Monday, October 21, 2013

Past Habitability of Mars

This post is one of two that was inspired by discussion questions I encountered from a week in which there weren't too many questions that caught my eye, but those that did have pretty complicated answers. This question was "Because Mars is just on the edge of the habitable zone of the Sun, is it possible that it did, at one time, fall within the boundaries of the habitable zone and, therefore, support life?"


Aerial image of an ancient Martian riverbed taken with ESA's Mars Express orbiter.
To the first part of the question, yes it did. There is ample evidence that Mars used to have liquid surface water, which is THE criterion for the astronomical habitable zone. Of course, Mars has no flowing water today, largely because its surface temperature is (on average) about -63 Celsius. So what changed? To start, Mars needed a thick atmosphere. A thick atmosphere allows the greenhouse effect to occur and keep the planet warm enough for it to be "habitable". Further, a low-pressure atmosphere makes it harder for liquid water to exist on Mars' surface because it would most likely just end up as water vapor in the Martian atmosphere. This seems weird if you don't remember your high school chemistry (I barely do), so I've included a phase diagram for water below. As you can see, at low pressures, water can go directly from a solid to a gas (in a process called sublimation).

Phase Diagram for water. Note that 1 bar corresponds to Earth's atmospheric pressure.
The Martian atmospheric pressure today is 160 times smaller than that of present-day Earth with a surface temperature too low for liquid water. So Mars must have had a thick atmosphere at some point, but where did it go? The most widely accepted solution is best explained through an analogy with Earth.

One of the single most underrated aspects of Earth that allows us to exist at all is Earth's magnetic field. Our magnetic field is created by the rotation of Earth's liquid metal outer core (famously portrayed as shutting down and being nuked into motion again in The Core). Earth's magnetic field has the important job of shielding Earth from the stream of plasma known as the solar wind. The solar wind specifically consists of a stream of protons and electrons emitted by the upper layers of the Sun's atmosphere (specifically the solar corona). This matters because ionized particles are generally bad news for both Earth's atmosphere and living organisms.
Diagram showing how a planet's magnetic field affects its interactions with the solar wind.
As was covered in class, Mars is small enough that it was able to cool completely. Mars' outer core solidified, so it appears to have lost its magnetic field about 4 billion years ago. Therefore, the atmosphere had no protection from the solar wind and has been eroded away ever since. In this scenario, the particles that make up the solar wind collide with the atoms and molecules in the atmosphere, and give them enough energy to escape Mars' gravity. This process also occurs naturally even with the protection of a magnetic field. In short, if an atom or molecule happens to have enough energy to escape the gravitational pull of the planet (in a thermal distribution, this will happen every once in a while). Not having a magnetic field just speeds up the process.

To summarize the above, it appears that Mars did have an atmosphere in the past. It also briefly had a magnetic field that allowed it to keep that magnetic field for some time. There is also abundant evidence that Mars had liquid water on its surface in the past. As such, we have every reason to believe that Mars used to be habitable.

While this seems straightforward, there's actually a big complicating factor known as the "Faint Young Sun problem". The Faint Young Sun problem was first noted by the famous Carl Sagan and George Mullan in 1972. In short, the Sun used to be less luminous than it is today, which would move the habitable zone closer in to the Sun, which would make young Earth, which we know from geologic evidence to have been quite hot, cooler than it should be. This applies equally well to Mars, and possibly more so because Mars is on the edge of the habitable zone today, which could put it entirely out of the habitable zone in the early solar system. I actually alluded to this idea towards the end of my previous post, Discussion Questions: 9/5.

Well that's problematic. We know that Mars had habitable conditions, but suddenly it's a lot harder for us to re-create them in our models. The typical solution to this problem is that Mars likely had a much thicker atmosphere than we initially thought. It is not unfeasible that Mars once had a thick atmosphere. Whether or not it could have lost that atmosphere over the past 4 billion years is still an open question, though it is an active area of research.

Wednesday, October 16, 2013

Discussion Questions: 9/26

Round three. Fight!

