Friday, May 22, 2009

At the mine

I'm writing this entry from 6800 feet below ground. I am wearing a baby blue jumpsuit (pictures to come, hopefully), safety glasses, steel toed boots, a hair net and a hard hat. At some point, my mom commented that hearing about working in the mine might be more interesting than posts on physics, and so I am going to give the human interest piece a try.

I have been working up in Sudbury, Ontario for the past two and a half weeks at the underground lab I mentioned in the overview posts (linked from the right of this blog). What is it like? Well, it's pretty cool, I have to admit. Life at the lab is in many ways defined by the cage schedule of the mine, as I'll explain. I get up before 7, because I have to catch the 7:30 cage underground. If I miss that cage, I'm pretty sure that I won't be able to go under on that day. So, I'm up at 7 (I don't have to shower, as you'll soon see), drive to the mine, go to my locker. I take off the civvies, and put on a mining jumpsuit (lots of reflective tape), hardhat, glasses, wellington boots. I get my head lamp (there's a slot on the hard hat for the head lamp to slide into), tag in (the mine has a lot of safety rules, but the main one is the tag-in and tag-out system. If you go underground, you have to be tagged in, and then when you come back up you tag out. That way, when the company wants to do some blasting, they can make sure no one is underground. If you forget to tag out, or tag out the wrong person, they are not allowed to blast. People do get calls at 4 in the morning about being tagged in, you do not want to be the person who forgets) and wait for the cage. When it arrives, we all pile in. The cage is very cage-like. It's maybe 5 ft wide and 15 ft deep, made all of beat-up metal, and the miners and lab workers pile in in rows of 4. Sometimes, when it's full, we'll be squeezed all the way in, and I hear stories that "in the old days, we used to put 5 in a row." Then we drop. A couple of people will put their lights on at this point, otherwise we'd just be going down in the dark. We stop at a few places along the way for people to get off at various levels (if we stop too many times, that's known as a "milk run"), and then finally, we arrive at the 6800 ft level.

Next, we have to hike about 1.5 km down a drift. The drift is 10 ft wide maybe, with screen or "shotcrete" helping to support the walls. We'll hike half the way down, and then we call ahead to the lab where someone has advanced ahead of us with an air monitor (the modern version of a canary) to make sure it's safe to proceed. Sometimes there will be water on the ground to tramp through, and there's evidence of mining all over the place. Eventually we arrive at the lab. At this point, we take off our clothes, and take the garbage bags off anything we've brought down with us. We shower (there's a built in shower every morning, which is nice when you're getting up so early [at least for a grad student]) and put on a clean jumpsuit and hair net, etc. The entire lab is a "clean room," which means that considerable effort has gone into making sure that all the dirt and dust picked up on the walk through the drift is cleaned off before we enter the lab. Hence the cleaning precautions.

So now we're in the lab. The walls are all whitewashed (but not straight, since it's a cave, essentially), and most of the ventilation and wiring is visible. It looks like the set of a sci-fi movie. So off I go to my experiment where I do the day's work (fiddling with high voltage power supplies, making sure the detector stays cold, that there is enough liquid nitrogen, doing various radioactive source calibrations, etc). Then, 45 minutes before the cage up time (again, there's a fixed schedule. I can't just come in and out whenever I want), we go through the reverse process, take off the lab clothes, put back on the mining gear, hike back out through the drift, etc. And you'd better make that cage.

So up we go back to the surface (there's a signal system for the cage, and you always know that when they signal 2 short pulses twice, the next stop is the surface), take off the mining gear, shower again (I love that the day is bracketed by showers), and voila, life underground at the lab.

It's a good thing I'm done this little summary, because a liquid nitrogen fill just completed so today's tasks are all done and the detector will survive the weekend, and I have to start cleaning up to catch the next cage out (I'm taking the early cage today).

Saturday, May 16, 2009

More philosophical meanderings

My mom writes in a comment:

I think I would like to know what the consequences are of discovering or measuring dark matter. Also, does what you are doing have any relation whatsoever to things like the Hubble telescope or general space travel that people seem to be doing more and more of? Might your discoveries, for instance, give us an idea of the future of the universe as we know it?
xox MOM


These are good questions. What would be the consequences of discovering dark matter? When people ask me this question, one of the first things that I have to emphasize is that there are no foreseeable applications to my research. Nothing obviously useful will come out of it, unlike, for example, research in quantum computing or more applied fields. Now, there's always the chance that something we develop in trying to detect dark matter could be useful to society (for example, there are a number of ideas to use technologies developed in this field for detecting nuclear weapons at border crossings), but I believe that justifying this research by appealing to possible applications is dishonest.

