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Diffstat (limited to 'content/know/concept/lorentz-force')
| -rw-r--r-- | content/know/concept/lorentz-force/index.pdc | 182 |
1 files changed, 65 insertions, 117 deletions
diff --git a/content/know/concept/lorentz-force/index.pdc b/content/know/concept/lorentz-force/index.pdc index 2362766..83357d2 100644 --- a/content/know/concept/lorentz-force/index.pdc +++ b/content/know/concept/lorentz-force/index.pdc @@ -5,6 +5,7 @@ publishDate: 2021-09-08 categories: - Physics - Electromagnetism +- Plasma physics date: 2021-09-08T17:00:32+02:00 draft: false @@ -27,6 +28,44 @@ $$\begin{aligned} \end{aligned}$$ +## Uniform electric field + +Consider the simple case of an electric field $\vb{E}$ +that is uniform in all of space. +In the absence of a magnetic field $\vb{B} = 0$ +and any other forces, +Newton's second law states: + +$$\begin{aligned} + \vb{F} + = m \dv{\vb{u}}{t} + = q \vb{E} +\end{aligned}$$ + +This is straightforward to integrate in time, +for a given initial velocity vector $\vb{u}_0$: + +$$\begin{aligned} + \vb{u}(t) + = \frac{q}{m} \vb{E} t + \vb{u}_0 +\end{aligned}$$ + +And then the particle's position $\vb{x}(t)$ +is found be integrating once more, +with $\vb{x}(0) = \vb{x}_0$: + +$$\begin{aligned} + \boxed{ + \vb{x}(t) + = \frac{q}{2 m} \vb{E} t^2 + \vb{u}_0 t + \vb{x}_0 + } +\end{aligned}$$ + +In summary, unsurprisingly, a uniform electric field $\vb{E}$ +accelerates the particle with a constant force $\vb{F} = q \vb{E}$. +Note that the direction depends on the sign of $q$. + + ## Uniform magnetic field Consider the simple case of a uniform magnetic field @@ -64,12 +103,12 @@ $$\begin{aligned} \end{aligned}$$ Where we have defined the **cyclotron frequency** $\omega_c$ as follows, -which is always positive: +which may be negative: $$\begin{aligned} \boxed{ \omega_c - \equiv \frac{|q| B}{m} + \equiv \frac{q B}{m} } \end{aligned}$$ @@ -78,7 +117,7 @@ the solution for $u_x(t)$ is given by: $$\begin{aligned} u_x(t) - = - u_\perp \sin\!(\omega_c t) + = u_\perp \cos\!(\omega_c t) \end{aligned}$$ Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity. @@ -87,13 +126,12 @@ Then $u_y(t)$ is found to be: $$\begin{aligned} u_y(t) = \frac{m}{q B} \dv{u_x}{t} - = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t) - = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t) + = - \frac{m \omega_c}{q B} u_\perp \sin\!(\omega_c t) + = - u_\perp \sin\!(\omega_c t) \end{aligned}$$ -Where $\mathrm{sgn}$ is the signum function. -This tells us that the particle moves in a circular orbit, -and that the direction of rotation is determined by $q$. +This means that the particle moves in a circle, +in a direction determined by the sign of $\omega_c$. Integrating the velocity yields the position, where we refer to the integration constants $x_{gc}$ and $y_{gc}$ @@ -101,10 +139,10 @@ as the **guiding center**, around which the particle orbits or **gyrates**: $$\begin{aligned} x(t) - = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc} + = \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + x_{gc} \qquad \quad y(t) - = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc} + = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + y_{gc} \end{aligned}$$ The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by: @@ -112,16 +150,18 @@ The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_ $$\begin{aligned} \boxed{ r_L - \equiv \frac{u_\perp}{\omega_c} + \equiv \frac{u_\perp}{|\omega_c|} = \frac{m u_\perp}{|q| B} } \end{aligned}$$ -Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$: +Finally, it is easy to integrate the equation +for the $z$-axis velocity $u_z$, which is conserved: $$\begin{aligned} z(t) - = u_z t + z_{gc} + = z_{gc} + = u_z t + z_0 \end{aligned}$$ In conclusion, the particle's motion parallel to $\vb{B}$ @@ -129,116 +169,24 @@ is not affected by the magnetic field, while its motion perpendicular to $\vb{B}$ is circular around an imaginary guiding center. The end result is that particles follow a helical path -when moving through a uniform magnetic field. - - -## Uniform electric and magnetic field - -Let us now consider a more general case, -with constant uniform electric and magnetic fields $\vb{E}$ and $\vb{B}$, -which may or may not be perpendicular. -The equation of motion is then: - -$$\begin{aligned} - \vb{F} - = m \dv{\vb{u}}{t} - = q (\vb{E} + \vb{u} \cross \vb{B}) -\end{aligned}$$ - -If we take the dot product with the unit vector $\vu{B}$, -the cross product vanishes, leaving: +when moving through a uniform magnetic field: $$\begin{aligned} - \dv{\vb{u}_\parallel}{t} - = \frac{q}{m} \vb{E}_\parallel -\end{aligned}$$ - -Where $\vb{u}_\parallel$ and $\vb{E}_\parallel$ are -the components of $\vb{u}$ and $\vb{E}$ -that are parallel to $\vb{B}$. -This equation is easy to integrate: -the guiding center accelerates according to $(q/m) \vb{E}_\parallel$. - -Next, let us define the perpendicular component $\vb{u}_\perp$ -such that $\vb{u} = \vb{u}_\parallel \vu{B} + \vb{u}_\perp$. -Its equation of motion is found by -subtracting $\vb{u}_\parallel$'s equation from the original: - -$$\begin{aligned} - m \dv{\vb{u}_\perp}{t} - = q (\vb{E} + \vb{u} \cross \vb{B}) - q \vb{E}_\parallel - = q (\vb{E}_\perp + \vb{u}_\perp \cross \vb{B}) -\end{aligned}$$ - -To solve this, we go to a moving coordinate system -by defining $\vb{u}_\perp = \vb{v}_\perp + \vb{w}_\perp$, -where $\vb{v}_\perp$ is a constant of our choice. -The equation is now as follows: - -$$\begin{aligned} - m \dv{t} (\vb{v}_\perp + \vb{w}_\perp) - = m \dv{\vb{w}_\perp}{t} - = q (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B} + \vb{w}_\perp \cross \vb{B}) -\end{aligned}$$ - -We want to choose $\vb{v}_\perp$ such that the first two terms vanish, -or in other words: - -$$\begin{aligned} - 0 - = \vb{E}_\perp + \vb{v}_\perp \cross \vb{B} -\end{aligned}$$ - -To find $\vb{v}_\perp$, we take the cross product with $\vb{B}$, -and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$: - -$$\begin{aligned} - 0 - = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B}) - = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2 - \quad \implies \quad \boxed{ - \vb{v}_\perp - = \frac{\vb{E} \cross \vb{B}}{B^2} + \vb{x}(t) + = \frac{u_\perp}{\omega_c} + \begin{pmatrix} + \sin\!(\omega_c t) \\ \cos\!(\omega_c t) \\ 0 + \end{pmatrix} + + \vb{x}_{gc}(t) } \end{aligned}$$ -When $\vb{v}_\perp$ is chosen like this, -the perpendicular equation of motion is reduced to: - -$$\begin{aligned} - m \dv{\vb{w}_\perp}{t} - = q \vb{w}_\perp \cross \vb{B} -\end{aligned}$$ - -Which is simply the case we treated previously with $\vb{E} = 0$, -with a known solution -(assuming $\vb{B}$ still points in the positive $z$-direction): - -$$\begin{aligned} - w_x(t) - = - w_\perp \sin\!(\omega_c t) - \qquad - w_y(t) - = - \mathrm{sgn}(q) \: w_\perp \cos\!(\omega_c t) -\end{aligned}$$ - -However, this result is shifted by a constant $\vb{v}_\perp$, -often called the **drift velocity** $\vb{v}_d$, -at which the guiding center moves transversely. -Curiously, $\vb{v}_d$ is independent of $q$. - -Such a drift is not specific to an electric field. -In the equations above, $\vb{E}$ can be replaced -by a general force $\vb{F}/q$ (e.g. gravity) without issues. -In that case, $\vb{v}_d$ does depend on $q$: - -$$\begin{aligned} - \boxed{ - \vb{v}_d - = \frac{\vb{F} \cross \vb{B}}{q B^2} - } -\end{aligned}$$ +Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$ +is the position of the guiding center. +For a detailed look at how $\vb{B}$ and $\vb{E}$ +can affect the guiding center's motion, +see [guiding center theory](/know/concept/guiding-center-theory/). |