Light

Geometric Optics

Parallax

\begin{equation} \tan{p} = \frac{1 \textrm{ AU}}{d} \end{equation} \begin{equation} d = \frac{1 \textrm{AU}}{\tan p} \simeq \frac{1}{p} \textrm{AU} \end{equation}

A major milestone in observational astronomy occurred when trigonometric parallax was first used to measure the distance to a celestial object. Although a successful observation wouldn't occur until 1838 by Friedrich Bessel for the star 61 Cygni, astronomers had known for years that if the earth was indeed orbiting the sun, parallax should be observable. In fact, the earlier geocentric astronomers used the lack of any stellar parallax observations to support their claim that the earth was indeed motionless. With today's instruments, measuring parallax for nearby objects is easy. However, the angles are so small that these measurement would have been difficult for early astronomers.

The star measured by Bessel has a parallax angle of 285.88 ± 0.54[5] mas. (mas = milliarcseconds). This puts 61 Cygni about 11 light years away. The closest star to us, Proxima Centauri, has a parallax of 0.7687 ± 0.0003 arcsec. This is equivalent to measuring something about an inch in diameter about 4 miles away. So, from City College, measuring a quater located at the Bronx zoo would be a similar level of precision.

The parsec

\begin{equation} d = \frac{1 \textrm{AU}}{\frac{1}{3600}*\frac{\pi}{180} } = 206,265 \; \textrm{AU} \end{equation}

Then, we'll make a new unit of distance called the parsec.
1 pc = 2.0625 × 10⁵ AU = 3.0856776 × 10¹⁵ m = 3.26 ly.

Def: the distance at which 1 AU subtends an angle of one arcsecond

No stars within 1 parsec

Light years

The distance light travels in vacuum in 1 year.

1 Light year = 9,460,730,472,580.8 km = 9 × 10¹⁵ m

Cosmic distances

Object	meters	AU	Light Year	Parsec
Sun	1.496 × 10¹¹	1	1.58 × 10^-5	4.85 × 10^-6
Pluto	7.5 × 10¹²	50.1	-	-
Voyager 1	2.089 × 10¹³	139.6	-	-
Proxima Centuri	4.014 × 10¹⁶	2.69 × 10⁵	4.2	1.301
Center of Milky Way	2.3 × 10²⁰	-	25,000	7,600
Andromeda Galaxy	2.4 × 10²²	-	2.5 × 10⁶	760000

Magnitude Scale

Apparent Magnitude

$m$ indicates how bright it looks to us. For the ancients, $m=1$ was the brightest star (not the sun). $m = 6$ was the dimmest visible star.

Describing Brightness

flux vs. luminosity — a) Luminosity is a measure of the entire output of the star. b) radiant flux is what we see as the brightness, since it depends on how far away the observer is.

Radiant Flux, $F$ is what we mean when we say brightness. Total amount of energy at all wavelengths that crosses a unit area perpendicular to the direction of the light's travel per unit time. Or, energy per second received from a star by 1 square meter. or Watts/Meter².

Luminosity, $L$, energy emitted per second. This is intrinsic to the star and doesn't depend on the observer's position.

$$\begin{equation} F = \frac{L}{4 \pi r^2} \label{eq:fluxluminosityr} \end{equation}$$ can be used to find the flux $F$ at a distance $r$ away from a source with luminosity $L$.

The radiant flux above that reaches the earth from the sun was measured to be on average about 1361 W/m² [ref] (This is above the atmosphere). What is the luminosity of the sun based on this measurement?

Absolute Magnitude

If we put every star at a distance of 10 pcs away from us, then ranked their magnitudes, we could have a scale that gave an absolute magnitude ($M$) .

