Problem #1: Elliptical galaxy XJS-16 has a spectrum which can be approximated as a 5500 K blackbody.
a) Calculate the wavelength of the peak of this galaxy's spectrum if it's nearby (i.e., redshift = 0).
b) Calculate the wavelength of the peak of this galaxy's spectrum if the galaxy has a redshift of 3.0.
c) If this galaxy has a redshift of 3.0, in what wavelength region (e.g., x-ray, ultraviolet, visible, infrared) is it brightest?
Solution -- Part a): This one goes back to the beginning of the class and provides a nice reminder of some of the things we did way back when. Hopefully, you've recalled that the peak of a blackbody spectrum is related to the the temperature by the following relation:
If so, then this part of the problem is simple:
Solution -- Part b): Now that we know at what wavelength the spectrum of the galaxy peaks if it's nearby and not moving, we can figure out what it might look like if it's far away. We're told that this galaxy has a redshift of 3.0. But what does that mean? Well,
so a redshift or 3.0 must mean that the change in wavelength is 3 times the rest wavelength. The rest wavelength is 545 nm, so the change in wavelength must be 3 times that, or 3 x 545 nm = 1635 nm.
So what's the wavelength that we observe?
where I knew to add the change in wavelength to the rest wavelength because the galaxy is receding from us. If an object is moving away, the waves it emits will be "stretched out" on their way toward us, and so we'll observe a wavelength longer than the wavelengths that were emitted.
Solution -- Part c): This wavelength is in the infrared portion of the electromagnetic spectrum. That's why many of the searches for very distant (and therefore highly redshifted) galaxies are done at infrared wavelengths rather than in the optical.
Problem #2: a) Calculate the value of the critical density
for a Hubble constant of 75 km/s/Mpc (Be careful with units! You should come out with an answer in units of kg/m3.)
b) Calculate the density of the visible mass in our Local Group of galaxies, if the Local Group contains about 4 x 1012 solar masses of material within a spherical volume of radius 3 Mpc.
c) If the universe has a density similar to the density you found for the Local Group in part b), is our universe open or closed?
Solution -- Part a): This problem can be really painful if you're not careful with units. To start with, I really recommend conversing the Hubble constant from its really stupid units of km/s/Mpc to the more sensible units of /seconds.
Here's how to do that:
Though /seconds may seem like an odd unit to you (since there's nothing "on top" of the seconds), it's really quite normal. You might remember that in lab, we calculated the age of the universe (the "Hubble time") from 1/H; this quantity had units of time, so it stands to reason that H must have units of 1/time.
Anyway, now we can calculate the critical density without having to worry about the unit conversions, and that will make things much simpler.
We're told that 4 x 1012 solar masses of material can be found within a spherical region of radius 3 Mpc. This is all we'll need to estimate a density, since density is nothing more than mass/volume. So the mass is
To get the volume in reasonable units (like m3), we'll need to convert the radius from Mpc to m:
So, the volume is
Now, density = mass/volume, so
Solution -- Part c): Since the density we found in part b) is less than the critical density, there isn't enough mass in the universe to overcome the Big Bang's expansion with gravity. Therefore, the universe will keep expanding forever, and we live in an open universe.