# Does the brightness of day follow simple harmonic 'motion'?

Is it really true that the brightness during the day on earth follows simple harmonic motion? My teacher mentioned this as an example but it doesn't feel obvious to me by any stretch of the imagination (at least for a tilted earth). So how can we work out whether the brightness of day is actually SHM or not?

My attempt: We need the dot product between the sun's direction and surface of the earth. if we start by assuming that the sun is sufficiently far, then the incident rays are parallel, and the intensity of the rays is constant as earth rotates. (since light follows gauss's law)

Taking the latitude angle to be $\phi$ and the azimuthal angle $\theta$. We can use standard polar coordinates to describe the position on earth as $x = r\cos(\omega t+c_0)\cos(\phi)$, $y = r\sin(\omega t+c_1)\cos(\phi)$, $z=r\sin(\phi)$. We can dot product this with the angle of the sun's rays w.r.t time of the year. Problem is, I don't know how work out how the coordinates transform to make the sun rotate around the earth at some angle.

But instead supposing that the angle is fixed at $d$, take it to be $z_r = I\sin(d)$ and $x_r = I\cos(d)$ where intensity is $I$. The dot product is $$rI(\cos(\phi)\cos(d)\cos(\omega t+c_0) + \sin(d)sin(\phi))$$

which is SHM. But have I got it right? What happens when the sun's rays are suddenly given a $y$ component?

This actually gives us the answer we need $cos(\theta''(L,t))$ on page 6. replacing terms, we find that $d = \epsilon$, the axis tilt of earth, $L = \phi$ and $\phi$ represents the rotation of 'sun around the earth'. I happened to set this $\pi/2$ which gave me a special case. Their equation implies a specific function for the direction of the sun. Can anybody explain where it comes from?

I should rename the question 'the sun's position in the sky'

I believe your equation is generally correct (hard to identify what all your angles mean without a diagram), where the $\cos(\omega t)$ term in the case above would take it from $25^o$ to $70^o$ and back in a full cycle.
• ok so $w$ is the rotational speed of earth around it's axis. I have rotated the solar system so that north is the north pole and setup the polar angle that way – lucky-guess Sep 21 '17 at 16:39
• It is pretty much SHM over the course of a day since it's the sum of a sine and a cosine. Also, quite neat that the whole idea of phase representing time zone is only true near the equator and for near zero tilt, where it ends up being $t_0 - mt$. where $m$ is the rotation rate of earth around the sun. – lucky-guess Sep 21 '17 at 18:03