I read an article recently about the supermassive black hole in the center of our galaxy. They discovered it had "erupted". Can you explain what this means?
It seems like you are referring to an article such as this one from Discovery.com.
When someone talks about a supermassive black hole "erupting", they're referring to the black hole accreting matter and emitting vast amounts of radiation as a result. Black holes are very sloppy eaters and always leave a mess. As I vaguely alluded to in my previous post, black holes that are in the process of eating have hot accretion disks formed by the material that was ripped apart by the massive gravity of the black hole. This material is, over time, consumed by the black hole. So a few things to note here. When a black hole is accreting, the material being accreted is emitting a lot of radiation.
An eruption is what happens when an object like a star or a gas cloud falls into a supermassive black hole. The object will be ripped apart and heat up as it accelerates inwards. The accreting material will give off a lot of radiation all across the electromagnetic spectrum. Not all of the emission will be radiation though. As I said before, black holes are very sloppy eaters, and will spew out a lot of gas as well (mostly hydrogen). This will only happen during the accretion process, so when the material is used up, the emission will cease, and the black hole will return to what we call its quiescent phase where it's mostly just sitting there.
Sag A*, the SMBH at the center of the Milky Way is currently quiet and not doing a whole lot. The linked article above mentions a nearby gas cloud that may get eaten within the year, which of course will be pretty exciting, so we'll have to keep our eyes and telescopes open.
If I had a green laser pointer and a red laser pointer, does the frequency of the light cause the green laser to shoot farther and be brighter than the red laser? Could you excite the green laser pointer so the beam starts to burn things on contact?
I chose this question initially to address a few misconceptions, but then I realized that there was actually some fun to this question. So let's get to it.
First off, the frequency of light has nothing to do with how bright it is. Green laser pointers will appear brighter than red laser pointers of the same power because of how our eyes work. Your rods, the receptors in your eye responsible for detecting light in general (without which you would be very bad at seeing in low-light environments) are rather good at detecting green light while they do not detect red light at all. (The cutoff wavelength for rods appears to be around 600 nanometers.) As such, the receptors in your eye combined (rods and cones) will detect more photokns from a green laser than they will from a red laser. Frequency only affects the amount of energy carried by an individual photon.
Can you burn things on contact with a laser if it is "excited" enough? If you make the beam intense enough (increase the number of photons per second), yes. At this point, the color doesn't matter too much (depending on the material). If you hit something with enough highly concentrated energy, it's going to burn so long as it's not purely reflecting all of the light you're shining on it.
As for the beam shooting farther... this is actually interesting, and there are two effects that need to be accounted for. First off, laser beams are not perfect. As the light travels, the beam itself spreads out, so the light is less concentrated at greater distances from the beam source. If we assume totally ideal conditions, then the beam spread is dependent directly on the wavelength of the light. Lower wavelengths diverge less. This means that the red laser would spread out more than the green laser over the same distance, so it won't make as good a beam.
The other effect to contend with on Earth is the atmosphere. Light doesn't just travel through air perfectly because air is made of stuff (good luck breathing if this wasn't the case). Photons travelling through air have a tendency to run into some of these molecules once in a while, and end up going elsewhere. This is what we call scattering, and it's kind of like a microscopic version of a reflection, except the end direction can't be predicted. Anyway, this process also depends on the wavelength of the light you're using. The effect here is reversed from the above though. Scattering has a stronger effect on photons with shorter wavelengths, so it has a stronger effect on the green laser.
Combining these two effects, which beam is the most visible farther away from the source of the two lasers? I haven't the (puts on sunglasses) faintest idea.
Looking at the Ring Nebula, which still has the remnant of its core visible, is that core truly dead and giving off no light so that the nebula can be viewed as an emission spectrum?
