When I have an expression in polar coordinates and I substitute the definition of $r$ and $\phi$, I get the corresponding expression in polar coordinates. I would like to go back and obtain the same expression now as a function of $r$ and $\phi$, but the code below doesn't simplify the ArcTan as I would like:
ρ[x_, y_] := Sqrt[x^2 + y^2];
Θ[x_, y_] := ArcTan[x, y];
$Assumptions = r > 0 && r ∈ Reals, ϕ ∈ Reals;
ϕ /. r -> ρ[x, y] /. ϕ -> Θ[x, y]
% /. x -> r Cos[ϕ] /. y -> r Sin[ϕ]
First of all, $Assumptions isn't used anywhere in the original code because you're just doing replacements. You have to use a function such as FullSimplify that actually checks the setting of $Assumptions. To decide what functions do this, look for the option Assumptions in the Options of that function.
With that said, here is how to do the inversion:
ρ[x_, y_] := Sqrt[x^2 + y^2];
Θ[x_, y_] := ArcTan[x, y];
$Assumptions =
r > 0 && r ∈ Reals && -Pi/2 < ϕ < Pi/2;
ϕ /. r -> ρ[x, y] /. ϕ -> Θ[x, y]
(* ==> ArcTan[x, y] *)
% /. x -> r Cos[ϕ] /. y -> r Sin[ϕ]
(* ==> ArcTan[r Cos[ϕ], r Sin[ϕ]] *)
FullSimplify[%]
(* ==> ϕ *)
The only other thing I had to change is to restrict the angle range in $Assumptions so that the ArcTan can in fact be uniquely inverted.
So you are trying to use coordinate transformation for a function. You can do it in this way. Say you have a function f[x,y] which you transform into g[r,q] and then use the reverse transformation to get back f[x,y].
f[x,y]
g[r_, q_] = f[x, y] /. {x -> r Cos[q], y -> r Sin[q]}
f1[x_, y_] = g[r, q] /. {r -> Sqrt[x^2 + y^2], q -> ArcTan[x, y]}
f[x, y]
f[r Cos[q], r Sin[q]]
f[x, y]
Is that what you are looking for?
Also notice that I don't use any form for f[x,y]. So it is true for both scalar and vectors.
In your example you get ArcTan[r Cos[ϕ], r Sin[ϕ]] instead of ϕ because ArcTan is bounded within -Pi/2 to Pi/2 but your ϕ doesn't have such bound. When you use it in another expression, it will give you right answer.
Tan[ArcTan[r Cos[ϕ], r Sin[ϕ]]]
Tan[ϕ]
