A very short post this week, as I am swamped - I will be away for almost 2 weeks and will explain in detail when I get back - but today was a good day in terms of my publishing.

Firstly, I got a letter accepted in the Sydney Morning Herald on science funding. Here's the letter.

But the bestest bit is that not only did I get the letter published (does this go on my CV), but I also got the letters' page cartoon dedicated to my letter.

I'm going to order a copy of the original cartoon to hang on my wall - I think it is excellent.

More when I have some more time.

## Monday, 29 October 2012

### Publishing Success!!!!

## Saturday, 20 October 2012

### Unearthing Foundations of a Cosmic Cathedral: Searching the Stars for M33's Halo

The Stellar Halo of a galaxy is a tenuous population of stars and other things surrounding the the bright spiral disk that characterises galaxies like our own Milky Way. Some of the oldest, most pristine, stars that we know can be found in the halo of our own galaxy, and so they provide clues to the processes that bought large galaxies into existence.

The problem is that they are, well, tenuous. When we look for halo stars in our own Milky Way, we have to sift through a sea of nearby fainter stars, to pick out the giant stars that are far away.

I've mentioned before about my work with the Pan-Andromeda Archaeological Survey, which has been mapping out the stars of our nearest large neighbours, the Andromeda and Triangulum galaxies.

Now that all the data is in, we're dissecting the various stellar populations in the vicinity of these tow galaxies, especially with regards to the large amount of substructure (the left over remnants of disrupted systems). But also we want to measure the shape and extent of the stellar halos.

But it is hard work. These things are still faint, and we still have to trawl through the contamination from our galaxy. Although it's important to remember that we are not hear to do the easy things!

And this is precisely what Rob Cockcroft of McMaster University has done. For the first time, we have measured the density of stars in the halo!

So here is an example of what we have to deal with. We see all the stars towards Triangulum (M33), but in the lots are in our galaxy (Disk and Halo), as well as there being misclassified galaxies in there as well (when something is small and faint, without a spectrum, it is hard to tell things apart).

Once we have cleaned all the contamination out of the way, we can actually try to measure the density drop off of what is the stellar halo. And this is it!

The important curves are the dashed ones with the data points on there. You can see that there are large error-bars, but this is because the measurement is hard. If we could do this and get really small error-bars, then someone would have done it years ago with crappier data and gotten large error bars :)

So, M33 has a faint halo of old stars, and now we need to think a little about just what it is telling us about the formation and evolution of this little galaxy (it is about 10th the mass of the Milky Way).

Just to give you an idea of what we are dealing with, now we have measured it, we can calculate that the total brightness of the halo represents less than 1% of the light emitted by the entire galaxy!

A stirling result! Well done Rob!

Unearthing Foundations of a Cosmic Cathedral: Searching the Stars for M33's Halo

Robert Cockcroft, Alan W. McConnachie, William E. Harris, Rodrigo Ibata, Mike J. Irwin, Annette M. N. Ferguson, Mark A. Fardal, Arif Babul, Scott C. Chapman, Geraint F. Lewis, Nicolas F. Martin, Thomas H. Puzia

The problem is that they are, well, tenuous. When we look for halo stars in our own Milky Way, we have to sift through a sea of nearby fainter stars, to pick out the giant stars that are far away.

I've mentioned before about my work with the Pan-Andromeda Archaeological Survey, which has been mapping out the stars of our nearest large neighbours, the Andromeda and Triangulum galaxies.

Now that all the data is in, we're dissecting the various stellar populations in the vicinity of these tow galaxies, especially with regards to the large amount of substructure (the left over remnants of disrupted systems). But also we want to measure the shape and extent of the stellar halos.

But it is hard work. These things are still faint, and we still have to trawl through the contamination from our galaxy. Although it's important to remember that we are not hear to do the easy things!

And this is precisely what Rob Cockcroft of McMaster University has done. For the first time, we have measured the density of stars in the halo!

So here is an example of what we have to deal with. We see all the stars towards Triangulum (M33), but in the lots are in our galaxy (Disk and Halo), as well as there being misclassified galaxies in there as well (when something is small and faint, without a spectrum, it is hard to tell things apart).

Once we have cleaned all the contamination out of the way, we can actually try to measure the density drop off of what is the stellar halo. And this is it!

The important curves are the dashed ones with the data points on there. You can see that there are large error-bars, but this is because the measurement is hard. If we could do this and get really small error-bars, then someone would have done it years ago with crappier data and gotten large error bars :)

So, M33 has a faint halo of old stars, and now we need to think a little about just what it is telling us about the formation and evolution of this little galaxy (it is about 10th the mass of the Milky Way).

Just to give you an idea of what we are dealing with, now we have measured it, we can calculate that the total brightness of the halo represents less than 1% of the light emitted by the entire galaxy!

A stirling result! Well done Rob!

Unearthing Foundations of a Cosmic Cathedral: Searching the Stars for M33's Halo

Robert Cockcroft, Alan W. McConnachie, William E. Harris, Rodrigo Ibata, Mike J. Irwin, Annette M. N. Ferguson, Mark A. Fardal, Arif Babul, Scott C. Chapman, Geraint F. Lewis, Nicolas F. Martin, Thomas H. Puzia

(Submitted on 15 Oct 2012)

We use data from the Pan-Andromeda Archaeological Survey (PAndAS) to search for evidence of an extended halo component belonging to M33 (the Triangulum Galaxy). We identify a population of red giant branch (RGB) stars at large radii from M33's disk whose connection to the recently discovered extended "disk substructure" is ambiguous, and which may represent a "bona-fide" halo component. After first correcting for contamination from the Milky Way foreground population and misidentified background galaxies, we average the radial density of RGB candidate stars over circular annuli centered on the galaxy and away from the disk substructure. We find evidence of a low-luminosity, centrally concentrated component that is everywhere in our data fainter than mu_V ~ 33 mag arcsec^(-2). The scale length of this feature is not well constrained by our data, but it appears to be of order r_exp ~ 20 kpc; there is weak evidence to suggest it is not azimuthally symmetric. Inspection of the overall CMD for this region that specifically clips out the disk substructure reveals that this residual RGB population is consistent with an old population with a photometric metallicity of around [Fe/H] ~ -2 dex, but some residual contamination from the disk substructure appears to remain. We discuss the likelihood that our findings represent a bona-fide halo in M33, rather than extended emission from the disk substructure. We interpret our findings in terms of an upper limit to M33's halo that is a few percent of its total luminosity, although its actual luminosity is likely much less.

Labels:
Astro-ph,
Galactic Cannibalism,
PAndAS

Location:
Sydney NSW, Australia

## Saturday, 13 October 2012

### Catching the bus.....

OK, it's been a very good week (for reasons that will become clearer in the near future) and so I am going to take a breather for half an hour for a little recreational mathematics. The question is all about catching the bus.

One good thing about living in Sydney, which I've noted before, is that it easy to get to see international rugby at the Olympic Park. An excellent free bus service is provided to bring people in from the far-reaches of Sydney, and then take them home again.

It is quite impressive that it works, with tens of thousands of people pouring out of the grounds and onto buses quite efficiently.

So, I've been thinking - If people turn up at a bus stop at a certain rate, and buses arrive at a certain rate, then what do we expect the number of people on each bus to be?

OK, the question is easy if the people arrive at a fixed regular intervals, as do the buses. But we are not here to do the easy things.

Being a physicist, we start by simplifying the problem. Let's assume that the average number of people who turn-up per hour is λ

Imagine you are on the gate of the park, watching the buses leave. What's the distribution for the number of passengers on each bus?

The problem requires two parts, both based on the Poisson distribution. I mentioned this in the recent discussion on the German tank problem, but it is an extremely powerful description of random events, from the drops coming from a tap, the number of photons arriving at a detector, and, quite famously, the number of men kicked to death by horses in the Prussian cavalry.

This means that there are lots of short gaps, and fewer long gaps between buses arriving. Now, real buses don't follow such a distribution in detail, but let's stick with this because the maths gets more funnerer.

Right, the next question is how many people accumulate at the bus stop between buses? It's over to the Poisson distribution again. In a time, t, the distribution for the number of people at the bus stop is given by

Why is there a distribution? Well, people are dribbling in at random and in the same time interval there might be one person, or two, or ten or even none.

