Week 5, Post 4

Worksheet 5, Problem 2

We are told that the intensity of radiation emitted from a blackbody is given by:

$I_{\nu}(T) = \frac{2\nu^2}{c^2} \frac{h\nu}{\exp(frac{h\nu}{kT} -1} \equiv B_{\nu}(T)$

Where:

• $B_{\nu}(T) is energy per time, per area, per frequency, per solid angle
• Boltzmann's constant $k = 1.4 \times 10^{-16} \text{ erg} \text{ K}^{-1}$
• Planck's constant $h = 6.6 \times 10^{-27} \text{ erg} \cdot \text{s}$.

a)

We're asked to integrate the blackbody flux $F_{\nu}(T)$ over all frequencies to obtain the bolometric flux emitted from a blackbody, which is $F(T)$.

From problem 1, we solved for $F_{\nu}(T)$:

$F_{\nu}(T) = \frac{2\pi\nu^2}{c^2}\frac{h\nu}{\exp{\frac{h\nu}{kT}}-1}$

The question suggests that we use integration by parts, by substituting the variable:

$u \equiv \frac{h\nu}{kT}$

Then, we can isolate $\nu$:

$\nu \equiv \frac{kT}{h}u$

So, we get:

$d\nu = \frac{kT}{h} du

Let's integrate using $u$ substitution:

${\displaystyle \int F_{\nu}(T) d\nu = \frac{2\pi}{c^2}\frac{k^3T^3}{h^2} \int_0^{\infty} \frac{u^3}{e^u -1} (\frac{kT}{h}) du}$

${\displaystyle= \frac{2\pi k^4 T^4}{c^2 h^3} \int_0^{\infty} \frac{u^3}{e^u - 1}du}$

Let's separate out the temperature-dependent term and isolate the term comprising an integral over all frequencies. This term $\sigma$ includes an integral and a bunch of constants:

${\displaystyle \sigma = \frac{2\pi k^4}{c^2 h^3} \int_0^{\infty} \frac{u^3}{e^u - 1}du \approx 5.7 \times 10^{-5} \text{ erg} \text{ s}^{-1} \text{ cm}^{-2} \text{ K}^{-4}}$

Finally, we can express $F(T)$:

$F(T) = \sigma T^4$

b) The Wien Displacement Law:

We are asked to convert the units of the blackbody intensity from $B_{\nu}(T)$ to $B_{\lambda}(T)$.

$\nu = \frac{c}{\lambda}$

Differentiate to get:

$\frac{d\nu}{d\lambda} = \frac{-c}{\lambda^2}$

Notice the negative sign. This is because $\nu$ and $\lambda$ are inversely proportional.

Recall:

$B_{\nu}(T) = \frac{2\nu^2}{c^2} \frac{h\nu}{\exp(\frac{h\nu}{kT})-1}$

$B_{\nu}(T) = B_{\nu}(T) \frac{d\nu}{d\lambda}$

Plug in $\frac{-c}{\lambda^2}$ for $\frac{d\nu}{d\lambda}$ and $\frac{c}{\lambda}$ for $\nu$:

${\displaystyle B_{\lambda}(T) = \frac{2(\frac{c}{\lambda})^2}{c^2} \frac{h(\frac{c}{\lambda})}{\exp(\frac{hc}{kT\lambda})-1}(\frac{-c}{\lambda^2})}$

${\displaystyle =-\frac{2c^2h}{\lambda^5}\frac{1}{\exp(\frac{hc}{kT\lambda})-1}}$

c)

We are asked to derive an expression for the wavelength $\lambda_{max}$ corresponding to the peak of the intensity distribution at a given temperature $T$. 

$B_{\lambda}(T) = \frac{2c^2 h}{\lambda^5}\frac{1}{\exp(\frac{hc}{kT\lambda})-1}$

We'll substitute $u \equiv \frac{h\nu}{kT}$. To do this, we'll replace $\lambda$ with $\frac{c}{\nu}$.

Also, we need to replace instances with $\nu$ with $u$.

$u = \equiv \frac{h\nu}{kT}$

$\nu = \frac{kTu}{h}$

Let's plug in these values and simplify:

${\displaystyle \frac{d\lambda}{dT}(B_{\lambda}(T)) = \frac{2c^2 h}{(\frac{c}{\nu})^5}\frac{1}{\exp(\frac{h\nu}{kT})-1}}$

${\displaystyle =\frac{2\nu^5 h}{c^3}\frac{1}{\exp(\frac{h\nu}{kT})-1} }$

${\displaystyle = \frac{2(\frac{kTu}{h})^5 h}{c^3}\frac{1}{\exp(u)-1} }$

${\displaystyle B_u(T) = \frac{2k^5 T^5 u^5}{h^4 c^3}\frac{1}{\exp(u)-1} }$

${\displaystyle B_u(T) = \frac{2k^5 T^5}{h^4 c^3}\frac{u^5}{e^u -1} }$

This is the maximum wavelength, and to find the peak of a function, we'll take its derivative and set it to 0, and find the corresponding value of the variable we are differentiating with respect to.

