Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I want to set a simulation for jet which has a precessional motion. The symmetry axis of jet is $z$ axis, i.e. jet is propagating along $z$ direction making angle $\theta$. I set the velocity component of jet as \begin{eqnarray*} v_z &=& v\cos \theta \\ v_x &=& -v\sin \theta \cos \phi \\ v_y&=& v\sin \theta \sin \phi \end{eqnarray*}

As the jet has precessional motion, it will rotate in $x-y$ plane. For this reason I have set the $-$ sign to the $v_x$ component to rotate the jet anti-clockwise(according to the Lissajous figures). But some one is arguing that $-ve$ sign will not appear in the $v_x$ component. Where $\phi (t)=\omega t$.

Could you please explain this matter ?

share|improve this question
    
jet is the emission from radio galaxy. You can compare it with a flow from a nozzle. –  kallo Jun 28 '12 at 1:41
    
In a large degree the active galactic nuclei and the jet are superfluous to question: this is simply about the conventional meaning of the azimuthal and polar angles. –  dmckee Jun 28 '12 at 4:12
1  
For future reference, when using two different coordiante systems (x,y,z $r,\theta,\phi$), it is good to say your convention. The easier way in this case is to write down the vector field in rectangular coordinates, not polar ones, so that there are no interpersonal imagery issues. I kept on seeing the hyperbolic flow as opposed to the circle you meant to write (and probably did write, in your conventions). –  Ron Maimon Jun 28 '12 at 5:14
add comment

2 Answers

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.

share|improve this answer
add comment

I don't see the time dependence. What you have written is a constant velocity. If you want precession about the z-axis, you will need to introduce time dependence in $\phi$. The easiest would be constant precession, with $\phi(t) = \omega t$.

Unless there is something else going on, the situation you have described is symmetric between the x- and y-axes. So you can arbitrarily say that the jet starts out coplanar with both the x- and z-axes, with a positive x component. If you did that, the negative sign would not be there. Putting the negative sign in $v_x$ is equivalent to starting the velocity with a negative x component.

share|improve this answer
    
I have clarified what I meant by that. I did not mean to rotate the axes; I meant that this would rotate the initial state of the flow. –  Colin McFaul Jun 28 '12 at 14:55
    
If there is no time dependence, then how is the jet precessing? I genuinely do not understand this. From your and @dmckee's comments, it sounds like I'm completely misunderstanding the question. If that's the case, then the question is badly written, and needs another edit. –  Colin McFaul Jun 28 '12 at 14:57
1  
The jet particles are going around and around the z axis along the integral lines of v--- the v is pointing around the z axis, so the particles are precessing, even though the velcocity at any point is not chaning in time. –  Ron Maimon Jun 28 '12 at 15:38
    
Yes,$\phi (t)=\omega t$. –  kallo Jun 28 '12 at 23:30
    
The flow of jet is precessing and $\phi (t)=\omega t$. –  kallo Jun 28 '12 at 23:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.