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I am currently working with an appropriate Lagrangian $L=\frac{1}{2m}(m\overrightarrow{v}+\frac{q}{c}\overrightarrow{A})^2-(\frac{q^2}{2mC^2})\overrightarrow{A}\cdot \overrightarrow{A}-q\phi$

I am trying to apply the Euler Langrange equation to this Lagrangian.

$\frac{d}{dt}(\frac{dL}{dv_x})-\frac{dL}{dx}=0$

So far I have managed to determine the first factor like so

$\frac{d}{dt}(\frac{dL}{dv_x})=m\frac{dv_X}{dt}+\frac{q}{c}(v_x\frac{dA_x}{dx}+v_y\frac{dA_x}{dy}+v_z\frac{dA_x}{dz}+\frac{dA_x}{dt})$

and this eventually gives

$\frac{d}{dt}(\frac{dL}{dv_x})=\frac{d}{dt}(mv_x+\frac{q}{c}A_x)=m\frac{dv_x}{dt}+\frac{q}{c}\frac{dA_x}{dt}$

$\overrightarrow{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$

However I am having difficulty attempting the second factor $\frac{dL}{dx}$ could I have some help on how I go about solving this?

  • 1
    What does $\mathbf A$ and $\phi$ depend on? – 2017-02-23
  • 0
    I have added A's dependence and $\phi$ is simply a scalar potential so I think it vanishes in differential due to its constant nature – 2017-02-23

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