Translate a vector field Imagine that you have a vector field $A = \frac{A_0}{r} e_{\theta}$ in cylindrical coordinates, where $A_0 \in \mathbb{R}$. Now you translate your coordinate system in $e_x$ direction by $x \mapsto x + d$ for some constant $d \in \mathbb{R}$. What happens to the field then?
I would say that $\frac{A_0}{r} \mapsto \frac{A_0}{\sqrt{(x-d)^2+y^2}}$, but what happens to the unit vector $e_{\theta}$?
So the question is: How can we express this vector field in the shifted coordinate system? If anything is unclear, please let me know.
 A: The coordinate change you wish to study is most natural in Cartesian terms. Therefore, change the given $A$ to standard Cartesians as a starting point:
$$ e_{\theta} = \langle -\sin \theta, \cos \theta \rangle  = \langle \frac{-y}{r}, \frac{x}{r} \rangle $$ 
Therefore, 
$$ A = \frac{A_o}{r}e_{\theta} = \frac{A_o}{x^2+y^2} \langle -y, x \rangle $$
Let $x' = x+d$ and $y' = y$ then $x = x'-d$ and $y = y'$. Thus,
$$ A = \frac{A_o}{r}e_{\theta} = \frac{A_o}{(x'-d)^2+(y')^2} \langle -y', x'-d \rangle $$
In the prime coordinates we also introduce $r', \theta'$ where these are defined implicitly by
$$ x' = r' \cos \theta', \qquad y' = r'\sin \theta'$$ 
hence $r' = \sqrt{(x')^2+(y')^2}$ and $\tan \theta' = y'/x'$. Returning to $A$ we find,
$$ A=\frac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \langle -r'\sin \theta', r' \cos \theta'-d \rangle $$
I suppose you probably want the end result in terms of the prime-polar frame $e_{r'}, e_{\theta'}$. Note $\nabla x' = \nabla x$ and $\nabla y' = \nabla y$ hence $e_x=e_{x'}$ and $e_y=e_{y'}$. Therefore,
$$ A=\frac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left( -r'\sin \theta'e_{x'}+ (r' \cos \theta'-d)e_{y'} \right) $$
Almost there, now just convert $e_x=e_{x'}$ and $e_y=e_{y'}$ to the prime polar frame. The cartesian and polar prime frames are related in the usual manner:
$$ e_{r'} = \cos \theta' e_{x'}+ \sin \theta' e_{y'} \qquad e_{\theta'} = -\sin \theta' e_{x'}+ \cos \theta' e_{y'} $$
which inverted yield
$$ e_{x'} = \cos \theta' e_{r'}-\sin \theta' e_{\theta'} \qquad 
 e_{y'} = \sin \theta' e_{r'}+\cos \theta' e_{\theta'}$$
Once more, return to $A$,
$$ A=\tfrac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left( -r'\sin \theta'[\cos \theta' e_{r'}-\sin \theta' e_{\theta'}]+ (r' \cos \theta'-d)[\sin \theta' e_{r'}+\cos \theta' e_{\theta'}] \right) $$
which simplifies nicely to:
$$ A=\tfrac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left[ (r'-d\cos \theta ') e_{\theta'}-d\sin \theta' e_{r'} \right] $$
which reminds me of a dipole field formula from my junior level electromagnetics course.