The only reason I have for searching for dark matter is to increase our ("Mankind's" with a capital M) understanding of the universe. 23% of the universe is dark matter, and 85% of all the matter is dark. There are two aspects to this. The first is humanity's standard musings over "why are we here? how did we get here?" Dark matter is a key component to the evolution of the universe, influencing the expansion rate of the universe and the way matter first clustered to form stars and then planets. If it didn't exist in the way that it does, the Earth would probably not exist and neither would this blog. I'm touching up on religion again, here, which interestingly enough seems to happen quite a lot in this blog.

The second aspect that interests me is that I just think it's cool to know more about the way the universe works. Why is there more matter than antimatter in the universe (another great physics question, as naively we might expect identical amounts in which case we would have all disappeared in a puff of energy a long long time ago)? What was the big bang? Does dark matter really take the form we think it does (I sort of like the fact that we can predict the existence of a particle and then go out and find it, which has happened many times in the past)?


To answer my mom's other questions, this is very closely related to the Hubble telescope in the sense that a lot of the evidence for dark matter comes from telescopes like Hubble, and that telescopes have a chance to detect dark matter in a completely different way from us. Not so much space travel, which in my mind isn't so interesting.

The picture is a simulation of structure formation in the universe. All the filaments and bright sports are made of dark matter (Courtesy http://www.casca.ca/ecass/issues/1997-DS/West/ and interestingly enough titled "hugh.gif")

First summary

In the last post, I finally finished the first "thread" about galaxy rotation curves. My dad (who apparently also reads this blog, although not as consistently as my mom) wasn't quite sure how everything tied together (I believe he missed some entries in the middle). So to briefly recap: I started by explaining the Doppler effect, which was then followed by a series of posts on the wave/particle nature of light. I then discussed the Bohr model of the atom, because it provides a nice framework for understanding the emission of light by atoms. Combining all those posts, we can now understand how to measure the speed of rotation of a galaxy - hydrogen in stars emits light at known wavelengths/frequencies which are then shifted by the Doppler effect. Knowing the math behind the Doppler effect, we can determine how fast the galaxy is rotating. Next, we talked about Newtonian gravity, which led to a prediction for what we expect the rotation of galaxies to look like. In the last post, I described what we actually see, providing evidence for dark matter.

This is the goal of this blog - to try and describe all the pieces that go into a physics argument in a way that's understandable. My hope is that interested readers will see that while specific parts of physics may be esoteric or complicated (i.e. high level math), in a general sense we're making deductions in a way that is very similar to those made in almost any other field of study.

Monday, May 11, 2009

Galaxy rotation curves

Ok, so finally I think we can look at rotation curves. We'll make the simplifying assumption that the objects we are interested in are in a perfectly circular orbit about the center of the galaxy, an assumption which doesn't really change anything so it's ok (another larger point about physics: quite often [in fact, almost always], we take a complicated problem and approximate it into something smaller that we can solve [often called the "spherical cow" approach - we would approximate a cow to be a sphere and go from there]. The question then often becomes "how good was the approximation?" In this case, there is no real difference between circular and elliptical orbits, so the approximation is fine and the conclusions are valid).

We know the equation of circular motion, F=mv2/r. And by hypothesis, the only force acting on the object in orbit is the force of gravity, F=G*m1*m2/r2. In this case, m1 is the mass of the galaxy, and m2 is the mass of the object. We equate the forces, so mv2/r = G*m1*m2/r2. Now, the mass from the circular motion equation is just the mass of the object in orbit, so m2 will cancel. All that remains is to solve for the velocity, since that's what we measure using the Doppler effect and red shift.

m2*v2/r = G*m1*m2/r2

First, divide both sides by m2

v2/r = G*m1/r2

Next, multiply through by r

v2 = G*m1/r

Now, take the square root of both sides

v = Sqrt(G*m1/r)

And that's it. We have derived that the velocity of an object in orbit about a galaxy should be proportional to the square root of the mass of the galaxy divided by the orbital radius. There is one more thing we should be aware of, which is that I haven't made any assumptions yet about the size of the mass of the galaxy. A galaxy is a very large thing, and what happens if you're inside part of it? For example, the Sun and the Earth are somewhere inside the Milky Way galaxy. We do orbit the Milky Way center, but part of the Milky Way is outside our orbit. The answer is that in this case, m1 refers the total mass inside the orbit. It doesn't matter how spatially extended it is in space, as long as the object in which we're interested is outside of the galaxy, the equations are fine. And since I'm particularly interested in the mass of the bright part of the galaxy, it's easy to know when we're outside that part, so everything holds.

