dE}{dl}\).
Taking into account the resistance force of the matter due to ionization, equation (2) can be represented in vector form as:
\[ \frac{d\vec p}{dt}=\frac{e}{c}[\vec v\vec B]-\left|\frac{dE}{dl}\right|\frac{\vec v}{v}, \tag{3} \]
where the square brackets denote the absolute value of the corresponding quantity, and \(v\) is the modulus of the particle velocity vector.
In the relativistic case, the left-hand side of equation (3) must be represented as follows:
\[ \frac{d\vec p}{dt} = \frac{d}{dt}(m\vec v) = \frac{d}{dt} \left( \frac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}\,\vec v \right), \tag{4} \]
where \(m_0\) is the rest mass of the particle.
In a three-dimensional rectangular coordinate system the relation holds
\[ v^2=\dot x^2+\dot y^2+\dot z^2, \tag{5} \]
where, for brevity of notation, the first derivatives with respect to time are denoted by dots over the corresponding coordinates.
Projection of the vector equation (4) onto the coordinate axes, taking (5) into account, makes it possible to write the equation of motion in the following form:
\[ \left. \begin{aligned} \frac{d}{dt} \left( \frac{m_0}{\sqrt{1-\dfrac{\dot x^2+\dot y^2+\dot z^2}{c^2}}}\,\dot x \right) &= \frac{e}{c}B_z\dot y-\frac{e}{c}B_y\dot z-\left|\frac{dE}{dl}\right|\frac{\dot x}{v}; \\[6pt] \frac{d}{dt} \left( \frac{m_0}{\sqrt{1-\dfrac{\dot x^2+\dot y^2+\dot z^2}{c^2}}}\,\dot y \right) &= \frac{e}{c}B_x\dot z-\frac{e}{c}B_z\dot x-\left|\frac{dE}{dl}\right|\frac{\dot y}{v}; \\[6pt] \frac{d}{dt} \left( \frac{m_0}{\sqrt{1-\dfrac{\dot x^2+\dot y^2+\dot z^2}{c^2}}}\,\dot z \right) &= \frac{e}{c}B_y\dot x-\frac{e}{c}B_x\dot y-\left|\frac{dE}{dl}\right|\frac{\dot z}{v}. \end{aligned} \right\} \tag{6} \]
Here \(B_x\), \(B_y\), and \(B_z\) are the components of the magnetic induction vector along the coordinate axes.
For brevity of notation, introduce the designation
\[ \varphi(v)=\left|\frac{dE}{dl}\right|\frac{1}{v} \tag{7} \]
and represent the right-hand sides of the equations of system (6) in the form
\[ \left. \begin{aligned} -\varphi(v)\dot x+\frac{e}{c}B_z\dot y-\frac{e}{c}B_y\dot z&=L; \\[4pt] -\frac{e}{c}B_z\dot x-\varphi(v)\dot y+\frac{e}{c}B_x\dot z&=M; \\[4pt] \frac{e}{c}B_y\dot x-\frac{e}{c}B_x\dot y-\varphi(v)\dot z&=N. \end{aligned} \right\} \tag{8} \]
Expanding the derivatives in the left-hand sides of system (6) and taking (8) into account, we obtain
\[ \frac{m_0}{\sqrt{1-\beta^2}}\,\ddot x + \dot x\,\frac{d}{dt} \left( \frac{m_0}{\sqrt{1-\dfrac{\dot x^2+\dot y^2+\dot z^2}{c^2}}} \right) = L; \]
\[ \left. \begin{aligned} \frac{m_0}{\sqrt{1-\beta^2}}\,\ddot{y} +\dot{y}\frac{d}{dt}\left( \frac{m_0}{\sqrt{1-\dfrac{\dot{x}^2+\dot{y}^2+\dot{z}^2}{c^2}}} \right)&=M;\\[6pt] \frac{m_0}{\sqrt{1-\beta^2}}\,\ddot{z} +\dot{z}\frac{d}{dt}\left( \frac{m_0}{\sqrt{1-\dfrac{\dot{x}^2+\dot{y}^2+\dot{z}^2}{c^2}}} \right)&=N. \end{aligned} \right\} \tag{9} \]
Here \(\beta=\dfrac{v}{c}=\dfrac{\sqrt{\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}}}{c}\) is the relativistic factor, and two dots denote the second derivative with respect to time. Expanding the derivative,
\[ \frac{d}{dt}\left( \frac{m_0}{\sqrt{1-\dfrac{\dot{x}^2+\dot{y}^2+\dot{z}^2}{c}}} \right) = \frac{m_0}{c^2}\, \frac{1}{\sqrt{(1-\beta^2)^3}}\, (\dot{x}\ddot{x}+\dot{y}\ddot{y}+\dot{z}\ddot{z}), \tag{10} \]
since all coordinates are functions of time. Substituting (10) into the system of equations (9) and carrying out simple transformations, we obtain
\[ \left. \begin{aligned} \ddot{x}\,[c^2(1-\beta^2)+\dot{x}^2]+\dot{x}\dot{y}\ddot{y}+\dot{x}\dot{z}\ddot{z}&=L_1,\\ \dot{x}\dot{y}\ddot{x}+\ddot{y}\,[c^2(1-\beta^2)+\dot{y}^2]+\dot{y}\dot{z}\ddot{z}&=M_1,\\ \dot{x}\dot{z}\ddot{x}+\dot{y}\dot{z}\ddot{y}+[c^2(1-\beta^2)+\dot{z}^2]\ddot{z}&=N_1. \end{aligned} \right\} \tag{11} \]
Here, for brevity of notation, the following designations have been introduced:
\[ \left. \begin{aligned} L_1&=\frac{c^2\sqrt{(1-\beta^2)^3}}{m_0}\,L;\\[4pt] M_1&=\frac{c^2\sqrt{(1-\beta^2)^3}}{m_0}\,M;\\[4pt] N_1&=\frac{c^2\sqrt{(1-\beta^2)^3}}{m_0}\,N. \end{aligned} \right\} \tag{12} \]
In modeling ordinary differential equations on analog computing machines (ACM), it is customary to solve these equations with respect to the highest derivatives (the method of lowering the order of the derivative).
We consider system (11) as a system of three equations with three unknown derivatives \(\ddot{x}\), \(\ddot{z}\), \(\ddot{y}\).
We compute the principal determinant of system (11)
\[ \Delta= \begin{vmatrix} c^2(1-\beta^2)+\dot{x}^2 & \dot{x}\dot{y} & \dot{x}\dot{z}\\ \dot{x}\dot{y} & c^2(1-\beta^2)+\dot{y}^2 & \dot{y}\dot{z}\\ \dot{x}\dot{z} & \dot{y}\dot{z} & c^2(1-\beta^2)+\dot{z}^2 \end{vmatrix} \tag{13} \]
Applying Cramer’s rule, or any other method of computing determinants, and taking (5) into account, after simple but cumbersome transformations we obtain
\[ \Delta=c^6(1-\beta^2)^2. \tag{14} \]
Now the expression, for example, for \(\ddot{x}\) will have the form
\[ \ddot{x}=\frac{1}{\Delta} \begin{vmatrix} L_1; & \dot{x}\dot{y}; & \dot{x}\dot{z};\\ M_1; & [c^2(1-\beta^2)+\dot{y}^{\,2}]; & \dot{y}\dot{z};\\ N_1; & \dot{y}\dot{z}; & [c^2(1-\beta^2)+\dot{z}^{\,2}] \end{vmatrix}. \tag{15} \]
Cumbersome algebraic transformations in the calculation of (15) make it possible, in the end, to obtain a comparatively simple expression for \(\ddot{x}\)
\[ \ddot{x}=\frac{1}{m_0}\left(\frac{e}{c}B_z\dot{y}-\frac{e}{c}B_y\dot{z}\right)\sqrt{1-\beta^2} -\frac{1}{m_0}\varphi(v)\dot{x}\sqrt{(1-\beta^2)^3}. \tag{16} \]
The expressions for \(\ddot{y}\) and \(\ddot{z}\) are computed analogously. As a result we obtain the system of equations of motion in the form:
\[ \left. \begin{aligned} m_0\ddot{x} &= \left(\frac{e}{c}B_z\dot{y}-\frac{e}{c}B_y\dot{z}\right)\sqrt{1-\beta^2} -\varphi(v)\dot{x}\sqrt{(1-\beta^2)^3};\\ m_0\ddot{y} &= \left(\frac{e}{c}B_x\dot{z}-\frac{e}{c}B_z\dot{x}\right)\sqrt{1-\beta^2} -\varphi(v)\dot{y}\sqrt{(1-\beta^2)^3};\\ m_0\ddot{z} &= \left(\frac{e}{c}B_y\dot{x}-\frac{e}{c}B_x\dot{y}\right)\sqrt{1-\beta^2} -\varphi(v)\dot{z}\sqrt{(1-\beta^2)^3}. \end{aligned} \right\} \tag{17} \]
For \(\beta=0\) we obtain the equations of motion of a charged particle in matter in the presence of a magnetic field for the nonrelativistic case
\[ \left. \begin{aligned} m_0\ddot{x} &= \frac{e}{c}B_z\dot{y}-\frac{e}{c}B_y\dot{z}-\varphi(v)\dot{x};\\ m_0\ddot{y} &= \frac{e}{c}B_x\dot{z}-\frac{e}{c}B_z\dot{x}-\varphi(v)\dot{y};\\ m_0\ddot{z} &= \frac{e}{c}B_y\dot{x}-\frac{e}{c}B_x\dot{y}-\varphi(v)\dot{z}. \end{aligned} \right\} \tag{18} \]
The obtained equations of motion have a form convenient for modeling on an analog computer. In a cylindrical coordinate system, the equations of motion in the relativistic case will have the form:
\[ \left. \begin{aligned} m_0(\ddot{r}-r\dot{\varphi}^{\,2}) &= \frac{e}{c}(r\dot{\varphi}B_z-\dot{z}B_{\varphi})\sqrt{1-\beta^2} -\varphi(v)\dot{r}\sqrt{(1-\beta^2)^3};\\ m_0\frac{1}{r}\frac{d}{dt}(r^2\dot{\varphi}) &= \frac{e}{c}(\dot{z}B_r-\dot{r}B_z)\sqrt{1-\beta^2} -\varphi(v)r\dot{\varphi}\sqrt{(1-\beta^2)^3};\\ m_0\ddot{z} &= \frac{e}{c}(\dot{r}B_{\varphi}-r\dot{\varphi}B_r)\sqrt{1-\beta^2} -\varphi(v)\dot{z}\sqrt{(1-\beta^2)^3}. \end{aligned} \right\} \tag{19} \]
In modeling on an analog computer it is also useful to take into account the relations between the kinetic energy of the particle and its velocity, in the form
\[ \left. \begin{aligned} \beta=\frac{v}{c} &= \frac{\sqrt{E(E-2E_0)}}{E+E_0};\\ E &= \frac{m_0c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_0c^2 = \frac{E_0}{\sqrt{1-\dfrac{v^2}{c^2}}}. \end{aligned} \right\} \tag{20} \]
Here \(E_0=m_0c^2\) is the rest energy of the particle.