Category Archives: Astronomy & Physics

Challenge Physics Question: Sliding off a Hemisphere

Here’s a challenge physics question, which you can solve with a working knowledge of first-year college physics (even of the non-Calculus variety), but which you might find a bit tricky. I will post the answer in a week or two.


A small object slides frictionlessly from the top of a hemisphere of radius R. (The size of the object is much less than R.) It will slide along the hemisphere for a while, but at some point it will fly off of the hemisphere.

First question: at what position does the small object leave the sphere? You can give this answer by giving the angle θ shown on the picture above where the small object leaves the sphere.

Second question: what distance d does the object land away from the edge of the sphere?

Answers should be in terms of R, the radius of the sphere, and g, the gravity of earth.

If you want to visualize what’s happening, here’s a quick simulation I rendered using Blender’s physics engine.

Vectors in Physics

Vectors and Scalars

There’s a lot of math in physics. It’s often said that mathematics is the language of science, and this is probably more true for physics than most any other science. Mathematics does an amazing job of representing and modelling reality. Of course, as you get to more complicated models of reality, and more advanced physics, you start needing new and fancier mathematical objects.

For most of high-school mathematics, and much of early college mathematics, you deal with mathematical objects that are just numbers, or that could be just numbers. There is one notable exception: geometry. In geometry, your primitive mathematical objects might be geometrical figures, and you learn some ways of making mathematical statements about them because of theorems about triangles and so forth. However, for much of the rest of the math you talk about, the primitives are just numbers. Sure, equations and formulae are mathematical objects too, but they aren’t the “primitive” objects, the basic things that have an independent identity; rather, they’re built from those primitive objects, and operations on those primitive objects.

“What about variables?” you might be asking at this point. Even in algebra, the variables are, in a sense, just numbers. Of course, to really get algebra, you have to learn how to think about variables abstractly, to be able to manipulate the expressions and understand what’s being said about the variables even if you don’t “solve” for them to figure out exactly what number they are. Indeed, sometimes a variable isn’t just one number at a time; consider the equation of a line:

y\ =\ m\,x\ +\ b

In this equation, m and b are constants; they are just numbers, and they’re supposed to be the same number all the time… although to talk about a line, you don’t necessarily have to know what those numbers are. y and x, however, are variables. For every value of x, there’s a corresponding value of y. In that sense, they aren’t “just numbers”… but every value that x could take is just a number.

When you start applying math to physics, the first wrinkle you add to “just numbers” is dimensionality. If you ask, “How far does the apple fall?”, then “three” is not an adequate answer. “Three meters”, however, would be. Most physical quantities have a dimensionality to them, which means that they have units. When you supply numbers in physics, if you leave the units out, then you haven’t really supplied a meaningful number. However, all of the math still works the same with numbers with units as they do just with numbers. You need to be very careful to bring the units with you. Indeed, your training in math courses have probably done you a disservice, as you’ve probably gotten used to plugging numbers into equations without bothering to write the units down, even when there are units in problems. When you’re doing this in physics, you should always keep the units with the numbers; otherwise, you aren’t even really doing it right, and aren’t talking about the right thing.

However, it turns out that not all useful quantities in physics can be represented by mathematical objects that are just numbers. I’m going to introduce the term “scalar” for a physical quantity that can be represented by just a number. For example, the mass of an object is a scalar: a single number, with units, can represent the mass of an object. However, you can’t represent the motion of an object with scalars. You have to introduce a new kind of mathematical object, the vector.

The sort of vector we’re talking about in physics is something that has both a magnitude and a direction. An example of a vector in physics is velocity. Velocity is not a synonym with speed. Rather, speed is the magnitude of velocity. If a car is going 30mph due north, then that car’s speed, the magnitude of the vector, is 30mph. But the velocity is not 30mph; by itself, “30mph” doesn’t have enough information to convey a vector quantity, because it doesn’t have a direction. The magnitude of a vector is a scalar; it’s just a number.

Here are some standard kinematic quantities:

Vector Quantity Magnitude SI Units
Displacement Distance m
Velocity Speed m/s
Acceleration Magnitude of Acceleration m/s2

Displacement is not the same as distance. Suppose you define a coordinate system, and there is a ball whose coordinates are x=5m, y=0, and z=0. It is not correct to say that the displacement of the ball from the origin is 5m. The distance of the ball from the origin is 5m. The displacement, however, is 5m along the x-direction; you have to specify a direction for a vector quantity. If another ball is at x=-5m, y=0, and z=0, then the displacement of that ball is 5m along the negative-x-direction. The distance of the second ball from the origin is also 5m. Notice that the magnitude (distance) has less information than the full vector quantity (displacement), as two things can have the same distance from the origin, but different displacements.

Because vectors are a different sort of mathematical object from scalars, when you write down a variable that is a vector quantity you need to do something to indicate that you’re talking about a vector. There are different conventions. One of the most common ones is to draw a little arrow over the vector. So, you might indicate displacement by:


Whether or not this renders properly inline will depend on whether your browser is properly supporting Unicode. However, if it does, then the letter after the colon should look like an r with an arrow over it, just like the letter set all by itself above this paragraph: r⃗. (If you don’t see the arrow, blame your browser. It might also be your font; you mignt try installing the Deja Vu fonts and configuring your browser to use them.) Hopefully you see this arrow, and won’t be too confused. If not, I’m also putting vectors in boldface, in hopes that you will be able to distinguish them from scalars.

Similar to displacement, you would write velocity as v⃗, or, setting it off and quoting it and making usre it really works regardless of browser:


The scalar magnitude of a vector is then written as the same letter, only without the boldface or the little arrow. The magnitude of velocity v⃗ is speed v. Sometimes we also use the notation of absolute value to talk about the magnitude of a vector, i.e.:

\left|\vec{v}\right|\ =\ v

Representing Vectors: Components, Little Arrows

Now that we’ve conceptually introduced a new mathematical object, the vector, you need to learn how to represent those objects. That is, what’s a way of expressing an object such that you could figure out what the quantity is, and such that you could do calculations with it. For purposes of visualization, a great way to represent a vector is as an arrow. The arrow points in the direction that is the direction of the vector, and the length of the vector represents the magnitude of that vector. On a graph, you can do this in a very straightforward manner with displacements. For instance, consider a ball that is at x=2.0m, y=3.0m, z=0. You could represent the displacement of that ball from the origin with an arrow as follows:


Notice that when I talk about displacement, I keep saying “relative to the origin”. If you say just “displacement” by itself, it should be assumed to be the displacement relative to the origin. (For this to make sense, of course, you must have defined a coordinate system so you know where the origin is.) You can also talk about the relative displacement between two objects. If there is a second ball at x=-1.5m, y=1.0m, z=0. You could talk about the displacement of the first (red) ball from the second (blue) ball, which would be represented by this arrow:


The arrow points just as described: from the blue ball to the red ball.

