1
$\begingroup$

I had a quiz in one of my physics classes the other day, and one of the questions is still bugging me.

Say we have a point in 3-dimensional cartesian space with coordinates $(0, 2m, 0)$. How you would represents a position vector pointing from the origin to this point in spherical coordinates: $(r, \theta, \phi )$.

Before writing down any equations, it's obvious to see that the magnitude of the position vector would be 2m, and it would be pointing in the $\hat y$ direction. The position vector in cartesian coordinates would be: $$\vec r=2m \hat y$$ Well if we define $\phi = 0$ as the positive x-axis and $\theta =0$ as the positive z-axis, then each angle should by $90^\circ$ or $\frac \pi2$ radians. Then I think the vector would be: $$\vec r=2m\hat r+\frac \pi2\hat \theta+\frac \pi2\hat \phi$$ Which shows the exact position of that point relative to the origin. I got this question wrong. I talked to my professor, and she said the real right answer should be $$\vec r=2m\hat r$$ Which doesn't specify the position. That's just the distance from the origin. That could be used to talk about an infinite amount of points. Her argument was that, since you can place the coordinate axes anywhere you want, the angles don't really matter. By that logic, since we're talking about a defined point in space with a specific location, we could just move the origin to that point. In that case, it would just be $\vec r=0$. It seems extremely wishy-washy and I feel kind of cheated. I trust my professor, but I didn't receive a very satisfying explanation for how my answer is incorrect. Can anyone give me a more reasonable explanation to clear up why writing the vector like this is okay? If the $\vec r=2m\hat r$ is correct, does that mean that $\vec r=2m\hat r+\frac \pi2\hat \theta+\frac \pi2\hat \phi$ is wrong?

$\endgroup$
2
  • $\begingroup$ An answer is not difficult, but should necessarily provide graphics, so it's not quick to put together, I will try, slowly, if nobody answers before me $\endgroup$ Sep 2, 2020 at 19:44
  • $\begingroup$ @enzotib Thanks for taking the time to help! $\endgroup$
    – Batclocks
    Sep 2, 2020 at 19:47

1 Answer 1

1
$\begingroup$

Look at the following graphics

enter image description here

the unit vector of the spherical coordinates are the unit vector tangent to the lines where two coordinate are constants and the third change.
For example, if you only change $\theta$, with $r=2m$ and $\phi=\pi/2$ fixed, you obtain the quarter of circle in the graphics, so $e_\theta$ is tangent to this circle.
If you only change $\phi$, with $r=2m$ and $\theta=\pi/2$ fixed, you obtain the circle that appears as an ellipse in the graphics, so $e_\phi$ is tangent to this circle.
If you only change $r$, with $\phi=\pi/2$ and $\theta=\pi/2$ fixed, you obtain the positive $y$ axis, so $e_r$ is tangent to this line.

You can see that the vector $r=2me_y$ has component only along $e_r$, so the result of the problem.

In general, the unit vectors in a generic position are given by \begin{align} e_r &= \sin\theta(e_x\cos\phi+e_y\sin\phi)+e_z\cos\theta, \\ e_\theta &= \cos\theta(e_x\cos\phi+e_y\sin\phi)-e_z\sin\theta, \\ e_\phi &= -e_x\sin\phi+e_y\cos\phi, \end{align}

and conversely \begin{align} e_x &= \cos\phi(e_r\sin\theta+e_\theta\cos\theta)-e_\phi\sin\phi, \\ e_y &= \sin\phi(e_r\sin\theta+e_\theta\cos\theta)+e_\phi\cos\phi, \\ e_z &= e_r\cos\theta-e_\theta\sin\theta. \end{align}

If you put $\theta=\phi=\pi/2$ in these equations, you get the $e_y=e_r$ in that point.

$\endgroup$
8
  • $\begingroup$ i believe that the confusion arose because the OP is using spherical unit vectors "the same way" as he would use cartesian ones to express some point, i.e. for the point with s.c. $(\rho, \theta, \phi)$ writing $\vec{r} = \rho e_{\rho} + \theta e_{\theta} + \phi e_{\phi}$ but these two things are actually different $\endgroup$
    – Javi
    Sep 2, 2020 at 20:25
  • $\begingroup$ Okay, so the unit vectors representing the basis for spherical coordinates are not really definite in the same way as cartesian coordinates. In other words, the three axis lines that we draw to visualize the 3d space do not necessarily represent the basis...? So to write a position vector in spherical coordinates from a point in cartesian coordinates, I would first find the components r, theta, and phi using equations such as z=rcos(theta) etc. Then I would use the above equations to determine the spherical unit vectors associated with the specific values of r, theta, and phi that I found? $\endgroup$
    – Batclocks
    Sep 2, 2020 at 22:04
  • $\begingroup$ @ enzotib OR the unit vectors $\hat x, \hat y,$ and $\hat z$ do form the basis, but the unit vectors $\hat r, \hat \theta,$ and $\hat \phi$ are just functions of those basis vectors. So $\hat r, \hat \theta,$ and $\hat \phi$ are not necessarily the same at every point? $\endgroup$
    – Batclocks
    Sep 3, 2020 at 2:51
  • $\begingroup$ @Batclocks it is a basis, but it changes from point to point: there is a different basis in each point of the space $\endgroup$ Sep 3, 2020 at 5:36
  • $\begingroup$ @enzotib Thanks for replying. So if we were describing two points in spherical coordinates (i.e. a source and field points in an electrostatics problem), would they both have a different set of unit vectors? Would the only thing mathematically linking those points together be the dependence of their individual unit vectors on the unit vectors $\hat x, \hat y,$ and $\hat z$? $\endgroup$
    – Batclocks
    Sep 3, 2020 at 21:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .