Summer Solstice in Nashville, Tennessee

Let us assume that the center of the Earth is at $(0, 0, 0),$ that axis of rotation is the $z$ -axis, and that the sunlight rays are parallel to the vector $\boldsymbol{v}=\langle \cos {\phi_0}, 0, \sin{\phi_0} \rangle ,$ where $\phi_0=\ang{90}-\ang{66.5}=\ang{23.5}.$ Notice that the angle formed by this vector and the $z$ -axis is $\ang{66.5}.$ Now, we are going to find the position of Nashville on the sphere given by its x, y and z coordinates at the moment of the sunrise on June 21st. We are going to use $\phi$ to represent the latitude of Nashville. So, $\phi= \ang{36.16}.$ We are going to assume Nashville is represented at that moment by the point $(x, y, z).$ Since this point is on the sphere then it must satisfy that $x^2+y^2+z^2=R^2,$ where $R$ is the radius of the Earth. Since the latitude of Nashville is $\phi,$ then $z=R \sin \phi.$ We can also get another equation for this point by noticing that since Nashville is on the boundary of the part of the Earth that is being illuminated by the sun at that moment. The boundary of the part of the Earth that is illuminated is formed by points of the sphere that are on the plane orthogonal to $\boldsymbol{v}$ and that passes through the origin. This plane has the equation $x\cos\phi_0+z\sin\phi_0=0.$ So, the coordinates of the point representing that city must form a solution of the system

$x^2+y^2+z^2=R^2,$ $z=R \sin \phi.$ $x\cos\phi_0+z\sin\phi_0=0.$

By solving this system we obtain that the corresponding point is $\left (-R\tan\phi_0 \sin\phi, -R\sqrt{1-\sin^2{\phi}(1+\tan^2\phi_0)}, R\sin \phi \right ).$ Now it easy to see that a vector pointing to east at Nashville is a vector perpendicular to the position vector of the previous point, has a positive first component, and is parallel to the $xy-$ plane. So this vector can be, for example, $\boldsymbol{w}= \left \langle R\sqrt{1-\sin^2{\phi}(1+\tan^2\phi_0)}, -R\tan\phi_0 \sin\phi, 0 \right \rangle .$ Now the angle form by the Sun and the east direction can be found using the formula $\arccos\left (\frac{\boldsymbol{v}\cdot\boldsymbol{w}}{|\boldsymbol{v}| |\boldsymbol{w}|} \right )$ Doing the corresponding substitutions and using a calculator in degree mode, we obtain that the value of the this expression in $\boxed{\ang{29.596}}$ .

Set up a reference frame with its origin at the center of earth such that the direction of the sun from the origin of the frame is along its x-axis. Set up another reference frame Ox'y'z' with its z' axis coincinding with the axis of rotation of earth, thus making an angle of $\theta = 23.5^{\circ}$ with the z-axis. a point on the surface of earth at latitude $\theta_L$ is given by

$(x', y', z') = R ( \sin(90 - \theta_L) \cos \phi, \sin(90 - \theta_L) \sin \phi, \cos(90 - \theta_L) )$

$= R (cos(\theta_L) \cos \phi, cos(\theta_L) \sin \phi, \sin \theta_L) )$

now $(x', y', z')$ and $(x, y, z)$ are related by a rotation about the y-axis by an angle of $\theta = 23.5^{\circ}$

$(x, y, z) = \begin{bmatrix} \cos \theta && 0 && \sin \theta \\ 0 && 1 && 0 \\ -\sin \theta && 0 && \cos \theta \end{bmatrix} (x' , y', z' )$

at sunrise, we have x = 0, hence

$\cos \theta \cos(\theta_L) \cos \phi + \sin \theta \sin(\theta_L) = 0$

so $\phi = -\cos^{-1} (- \dfrac{\sin \theta \sin(\theta_L) }{ \cos \theta \cos \theta_L} ) = - \cos^{-1} (- \tan \theta \tan \theta_L )$

the east direction is the direction of motion of the point on the sphere and is obtained by the derivative of (x', y', z')

with respect to $\phi$

$p' = ( \cos(\theta_L) (-\sin \phi) , \cos(\theta_L) \cos \phi, 0 )$

hence the direction of the east in the absolute frame is

$q' = (\cos \theta \cos(\theta_L)(-\sin \phi) , \cos(th_L) \cos \phi , -\sin \theta \cos(\theta_L) (-\sin \phi) )$

the angle $\psi$ between this direction and direction of the sun which is $(1, 0, 0)$ is

$\psi = \cos^{-1} ( \cos \theta \cos(\theta_L) (-\sin \phi ) / | q' | ) = \cos^{-1} ( - \cos \theta \sin \phi)$

Plugging in the given values for $\theta_L$ and $\theta$ , we can calculate $\phi$ and then we calculate the angle between the direction of East and the direction of the Sun. Starting with the last equation, and with a little bit of manipulation, one can show that an alternate formula for the angle $\psi$ is

$\psi = \displaystyle \sin^{-1} \left( \dfrac{ \sin \theta}{\cos \theta_L} \right)$

Let's place Earth at the centre of a cartesian coordinate system, scaled such that its radius equals 1. Let the sun be in the positive x-direction, and the earth axis in the x-y-plane. The north pole $\vec{p}$ then is at $\vec{p}=(\cos{66.5°},\sin{66.5°},0)$ Nashville $\vec{n}$ is at the day/night terminator, so it has x=0. Also, $\vec{n}\cdot\vec{p}=\sin 36.16°$ , so $\vec{n}=(0,y,\sqrt{1-y^2}) \text{ with } y=\frac{\sin {36.16°}}{\sin {66.5°}}$ For Nashville, the horizon is made up of vectors perpendicular to $\vec{n}$ , and in particular, the East direction is also perpendicular to $\vec{p}$ . So the unit vector in their east direction can be found as $\vec{e}=\frac{\vec{p}×\vec{n}}{|\vec{p}×\vec{n}|} = \frac{\vec{p}×\vec{n}}{\cos{36.16°}}$ Since we want the angle between the sun direction $(1,0,0)$ and east, we only need the first coordinate of this: $\cosφ=e_1=\frac{p_2n_3-p_3n_2}{\cos {36.16°}}=\frac{\sqrt{\sin^2{66.5°}-\sin^2{36.16°}}-0}{\cos {36.16°}}$ And therefore $φ=\boxed{29.596°}$