Arrows are fine for graphically representing vectors, and are great ways to visualize them. You can build intuition about vectors by visualizing them this way. For doing actual calcualtions, however, it’s easier to represent the vector in terms of its components. The components of the vector are just how far the vector points along each axis. If the displacement of the red ball from the origin is the vector r⃗, then the components are rx=2m, ry=3m, and rz=0. Components give us another way of writing a vector, in notation as a “column vector”:

\vec{r}\ =\ \left[\begin{array}{c} 2\mathrm{m} \\ 3\mathrm{m} \\ 0 \\ \end{array}\right]

In this example, the three numbers are just lined up above each other. This is not meant to indicate a fraction, but the three separate numbers. You can also write vectors as a row vector; for some more advanced math that goes with physics, it’s important to distinguish between row and column vectors, but for what we’re doing here, it doesn’t matter.

\vec{r}\ =\ \left[\begin{array}{ccc}2\mathrm{m} & 3\mathrm{m} & 0 \\ \end{array}\right]

Again, the vector is written as three separate numbers, in brackets, with space between them. You should not take this to mean that the numbers should be multiplied together. (When using this notation in your writing, make sure to leave enough space so it’s clear what you mean.)

Notice that vectors can have units! This should lead you to worry about how long you draw the arrow to represent a vector that isn’t displacement, since the graph above have axes that are scaled in meters (i.e. as lengths). The answer is, because a speed is just a different kind of unit than distance, there isn’t really a right way to do it. As such, you can make the lengths whatever you want. For consistency, though, you should make the relative lengths of velocity arrows are scaled along with the magnitudes of the two vectors. Suppose, for example, that the red ball has velocity v⃗1=[-2m/s, 1m/s, 0m/s], and the blue ball has velocity v⃗2=[4m/s, -2m/s, 0m/s]. You could then indicate the two velocity vectors of the two balls as follows:


With velocity vectors, it’s convential to put the tail of the arrow at the position of the object whose velocity you’re representing. Notice how the relative lengths and the directions of the two vectors are correct relative to each other, but the absolute scaling of the two vectors compared to the labels of the graph isn’t set to anything in particular.

Calculating the magnitude from components

If you have a vector v⃗ represented by components vx, vy, and vz, then it’s very straight forward to calculate the magnitude of the vector by extension of the Pythagorean theorem:

\left|\vec{v}\right|\ =\ \sqrt{{v_x}^2\,+\,{v_y}^2\,+\,{v_z}^2}

For two dimensions, this is not terribly surprising. Consider the red and blue balls above, and the displacement from the blue ball to the red ball. If you call that displacement r⃗, then you could visualize the vector and its components as follows:


Looking at this, you see that indeed you have a right triangle, and the magnitude of the vector is the hypotenuse, so it’s no surprise that here

r\ =\ \left|\vec{r}\right|\ =\ \sqrt{{r_x}^2\,+\,{r_y}^2}

(Note that in this specific case, rz=0.)

Vector operations

So what can you do with vectors? There are a handful of operations that are defined for vectors:

  • You can add two vectors together
  • You can subtract one vector from another
  • You can multiply a vector by a scalar
  • You can take the dot product of two vectors
  • You can take the cross product of two vectors

Notice that there are some things that are not on this list. You can not add a scalar to a vector! That’s simply not a defined mathematical operation. Also, you cannot divide by a vector; that, too, is not a defined mathematical operation. Notice too that there are two different ways to muliply vectors, the dot product and the cross product!

The first three vector operations are pretty easy. To figure out the sum of a vector, just add its components together. That is,

\vec{a}\,+\,\vec{b}\ =\ \left[\begin{array}{c} a_x\,+\,b_x \\ a_y\,+\,b_y \\ a_z\,+\,b_z \\ \end{array}\right]

In terms of drawing little arrows, put the tail of the second vector at the tip of the first vector, and then draw an arrow from the tail of the first vector to the tip of the second vector. When you put the tail of the second at the tip of the first, make sure that you pick up and move the second vector without stretching and rotating it. For example:


To multply a vector by a scalar, multiply each of the vector’s components by that scalar. Three examples:

3\,\left[\begin{array}{c} 2\mathrm{m/s} \\ 3\mathrm{m/s} \\ -1\mathrm{m/s} \\ \end{array}\right]\ =\ \left[\begin{array}{c} 6\mathrm{m/s} \\ 9\mathrm{m/s} \\ -3\mathrm{m/s} \\ \end{array}\right]

-\frac{1}{2}\,\left[\begin{array}{c} 2\mathrm{m/s} \\ 3\mathrm{m/s} \\ -1\mathrm{m/s} \\ \end{array}\right]\ =\ \left[\begin{array}{c} -1\mathrm{m/s} \\ -1.5\mathrm{m/s} \\ 0.5\mathrm{m/s} \\ \end{array}\right]

c\,\vec{a}\ =\ \left[\begin{array}{c} c\,a_x \\ c\,a_y \\ c\,a_z \\ \end{array}\right]

In terms of representing vectors as arrows, when you multiply a vector by a scalar, you stretch the length of the arrow by the same factor as the scalar. If the scalar is negative, you turn the arrow arround. The simplest example is when you multiply a vector by -1; in that case, it would be represented by an arrow of exactly the same length, but pointing in the opposite direction.

For subtracting vectors, I find it easiest to think of adding the negative of the second vector to the first, that is:

\vec{a}\,-\,\vec{b}\ =\ \vec{a}\,+\,(-\vec{b})\ =\ \vec{a}\,+\,(-1)\vec{b}

For example, below, I have two vectors a⃗ and b⃗. To figure out a⃗b⃗, first I multiply b⃗ by -1 (i.e. flip it around), and then I do the standard tip-to-tail addition:


Dot Products

We’ll leave cross products for another time. However, it’s worth knowing how to take the dot product of two vectors. When you take the dot product of two vectors, the result is actually a scalar. The dot product of two vectors is defined as:

\vec{a}\,\cdot\,\vec{b}\ =\ |\vec{a}|\,|\vec{b}|\,\cos{\theta}

That is, it is the magnitude of the two vectors multiplied together, times the angle between the two vectors. (In this case, we are using the greek letter θ to represent the angle between the two vectors.