OK, we cans stick the two equations together and ask the question "what is the distribution of number of people leaving on the buses?" What we end up with is an expression that looks like this.

Why is there an integral in there? Well, multiplying the two probabilities gives us the distribution of the number of people on a bus, after a waiting time between t and t+dt. A little algebra, this becomes

What is that? The final probability distribution depends upon a rather mysterious quantity called the Gamma function. I could write a long post on the gamma function, but seeing that we are only interested in integer values (because we can only have whole numbers of passengers on the bus), then we know that

and so the distribution for the number of passengers on the bus becomes

Isn't that lovely? Well, I think so. So, let's take a simple case. Let's assume that the average number of buses arriving per hour is the same as the average number of passengers arriving. As you can imagine, the number of people on the bus will fluctuate, with some buses having 1, or 2, and we should expect a number of buses should have no people on there at all.

The distribution of the number of passengers becomes

Cool huh? Let's ask the question, what is the probability that a bus has no passengers? Popping n=0 into the above, the chance of this is 1/2. And the chance that a bus has one person on it is 1/4, and 2 people is 1/8.

In fact, we have a pretty interesting progression here, and it has been known from antiquity that 1/2+1/4+1/8+.... = 1, which is what you want all of the probabilities to add up to.

Let's consider a more realistic scenario, with 1000 people per hour arriving at the bus stop, and 20 buses per hour. What does the distribution of bus occupancy look like? You might think that it is most likely that there are about 50 people on each bus. Here's the distribution

The most likely number of people on a bus is zero! And lots of the buses have only a hand full of people of them, while some are jam-packed with more than 150 on the bus - Standing room only!

You might think that chucking more buses per hour is a solution to the problem, but we always have this shape. The most likely number of people on a bus is zero. Throwing less buses doesn't help either, the most probable number of passengers is zero. Luckily this is not how real buses at Sydney Olympic Park operate, or we would never get home!

One last question then, as I am cooking pizzas tonight and need to get started. Let's imagine that you are not the gate guards, but a passenger, just a random passenger. What's the most likely number of passengers on the bus (including you)?

While the most probable number of passengers on a bus is zero, there is no one there to see that. And while there are lots of buses with 1 person on, it is actually unlikely to be you (I know this might sound a little paradoxical, but think about it).

So, we effectively have to weight each number of passengers probability but the number of passengers there to see it. What does that do?

This is the case that we looked at above, with 1000 passengers per hour, and 20 corresponding buses. Most people will report that they were comfortably on a bus with a total of 50 passengers, while only a few will report they were on really empty or really full buses, even though the guards on the gate say that many buses were leaving with virtually no one on them, and a few very very crowded.

And while you are chewing that over, I'm off to do some cooking.

One good thing about living in Sydney, which I've noted before, is that it easy to get to see international rugby at the Olympic Park. An excellent free bus service is provided to bring people in from the far-reaches of Sydney, and then take them home again.

It is quite impressive that it works, with tens of thousands of people pouring out of the grounds and onto buses quite efficiently.

So, I've been thinking - If people turn up at a bus stop at a certain rate, and buses arrive at a certain rate, then what do we expect the number of people on each bus to be?

OK, the question is easy if the people arrive at a fixed regular intervals, as do the buses. But we are not here to do the easy things.

Being a physicist, we start by simplifying the problem. Let's assume that the average number of people who turn-up per hour is λ

_{1}, and the average number of buses per hour that turn up is λ_{2}. Let's also assume that the bus instantaneously picks up all the passengers who are waiting, and then heads off.Imagine you are on the gate of the park, watching the buses leave. What's the distribution for the number of passengers on each bus?

The problem requires two parts, both based on the Poisson distribution. I mentioned this in the recent discussion on the German tank problem, but it is an extremely powerful description of random events, from the drops coming from a tap, the number of photons arriving at a detector, and, quite famously, the number of men kicked to death by horses in the Prussian cavalry.

