The $x$-component of $B$ is:

$B_x=\dfrac{\partial {A_z}}{\partial y}-\dfrac{\partial {A_y}}{\partial z} =\dfrac{\partial {(A_z+C_1)}}{\partial y}-\dfrac{\partial {(A_y+C_2)}}{\partial z}$

where $C_1$ and $C_2$ are constants.


$A_z$ can be replaced with $A_z+C_1$

$A_y$ can be replaced with $A_y+C_2$


$A_x$ can be replaced with $A_x+C_3$

Thus can we replace $(\vec{A})$ with $(\vec{A}+\vec{C})$?

where $\vec{C}$ is an arbitrary constant vector.

  • 2
    $\begingroup$ ...and the question is? It's unclear what you are confused about - you just showed that you can add a constant without changing the B-field, what do you want to know about that? $\endgroup$
    – ACuriousMind
    Sep 24, 2017 at 10:09

1 Answer 1


Yes, we can replace $\vec A(\vec x)$ with $\vec A'(\vec x) = \vec A(\vec x) + \vec C$. This is similar to electrostatics, where we can replace the electric potential $\Phi(\vec x)$ by $\Phi'(\vec x) = \Phi(\vec x) + c$, where $c$ is a constant energy shift.

What you found is an example for the gauge freedom that exists in magnetostatics: In general, replacing $$ \vec A(\vec x) \quad \rightarrow \quad \vec A'(\vec x) = \vec A(\vec x) + \nabla \chi(\vec x) $$ where $\chi(\vec x)$ is an arbitrary function, does not change the magnetic field (exercise for the reader: check that this is true).

In electrodynamics, the gauge freedoms for $\Phi(\vec x, t)$ and $\vec A(\vec x, t)$ combine to a single gauge freedom for the 4-vector $(\Phi, \vec A)$. See e.g. the linked Wikipedia article.


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