If you remember your trigonometry and the definition of cosine, you may realize that b cos(θ) is the same as the projection of b⃗ on to a⃗:


So,the dot product of two vectors is the magnitude of the two vectors multiplied together, then modified by how much the two vectors are along each other. If they’re pointing in almost the same direction (θ is close to 0), then cos(θ) will be close to 1, and the dot product will be close to the product of the two magnitudes. If the two vectors are almost perpendicular (θ is close to 90°), then cos(θ) will be close to 0 and the dot product will also be close to zero. Finally, if the two vectors point in opposite directions, θ will be bigger than 90° (with θ being exactly 180° if the two vectors are in exactly the opposite direction) and the dot product will be negative.

Calcuating the dot product from components of vectors is straightforward. Just multiply the same components of each vector by each other, and add them all up. That is,

\vec{a}\,\cdot\,\vec{b}\ =\ a_x\,b_x\ + a_y\,b_y\ +\ a_z\,b_z

Unit Vectors

There is one more concept that’s worth mentioning. When you define a coordinate system, you define unit vectors to go along with them. Unit vectors are vectors that have length 1 (“unit length”, hence the name), and that point right along the axes. Note that unit vectors are unitless! There are lots of ways to notate unit vectors. One is to use the variable e⃗ with a subscript indicating which vector you’re talking about. So, the three unit vectors would then be:

\vec{e}_x\,=\,\left[\begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array}\right]\ \ \ \ \vec{e}_y\,=\,\left[\begin{array}{c} 0 \\ 1 \\ 0 \\ \end{array}\right]\ \ \ \ \vec{e}_z\,=\,\left[\begin{array}{c} 0 \\ 0 \\ 1 \\ \end{array}\right]

Putting together multiplying vectors by scalars and adding vectors together, you can compose any vector by multiplying its components by unit vectors and adding them all together:

\vec{a}\ =\ a_x\,\vec{e}_x\ +\ a_y\,\vec{e}_y\ +\ a_z\,\vec{e}_z

The Big Bang Wasn’t All At One Point (Cosmos Commentary)

I finally got around to watching the first episode of Cosmos. I quite enjoyed it, although probably not as much as I would have were I still 9 (which is the age I was back when I used to watch Carl Sagan doing Cosmos… and that, indeed, is probably a nontrivial part of why I’m an astronomer today rather than a paleontologist). I think it’s awesome that once again we’ve got a very charismatic astronomer on TV sharing the wonders of the Universe with us. Alas, I doubt it will have anywhere near the cultural impact that the original Cosmos did, simply because there is so much more out there to pay attention to now. (Not only is there more out there to pay attention to, but over time American society has become more and more ADHD.) Back in the late 1970s, there was little more than three networks of TV to choose from; it was the rare household that had cable. Now, most people have many more options for TV, never mind the ability to download stuff off the Internet on demand. Even if it’s just as high-quality, just as cool, and just as engaging as the old Cosmos was, I fear that the new Cosmos will not be noticed by as large a fraction of the population, and will be more quickly forgotten as people move on to the next shiny thing.

As for the show itself: it all seemed pretty basic to me, but then again, I’m a PhD physicist and professional astronomer who does a fair amount of astronomy outreach, so I was not the primary target audience. I liked the homage to the old show– not just the explicit one at the end (which brought a tear to my eye), but the “we are starstuff” comment, and the Ship of the Imagination (which, as Tyson points out, allows you to travel much faster the speed of light, something I’m doing all the time when I teach astronomy classes).

I did have a couple of quibbles, though. My first was when he was flying through the Solar System’s asteroid belt. The asteroid belt was thick with rocks, creating a massive hazard. The real asteroid belt is not like that. There is less mass, total, of asteroids, than there is in any single planet, and they’re spread out over a huge area in the disk of the solar system. This is why we can fly spacecraft through the asteroid belt without worrying about weaving and dodging. There, asteroids just aren’t that thick.

What, is this Cosmos, or The Empire Strikes Back?

To be fair, when he was out in the Oort cloud, although yet again they were shown too thick (I know, for purposes of actually being able to see something), he did mention that the Oort cloud objects are typically as far apart as Earth is from Saturn. Still, the visual image will stick with people more than the words.

My primary quibble with the show, though, is the title of this post. One sentence of what he said promulgated one of the primary misconceptions about the nature of the Big Bang. “Our entire universe emerged form a point smaller than single atom.” GAH! No! Indeed, Tyson was (perhaps deliberately) cagey about the difference between our Universe and our Observable Universe. He did use the term “Observable Universe”, with a good description. (It’s as far away as we can see, a horizon defined by the speed of light and the 13.8-billion-year-old age of the Universe.) However, thereafter, he seemed to be conflating the Universe with the Observable Universe. While there are some good reasons why one might do this, the way in which he did it fed into a very common misconception about the Big Bang.

Even though our observable universe is finite, the whole universe is much bigger– indeed, perhaps (probably?) infinite.

Here’s the real story, given the Big Bang model as we best understand and use it in astronomy: the Big Bang didn’t happen all at one point. Rather, the Big Bang happened everywhere. The problem with describing it as happening at one point is that it gives you the misconception that we could identify a point in space away from which everything is rushing. This is not the description of our Universe that shows up in modern cosmological models. Every point in the Universe is equivalently the center. Any point in space you can identify: that is where the Big Bang happened. Everything is rushing away from everything else. It’s really not like an explosion, where there’s a center everything rushes away from. (I wrote about this years ago in my blog post “Big Bang”: A terrible name for a great theory.)

Strictly speaking, it is true that our observable universe was once upon a time compressed into a size smaller than the size of an atom. However, saying that by itself implies a misconception: that that compressed, less-than-an-atom size of extremely dense, extremely exotic matter is all there was. In fact, that’s not right. Our Observable Universe was that small… but just as today there is other Universe (filled with galaxies) outside the boundaries of our Observable Universe, at that early epoch there was more extremely exotic dense-matter Universe outside the atom-sized ball that would one day expand and become today’s Observable Universe. Indeed, if the Universe today is infinite, it was always infinite… even back at that early epoch we’re talking about.

The Observable Universe (or a 2d projection thereof) at a period a tiny fraction of a second later than what I’m talking about in the text.
The whole Universe (or a 2d projection thereof) at the same epoch.

This may seem like a minor quibble, but the notion of the Big Bang as an explosion, something everything is rushing away from, is a very tenacious misconception that leads to other misconceptions about our Universe amongst many people I run into. It’s a little difficult to wrap your head around the real model– indeed, people find talks about cosmology that try to describe the real situation (and also the cosmology section of my current ongoing astronomy class) very brain-hurty. But, to my point of view, that’s part of the fun!