The other side of the coin is that you can use this distribution to calculate the time between events, which is what we want to use to describe the time between buses arriving. It turns out that this is an exponential distribution, and the probability of a bus arriving between t and t+dt after the last one is

This means that there are lots of short gaps, and fewer long gaps between buses arriving. Now, real buses don't follow such a distribution in detail, but let's stick with this because the maths gets more funnerer.

Right, the next question is how many people accumulate at the bus stop between buses? It's over to the Poisson distribution again. In a time, t, the distribution for the number of people at the bus stop is given by

Why is there a distribution? Well, people are dribbling in at random and in the same time interval there might be one person, or two, or ten or even none.

OK, we cans stick the two equations together and ask the question "what is the distribution of number of people leaving on the buses?" What we end up with is an expression that looks like this.

Why is there an integral in there? Well, multiplying the two probabilities gives us the distribution of the number of people on a bus, after a waiting time between t and t+dt. A little algebra, this becomes

What is that? The final probability distribution depends upon a rather mysterious quantity called the Gamma function. I could write a long post on the gamma function, but seeing that we are only interested in integer values (because we can only have whole numbers of passengers on the bus), then we know that

and so the distribution for the number of passengers on the bus becomes

Isn't that lovely? Well, I think so. So, let's take a simple case. Let's assume that the average number of buses arriving per hour is the same as the average number of passengers arriving. As you can imagine, the number of people on the bus will fluctuate, with some buses having 1, or 2, and we should expect a number of buses should have no people on there at all.

The distribution of the number of passengers becomes

Cool huh? Let's ask the question, what is the probability that a bus has no passengers? Popping n=0 into the above, the chance of this is 1/2. And the chance that a bus has one person on it is 1/4, and 2 people is 1/8.

In fact, we have a pretty interesting progression here, and it has been known from antiquity that 1/2+1/4+1/8+.... = 1, which is what you want all of the probabilities to add up to.

Let's consider a more realistic scenario, with 1000 people per hour arriving at the bus stop, and 20 buses per hour. What does the distribution of bus occupancy look like? You might think that it is most likely that there are about 50 people on each bus. Here's the distribution

The most likely number of people on a bus is zero! And lots of the buses have only a hand full of people of them, while some are jam-packed with more than 150 on the bus - Standing room only!

You might think that chucking more buses per hour is a solution to the problem, but we always have this shape. The most likely number of people on a bus is zero. Throwing less buses doesn't help either, the most probable number of passengers is zero. Luckily this is not how real buses at Sydney Olympic Park operate, or we would never get home!

One last question then, as I am cooking pizzas tonight and need to get started. Let's imagine that you are not the gate guards, but a passenger, just a random passenger. What's the most likely number of passengers on the bus (including you)?

While the most probable number of passengers on a bus is zero, there is no one there to see that. And while there are lots of buses with 1 person on, it is actually unlikely to be you (I know this might sound a little paradoxical, but think about it).

So, we effectively have to weight each number of passengers probability but the number of passengers there to see it. What does that do?

This is the case that we looked at above, with 1000 passengers per hour, and 20 corresponding buses. Most people will report that they were comfortably on a bus with a total of 50 passengers, while only a few will report they were on really empty or really full buses, even though the guards on the gate say that many buses were leaving with virtually no one on them, and a few very very crowded.

And while you are chewing that over, I'm off to do some cooking.

## Monday, 8 October 2012

### Open-access science: be careful what you wish for

A very quick post tonight, but I've had an article published in The Conversation titled Open-access science: be careful what you wish for.

This is a word of caution on the push to make all science, bother the publications and data paid for with public money, available to all.

To cut a long story short, I fully support this move. I would love my science to be read by all (well, at least those who are interested :).

The caution is, however, that doing this costs. The current funding model is a bit busted; scientists need to publish in established journals, as articles that do not appear in the "highest impact journals" are not considered as important.

But the journals are owned by big publishing houses, and so libraries need to pay for access to the papers, often to see the science generated by their own researchers. It's a bit of a complex mess.

Anyway, my caution is that while the funding for open source science has to come from somewhere, it is not good to raid existing science budgets, when those funds could be used to support more science and innovation.