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.
The Ring Nebula is the remnant of a star that was probably more massive than the Sun, but not massive enough to explode as a supernova. Rather, in its old age, the star's core became unstable from fusing hydrogen into helium and helium into carbon in shells around an inactive carbon/oxygen core. This is unstable because helium burning is ignited intermittently, resulting in bursts of energy being released from the core. This causes the star to rapidly pulse, varying in luminosity and size on relatively short timescales (yeah, we astronomers consider 10,000-100,000 years to be short). This phase of a star's life is called the thermally pulsing asymptotic giant branch (TP-AGB), and more information can be found here if you're interested. Ultimately, these pulses become strong enough that the outer layers of the star are lost to the interstellar medium, leaving only the hot, inert former stellar core behind.

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)

Wednesday, October 2, 2013

Discussion Questions: 9/17

Round two of Suvrath's discussion questions. This time, I didn't even try to answer some of the questions because they involved discussions of ethics that I am definitely not qualified to do. I will stick to the scientific questions that I can actually answer without projecting my opinion into the answer (too much).

Why is an incandescent bulb a good example of a blackbody where a fluorescent light bulb is not?
This all comes down to how the two different bulbs work.

An incandescent bulb works by running a current (moving electrons) through a filament that acts as a resistor. A resistor is just any material that doesn't allow charges to move through it quite so easily. Electrons moving through a resistor will lose energy as they go. This energy heats up the resistor, which will eventually get hot enough to glow. Because the filament is a hot, dense object, it will emit like a blackbody, which means it emits light all across the entire visible spectrum. Of course, this also means it is emitting light at non-visible wavelengths also, which is a huge part of why they are so inefficient.

Fluorescent bulbs work by running a current through a diffuse (generally) mercury gas. And, as you know by now, a hot, diffuse gas gives off an emission spectrum. Why do we use mercury though? (Hint, it's not because someone wants to poison you, though it's a good reason not to inhale too deeply right after you break a fluorescent bulb.) Here's an emission spectrum of mercury.
Mercury is a good choice because it emits light at colors that span the whole visible spectrum. Unfortunately, it also emits an awful lot of UV light as well. But we can fix this! We have the technology The coating on a flourescent bulb (this coating seems to be what causes them to appear that milky whitish color) specifically absorbs UV photons and spits the energy back out as visible light, covering even more of the visible spectrum. So you can get all of your basic colors looking like they're supposed to and get the same amount of light far more efficiently than incandescent bulbs. Sounds good to me.

This is not to say that fluorescent bulbs produce no waste heat. They just produce less of it than equivalent (visual) brightness incandescent bulbs.

Are there any stars whose peak emission is not invisible light. If so, would they still be capable of supporting life?
Absolutely! In fact, we can calculate exactly when the surface temperature of a star makes the peak wavelength of its spectrum no longer corresponding to visible light using Wein's law!
Where T is the surface temperature of the star and λ is the wavelength (color) at which the star emits the most light. The violet edge of the visible spectrum occurs at about 400 nanometers, or 0.4 microns (in the units given above) while the red edge occurs at about 700 nm (0.7 microns). Wein's law tells us that stars with temperatures higher than 7250 K and lower than 4100 K will have peak colors outside of the visible spectrum.

Can these stars have habitable zones? Absolutely! Whether or not a star has a habitable zone doesn't depend on its peak color, but on its total energy output. For main sequence stars, a star's temperature, mass, luminosity, and size are all correlated to one another, so bigger stars tend to be hotter and more luminous. The only difference will be that the cooler stars will have smaller habitable zones that are closer in to the star, while hotter stars have larger habitable zones that are farther away. For more than you probably ever wanted to know on habitable zones, check out my previous post specifically on the topic of habitable zones.

There is one critical difference between hotter and cooler stars that would affect the potential for life to develop. Hotter stars live shorter lives. Why does this matter? Life as we know it on Earth took 4.6 billion years from the formation of the solar system to develop. Could intelligent life develop faster? Possibly, but it may also take longer than it did for us. With a sample size of 1, there's no way to tell. But it's probably a safe bet that life needs a decent amount of time to come into existence in the first place, thereby favoring smaller, cooler stars.

What do the spectra of non-blackbody radiators look like?
Conveniently, I was able to cover some of these in class when I subbed for Suvrath. Of course, I only really covered the basics that we know of from (you guessed it) Kirchoff's Laws of Spectroscopy. There is another type of spectrum that isn't covered in typical undergrad astronomy: the power law. A "power law" spectrum is a spectrum whose shape is given by the mathematical relationship
The symbol Iν stands for the specific intensity, or the amount of light emitted at a given frequency, while the symbol ν is generally used to represent frequency, for some reason. α here is just used as a stand-in for some positive number Ultimately, a power-law spectrum will look like Figure (number this appropriately).
General shape of a power law shown on a linear plot. Image courtesy Wikipedia.
Now, the math-savvy readers may note that the equation above should asymptote (approach infinity) as ν approaches zero. This is generally fixed by limiting the range of the power law at low values of whatever variable you're considering. In typical jargon, this is to say that the power law "turns over". Power laws occur an awful lot in nature, but that's a subject for another time.

So what kind of weird objects produce a continuous spectrum that isn't a blackbody? Anything with jets. A "jet", in astronomical terms, is a beam of radiation and energetic particles travelling in the same direction. Generally, objects that emit jets are emitting two jets in opposite directions and are actively accreting matter from a disk. The most interesting of these are black holes. A stellar-mass black hole that is stealing matter from a nearby star with which it shares an orbit (low-mass x-ray binary) or a supermassive black hole that is eating anything unfortunate enough to come too close will have a disk of rather hot material orbiting it. As shown in the artist's depiction below, the two jets will be emitted perpendicular to the hot accretion disk.

While the precise origin of jets is still a bit uncertain, there is reason to believe that they, to some degree, are related to the magnetic fields created from the rotating disk of material around the black hole. The evidence for this is in the spectrum of the radiation. Synchotron radiation is emitted from charged particles rotating in a magnetic field. We expect that particles in the jet will, themselves, have a power-law distribution of energies, which gives the emitted radiation a power-law spectrum as well, where the emitted power (energy emitted every second) at a given frequency goes as
P just stands for the emitted power and p is a variable whose value depends exactly on the source of the synchotron radiation. In general, synchotron emission comes out at radio wavelengths.

Edit: We also get power law spectra from what is known as inverse Compton scattering. In inverse Compton scattering, the hot accretion disk emits photons thermally that bump into very energetic electrons that are flying around in the accretion disk's corona. The corona in this case is just the region above and below the accretion disk which contains very high energy particles flying around. The photons will crash into a relativistic electron and actually gain energy from the collision, which makes them into x-ray photons. However, this has nothing to do with the jets. It's just another source of continuous power-law emission.

Man, I had to dig up my 502 notes for that question. If any of my friends see fit to correct me on what I've said about jets, I'll post updates.

What is it about "green" stars that causes them to appear white to the human eye? Also, why aren't there purple stars?
A star whose peak color (see Wein's Law above) occurs at green wavelengths (~550 nanometers) isn't just emitting green light. In fact, it isn't even emitting mostly green light. It just happens to emit more green light than it does light at any other individual color, but not by much. To illustrate this, I've created a little plot in Excel that shows the blackbody spectrum of an object whose emission peaks at 550 nm (about 5270 K) that spans the whole visible spectrum of light.
That's not a huge difference. While there is definitely a maximum at 550 nm, emission at other colors isn't too far behind.

Now we get into how the human eye works a bit. We have two different types of receptors: rods and cones. Your rods just detect light while your cones are responsible for detecting color. The cones in human eyes, in particular, detect light at red, green, and blue wavelengths, and blend the combination of these three into any color we can perceive. This is why old cathode tube TVs and computer monitors have pixels comprised of red, green, and blue bars, or why Photoshop or MS Paint create colors as a mixture of red, green, and blue. If you have roughly equal amounts of red, green, and blue, the object will look white. This is what happens for "green" stars.

This is also why we can't see purple stars. A given star's spectrum could very well peak at purple wavelengths, but the detectors in our eyes will only see them as blue.