There was one throwaway comment about the Big Bang that Tyson made in Cosmos that I really liked. Just before the comment about the atom-sized Universe that got me worked up to make this post, he said about this early Big Bang epoch that “It’s as far back as we can see in time… for now.” That “for now” is great, and spot on. If you read A Brief History of Time by Stephen Hawking, he’ll talk about how the Big Bang was the beginning of time, and how it’s not even really meaningful to ask what was “before” the Big Bang. While that’s true in a purely classical General Relativity description of the Big Bang, we know that such a description can’t be right… because our Universe also has Quantum Mechanics in it, and we have huge amounts of experimental evidence telling us that we need to take Quantum Mechanics seriously. The real story is that there is an extremely early epoch in the Universe (what I tend to think of as “the beginning” nowadays) about which we can make supportable statements based on our understanding of physics. However, we also know that we don’t understand physics well enough to really know what the Universe was like before that early epoch. So, it is meaningful to talk about a before, it’s just that that before is a “known unknown”.

For now.

The Astronomical Magnitude System

Background: What and Why?

Magnitudes are a system astronomers use to talk about the brightness of objects. They’re often a headache for astronomy students, because some things about them are counterintuitive. They also leave physicists who come to astronomy scratching their heads, because said physicists may not appreciate the historical reason why it actually makes sense to use magnitudes. I go back and forth myself on whether or not I should teach magnitudes in an introductory astronomy class. Sometimes I decide not to, because they really don’t add much to the understanding of the physics of astronomical systems, they’re just one more complication when it comes to dealing with numbers. However, at the moment, my opinion has vacillated towards thinking it is worth teaching, for two reasons. First, because they are ubiquitous in astronomy, you will see them referenced a lot. For instance, if you click on an object in Stellarium, you will be told the magnitude of that object. Second, and perhaps more importantly, it’s a worthwhile intellectual exercise. The magnitude scale is a logarithmic scale, and I think it’s good brain exercise to make students wrestle with that.

The two most important things to know about magnitudes— the first of these being one of the things that makes the difficult to work with— are as follows:

  • A larger value of the magnitude represents a dimmer object
  • The magnitude scale is a logarithmic scale, so a given difference of magnitudes represents a given multiple of brightness.

I’ll explore both of these in greater depth below as I give the definition of the magnitude system.

Historically, the ancients categorized stars in brightness classes. The brightest stars in the sky (all named) were called “first magnitude” stars. The ones that appeared a step down in brightness were “second magnitude” stars, and so forth. Modern astronomers reverse-engineered this historical listing of stars to give a numerical “magnitude” that corresponds to the measured brightness of the stars such that stars would more or less have the same magnitude as they had when classified by the ancients. Because the response of our eye is much more approximately logarithmic than it is linear, this led to the magnitude scale being a logarithmic scale.

Flux and Apparent Magnitudes

When an astronomer talks about “the” magnitude of an object, she is usually referring to the apparent magnitude. This corresponds to a measurement of how bright an object is to a given observer (almost always an observer at Earth), not to the intrinsic energy output of the object. To quantify brightness, we use the concept of flux or energy flux, which is defined as the rate at which energy is collected by a telescope with a 1 m2 aperture. Flux comes in units of W/m2 (where a W, of course, is a J/s). Notice the m2 in the denominator. This is what makes flux not depend on the telescope that’s looking at the object. If two telescopes look at the same object, the one with the larger aperture will collect more energy from the same object. To figure out the rate of energy collection, you have to multiply the flux by the collecting area of the telescope. (Real telescopes also have an efficiency that combines the reflectively of their mirrors and the intrinsic quantum efficiency of their detector, but we’ll not worry about that here.)

Suppose you’re looking at two stars, one with flux f1, the other with flux f2. The difference between the magnitudes of these two objects is then defined by:

m_1\ -\ m_2\ =\ -2.5\,\log\frac{f_1}{f_2}

Notice that this is a relative definition; it defines the difference in magnitudes in terms of the quotient of the fluxes of the two objects. The logarithm here is a base-10 logarithm. Warning: while the button on many calculators that performs this function is often labelled “log”, the log() function in many computer languages and applications actually does a natural logarithm (usually called “ln”) rather than a base-10 logarithm. To find out which you have, try taking the log of 10. If you get 1, then you’re doing a base-10 logarithm. If you’re doing a natural logarithm, you will get 2.3.

What is a log?

A logarithm is the inverse operation of exponentiation. This means that if you have

x\ =\ 10^a

then it’s also true that

a\ =\ \log(x)

You can think of a logarithm as being the operator that returns whatever you have to raise 10 to in order to get the argument of the logarithm. The natural logarithm (ln) is the same thing, only it’s what you raise e to rather than 10. Because logarithms are exponents, a few interesting properties apply to logs, which are useful when dealing with magnitudes:

\log(ab)\ =\ \log(a)\ +\ \log(b)
\log(\frac{a}{b})\ =\ \log(a)\ -\ \log(b)
\log(\frac{1}{a})\ =\ -\,\log(a)
\log(1)\ =\ 0
\log(a^b)\ =\ b\,\log(a)

Redux: Flux and Apparent Magnitude

Returning to our definition of the relative magnitude of two objects:

m_1\ -\ m_2\ =\ -2.5\,\log\frac{f_1}{f_2}

this also tells us how to get the ratio of fluxes from the magnitudes:

\frac{f_1}{f_2}\ =\ 10^{(m_1-m_2)/-2.5}\ =\ 10^{(m_2-m_1)/2.5}

The -2.5 in front of the logarithm is very important. The negative sign gives the feature that brighter objects have lower magnitudes. For instance, if object 1 has 10 times the flux of object 2, then:

m_1\ -\ m_2\ =\ -2.5\,\log(\frac{f_1}{f_2})\ =\ -2.5\,\log(10)\ =\ -2.5

Because the difference is negative, m1 is less than m2, given that object 1 has a higher flux. The 2.5 means tells you how many times one object must be brighter than another to have a certain difference in flux. It’s chosen so that a factor of 100 in flux corresponds to a difference of five magnitudes— it works out, because log(100)=2.

So far, however, I’ve only told you how to compare magnitudes. How, then, can we talk about “the” magnitude of an object? For that to happen, we must have some agreed-upon reference that is the standard of comparison. Here’s where things get sad. There two different standards in widespread usage. A more modern system, known as “AB” magnitudes, defines an object of a given flux as having magnitude 0. The more traditional system, still in widespread use by a lot of astronomers, and the system used by programs such as Stellarium, uses “Vega-based” magnitudes. In this system, the star Vega is defined to have magnitude 0. If you compare the flux of an object to the flux of Vega, you get “the” apparent magnitude for that object. The difference between Vega magnitudes and AB magnitudes matters when you’re talking about magnitudes through different filters. Note, however, that if you are comparing to objects, the difference in magnitude will be the same regardless of which system you’re on. The system matters when you cite the magnitude of an object in a given bandpass or filter.