Raid existing budgets? Surely you think this is not possible, but check out what's recently happened in the UK. While I support the move to open source, I'd rather see money being spent on science being done, and we think of clever ways of making it accessible to all!

This is a word of caution on the push to make all science, bother the publications and data paid for with public money, available to all.

To cut a long story short, I fully support this move. I would love my science to be read by all (well, at least those who are interested :).

The caution is, however, that doing this costs. The current funding model is a bit busted; scientists need to publish in established journals, as articles that do not appear in the "highest impact journals" are not considered as important.

But the journals are owned by big publishing houses, and so libraries need to pay for access to the papers, often to see the science generated by their own researchers. It's a bit of a complex mess.

Anyway, my caution is that while the funding for open source science has to come from somewhere, it is not good to raid existing science budgets, when those funds could be used to support more science and innovation.

Raid existing budgets? Surely you think this is not possible, but check out what's recently happened in the UK. While I support the move to open source, I'd rather see money being spent on science being done, and we think of clever ways of making it accessible to all!

## Thursday, 4 October 2012

### Flipping Bad Physics: David Blaine

The newspapers are ringing with the latest exploits of "magician" David Blaine. To a physicist, hopefully those who recently completed my first year course on electromagnetism, it is obvious that this story is little more than "man sits in a fancy Faraday cage for a while".

But that's not the point of the post. It's the text accompanying the image. And specifically the phrase "a million volts of fiery electric current"! ARGH!!!

I know Joe Public does not care, but such a sentence causes the inner guts of those with a passing knowledge of physics to twist in an awful pain. Put simply, voltage (measured in volts) is not the same as current (measured in amps).

If you are being electrocuted, the difference may not bother you, but the fact that smoke is coming out of your ears is not an excuse for physics illiteracy.

Right. This is going to be a short post, as I have lots to do, but let's use a gravitational analogy.

Let's imagine you are on the top of a very tall building. It is a long, long way down. You have a certain amount of gravitational potential energy which is proportional to your height off the ground. Simply put, this is your gravitational "voltage".

You step off the building. Let's ignore the acceleration time, but assume you instantaneously fall at the terminal velocity in air. This is almost 200 km/h, and will probably result in a messy end. The speed at which you fall is the gravitational "current" (in reality, you want a line of people jumping off to get a real current, but let's not have that disturbing picture).

Let's consider changing the experiment slightly. We keep the height the same, so the gravitational "voltage" stays fixed, but now you jump into a tube on the side of the building filled with treacle. Again, you fall at the terminal velocity, but this is much lower.

You gravitational "current" has gone down because the "resistance" to your motion has gone up.

Is it really that difficult? If you have a journalist friend, get them to watch this all explaining video

The penguins going up the steps are the electrons inside a battery. The height they go to is the voltage. The rate at which they slide down the path is the current. Simple.

But that's not the point of the post. It's the text accompanying the image. And specifically the phrase "a million volts of fiery electric current"! ARGH!!!

I know Joe Public does not care, but such a sentence causes the inner guts of those with a passing knowledge of physics to twist in an awful pain. Put simply, voltage (measured in volts) is not the same as current (measured in amps).

If you are being electrocuted, the difference may not bother you, but the fact that smoke is coming out of your ears is not an excuse for physics illiteracy.

Right. This is going to be a short post, as I have lots to do, but let's use a gravitational analogy.

Let's imagine you are on the top of a very tall building. It is a long, long way down. You have a certain amount of gravitational potential energy which is proportional to your height off the ground. Simply put, this is your gravitational "voltage".

You step off the building. Let's ignore the acceleration time, but assume you instantaneously fall at the terminal velocity in air. This is almost 200 km/h, and will probably result in a messy end. The speed at which you fall is the gravitational "current" (in reality, you want a line of people jumping off to get a real current, but let's not have that disturbing picture).

Let's consider changing the experiment slightly. We keep the height the same, so the gravitational "voltage" stays fixed, but now you jump into a tube on the side of the building filled with treacle. Again, you fall at the terminal velocity, but this is much lower.

You gravitational "current" has gone down because the "resistance" to your motion has gone up.

Is it really that difficult? If you have a journalist friend, get them to watch this all explaining video

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