Consider $\mathbb S_R^2 \subseteq \mathbb R^3$ be a sphere of radius $R$. Let $N$ denote the north pole and $0 < r < R \pi$. Define the metric circle with center N to be all points, $x \in \mathbb S_R^2$ that $$d_{\mathbb S_R^2}(x,N) = r$$ Here $d_{\mathbb S_R^2}$ is the induced metric on the sphere. Parameterize $\mathbb S_R^2$ with the usual spherical coordinates:

$$\Gamma:[0,2\pi) \times [0,\pi]\to\mathbb{R}^3 \quad \quad (\phi, \theta)\mapsto (R\sin\theta\cos\phi,R\sin\theta\sin\phi, R\cos\theta),$$

My hunch is that the metric circle with center N is spherical circle: $$\Gamma ( [0,2\pi) \times \{ r / R \} ) \subseteq \mathbb S_R^2$$ However, I cannot justify this claim using only elementary arguments using only elementary techniques.

Edit: Bump!

  • $\begingroup$ Your definition of a spherical circle is unclear. $\endgroup$ Feb 23 at 19:02
  • $\begingroup$ @MoisheKohan I have now edited the post. Hopefully, it is clear now. $\endgroup$
    – user82261
    Feb 24 at 1:32
  • $\begingroup$ Hmm, to me the induced metric on sphere is the angular metric. Do you mean the chordal metric? (The restriction of the Euclidean distance.) $\endgroup$ Feb 24 at 2:02
  • $\begingroup$ @MoisheKohan I think so. I am assuming the chordal metric and the induced metric are the same thing. $\endgroup$
    – user82261
    Feb 24 at 3:19
  • $\begingroup$ The answer to your question is positive and one does not need calculus to prove this, just elementary geometry. The key is that the chordal distance determines the angular distance and vice versa. $\endgroup$ Feb 24 at 3:44

1 Answer 1


Consider sphere $S(0,R)$ of radius $R$ in the Euclidean 3-space $E^3$ and centered at the origin $0$. There are two metrics one defines on $S(0,R)$:

  • The chordal metric $d(P,Q)=a=|P-Q|$, the restriction of the Euclidean distance function to the sphere.

  • The angular metric $\alpha=\angle(P,Q)$ which is the angle of the triangle $P0Q$ at $0$ (if you like, you can multiply it by the radius $R$).

Let us express one metric in terms of the other. For this we use the cosine formula: $$ a^2= 2R^2 - 2R^2 \cos(\alpha)= 2R^2(1- \cos(\alpha)). $$ Since $0\le \alpha\le \pi$, it follows that $\alpha$ determines $a$ and vice versa: $$ \alpha= \arccos(1-\frac{a^2}{2R^2}) $$ Now, consider the circle $C(P,\alpha)\subset S(0,R)$ of radius $\alpha$ centered at $P$, where the radius of the circle is computed in terms of the angular metric. Then a point $Q\in S(0,R)$ lies on $C(P,\alpha)$ if and only if $$ a:=d(P,Q)= R\sqrt{2(1-\cos(\alpha))}. $$ In other words, $C(P,\alpha)$ is the same as the circle in $S(0,R)$ of the chordal radius $a$ and centered at $P$, answering your question. (Your question has a typo, I think, you meant $r=p$, but, regardless, your conjectural formula for the angular radius in terms of the chordal radius is wrong.)

  • $\begingroup$ I have corrected the typos. Thanks. $\endgroup$
    – user82261
    Feb 25 at 10:33

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