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Given an electric charge $q$ of mass $m$ moving at a velocity ${\bf v}$ in a region containing both electric field ${\bf E}(t,x,y,z)$ and magnetic field ${\bf B}(t,x,y,z)$ (${\bf B}$ and ${\bf E}$ are derivable from a scalar potential $\phi(t, x, y, z) $and a vector potential ${\bf A}(t,x,y,z)$),

knowing that

  1. ${\bf E}=- \nabla \phi - \frac{\partial {\bf A}} {\partial t}$
  2. ${\bf B}= \nabla \times {\bf A} $
  3. $U=q \phi - q {\bf A} \cdot{\bf v}$ ($U$ is the velocity-dependent potential)

the Lagrangian is $$L=(1/2) m v^2-U=(1/2) m v^2- q\phi + q{\bf A} \cdot{\bf v}.$$

Considering just the $x$-component of Lagrange's equation, how can I obtain $$m \ddot{x}=q\left (v_x \frac{\partial A_x}{\partial x} + v_y \frac {\partial A_y}{ \partial x} + v_z \frac{\partial A_z} {\partial x}\right )-q\left (\frac{\partial \phi }{\partial x} + \frac{d A_x}{dt}\right) ~?$$

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2 Answers 2

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Use the Euler-Lagrange equations (motivated in a previous post of mine). These are

$$ \partial_x L - \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = 0 $$

We therefore need $$ \partial_x L = -q \partial_x \Phi + q \left( \partial_x A_x v_x + \partial_x A_y v_y + \partial_x A_z v_z \right) $$ $$ \partial_{v_x} L = m v_x + q A_x $$ $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = m \dot v_x + q \dot A_x $$

Putting this all together into the first equation and adding $\dot v_x m$ gives the equation for which you are looking.

Note especially that each component of $A$ is a function of all three coordinates: $A_i = A_i(x,y,z)$; the same goes for $\Phi = \Phi(x,y,z)$ (where usually $A_0$ is identified with $\Phi$ in relativistic electrodynamics).

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  • $\begingroup$ Thank you :) How can I express $\dot{A_x}$? I have some questions.. 1) $A(t,x, y,z)$ returns a 4-dimensional vector and each component of it is function of x, y,z (Is time component function of x,y,z?) 2)$\phi(t,x,y,z)$ returns a number, a 1-dimensional vector that has only one component, function of x, y,z (and t?) .. thanks again $\endgroup$
    – sunrise
    Oct 14, 2012 at 10:03
  • $\begingroup$ $\dot A_x$ is simply $\frac{\textrm{d}}{\textrm{d}t} A_x(x,y,z,t)$. @1: Are you sure that your $A$ is four-dimensional rather than 3-dimensional? Your usage of $\Phi$ independently of $A$ indicates that $A$ is only three-dimensional (because oterwise $\Phi$ would be one of the components of $A$). @2: $\Phi(x,y,z,t)$ is indeed a scalar (1-dimensional). Whether or not it depends on $t$ depends on the problem at hand, in general, it does depend on $t$. However, this is not relevant for the Euler-Lagrange equations as we don’t have to partially derive $\Phi$ with regard to $t$. $\endgroup$
    – Claudius
    Oct 14, 2012 at 10:20
  • $\begingroup$ I know that I said some foolish things ;) but I couldn't write here before... Thanks for your answers! $\endgroup$
    – sunrise
    Oct 16, 2012 at 20:47
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Use the Euler-Lagrange equation:

$$\frac{\partial L}{\partial x} = \frac{d}{dt}\frac{\partial L}{\partial v_x}$$ $$-q \frac{\partial \phi}{\partial x} + q\frac{\partial}{\partial x}\left(\mathbf{A} \cdot \mathbf{v}\right) = m \dot{v_x} + q \frac{dA_x}{dt}$$

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