Since a star that is 100 times brighter will have an apparent magnitude difference of 5, we can write as a ratio between the flux from two stars:

$$\begin{equation} \frac{F_2}{F_1} = 100^{\left( m_1-m_2\right) / 5 } \end{equation}$$

$$\begin{equation} \frac{F_2}{F_1} = 100^{\left( m_1-m_2\right) / 5 } \label{eq:ratiooffluxes} \end{equation}$$ or, if we take the log of both sides: $$\begin{equation} m_1 - m_2 = -2.5 \log_{10}\left( \frac{F_1}{F_2}\right) \end{equation}$$ The we combine eqs. \ref{eq:fluxluminosityr} and \ref{eq:ratiooffluxes} to obtain: $$\begin{equation*} 100^{\left(m-M\right)/5} = \frac{F_{10}}{F} = \left(\frac{d}{10\;\textrm{pc}}\right)^2 \end{equation*}$$ Solving for $d$ $$\begin{equation} d = 10^{\left(m-M+5\right)/5} \textrm{pc} \end{equation}$$ $m-M$ is effectively a measure of the distance to the star and is called the distance modulus: $$\begin{equation} \label{eq:m-M} m-M = 5 \log_{10}(d)-5 = 5 \log_{10}\left( \frac{d}{10 \;\textrm{pc}}\right) \end{equation}$$

Distance Modulus

$$\begin{equation} m-M = 5 \log_{10}(d)-5 = 5 \log_{10}\left( \frac{d}{10 \;\textrm{pc}}\right) \end{equation}$$

What is the absolute magnitude of the sun?→

For other bodies we'll obtain: $$\begin{equation} M = M_\textrm{Sun} - 2.5 \log_{10} \left( \frac{L}{L_\unicode{x2609}}\right) \end{equation}$$

Light as a Wave

$$\begin{equation} c = \lambda \nu \end{equation}$$

Region	Min $\lambda$	Max $\lambda$
Gamma Rays	-	1 nm
X-Ray	1 nm	10 nm
Ultraviolet	10 nm	400 nm
Visible	400 nm	700 nm
Infrared	700 nm	1 mm
Microwave	1 mm	10 cm
Radio	10 cm	-

Double Slit

grating — Drawing showing how blue light will have a first maximum closer to the optical axis.

em-waves — Electromagnetic Radiation is explained by oscillating electric and magnetic fields

The Poynting Vector

$$\begin{equation} \mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B} \end{equation}$$

Radiation Pressure

Absorption

$$\begin{equation} F_\textrm{rad} = \frac{\langle S \rangle A}{c}\cos \theta \label{eq:rad-pressure-absorption} \end{equation}$$

Reflection

$$\begin{equation} F_\textrm{rad} = \frac{2 \langle S \rangle A}{c}\cos^2 \theta \end{equation}$$

Design a solar sail based propulsion system. The spacecraft has a mass of 100 kg and we would like an acceleration of 5 m/s².

Using the second law: $F = ma$. If we have a 100 kg spacecraft and we want an acceleration of 5 m/s², then our force needs to be 500 N. Setting this equal to the desired radiation pressure: (using eq. \eqref{eq:rad-pressure-absorption}) $$\begin{equation} 500 \; \textrm{N}= F_\textrm{rad} = \frac{\langle S \rangle A}{c}\cos \theta \end{equation}$$ We can assume the angle between the Poynting vector and the sail's surface normal is 0 degrees. The $\langle S \rangle$ is just the average energy/time per square meter delivered by the sun: $$\begin{equation} \langle S \rangle = 1365 \; \textrm{W} / \; \textrm{m}^2 \end{equation}$$ and $A$ is the area of the sail, which assuming a circular sail would be given by: $a = \pi r^2$. Putting all this together and solving for $r$: $$\begin{equation} r = \sqrt{\frac{c m a}{\pi \langle S \rangle }} = 5.9 \; \textrm{km} \end{equation}$$ That's probably too large...

Solar Sail Examples

Near_Earth_Asteroid_Scout — Near-Earth Asteroid Scout (NEA Scout), mission concept. Expected to launch in Dec 2019

The IKAROS mission.

By Andrzej Mirecki - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=14656159

LightSail2 — Light Sail 2.

https://www.planetary.org/space-images?imgkeywords=lightsail-2-image-from-space

Blackbody Radiation

$$\begin{equation} \lambda_\textrm{max} T = 0.0028977729 \;\textrm{m} \cdot \textrm{K} \label{eq:wiensdisplacement} \end{equation}$$

Orion Constellation

Stefan-Boltzmann equation

$$\begin{equation} L = A \sigma T^4 \end{equation}$$

or for a spherical star of radius $R$:

$$\begin{equation} L = 4 \pi R^2 \sigma T_e^4 \label{eq:sb-equation-star} \end{equation}$$

Find the peak of the sun's spectrum using its Luminosity and Radius.

What's the blackbody radiation of a person? What color light do people emit? (i.e. find the peak wavelength for an average human.)

Quantization

Rayleigh-Jeans $$\begin{equation*} B_\lambda(T) \simeq \frac{2 c k T}{\lambda^4} \end{equation*}$$ Wien's approximation $$\begin{equation*} B_\lambda(T) \simeq \frac{2 h c^2}{\lambda^5}e^{\frac{- h c}{\lambda k T}} \end{equation*}$$

All was not well in physics around 120 years ago.

There were two functions that physicists had to describe blackbody radiation. One known as Rayleigh-Jeans $$\begin{equation} B_\lambda(T) \simeq \frac{2 c k T}{\lambda^4} \label{eq:rayleighjeanslaw} \end{equation}$$ and the other called Wien's approximation $$\begin{equation} B_\lambda(T) \simeq \frac{2 h c^2}{\lambda^5}e^{\frac{- h c}{\lambda k T}} \label{eq:wiensapprox} \end{equation} $$

The Rayleigh-Jeans law was not perfect because as the wavelength decreases towards zero, this relation shows that the radiance energy would increase to infinity - obviously problematic.

Wien's Approximation was unable to predict correctly the energies at large wavelengths.

And so there were two different relations, one for short wavelengths, and one that worked for long wavelengths. This is never desirable for physicists. One equation that works for all wavelengths is better!

Experimentally, it was known that the following limiting conditions must apply.

At a fixed temperature, the radiance approaches zero as the wavelength goes to zero.
At a fixed temperature, the radiance approaches zero as the wavelength goes to infinity.
At a fixed wavelength, the radiance approaches zero as the temperature goes to zero.
At a fixed wavelength, the radiance approaches infinity as the temperature goes to infinity.

Wien's approximation satisfies the first three but not the last. Rayleigh-Jeans satisfies the last three, but not the first.

Planck function

$$\begin{equation} B_\lambda (T) = \frac{2 h c^2 / \lambda^5}{e^{hc / \lambda k T}-1} \end{equation}$$

planck rayleigh jeans wien — Comparing the three functions for a radiator with temperature of 0.008 Kelvin. Wien's approximation fails in the low frequency limit, while Rayleigh-Jeans fails in the high-frequency. Planck's function fits both!

Show that the maximum of the Planck function gives Wien's displacement

The Planck function is: $$\begin{equation*} B_\lambda (T) = \frac{2 h c^2 / \lambda^5}{e^{hc / \lambda k T}-1} \end{equation*}$$ We need to take the derivative w.r.t. $\lambda$: $$\begin{equation} \frac{dB_\lambda}{d\lambda} = (-5) \left( \frac{2 h c^2}{\lambda^6}\right)\frac{1}{e^{\frac{hc}{k \lambda T}}-1} + \frac{2 h c^2}{\lambda^5}\frac{-1}{\left(e^{\frac{hc}{k \lambda T}}-1\right)^2}\cdot e^\frac{hc}{k\lambda T}\left( \frac{-1}{\lambda^2}\right)\left( \frac{hc}{kT}\right) \end{equation}$$ and then set it equal to 0. $$\begin{equation} \frac{dB_\lambda}{d\lambda} = 0 \end{equation}$$ Doing this and simplifying: $$\begin{equation} -5 + \frac{1}{e^{\frac{hc}{k \lambda T}}-1} e^\frac{hc}{k\lambda T} \left( \frac{hc}{k \lambda T} \right) = 0 \end{equation}$$ Let's set $$\begin{equation} \frac{hc}{k \lambda T} = x \end{equation}$$ Then we get for the above: $$\begin{equation} 5 \left(e^x - 1 \right) = x e^x \end{equation}$$ which is not solvable using algebraic means. We must use computers!

Some mathematica code to do a numerical solution:

  NSolve[5 (Exp[x] - 1) == x Exp[x], x]

yields $x = 4.965$ Thus, returning to the constants: $$\begin{equation} \frac{h c}{k \lambda T} = 4.965 \end{equation}$$ or $$\begin{equation} \lambda T = \frac{h c}{4.965 \; k} \end{equation}$$ Plugging in the values for $h$, $c$, and $k$ gives us: $$\begin{equation} \lambda T = 0.002897 \; \textrm{m} \; \cdot \; \textrm{K} \end{equation}$$