As a few examples, on the Vega system the Sun has an apparent magnitude of -26.74. Sirius, the brightest star in the night sky, has a magnitude of -1.46. Vega has a magnitude of 0.03. (What? you cry. Why not exactly 0? It turns out that “Vega magnitudes” are based on a historical idealized Vega rather than on the real thing.) The North Star, Polaris, has a magnitude of 2.02. The dimmest stars you can see under a good dark sky, on a moonless night and away from city lights, will be magnitude 5 or 6. (Numbers are visual magnitudes, and are from SIMBAD, except for the Sun’s magnitude which is from this page from GSFC.)

Filters and “Color Index”

Up to now, we’ve been talking about the flux of an object, implicitly including all the flux at all wavelengths. The technical term for this is the bolometric flux. In reality, we generally do not collect the flux at all wavelengths. Every detector we use is sensitive only to a finite range of wavelengths. For example, our eyes are only sensitive to wavelengths in about the range 450-650 nm, but stars emit light at wavelengths outside that range as well as inside that range. What’s more, there is a benefit to talking about the flux in even smaller ranges of wavelengths; this allows us to quantify the color of an object, by comparing (say) its flux at red wavelengths to its flux at green wavelengths.

In order to talk about colors and fluxes through different filters, we have to choose a filter set to use. A filter is defined by its transmission function; that is, what fraction of the photons it lets through as a function of wavelength. Two of the common filter systems in use are the Johnson-Cousins filters and the Sloan Digital Sky Survey filters. This image shows the Johnson-Cousins passbands:

UBVRI passbands from Bessel, 1990, PASP, 102, 1181

This system is sometimes called the “UBVRI” system. U stands for Ultraviolet, B for blue, V for visual, R for red, and I for infrared. All of these letters make sense except for V; V is not violet. V is more like green or yellow-green. It corresponds to more or less the peak of the eye’s sensitivity.

You can then define the flux through a filter as the flux coming from an object including just the photons transmitted through that filter. This will be less than the flux of the object as a whole (unless the object is very perverse and all of its light is emitted exactly at the wavelength where the transmission of the filter is exactly 1.0). A color, then, corresponds to the ratio of fluxes an object shows through different filters. For a given object, fB/fV would be the ratio of its B-band flux to its V-band flux. This ratio will be higher for bluer objects, because more of the light will be emitted at shorter wavelengths. If f0V and f0B are the fluxes of the reference object (i.e. Vega), then we have:

m_B\ -\ m_{0B}\ =\ -2.5\,\log\frac{f_B}{f_{0B}}
m_V\ -\ m_{0V}\ =\ -2.5\,\log\frac{f_V}{f_{0V}}

However, remember that the magnitude of Vega is defined to be 0, so:

m_B\ =\ -2.5\,\log\frac{f_B}{f_{0B}}
m_V\ =\ -2.5\,\log\frac{f_V}{f_{0V}}

If we subtract these two numbers, we get:

m_B\ -\ m_V =\ -2.5\,\log\left(\frac{f_B}{f_{0B}}\,\frac{f_{0V}}{f_V}\right)

That’s a bit of a mouthful, but it corresponds to the “color index” B-V of an object, defined by:

B\,-\,V\ =\ m_B\ -\ m_V

Notice that because a higher magnitude is a dimmer object, as B-V gets larger, it means that the flux in the B filter gets lower compared to the flux in the V filter. This is backwards from what might be intuitive.

Astronomers will very frequently cite the color index (something like B-V, B-R, V-I, or similar) as a way of talking about the color of stars or other astronomical objects.

Absolute Magnitudes and Distance Modulus

Apparent magnitude (or just “magnitude”) corresponds to flux, or the observed brightness of an object. We also have a magnitude defined that corresponds to the luminosity (intrinsic energy output) of an object, and we call that “absolute magnitude”. The absolute magnitude is defined as the magnitude that would be observed for an object by an observer 10pc away from the object. (pc means parsecs; one parsec is equal to 3.262 light-years, or 3.086×1016 meters.) This may seem arbitrary… and it is. It’s just the reference distance. We had to pick something. So, hey, why not 10pc?

Imagine we’re looking at two stars, the Sun, and another star just like the Sun that is exactly 10pc away. The flux of the Sun is f, and the flux of the other star we shall call fG. Both have the same luminosity (which we’ll just call L). We know that flux depends on distance by:

f\ =\ \frac{L}{4\pi\,d^2}

So, we have:

\frac{f_\odot}{f_G}\ =\ \left(\frac{L}{4\pi\,{d_\odot}^2}\right)\,\left(\frac{4\pi\,{d_G}^2}{L}\right)


\frac{f_\odot}{f_G}\ =\  \left(\frac{d_G}{d_\odot}\right)^2

From the definition of magnitudes, we also have:

m\ -\ M\ =\ -2.5\,\log\frac{f_\odot}{f_G}

where m is the magnitude of the Sun and M is the magnitude of the Sun as observed 10pc away. Because the magnitude of the Sun as observed 10pc away is the definition of absolute magnitude, M is also the absolute magnitude of the Sun. By convention, we use capital letters for the absolute magnitudes of stars. (However, be careful! We’ll often use capital letters for color indexes even when we’re talking about fluxes. It turns out that (discounting redshift effects) the absolute and apparent color index of an object will be exactly the same, so we can afford to be sloppy.)

Putting these two equations together, we have:

m\ -\ M\ =\ -2.5\,\log\left[\left(\frac{d_G}{d_\odot}\right)^2\right]
m\ -\ M\ =\ 5\,\log\frac{d_\odot}{10\,\mathrm{pc}}

This last equation defines the distance modulus for the Sun. To get there from the previous equation, I’ve used a couple of the properties of logarithms given earlier. In general, for an object a distance d away from us, its distance modulus is the difference between its apparent magnitude m and absolute magnitude M, and is:

m\ -\ M\ =\ 5\,\log\frac{d}{10\,\mathrm{pc}}

For the Sun, the distance from the Earth is 1AU. Putting this in to the distance modulus equation, together with a unit conversion from AU to pc, we can figure out the absolute magnitude of the Sun:

m\ -\ M\ =\ 5\,\log\frac{d}{10\,\mathrm{pc}}
M\ =\ m\ -\ 5\,\log\frac{d}{10\,\mathrm{pc}}
M\ =\ -26.74\ -\ 5\,\log\left[\left(\frac{1\,\mathrm{AU}}{10\,\mathrm{pc}}\right)\,\left(\frac{1\,\mathrm{pc}}{206265\,\mathrm{AU}}\right)\right]
M\ =\ 4.83

