In 3D Euclidean space, we know that distance between 2 points: $a=(x_1,y_1,z_1)$ and $b=(x_2,y_2,z_2)$ is $s^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$from metric $ds^2=dx^2+dy^2+dz^2$. But I was thinking if we have the misfortune of working in spherical coordinates, i.e. only given $a=(r_1,\theta_1,\phi_1)$ and $b=(r_2,\theta_2,\phi_2)$, how can I work out $s^2$?

I suppose I need to do the integration $$\int_a^b (dr)^2+r^2(d\theta)^2+r^2\sin^2\theta(d\phi)^2$$which I tried and got $$\left(\int_a^b dr\right)^2+\left(\int_a^b rd\theta\right)^2+\left(\int_a^b r\sin\theta d\phi\right)^2=(r_2-r_1)^2+(r_2\theta_2-r_1\theta_1)^2+(r_2\sin\theta_2\phi_2-r_1\sin\theta_1\phi_1)^2$$

Is this correct??

  • $\begingroup$ Nope. When $r_1=r_2$ you must get something that depends only on $r$ and $\cos(\theta_1-\theta_2)$ and $\cos(\phi_1-\phi_2)$, by en.wikipedia.org/wiki/Spherical_law_of_cosines. $\endgroup$ – Jack D'Aurizio Dec 23 '14 at 0:21
  • $\begingroup$ Why not just to convert $(r_i,\theta_i,\phi_i)$ into cartesian coordinates and compute the distance by the usual way? $\endgroup$ – Jack D'Aurizio Dec 23 '14 at 0:22
  • $\begingroup$ well, yes I suppose we can just convert to cartesian but I was just wondering if there was a way to do the integral just using spherical polar $\endgroup$ – Tobyhas Dec 23 '14 at 0:24
  • $\begingroup$ For sure. But just notice that the integral of the square is not the square of the integral. $\endgroup$ – Jack D'Aurizio Dec 23 '14 at 0:26
  • $\begingroup$ @JackD'Aurizio yeh I was a bit suspicious about doing the square of integral...but I was thinking: in cartesian coordinates, $s^2=\int_a^b (dx)^2+(dy)^2+(dz)^2$ and in order to recover from this the familiar $s^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$ I did $(\int_{x_1}^{x_2} dx)^2+(\int_{y_1}^{y_2} dy)^2+(\int_{z_1}^{z_2} dz)^2$ and used the similar method in the case of the spherical polar coordinates $\endgroup$ – Tobyhas Dec 23 '14 at 0:35

Assuming that $\theta$ is the latitude and $\phi$ is the longitude we have that the cartesian coordinates of the first point are: $$ (r_1 \cos\theta_1 \cos\phi_1, r_1 \cos\theta_1 \sin\phi_1, r_1\sin\theta_1),$$ so the distance between the two points is given by: $$ \sqrt{r_1^2+r_2^2-2r_1r_2\left(\cos\theta_1\cos\theta_2\cos(\phi_1-\phi_2)+\sin\theta_1\sin\theta_2\right)}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.