Well, someone asked about the Ring Nebula, so I have to put a picture of it in this post because damn, that thing is gorgeous.
|Hubble Space Telescope image of the Ring Nebula combined from images taken in multiple filters.|
This is where we find the Ring Nebula. The little white dot seen in the center of the nebula is the remnant of the stellar core that will one day become a white dwarf (when the outer layers finally drift away entirely). Until then, it is certainly a hot, dense object emitting thermally. So why do we mostly see an emission spectrum from the Ring Nebula (as was asserted in class)? First, the white dwarf's surface temperature is estimated at roughly 125,000 K. That corresponds to peak emission at about 23 nanometers, which is just a factor of two below the UV/X-ray boundary. The VAST majority of the thermal radiation produced by this object is not only not visible light, but is also energetic enough to ionize pretty much anything near it.
Why does that matter? The emission that we see from the gaseous part of the Ring Nebula is a result of the material ionized by the radiation from the white dwarf. Specifically, it comes from free electrons re-joining nuclei to form neutral atoms again. While the radiation doesn't directly come from the recombination of the electron with a nucleus, it comes from the electron rejoining into an excited state (which probably emits a photon that we can't see), then falling down toward the ground level (which emits a photon that we can).
To get to the heart of the question, the emission spectrum of the Ring Nebula far outweighs the visible blackbody emission of the central white dwarf because the VAST majority of the radiation from the white dwarf is high enough energy to ionize the gas surrounding it. As that gas moves away from the white dwarf, it cools significantly, de-ionizes, and the electrons fall down to lower energy levels, emitting visible light photons as they go. So while there is definitely thermal visible light coming from the white dwarf, we mostly just see the emission lines from the nebula gas spectroscopically.
Assuming the Earth is a blackbody since it's solid and opaque with a cold gas (atmosphere) around it, is it possible to see the absorption spectrum of Earth?
Absolutely! You'd probably be surprised at how hard this is to measure though.
First, I should note that the atmospheric spectrum of Earth is actually astrobiologically interesting because it gives us an idea of what we should look for in exoplanet atmospheres if we're looking for some familiar kind of life. Unfortunately, any spectra we would see of distant planets would be from unresolved sources, meaning we wouldn't be able to distinguish individual locations on that planet. So we need a way to look at the Earth as if it were a single point. Because every telescope we have is awfully close to Earth (even the space telescopes), this is not trivial. Fortunately some rather smart people have thought about this and have come up with a very interesting solution: look at light that is reflected from the Moon!
No, I'm not making that up. Yes, it works. The Moon itself has a very featureless spectrum, and is very good at reflecting light. There are two different types of atmospheric spectra you can observe this way: a reflection spectrum and a transmission spectrum. The reflection spectrum is sunlight that reflects off of Earth's surface, and onto the Moon, then back to telescopes on Earth. The reflection spectrum is best observed during a solar eclipse, but can also be generally observed during a new Moon, you just have to be more careful about the atmosphere. The transmission spectrum comes from light that only passes through Earth's atmosphere in a grazing fashion on its way to the Moon. This spectrum is best observed during a lunar eclipse.
Each of the above is a way that we have measured the atmospheric spectra of exoplanets, mostly Hot Jupiters at this point because they're the easiest to observe in many respects. We would love to make such measurements for more Earth-like planets, but this requires some specialized instruments for which there were, at some point, plans that have since been scrapped (much to the dismay of Jim Kasting).
Of course, you probably want to see what such a spectrum looks like in the infrared. The following graph was adapted from Christiansen & Pearl (1997), and the x-axis is given in the god-awful units of wavenumber, which I hate with a fiery passion. For whatever reason, some people like to talk about spectra with wavenumber, which is just the inverse of the wavelength. For reference, wavenumber of 200 corresponds to 50 microns and wavenumber of 1000 corresponds to 10 microns, so wavelength actually increases to the left on this plot. Yes, I know. It's stupid and unintuitive. Sorry.
|Infrared spectrum of the Earth from 6 microns (far right) to 50 microns (far left)|