As a final note, you can, of course, turn around the distance modulus equation to write the distance in terms of the absolute and observed magnitudes of an object:

d\ =\ (10\,\mathrm{pc})\,10^{(m-M)/5}


The magnitude system is a system astronomers use to talk about the brightnesses of stars. It is a logarithmic system, so a step of magnitude corresponds to a factor of energy output or energy detected. It is also backwards, so that larger magnitudes correspond to dimmer objects. Magnitudes can be “bolometric magnitudes” (taking into account all flux at all wavelengths), but more commonly we talk about magnitudes through a given filter. The color index is the difference of two magnitudes; by convention, we subtract the magnitude through the redder filter from the magnitude through the bluer filter. Absolute magnitudes are defined as the magnitude that would be observed by an observer 10pc away from the object; absolute magnitudes correspond to the luminosity of an object, while apparent (or observed) magnitudes correspond to the flux of an object. Finally, the distance modulus is defined as the difference between the observed and absolute magnitudes of an object, and corresponds directly to the distance between you and that object.

A computer animation of a thermonuclear supernova

A year ago, I taught a 3D Computer Modelling and Animation class. Most of the class was focused on the students working on projects in groups of 1-3. During that time, I did a small project myself as well. I posted a still image from the project a year ago, and promised to post the movie. I’m only now getting around to doing that….

Here is a direct link to the movie. The text in this blog post, and the movie, are also available on the web here. The movie is currently in Ogg Theora format. At some point, I may also put online on that web page a file in another format.

Click to embiggen

Thermonuclear Supernovae

A thermonuclear supernova, also called a Type Ia supernova, occurs when a white dwarf star passes a critical mass (the Chandrasekhar mass). Too massive to support itself under the influence of gravity it starts to compress. This compression triggers runaway nuclear fusion, and the entire star blows itself away in a massive thermonuclear explosion.

White dwarf stars are whats left over when a moderate mass star (less than about 8 times the mass of the Sun) ends its life. Towards the end of its life, such a star will slough off its outer layers, which briefly (for a few ten thousand years) glow as a planetary nebula. The core of the star, which is probably somewhere between 0.4 and 1.4 times the mass of the Sun but only about the size of the Earth, is left behind. It’s made of Carbon and Oxygen, but given its mass and size is incredibly dense. It is supported by “Fermi degeneracy pressure”. To those who know some Quantum Mechanics, the electrons in the white dwarf are in a degenerate fermi gas. If you don’t know what that means, suffice to say that the electrons (and thus the nuclei that go along with them) are packed together absolutely as close together as the fundamental laws of physics (the same basic things that give us the Heisenberg Uncertainty Principle) will allow them to be.

Such a configuration in a star is only stable up to 1.4 times the mass of the Sun (which is the aforementioned Chandrasekhar mass). If it starts smaller than that, how does it get to the necessary size? It must have a source of mass somewhere. There are two possible ways for a white dwarf to reach the Chandrasekhar mass. First, if the white dwarf star has another regular star as a companion, and if it’s orbiting that star closely enough, it’s possible that the gravity of the white dwarf will be able to slowly pull some of the gas off of the surface of the other star and accete it on to itself. If the mass builds up to the critical mass, the white dwarf starts to collapse, and, boom, thermonuclear bomb 1.4 times the mass of the Sun. This is called the “single degenerate” scenario, beacuse there is only one white dwarf (the degenerate object).

The second possibility is called the “double degenerate” scenario. In this case, two white dwarfs, both of them less than the Chandrasekhar mass, come together. Neither one by itself has enough mass to explode. But, if the two come together and merge, the result can be a degnerate object that’s above the Chandrasekhar mass, and boom, supernova.

This movie depicts the single-degenerate scenario, where a white dwarf has a regular star (or perhaps a subgiant or giant star) as a companion. The mass pulled off of the companion builds up in an swirling accretion disk around the white dwarf. Mass from the inner part of the disk falls in on to the white dwarf until it reaches the critical mass and explodes.

Stages and Timescales

So that the movie can complete in a reasonable period of time, I play a little fast and loose with timescales. I’m going to describe the major steps of the movie, and talk about how everything evolves too fast in the movie as compared to in real life.


A star that is several times the mass of the Sun will live a few hundred million years before it becomes a white dwarf. If it’s close enough to its companion (which will be of lower mass than the star that left behind the white dwarf started as), it might start accreting matter from it right away. If it’s farther, it might not start accreting matter until the companion star approaches the end of its life and starts to swell. This can mean a delay of anywhere from millions to hundreds of millions of years after the first star becomes a white dwarf before it has accreted enough mass of its companion to reach the very final stages depicted in the movie here. It’s possible while this is happening that there might be sub-supernova explosions, as some of the hydrogen gas collected on to the white dwarf undergoes a (smaller, but still huge) fusion explosion; we might observe such an event as a nova.

You might object that the camera is moving through the system faster than the speed of light, and you would be right. But, what the heck, it’s an animation! I’m showing you what’s there, not what it would look like if you were really flying through the system.


The explosion itself is instantaenous on astronomical timescales. There is some debate amongst theorists who model the explosions exactly how it happens, but even those arguing for a slower explosion still calculate that the explosion itself is over in about a second (e.g. Ciaraldi-Schoolman, Seitenzhal, and Röpke, 2013). The movie doesn’t really depict the thermonuclear fusion itself; it depictes the expanding blast wave of material blasted away and expanding as a result of the explosion. (Pictures of nuclear explosions we’ve created with our bombs on Earth are the same; the actual nuclear event is over instantly, and then the “explosion” is the expanding blast wave.) Watching the movie, you may think that the expanding blast wave is awfully sedate for such an extreme explosion. In fact, if anything it’s expanding too fast compared to how long it should take to expand in reality! If the star depicted in the video is a subgiant star, it probably has a radius that is at least several times the radius of the Sun. The white dwarf is a similar distance away from the surface of the companion star. If the white dwarf is 10 Solar Radii away from the surface of the star, that’s a distance of about 6,000,000 km. The blast from a supernova expands fast, but not at the speed of light. From memory (and I should really check this), we expect the blast wave to expand out at something like a third the speed of light. At that speed, it would take the blast wave a full minute to reach the nearby star! Things are far apart in astronomy. The 4-5 seconds it takes the blast wave to reach the companion star in the movie is almost certainly too fast.