${\displaystyle\frac{du}{dT}(B_u(T) = 0 = \frac{2k^5 T^5}{h^4 c^3}(\frac{5(e^u - 1)u^4 - u^5(e^u)}{(e^4 - 1)^2}) }$

${\displaystyle 0 = \frac{2k^5 T^5}{h^4 c^3} (\frac{5 e^u u^4 - 5u^4 - e^u u^5}{e^2u - 2e^u +1})}

Plugging this into Wolfram Alpha, we solve for $u = 4.96511 \approx 5$

Recall the substitution we used:

$u = \frac{h\nu}{kT} = \frac{hc}{k \lambda T}$

Rearrange this to isolate $\lambda$:

$\lambda_{max} = \frac{hc}{kuT} = \frac{hc}{5kT}$

d) The Rayleigh-Jeans Tail:

Recall:

$B_{\nu}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{\exp(\frac{h\nu}{kT})-1}$

The first-order Taylor expansion for $e^x$ is $1 + x$. Let's plug this in and solve:

$B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{(1 + \frac{h\nu}{kT})-1} = \frac{2\nu^2}{c^2}\frac{h\nu}{\frac{h\nu}{kT}}$

$= \frac{2\nu^2 kT}{c^2}$

e)

In part a), we integrated $F_{\ nu}(T)$ over all frequencies to obtain the bolometric flux emitted from a blackbody, $F(T)$.

$F(T) = \sigma T^4$, where $\sigma = \frac{2\pi k^4}{c^2 h^3}$

A spherical blackbody with radius $R$ has surface area $4 \pi R^2$.

Integrating the flux over the surface of the blackbody, we get:

$L(T) = 4\sigma\pi R^2 T^4$

f)

We are told that there are two stars, one is blue and one is yellow. The yellow star is six times brighter than the blue star.

Recall luminosity:

$L(T) = 4\sigma\pi R^2 T^4$

Recall $\lambda_{max}$:

$\lambda_{max} = \frac{hc}{5kT}$

Rearrange to isolate $T$:

$T = \frac{hc}{5k\lambda_{max}} = \frac{h\nu_{max}}{5k}$

Qualitatively, luminosity goes up with the square of radius, so the yellow star is larger because it is brighter. However, the blue star is hotter because temperature is proportional to frequency.

Let's compare their radii quantitatively. Let's use subscript $B$ to represent the blue star, and subscript $Y$.

$L_Y(T) = 6 L_B(T)$

$4\sigma\pi R_Y^2 T_Y^4 = 6\cdot 4\sigma\pi R_B^2 T_B^4$

$R_Y^2 T_Y^4 = 6 R_B^2 T_B^4$

$R_Y^2 (\frac{hc}{5k\lambda_{Y}})^4 = 6 R_B^2 (frac{hc}{5k\lambda_{B}})^4$

$R_Y^2 (\frac{1}{\lambda_Y})^4 = 6 R_B^2(\frac{1}{\lambda_B})^4$

$\frac{R_Y^2}{R_B^2} = 6(\frac{\lambda_Y}{\lambda_B})^4$

$\frac{R_Y}{R_B} = \sqrt{6}(\frac{\lambda_Y}{\lambda_B})^2$

The yellow part of the spectrum is 590-560 nm, so we'll take it at the middle, 575 nm. The blue part of the spectrum is 490-450 nm, so the middle is 470 nm.

$\frac{R_Y}{R_B} = \sqrt{6}(\frac{575 \text{ nm}}{470 \text{ nm}})^2 = 3.67$

The yellow star is 3.67 times larger than the blue star.

General Post 1

https://www.businessinsider.com/spacex-crew-dragon-capsule-nasa-demo1-mission-iss-docking-2019-3

https://www.space.com/spacex-crew-dragon-success-commercial-spaceflight-era.html

SpaceX became the first private company to dock a commercial spaceship for astronauts to the International Space Station. The spaceship, named the Crew Dragon, was launched on Saturday, March 2 at 2:49 EST, and reached the International Space Station by Sunday morning. The docking process started with a set of safety checks, after which the Crew Dragon was carefully maneuvered to the front of the ISS. Then, the ship made soft contact with a docking node and latched on with six arms. Shortly after, the spaceship fastened itself tightly to Node 2 of the ISS. On Friday, March 8, the six-day flight test concluded with a splashdown in the Atlantic Ocean.

Although the Crew Dragon can carry seven passengers, it was unmanned for this demonstration, carrying only 400 pounds of cargo and a crash-test dummy. However, this event was described by NASA as a “critical first step” in re-establishing US-manned access to space. The purpose of this test was to show that the Crew Dragon could safely transport US astronauts. The Crew Dragon is meant to be a private successor to the Space Shuttle program, which ended in July 2011 after the last flight of the space shuttle Atlantis. Since then, the only way to send astronauts to the ISS was via Russian Soyuz spacecraft. In 2014, NASA signed contracts with SpaceX and Boeing to develop private American solutions. This test flight brings SpaceX closer to carrying out contracted missions for NASA.

General Post 2

http://www.astronomy.com/news/2019/03/a-map-to-planet-nine-charting-the-solar-systems-most-distant-worlds


A new dwarf planet was recently discovered by a team of astronomers, and is now the most far out object in the solar system, earning it the name “Farfarout”. This followed the same team’s discovery of the previous most distant object in the solar system back in December, which was named “Farout”. Farfarout is two billion miles farther away than Farout, a total of 140 AUs from the Sun. The discovery of these very far planets is significant because they could contribute evidence for a large planet on the outer reaches of the solar system, referred to as “Planet Nine”

The orbits of these planets are difficult to track because they are very far away. To collect enough data to determine the orbits and compositions of the planets, the researchers are using powerful earth-based telescopes. As more powerful telescopes come online and better software to process data is developed, we are able to better examine the far reaches of the solar system. However, there may be more than just small icy planets.

There is more evidence to suggest that there is a “super-Earth” planet about 5 to 15 times the mass of Earth which influences the orbits of these small trans-Neptunian Objects (TNOs). These frozen dwarf planets, located beyond Neptune, make their closest approaches to the Sun at around the same location. This is interesting because they all have different orbits, which suggests that a much heavier planet is pulling them into a cluster.