Great circle between two points on the Earth's surface

[brotherbobby]

Homework Statement

Let vectors from the center of the earth to two points on the surface be $\vec{r}_1,\vec r_2$. If the angle between them is $\theta$, find the expression for $\theta$ in terms of the co-ordinates of the two points, them being their latitudes $\lambda_1,\lambda_2$ and $\phi_1,\phi_2$. (See sketch below)

Relevant Equations

1. A vector $\vec a = a(\cos\alpha\hat i+\cos\beta\hat j+\cos\gamma\hat k)$, where the angles are well known. Here, $\alpha = \phi$ and $\gamma = \pi/2-\lambda$.
2. If the direction cosines written in (1) above be given by the symbols $l,m,n$, then $\cos\theta=l_a l_b+m_a m_b+n_a n_b$.
3. We also have $l^2+m^2+n^2=1$ as the relation between the direction cosines for a vector.

Problem statement : I copy and paste the problem as it appeared in the text. [Kleppner, D., Kolenkow, R. (2014). An Introduction to Mechanics. Cambridge University Press (2E)]. The statement itself is rather long-winded. The question to be answered appeared towards the end of the passage :
$\color{red}{\text{Find an expression for}\;\theta\;\text{in terms of the co-ordinates of the two points.}}$
By co-ordinates, I am assuming the latitudes and longtidues of the two points, $[(\lambda_1,\lambda_2),(\phi_1,\phi_2)]$.

Attempt : I took the liberty to draw a better image which show the latitudes and longitudes of the two places.

The image shows places $P_1$ and $P_2$ drawn in purple and green and green, respectively, as well as their latitudes $\lambda_i$ and longitudes $\phi_i$ for $i = 1,2$. The angle between their position vectors is $\theta$, shown in red.
Cluttered though the symbols are, I hope they can be distinguished.

We know $\cos\theta = l_1 l_2+m_1 m_2 + n_1 n_2\qquad (1).$

Here, $l_i = \cos\phi_i$, $n_i = \cos\gamma_i = \sin\lambda_i$, where $\gamma$ is the (well-known) angle that a vector makes with the $z-$axis. The values $m_i=\cos\beta_i$, the angle with the $y-$axis. These can be calculated using $l^2+m^2+n^2=1\Rightarrow m_i = \sqrt{1-\cos^2\phi_i-\sin^2\lambda_i}$, the index $i$ taking values 1 and 2 throughout.

We know, $1-\cos^2\phi_i = \sin^2\phi_i$.

From $(1)$ above, all this yields $\boxed{\cos\theta = \cos\phi_1 \cos\phi_2+\sqrt{\sin^2\phi_1-\sin^2\lambda_1}\sqrt{\sin^2\phi_2-\sin^2\lambda_2}+\sin\lambda_1 \sin\lambda_2}$.

I'd check the answer presently, but am I on the right track?

Many thanks.

 

[pasmith]

I would rotate the sphere so that one point is at the north pole (0,0,1). Then $\theta$ is just the colatitude of the image of the second point.

 

[kuruman]

I would write two unit vectors $$\mathbf{\hat r}_i=\cos\phi_i\cos\lambda_i~\mathbf{\hat x}+\sin\phi_i\cos\lambda_i~\mathbf{\hat y}+\sin\lambda_i~\mathbf{\hat z}~~~~(i=1,2)$$ then take their dot product.

 

[brotherbobby]

Sorry for coming in late. I would use the method you proposed @kuruman presently, but please tell me if there's something wrong with my solution above. (Post #1)

I think it would be better if I put the problem statement again.

Of course it's crucial what I am assuming as "co-ordinates of the points". The ones given are $\vec r_1,\vec r_2$, but these are vectors. In their explanation of the problem, see image to the right, they talk of the latitude $\lambda$ and longtidue $\phi$ of a place on the surface of the earth.
I assumed, by this, the coordinates of the two places are $(\lambda_i,\phi_i)$ for $i=1,2$ and obtained this answer for the angle $\theta$ between the two places :
$\small{\boxed{\cos\theta = \cos\phi_1 \cos\phi_2+\sqrt{\sin^2\phi_1-\sin^2\lambda_1}\sqrt{\sin^2\phi_2-\sin^2\lambda_2}+\sin\lambda_1 \sin\lambda_2}}$.

I'd like to know where's the mistake.

 

[kuruman]

I'd like to know where's the mistake.

$\mathbf{\hat r}=\cos\phi\cos\lambda~\mathbf{\hat x}+\sin\phi\cos\lambda~\mathbf{\hat y}+\sin\lambda~\mathbf{\hat z}.$

The directional cosines are
$l=\mathbf{\hat r}\cdot \mathbf{\hat x}=\cos\lambda\cos\phi$
$m=\mathbf{\hat r}\cdot \mathbf{\hat y}=\cos\lambda\sin\phi$
$n=\mathbf{\hat r}\cdot \mathbf{\hat z}=\sin\lambda$

It follows that
$l^2+m^2+n^2=1\implies m^2=1-l^2-n^2=1-\cos^2\lambda\cos^2\phi-\sin^2\lambda$
Therefore,
$m_i=\sqrt{1-\cos^2\lambda_i\cos^2\phi_i-\sin^2\lambda_i}.$

Your derivation, without showing any work, asserts that

(Edited for typos)