Intuition for spherical coordinates of a highly symmetric point

Question

Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought "oh, that point lies on the "dead center" of the sphere as far the first octant is concerned", so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions: $$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$ $$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$ $$z(r,\theta)=r\cos(\theta)$$ I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?

Answer 1

You're right of course that the point $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ is symmetric with respect to permutations of the $x,y,z$ axes, and so e.g. lies on the line of symmetry about which you can rotate $x,y,z$. But you're neglecting the fact that $(r,\theta,\varphi)$ coordinates do not transform nicely under rotations that move the $z$-axis. So you cannot reason that the angle $\theta$ must take some special value by symmetry.

You can reason this way for $\varphi$; considering the fact the point is unchanges by reflection in the plane $x = y$ guarantees that $\varphi \equiv \pi/2 - \varphi$ is either $\pi/4$ or $5\pi/4$. Clearly it's the former as it lies in the first $(x,y)$ quadrant.

But $\theta$ must be expressed differently. Note that $\theta$ is the angle made with the $z$ axis, which can equivalently be expressed in terms of the angle with the $(x,y)$ plane. Since $$\sqrt{x^2+y^2} = \sqrt{2} \times \frac{1}{\sqrt{3}} \qquad \text{ but }\qquad \sqrt{z^2} = \frac{1}{\sqrt{3}}$$ we see that in an appropriate sense you should think of your point as being "$\sqrt{2}$ times more in the $(x,y)$ plane than in the $z$ direction". This is made precise by noting $$ \theta = \arctan\frac{\sqrt{x^2+y^2}}{\sqrt{z^2}} = \arctan \sqrt{2} $$ rather than $\arctan 1 = \pi/4$. (This value of $\theta \approx 54.7^\circ$ is the so-called magic angle.)

So to correct your intuition, I think you should internalize the idea that we are in 3 dimensions; the point has equal magnitude in each direction; and $\theta$ reflects how close you are to one of three possible directions; hence, since only one third of the magnitude [squared] is concentrated in that particular direction, the angle made $\theta$ will be larger than $\pi/4$.

Answer 2

Your intuition is failing you. If you do the conversion, the point $(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$ you first need $\cos (\theta) =\frac 1{\sqrt 3}$, which says $\theta\approx 0.955$. You do have $\phi=\frac \pi 4$

Answer 3

Your intuition is wrong. $\cos\left(\frac{\pi}4\right) = \frac{1}{\sqrt{2}}$.

The correct angle here is $\arccos\left(\frac 1{\sqrt{3}}\right)$ which does not have a closed form solution I am aware of. If $\cos(\theta) = \frac 1 {\sqrt{3}}$, then $\sin{\theta} = \sqrt{\frac{2}{3}}$. Thus $\sin(\varphi )= \frac 1{\sqrt{2}}$. Thus $\varphi = \frac \pi4$.

Answer 4

The sphere intersects the $x$, $y$, and $z$ axis at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. To see if a point is in the "dead center" you can find the distance to each of these points and see if they are the same. For $(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}})$, the distances are $1$, $1$, $\sqrt{2-\sqrt{2}}$, so it isn't in the "dead center". However, for $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$, the distances are all $\sqrt{2-\frac{2}{\sqrt{3}}}$.

Answer 5

Direction cosines

$$l^2 + m^2+n^2= \cos^2 \alpha+ \cos^2 \beta+ \cos^2 \gamma=1; \text {if } l=m=n, \text{then } l=m=n=\frac{1}{\sqrt 3}$$

The vector makes this angle to the coordinate axes including the $z$ axis. The parallel circle angle

$$ \theta=\pi/2 - \gamma = \pi/2 - \cos^{-1} \frac{1}{\sqrt3 }=\sin^{-1} \frac{1}{\sqrt3 }\approx 35.264^{\circ}; $$

Due to equal inclination to $x-, y-$ axes and projection on $x,y$ plane directly the azimuth (polar coordinate) angle $ \varphi= \dfrac{\pi}{4}.$