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un:em-gyration [2024/10/12 08:53] – asadun:em-gyration [2024/10/20 09:54] (current) – asad
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 ====== Electromagnetic gyration ====== ====== Electromagnetic gyration ======
  
-Gyration (helical rotation) of a particle of charge $q$ inside an electric ($\mathbf{E}$) and magnetic ($\mathbf{B}$) field is due to the combined effect of the Coulomb force and the Lorentz force.((The main reference for this article is Baumjohann & Treumann, //Basic Space Plasma Physics//, Imperial College Press, 1999.)) The equation of motion for this particle can be written as:+The gyration (helical rotation) of a particle with charge $q$ within electric ($\mathbf{E}$) and magnetic ($\mathbf{B}$) fields occurs due to the combined effect of Coulomb force and Lorentz force.((The primary reference for this article is Baumjohann & Treumann, //Basic Space Plasma Physics//, Imperial College Press, 1999.)) The equation of motion for this particle can be expressed as:
  
 $$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$ $$ m\frac{d\mathbf{v}}{dt} = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$
  
-where $m$ is the mass of the particle and $\mathbf{v}$ is its velocity. If there is no electric field, this equation will contain only the Lorentz force part. Then taking the [[vector|dot product]] of the velocity with the two sides gives:+where $m$ is the mass of the particle and $\mathbf{v}$ is its velocity. In the absence of an electric field, only the Lorentz force term remains in this equation. By taking the [[vector|dot product]] of both sides with the velocity, we obtain:
  
-$$ m\frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = q (\mathbf{v}\times\mathbf{B}) \cdot \mathbf{v} \ \Rightarrow \ frac{d}{dt}\left(\frac{1}{2}mv^2\right) = 0 $$+$$ m\frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = q (\mathbf{v}\times\mathbf{B}) \cdot \mathbf{v} \ \Rightarrow \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = 0 $$
  
-Because $\mathbf{v}\cdot (\mathbf{v}\times\mathbf{B})=0$ and both sides are divided by two. That means the kinetic energy ($mv^2/2$) and the magnitude of the velocity (speed) of this particle are both constant. A static magnetic field can never change the kinetic energy of a particle.+Since $\mathbf{v}\cdot (\mathbf{v}\times\mathbf{B})=0$ and both sides are divided by 2, it implies that the kinetic energy ($mv^2/2$) and the magnitude of the velocity (speed) remain constant. A static magnetic field can never change the particle’s kinetic energy.
  
-In a constant magnetic field along the z-axis, B=Bhatk and the three components in the equation of motion are then+If the magnetic field is aligned along the $z$-axis, $\mathbf{B}=B\hat{k}$, the components of the equation of motion become:
  
 \begin{align*} \begin{align*}
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 \end{align*} \end{align*}
  
-where $\dot{v}_x=dv_x/dt$ is the first derivative; Same for three components. The last equation states that the $z$-component of the velocity in the direction parallel to the magnetic field is constant. Differentiating the first equation again gives the second derivative like this.+where $\dot{v}_x=dv_x/dt$ represents the first derivative. The same holds for the other components. The last equation states that the $z$-component of the velocity, parallel to the magnetic field, remains constant. Differentiating the first equation yields the second derivative:
  
 \begin{align*} \begin{align*}
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 \end{align*} \end{align*}
  
-where $\omega_g=(qB/m$) is the gyrofrequency or cyclotron frequency, whose sign is opposite for positive and negative charges. Similarly it can be shown that $\ddot{v}_y = -\omega^2_g v_y$ for the other component. These two second derivative equations are actually harmonic oscillator equations whose solutions are:+where $\omega_g=(qB/m)$ is the gyrofrequency or cyclotron frequency, whose sign is opposite for positive and negative charges. Similarly, it can be shown that $\ddot{v}_y = -\omega^2_g v_y$ for the other component. These second-order differential equations are those of a harmonic oscillator, with solutions:
  
 \begin{align} \begin{align}
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 \end{align} \end{align}
  
-where the sign of the two components of displacement will be opposite for electrons and ions. and gyroradius+The sine terms in the displacement components will be opposite for electrons and ions. The gyro-radius is given by:
  
 $$ r_g = \frac{v_\perp}{|\omega_g|} = \frac{mv_\perp}{|q|B} $$ $$ r_g = \frac{v_\perp}{|\omega_g|} = \frac{mv_\perp}{|q|B} $$
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 where $v_\perp = (v_x^2+v_y^2)^{1/2}$ is the constant speed in the plane perpendicular to the magnetic field. where $v_\perp = (v_x^2+v_y^2)^{1/2}$ is the constant speed in the plane perpendicular to the magnetic field.
  
-Now the trajectory obtained according to the equations $\ref{ho1}$ and $\ref{ho2}$ is actually a circular orbit. The center of this orbit $(x_0,y_0)$ is called the guiding center. Around this center there is a circular current along the orbit. The internal magnetic field created by this current acts in opposition to the external magnetic field that creates this current. Its name is diamagnetic effect.+According to equations $\ref{ho1}$ and $\ref{ho2}$, the particle follows a circular orbit. The center of this orbit, $(x_0,y_0)$, is known as the guiding center. A circular current flows along the orbit, and the internal magnetic field generated by this current opposes the external magnetic field that created it. This phenomenon is known as the diamagnetic effect.
  
-If the particle's velocity has no component parallel to the magnetic field, the particle will move in a circular path. But if there is a component in parallel ( $v_z$ or $v_\parallel$ ), then the particle trajectory will be twisted like a helix as shown below.+If the particle has no velocity component parallel to the magnetic field, it will continue to move in a circular path. However, if a parallel component exists ($v_z$ or $v_\parallel$), the particle will follow a helical trajectory, as illustrated below.
  
 {{:bn:un:helical.png?nolink&500|}} {{:bn:un:helical.png?nolink&500|}}
  
-The distance from one orbit to another in the direction of the magnetic field is called the pitch, which we usually measure by the pitch angle ($\alpha$):+The distance between successive orbits along the magnetic field direction is called the pitch, usually measured by the pitch angle ($\alpha$):
  
 $$ \alpha = \tan^{-1} \frac{v_\perp}{v_\parallel} $$ $$ \alpha = \tan^{-1} \frac{v_\perp}{v_\parallel} $$
  
-That is, the pitch angle depends on the ratio of the two components of the particle's velocity.+Thus, the pitch angle depends on the ratio of the two velocity components.
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