The movie starts outside the explosion, but eventually the blast wave overtakes the camera and we see it from the inside… at about the same time that the blast wave is overtaking the companion star. What happens to the companion star? You might think it would suck to be next to a thermonuclear bomb one and a half times the mass of the Sun. And, it would. You might think you would be completely blown away. And, you would be. But a star wouldn’t. The gravitational force holding together the companion star is strong enough to allow it to survive despite the tremendous amount of energy deposited into it by being next to an exploding white dwarf. That being said, it is a lot of energy, and so we expect some fraction of the outer layers of the star to be stripped away. That indeed happens in this movie.


Finally, the movie zooms out, so that you can see the supernova in the context of its host galaxy. This supernova in the movie is really on the outskirts of the galaxy. It’s true that you find thermonuclear supernova more often in the outskirts of the galaxy than you do core-collapse supernovae (the other type), so it’s not unreasonable for the supernovae to be where I’ve shown it. I have made it rather optimistically far out, however. It’s also true that when it reaches maximum brightness, a supernova can be as bright as its whole host galaxy, which is qualitatively what you see in the movie. However, here is perhaps the greatest acceleration of time. It takes about 20 days for a supernova to reach maximum light after it explodes… not the mere handful of seconds that you see in the movie. However, the movie would be really boring if you had to sit and watch it for 20 days for the supernova to get as bright as the galaxy!

(You might be surprised by how long it takes for the supernova to reach maximum light; why isn’t it brightest right at the explosion? Parts of the exploding gas cloud are opauqe. Not all of it, and indeed as it expands, more and more of the outer layers become transparent. However, because the inner parts are opaque, the energy is effectively trapped inside the expanding cloud, and the rate at which it can be radiated away is limited by the surface area of the opaque part. So, as the supernovae gets bigger, it has more and more surface area, and so can get brighter. However, that’s only part of the story. The profile of rising and falling on the lightcurve is also driven by the detailed physics of what’s going on in the supernova. Some of the energy of the explosion goes into creating unstable nuclei that aren’t entirely stable, which then decay over time, releasing their energy into the expanding gasses.)

After maximum light, a supernova fades. Depending on how far away it is and how sensitive a telescope you use, it will be visible for weeks or months, or perhaps even years. In fact, if it’s close enough, it will be visible for thousands of years. At that point, though, we no longer call it a supernova explosion, but rather call it a “supernova remnant”. As an example, the X-ray and radio source Cassiopeiae A is the left over remnant of a thermonuclear supernova in our Galaxy that exploded in 1572.

Galaxy Image: NGC 1309, imaged by the Hubble Space Telescope. Hubble Legacy Archive, ESA, NASA; processing by Martin Pugh.

Music: Symphony No. 5 in C Minor by Ludwig van Beethoven, performers unknown; from the public-domain music site

This work is licensed under a
Creative Commons License.

Friday Galaxy Blogging: NGC 5278

This is NGC 5278, as imaged by the Sloan Digital Sky Survey. This color image was built by putting together broadband ugriz images, with the g filter mostly mapped to the blue channel of the image, r to the green channel, and i to the red channel. (u influenced blue and z influenced red as well.)

This was one of the galaxies in Chloe Wightman’s keystone project at Quest in 2013; she was looking at Galaxy Zoo-identified merging galaxies, and comparing morpological features to optical emission lines.


Thermonuclear Supernova in M82

I discovered (via random folks posting to my twitter feed) this morning that we (as in “humanity”) discovered a new supernova in the galaxy M82 in the last day or so. This is very cool for a lot of reasons.

The Galaxy

So what is M82? M82 is a nearby galaxy, as galaxies go. It’s not quite a spiral or an elliptical; it’s usually categorized as an irregular galaxy. It’s very close to the galaxy M81; the two galaxies are less than a degree apart on the sky, and are found not far from the tip of the bowl of the Big Dipper.

M81 and M82. Image from SDSS via SkyView, stacked by me to make a color image.

In the image above, M81 is the face-on spiral galaxy towards the bottom, and M82 is the thin line of a galaxy above it. If you look at this system with a radio telescope, you see that the two galaxies actually share a stream of gas; it’s clear that the two of them are interacting with each other.

M82 is also what we call a starburst galaxy. This is a galaxy that’s undergoing a burst of star formation. Stars are being formed in it much faster than they are in a typical everyday star-forming galaxy like our own. When stars are being formed rapidly, this will include some high-mass, short-lived stars. Higher mass stars may only live a few million or a few ten million years, which is a very short lifetime for a star. (For comparison, our star will live nearly ten billion years.) These high-mass stars also die with a bang: they explode in acore-collapse supernovae. So, you’d expect core-collapse supernovae to be more common in starburst galaxies than in regular everyday galaxies, simply because the high-mass stars that are the progenitors of core-collapse supernovae are being made at a faster rate there than they are elsewhere.

Unfortunately, often these starburst galaxies are very dusty, and the sites of most active current star formation are buried deep inside dust. As such, some of those supernovae may explode but never be caught because the light of the explosion is shrouded by dust.

M81 and M82 are very nearby galaxies, on the cosmic scale. According to NED, M82 is between 3 and 5 Mpc away; I’ll use NED’s “average” value of 3.8 Mpc. That means that M82 is 12 million light-years away. Whereas the supernova was just seen here on Earth in the last couple of days, it actually exploded 12 million years ago.

The Supernova

The new supernova was reportedly discovered by amateur astronomers in Russia (please suggest a better link or citation to said amateurs in the comments if you have one!), and has been confirmed by numerous people. As the image in the buzzfeed “discovery” link I just gave you shows, this supernova is actually on the outskirts of M82. What’s more, this supernova has been shown by its spectrum to be a thermonuclear supernova rather than a core-collapse supernova.

Very massive stars (more than eight times the mass of the sun) end their lives in a core-collapse supernova, blowing themselves all to hell and leaving behind a neutron star. Stars of lower mass end their lives in a much more sedate (but still pretty awesome) planetary nebula and leave behind a white dwarf. A white dwarf is a star that’s about the mass of the Sun, but only the size of the Earth. It is hot just because of left over heat from its formation; it’s no longer actively generating energy inside it. It just sits there and cools off. A typical white dwarf is made up of carbon and oxygen (although in a rather exotic state that I won’t go into right now). It’s not as mind-bogglingly dense as a neutron star, but it’s still amazingly dense. There is an upper limit to the possible mass of a white dwarf star (at 1.4 times the mass of the Sun); one over the size of that would collapse under gravity. If a real white dwarf does grow to this size (either by colliding with another white dwarf, or by pulling matter off of a companion star over time), it starts to collapse under gravity, but doesn’t complete that collapse. It gets dense enough to trigger runaway nuclear fusion of the carbon that composes it. Basically all of the carbon in the white dwarf undergoes fusion, and a chain reaction of fusion continues until all the atoms have combined to make iron (or something right about the same size as iron). In other words, the white dwarf becomes… a thermonuclear bomb that is one and a half times the mass of the sun.


