ScienceNostalgia101 2 Posted July 20 From my understanding, it is not only possible to have a rotating vector field with a point in its centre where the velocity vector is precisely zero, but that the reasons to expect one follow from the fact that if the surroundings are rotating about one point, they are not moving at that point. But what if you had a moving vector field? For instance, let's say you had the centre of some galaxy, about which all the stars around it were orbiting, moving through space. Obviously in the reference frame that is that moving galaxy, the velocity of that centre is zero. But in some external reference frame not moving parallel to that galaxy, that centre is not zero. Does there have to be some other point in that galaxy, then, at which the velocity is zero in the other reference frame, or is there some means by which literally no point in that galaxy has a zero point? Not sure whether this is better suited to the math or physics board; thought I'd put it here since I assume chemistry and geology have used for the math of rotational dynamics as well. 0 Share this post Link to post Share on other sites

timo 569 Posted July 20 I'll try to express what I understood in mathematical terms. From the title of your thread I was assuming a rotating vector field would be f(x, y, z, t) = (sin(t), cos(t), 0). That is a homogeneous field that rotates around the z-axis. But your first sentence seems to be considering a time-static, rotationally-symmetric field like f(x, y, z) = (y, -x, 0), instead. I do not understand why f(0, 0, 0) had to be (0, 0, 0) in that case (even though it is in my example). Mathematically, the function could have any value. Physically, the magnetic force around a wire would be a counter-example, since it "rotates around the wire" and diverges at the origin (i.e. is undefined). Without looking it up I'd expect that the velocities of circular planetary orbits also diverge at the origin. So I think there are some implicit assumptions you make in your first sentence that you are omitting, e.g. something like "the function is continuous and defined everywhere". Your moving example would probably be something like f(x,y,z,t) = (y-t, -x, 0), I think. So now, time is back in the picture, which is a significant difference. You can quickly solve that for (0, 0, 0) == (y-t, -x, 0) and see that for all (x=0, y=t, z=arbitrary) f(x, y, z, t) = (0, 0, 0). So in that particular example, the answer would be "at each time, there are points at which the function is zero" and "there is no point at which the function is zero at all times". If you constrain the domain of the function, the first statement can become invalid, of course (if the calculated points are outside of the domain). Not sure I really understood what you were asking. But I hope this helps. 0 Share this post Link to post Share on other sites

studiot 2018 Posted July 20 Rotation and translation are separate transformations. There is no rotation that produces a pure translation and no trnaslation that produces a pure rotation. 0 Share this post Link to post Share on other sites

HallsofIvy 6 Posted 16 hours ago (edited) In order to combine rotation and translation you need to use "homogeneous coordinates". That is, we represent the point (x, y, z) by the vector (x, y, z, 1) while identifying the vector (ax, ay, az, a), for any non-zero a, with (x, y, z, 1). Then the rotation around the z- axis, through angle [math]\theta[/math] together with translation by (dx, dy, dz), is given by the 4 by 4 matrix [itex]\begin{bmatrix}cos(\theta) & -sin(\theta) & 0 & 0 \\ sin(\theta) & cos(\theta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ dx & dy & dz & 1\end{bmatrix}[/itex]. [itex]\begin{bmatrix}cos(\theta) & -sin(\theta) & 0 & 0 \\ sin(\theta) & cos(\theta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ dx & dy & dz & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ 1\end{bmatrix}= \begin{bmatrix}xcos(\theta)- y sin(\theta)+ dx \\ xsin(\theta)+ ycos(\theta)+ dy \\ z+ dz \\ 1\end{bmatrix}[/itex]. Edited 16 hours ago by HallsofIvy 0 Share this post Link to post Share on other sites