Is it possible to define a coordinate system with 3 angles and no radii?

Question

In Cartesian coordinates, a point or vector is defined with an x distance or magnitude, a y distance or magnitude, and in 3 dimensions, a z distance or magnitude.

In 2 dimensional polar coordinates, a point or vector is defined by a radius and an angle.

In 3 dimensions, we have cylindrical coordinates and spherical coordinates.

In cylindrical coordinates, we have a vertical distance for a point or vector, and an angle and radius for the "circle" in the cylinder.

In spherical coordinates, we have two angles and one radius.

But is it possible for a coordinate system to be defined with three angles and no radius?

Answer 1

The question is a bit vague, coordinate system in particular would need a proper definition to give a rigorous answer.

But essentially the answer is yes. Imagine your space is a circle. Then a coordinate system is just an angle. You can generalise this to $n$ dimensions in at least two ways. You can take the product of $n$ circles, an $n$-torus, where natural coordinates are $n$ angles, or you can take the $n$-sphere, where again you can use $n$ angles as coordinates (that may be hard to visualise, but a 2-sphere is easy, and the angle are latitude and longitude).

Or you could decide you are work with conformal quantities - essentially you are defining an equivalence relation so that only angle matters and points along the same direction are identified (this is pretty loosely stated, look up conformal geometry or conformal transformation of a metric tensor if you are interested).

Answer 2

Working on $\mathbb{R}^3$:

Yes, but not with them all being angles around the same point.

For example, fix some point $y$ in $\mathbb{R}^3$ other than the origin $O$. Then the following data uniquely define a point $x$ in $\mathbb{R}^3$:

1. The standard two angles $\theta(x)$ and $\varphi(x)$ from spherical geometry.
2. The angle $\psi(x)$ between the line through $O$ and $y$ and the line through $x$ and $y$.

Proof of the above:

The angles $\theta(x)$ and $\varphi(x)$ pin $x$ to being on some ray with endpoint $O$, as in spherical coordinates. We can relate the distance from $x$ to the origin to the angle $\psi(x)$ bijectively (consider the triangle between $O$, $x$, and $y$).

Proof that you cannot have all three angles around the same point:

Without loss of generality, that point is $O$ (else translate so that it is). If $x$ is a scalar multiple of $y$, then for all points $z$, the angles $xOz$ and $yOz$ are equal, so no such arrangement can distinguish $x$ and $y$.

Answer 3

I propose a "goniometric" solution obtained by placing "observers" on the 3 axes. See figure below.

Let $O$ be the origin of coordinates.

Let us install "observers" in $X(1,0,0), Y(0,1,0), Z(0,0,1)$ resp.

Let us name

$$\alpha_x:=OXM, \ \ \alpha_y:=OYM, \ \ \alpha_z:=OZM$$

the deflection angles (wrt to origin $O$) under which a point $M(x,y,z)$ is seen from points $X,Y,Z$ resp.

Let $c_x:=\cos(\alpha_x), \ \ c_y:=\cos(\alpha_y), \ \ c_z:=\cos(\alpha_z).$

Let us consider the case of angle $\alpha_x$. We have:

$$\vec{MO}\begin{pmatrix}1\\0\\0\end{pmatrix}, \ \ \ \ \vec{MX} \begin{pmatrix}x-1\\y\\z\end{pmatrix}$$

Let us express in two ways the dot product $\vec{MO}.\vec{MX}$ :

$$1.(x-1)+0.y+0.z=c_x\sqrt{(x-1)^2+y^2+z^2} \tag{1}$$

Squaring (1), we obtain:

$$(c_x^2-1)(x-1)^2+(y^2+z^2)c_x^2=0 \ \iff \ y^2+z^2=t^2_x (1-x)^2 $$

where $t_x=\tan \alpha_x.$

What has been done with angle $\alpha_x$ can be done in exactly the same way for the two other angles, yielding a system of 3 equations with 3 unknowns:

$$\begin{cases}&&y^2&+&z^2&=&t_x^2(1-x)^2 \ \ (a)\\x^2&+&&&z^2&=&t_y^2(1-y)^2 \ \ (b) \\x^2&+&y^2&&&=&t_z^2(1-z)^2 \ \ (c)\end{cases}\tag{2}$$

Let us now add resp. $x^2$ to (a), $y^2$ to (b), $z^2$ to (c), giving system (3), equivalent to system (2):

$$x^2+y^2+z^2=p(x)=q(y)=r(z) \ \ \text{with} \ \ \begin{cases}p(x)&:=&x^2+t_x^2(1-x)^2\\ q(y)&:=&y^2+t_y^2(1-y)^2\\ r(z)&:=&z^2+t_z^2(1-z)^2 \end{cases}\tag{3}$$

Geometrical interpretation of (3): As

$$p(x)=q(y), \ \ \ \ q(y)=r(z), \ \ \ \ r(z)=p(x)$$

are the equations of 3 hyperbolic cylinders (see figure) with orthogonal directing lines, (3) means that $M(x,y,z)$ is at the intersection of these cylinders.

One obtains in this way a finite number of solutions $(x,y,z)$ (more exactly at most $8$ solutions).

Spurious solutions have been introduced by squaring(s). Therefore, it remains to find out which of these solutions fulfill the original equations, such as (1). In this way one gets a unique solution.

Fig. 1: The 8 cylinders' triple intersections are visible at the top and bottom of the image. Only one of them (point $M$) is valid.

Appendix: For readers interested by the Matlab program I have written for this figure:

clear all;close all;hold on;axis equal off;
set(gcf,'color','w');
a=3;I=-a:0.2:a;
[x,y,z]=meshgrid(I);
axis([-a,a,-a,a,-a,a]);
M=[2.4,2.69,3];plot3(M(1),M(2),M(3),'ok');
tx=1.3;ty=0.8;tz=0.1;
p=x.^2+tx^2*(1-x).^2;
q=y.^2+ty^2*(1-y).^2;
r=z.^2+tz^2*(1-z).^2;
for k=1:3;
   if k==1;s=p-q;end;
   if k==2;s=q-r;end;
   if k==3;s=r-p;end;
   [faces,verts,colors] = isosurface(x,y,z,s,0,abs(z));
   patch('Vertices',verts,'Faces',faces,'FaceVertexCData',colors,...
  'FaceColor','interp','EdgeColor','none');
end;
view(55,16);alpha(0.2);
plot3([1,0,0,0,0],[0,0,1,0,0],[0,0,0,0,1]); 
b=1.2;c=-0.25;d=0.2;text([c,b,0,0,M(1)+d],[c,0,b,0,M(2)+d],...
[0,0,0,b,M(3)],{'O','X','Y','Z','M'});