Now, let's look at a plot. If all the mass were in the bright part of the galaxy, then outside the bright part (say a radius of 100, just to make the plot look right), from the last equation, we would expect the velocity to fall like 1/Sqrt(r) (G and m1 would be constant). That would look like this:


Instead, we measure a flat line, like this:


Therefore, by deduction, we know that either Newtonian gravity is wrong (a possibility, I'll admit), or that there is more mass than we thought, mass that is not contained in the bright part of the galaxy. In fact, we know the distribution of that mass, as it has to increase like 1/Sqrt(r) or else the the velocity would not be flat.

This is what some of the actual data looks like:

These are measurements of galaxy rotation curves (Begeman, Broeils and Sanders, Mon. Not. R. astr. Soc., 1991, 249, 523). I apologize for the image quality, but velocity is on the y-axis and radius is on the x-axis, and all the black points are actual measurements. You can see at small radius the velocity increases. This is where the bright part of the galaxy is, and as the radius increases, we are containing more mass in the orbit. At larger radius, we would expect to see the velocity drop. But instead, it stays constant. The dashed and dotted lines are the the components of the mass, the bright part and the dark part. This data provides evidence that there is matter that we are not seeing, that is not interacting with light, but is dark matter.

Saturday, May 9, 2009

Newton's theory of gravity (part 2)

I have not posted (I can't quite bring myself to use "blogged" as a verb in the past tense, but I should probably get over such squeamishness) in almost a month now, for which I apologize. My excuses are standard - my work intruded. In the last month I have started looking at data from a new run and been to a conference (I've also been to my 10th high school reunion), and I am now visiting the mine in Canada I mentioned in one of the original posts to help run an experiment underground. Enough about my current activities, however, as it's time that I finally finished off the series on galactic rotation curves and Newtonian gravity. Unfortunately, I've realized that I still can't finish this in one post, but we'll get there eventually.

Mother, I'm sorry to say that we're going to need equations, but hopefully it won't be so bad. We'll start with Newton's Second Law of Motion, which says that F=ma. That's it. Force equals mass times acceleration. This law governs almost all macroscopic kinematics (a fancy word for the physics of motion). When you apply a force to something (say by pushing a book across a table), the book will accelerate with a particular acceleration depending on its mass. As with a lot of physics, this is intuitive, especially if we rewrite the equation as a=F/m. If I push a very light object, I can accelerate it rapidly (m is small, so the acceleration is big). If I try to push a heavy object like a car, I can barely move it at all (m is very large, so the acceleration is small). Likewise, the harder I push (more force), the faster it will accelerate.

Ok, next we'll look at circular motion where I'm presented with a pedagogical dilemma. Ultimately, I only really want the equation of circular motion (F=mv2/r), but I can't very well just state the equation without trying to explain it. I condescendingly fear, however, that any attempts at explaining it with a single paragraph in this blog will only result in confusion (which raises a question concerning the wisdom of the entire endeavor, but I'll ignore that). But here goes. In a circle, there are two important directions: the radial direction, which points from the center of the circle to a point on the circle itself, and the tangential direction, which is perpendicular to the radial direction. At any point on the circle, these directions have the same relationship to each other.

The key to circular motion (i.e. when an object just travels in a perfect circle forever and ever) is that the instantaneous velocity is always tangential to the circle while the acceleration is always radially inward. Imagine that my mother is on a ferris wheel. At point "A" on the ferris wheel, she is travelling straight to the left. At point "B" on the wheel, she is travelling straight down (for example, if she were to jump off the ferris wheel at that point, she would drop straight down). Likewise at points "C" and "D," she is travelling to the right and straight up, respectively. The point being that she is always moving tangentially and never radially.

The strange thing is that the force she experiences (the "centripetal force") is always radially inward. At this point, I think I may just have to revert to a "trust me." It's a matter of mathematical fact that to keep her motion circular, she must experience only radial forces. And if you work out the math, the necessary force is F=mv2/r; in other words, the force required to keep her in circular motion is proportional to the square of her velocity divided by the radius. This does make sense intuitively. The faster she is moving, the more force is required to get her to go around the circle. And, the larger the circle, the less change is required to keep her in orbit.

I'm very unsatisfied with that explanation, but I'll leave it for now. The final equation we need is Newton's equation of gravity. This is an empirical law (keeping in mind the discussion in the prior post). Newton discovered that two masses (call them m1 and m2) separated by a distance (call it r) will exert an attractive force on each other, F=G*m1*m2/r2, where G is a constant that is experimentally determined and set by nature. Every mass in the universe exerts such a force on every other mass and vice versa. However, the value of G is very small, so it takes either an extraordinarily large mass (like the Earth or the Sun) or an extremely small separation to observe this force. But that's all there is to Newtonian gravity.