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I'm reading "Classical Mechanics" (5ed) by Berkshire and Kibble, in the example for uniform magnetic field on pg.243 (Chapter 10 Lagrangian Mechanics) I came across this

A charged particle moves in a uniform static magnetic field B. Find the solutions of the equations of motion in which ρ (axial radius, cylindrical coordinates) is constant.

For a uniform magnetic field, we may take $$\phi=0 \text{ and } \boldsymbol{A}=\frac{1}{2}\boldsymbol{B}\times\boldsymbol{r}$$

The authors did not explain where these come from and I cannot understand why such conditions are imposed.

I'm particularly confused about the first condition (scalar potential=0),neither of the four Maxwell's equations require $\phi=0$ for when $\partial_tB^i=0$.

Is this purely a choice or am I missing something?

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The authors are just helping you out. It should take only a few moments to verify that those choices for $\phi$ and $\mathbf A$ yield the correct electric and magnetic fields.

Electromagnetism exhibits gauge invariance, so there are an infinity of other choices of $\phi$ and $\mathbf A$ which would also yield the correct fields, but the one they give you is simple and convenient.

If you want to be less convenient, pick any scalar function $\chi(x,t)$ and add $\frac{\partial \chi}{\partial t}$ to $\phi$ and subtract $\nabla \chi$ from $\mathbf A$. These potentials correspond to the same electric and magnetic fields as before so your final answer would be the same, but in the absence of additional motivation there's no reason to do this to yourself.

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  • $\begingroup$ I understand that $\phi=0 \to \boldsymbol{E}=0$, $\boldsymbol{B}=\frac{1}{2}\boldsymbol{A \times r}$ $\to$ $\partial_t\boldsymbol{B}=0$ but how can I work out $\boldsymbol{A}$ from $\boldsymbol{B}$ ($\phi=0$ only concerns the electric field?), also I'm quite new to EM. $\endgroup$
    – Chern-Simons
    Commented Feb 22, 2021 at 8:44
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    $\begingroup$ @ChernSimons Firstly, $\mathbf E = -\nabla \phi - \frac{\partial}{\partial t} \mathbf A$, so it generally involves both the scalar and vector potentials. Secondly, $\mathbf B = \nabla \times \mathbf A$, not $\frac{1}{2}\mathbf A \times \mathbf r$. If you use the $\phi$ and $\mathbf A$ that the authors recommend, you can calculate in a few lines that $\mathbf E=0$ and $\nabla \times \mathbf A = \mathbf B$ (where $\mathbf B$ is taken to be a constant). Therefore, this choice of $\phi$ and $\mathbf A$ produces the fields you want. $\endgroup$
    – J. Murray
    Commented Feb 22, 2021 at 13:49
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    $\begingroup$ @ChernSimons There is no unique way to determine $\mathbf A$ from $\mathbf B$ because as I said, there are an infinite number of choices of $\phi$ and $\mathbf A$ which produce exactly the same electric and magnetic fields. What one does is make a choice of gauge-fixing condition (like $\nabla \cdot \mathbf A=0$, or $\nabla \cdot \mathbf A = \partial \phi/\partial t$) to narrow down this freedom and then use the Maxwell equations to calculate $\mathbf A$. This is all covered in any intro E&M text, but in this case you don't need it; the authors provide a valid choice of potentials for you. $\endgroup$
    – J. Murray
    Commented Feb 22, 2021 at 13:54

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