# Stream plotting in polar coordinates [duplicate]

I got to a scalar field that really looks just fine to me in polar coordinates. "looks just fine" = all the physical limits make sense. The scalar field in polar coordinates r and θ is:

ϕ[r_, θ_] := -v0*Cos[θ + α] (r + R^2 (1 - 2*ϵ*Sin[θ]^2)/r)


Now using equation

v = -Grad[ϕ[r, θ], {r, θ}, "Polar"]


I want to get the desired vector field. The equation above leaves me with this

v={v0 Cos[α + θ] (1 - (
R^2 (1 - 2 ϵ Sin[θ]^2))/r^2), -(((
4 R^2 v0 ϵ Cos[θ] Cos[α + θ] Sin[θ])/r +
v0 (r + (R^2 (1 - 2 ϵ Sin[θ]^2))/
r) Sin[α + θ])/r)}


All nice, whatever this is. BUT I want to plot that vector field. And in order to do it, I did:

vx =
TransformedField["Polar" -> "Cartesian",
v0 Cos[α + θ] (1 - (
R^2 (1 - 2 ϵ Sin[θ]^2))/
r^2), {r, θ} -> {x, y}] // Simplify

(*  (v0 ((x^2 + y^2)^2 -
R^2 (x^2 + y^2 (1 - 2 ϵ))) Cos[α +
ArcTan[x, y]])/(x^2 + y^2)^2          *)

vy =
TransformedField[
"Polar" ->
"Cartesian", -(((
4 R^2 v0 ϵ Cos[θ] Cos[α + θ] Sin[θ])/r +
v0 (r + (R^2 (1 - 2 ϵ Sin[θ]^2))/
r) Sin[α + θ])/r), {r, θ} -> {x, y}] //
Simplify

(*  -((
v0 (4 R^2 x y ϵ Cos[α +
ArcTan[x, y]] + ((x^2 + y^2)^2 +
R^2 (x^2 + y^2 - 2 y^2 ϵ)) Sin[α +
ArcTan[x, y]]))/(x^2 + y^2)^2)      *)


Which for parameters:

v0 = 1;
R = 1;
ϵ = .02;
α = Pi/4;

Show[StreamPlot[{vx, vy}, {x, -5, 5}, {y, -5, 5}],
PolarPlot[f[θ], {θ, 0, 2*Pi}]] Brings me to a very strange and unexpected vector field. It's a vector field that has no sense at all.

And I have a strong feeling that this method of plotting a vector field is wrong, because i am 99.9% sure that the scalar field I got is ok. So, my question to is: What is wrong and how do I change it?

(BTW: scalar field is only defined for r that are equal to or bigger than f[θ_]:=R(1-ϵSin[θ]^2).)

Any help will be highly appreciated.

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Try this:

 v0 = 1;R = 1;ϵ = .02;α = Pi/4;
ϕ[r_, θ_] := -v0*Cos[θ + α] (r + R^2 (1 - 2*ϵ*Sin[θ]^2)/r);
v = -Grad[ϕ[r, θ], {r, θ}, "Polar"];
gxy = TransformedField["Polar" -> "Cartesian",v, {r, θ} -> {x, y}] // Simplify;

StreamPlot[gxy, {x, -5, 5}, {y, -5, 5}];


Or use Cartesian coordinate from the beginning:

f = TransformedField[
"Polar" -> "Cartesian", ϕ[
r, θ], {r, θ} -> {x, y}]; 