That’s what happened in M82.

Because these supernovae are so amazingly bright, they can be seen out to extreme distances. I’m not talking M82; I’m talking many, many, many times farther away. Whereas M82’s supernova exploded just 12 million years ago, we’ve seen supernova out to distances that means they exploded 10 billion years ago. This property, together with other properties, of this type of supernova makes them lighthouses useful for measuring the history of the Universe. Because we can see them so far away, we can see supernova from when the Universe was much younger. Indeed, it was measurements of these supernovae that allowed us to measure the history of the expansion rate of the Universe and discover that the Universe’s expansion is actually accelerating.

It’ll be fun to watch over the next weeks and months as this new nearby supernova gets brighter and then fades away.

The Higgs Boson: a talk in Second Life tomorrow morning (April 6)

It’s been a year since I’ve given a public outreach physics and astronomy talk in Second Life. I used to do these things fairly regularly as a part of MICA (the Meta-Institute of Computational Astronomy). However, the MICA project has completed, its island in Second Life has gone online, its Second Life groups have been disbanded, and MICA no longer really exists. (Its website is still up, and should stay up for at least a little while. If I were smart, I’d probably make sure to download and archive elsewhere all of the audio recordings of my own talks….) A write-up of what MICA did and was all about is available at, and was published in the conference proceedings of a SLActions conference on virtual worlds

I’ve always meant to find other venues for continuing to do popular talks in virtual worlds. Someday, I’d like to escape from Second Life’s walled garden and start doing these talks in an OpenSim grid, and even did the first steps for trying to get set up to do them in my own region on OSGrid. However, of course, the audience in Second Life for now is still far bigger.

Fortunately, the Exploratorium, the excellent science museum in San Francisco, has a presence in Second Life. This Saturday (tomorrow, 2013 April 6) at 10AM pacific time (17:00 UT) I’ll be giving a talk about the Higgs boson in the Exploratorium region in Second Life. Remember, basic Second Life accounts are free. Drop by if you’re interested.

Dark Matter found? Don’t break out the champagne just yet.

You may have seen announcements that Dark Matter has been “found”. I don’t believe there’s a publicly available scientific paper on this yet, so the original source for this is two press releases from CERN: One from four days ago and one from today.

First, I want to say what is meant by dark matter being “found” here. It’s not evidence that previously-uncertain Dark Matter exists. We already know that Dark Matter exists; the Bullet Cluster observations several years ago was unambiguous confirmation that non-baryonic dark matter exists. We don’t know what it is from that, but we know it exists. (Here is a podcast I did three years ago about the evidence for the existence of Dark Matter, including the Bullet Cluster.) So what does it mean to say that these new CERN results may have “found” Dark Matter?

Although we know Dark Matter exists, there remain a huge number of mysteries about it. Many of these can be summarized under: what is it? All we really know is that it’s not made out of baryons, that is, protons and neutrons. So, it can’t be an excess of dim stars or rogue planets (a model that was once considered a real possibility for our Galaxy’s dark matter). Thus far, we’ve observed it because of the effects of its gravity. We’ve seen it in comparisons of the structure in the Universe to models of structure growth from early-Universe conditions; in the dynamics of galaxy clusters and galaxies; and through gravitational lensing. It would be nice to observe it in other ways.

To “find” Dark Matter, we’d like to do one (or more) of two or three things. Either, we’d like to see the results of decay products in our atmosphere or in space because of interactions of Dark Matter particles out in space. Or, we’d like to have an actual Dark Matter particle interact with a particle detector we have on Earth (analogous to how we see neutrinos from the Sun). Or, we’d like to actually make some of the stuff in a collider like the LHC at CERN in Switzerland, and see its decay products or signature there.

The current announcement from CERN is potentially of the first type. There is a detector, the “Alpha Magnetic Spectrometer” or AMS, on the International Space Station. This spectrometer is measuring electrons and positrons (the antiparticles of electrons) coming from space— that is, cosmic rays that are electrons and positrons. They see too many positrons for what we’d expect. One possible reason for the excess of positrons is that they are the result of very rare Dark Matter annihilations in our Galactic halo. (Although such annihilations, if they are happening, would be rare, there is so much bloody Dark Matter out there that if it’s doing this, it would produce enough excess positrons for us to observe.)

What’s really been detected is a positron excess, which is interesting all by itself. Whatever it turns out that this positron excess is coming from, it’s going to be at least new astronomy, and potentially also new physics. It may not be as sexy and headline-worthy as “WE FOUND TEH DARK MATTER!!!1!!one!”, but it will still be interesting, and will tell us something about nature. What’s been seen is consistent with it coming from the Dark Matter halo of our galaxy, but other sources can’t be ruled out yet. As more data is collected, the investigators running this experiment will be able to test whether the details of what is seen remain consistent with what would be expected from Dark Matter, versus other possible sources.

Image of a Thermomnuclear Supernova Progentior

Holy cow, it’s been a long time since I blogged.

The class I’m teaching right now is 3d Computer Modelling and Animation. Perhaps the hardest thing about it is figuring out if the word Modelling has one or two l’s in it… it depends on whether you’re in the USA or Canada, I think.

For this class, I’m making all of the students do a major project. Some of them are doing some pretty interesting things, and already several of them have figured out things about Blender (the 3D software we’re using, a quite powerful free package that you should check out yourself) that I don’t know myself. A couple are playing around with motion tracking, in order to add 3D rendered elements into a live action video scene. One is building a game using the Blender game engine. Others are doing various other animations.

I’ve decided to take on a project myself. For this project, I am going to model a white dwarf in a mutual orbit with a main sequence or red giant star, pulling matter off of it into an accretion disk. During the animation, the white dwarf will go critical, and explode in a supernova, blowing itself way, and blowing off some of the outer layers of the companion star.

So far, I’ve managed to create the basic progenitor model, and do a little bit of animation of the textures so that the disk is spinning, the star’s surface is roiling, and the gas bridge between the star and the disk looks a little like it’s streaming. Here’s a rendered frame from what I’ve done so far:

Click to embiggen (CC-BY-3.0)

I’ll certainly post the full animation once I’ve completed it. Next, I’m going to have to start worrying about how to deal with the supernova. Eventually, I’ll set the whole thing to music.