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You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$

is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, and if you look very close to the z-axis (as I assume you are doing with a jet), the spiral trajectories have the same period to leading order, so the motion is asymptotically rigid in this near-z-axis region.

Your miscommunication is probably caused by having flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

(this is different from the previous thing). That's why you had a difference of opinion.

You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$

is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, and if you look very close to the z-axis (as I assume you are doing with a jet), the spiral trajectories have the same period to leading order, so the motion is asymptotically rigid turning around the z-axis in this near-z-axis region (you can see this by making x,y small, so that r=z--- then the vector field reduces to the standard rigid rotation vector field:

$$ v_x = - Cy $$ $$ v_y = Cx $$

Where C is ${1\over z}$. This is the vector field that generates rigid rotations around the z axis.

Your miscommunication is probably caused by having flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

That's why you had a difference of opinion. It helps to write out the "obvious" formulas when using different coordinate systems, not because people don't know what polar coordinates are, but because it fixes your conventions so that these types of miscommunications don't happen.

By the way, I would be really surprised if real active galactic nuclei jets turn rigidly around the axis, or that the velocity would be constant magnitude like this. I think a better model is:

$$ v_z = f(\theta) $$ $$ v_x = -{y\over \rho} g(\rho) $$ $$ v_y = {x\over \rho} g(\rho) $$

Where $\rho=\sqrt{x^2+y^2}$ (I sidestepped the polar convention ambiguities by avoiding using $\phi$). where f is a decaying function of $\theta$, like a Gaussian, with a peak at z=0, and g is some function of $\sqrt{x^2 + y^2}$ that depends on the source of the precession.

You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$

is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, asymptotically rigidly with a constant period right near the z-axis.

I think you flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

(this is different from the previous thing). That's why you had a difference of opinion.

You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$

is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, and if you look very close to the z-axis (as I assume you are doing with a jet), the spiral trajectories have the same period to leading order, so the motion is asymptotically rigid in this near-z-axis region.

Your miscommunication is probably caused by having flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

(this is different from the previous thing). That's why you had a difference of opinion.

You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_x(r) = r \cos\theta$$ $$ v_y(r) = r \sin\theta\cos\phi $$ $$ v_z(r) = r \sin\theta\sin\phi $$

is radially outward. Then you flip the sign on v_x, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward to the y-z plane, it is a saddle flow. What you want to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. So your convention for coordinates flips the normal ones:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

(this is different from the previous thing). In this situation, the gas will precess around the axis as you intend, the net speed will be constant, and the trajectories will be spirals around the z axis.

You use two different coordinates, polar and rectangular, without specifying how they are related. The relation I will assume is the textbook one, where

$$ z = r\cos\theta$$ $$ x = r\sin\theta\cos\phi$$ $$ y = r \sin\theta\sin\phi$$

So that the vector field

$$ v_z(r) = \cos\theta$$ $$ v_x(r) = \sin\theta\cos\phi $$ $$ v_y(r) = \sin\theta\sin\phi $$

is unit length radially outward. Then you are only flipping the sign on $v_x$, so that the field is radially outward in the y-z plane, but inward in x. Another way to describe this field is:

$$ v_z = {z\over r} $$ $$ v_x = -{x\over r} $$ $$ v_y = {y\over r} $$

this thing sucks the flow inward in x to the y-z plane, it is a saddle flow, with the x-y-plane repelling and the y-z plane attracting. What you certainly intended to do is

$$ v_z = {z\over r} $$ $$ v_x = -{y\over r} $$ $$ v_y = {x\over r}$$

In other words, you need to flip x and y. This vector field describes a gas which precesses around the axis as you intend, with a net speed which will be constant, and the trajectories spiral around the z axis, asymptotically rigidly with a constant period right near the z-axis.

I think you flipped the x and y axes in your polar coordinate convention compared to the usual one, so you used:

$$ z = r\cos\theta$$ $$ x = r\sin\theta\sin\phi$$ $$ y = r \sin\theta\cos\phi$$

(this is different from the previous thing). That's why you had a difference of opinion.