# Why does feedback hurt sometimes?

Research is hard. And not for the reasons you might expect! Sure, my daily life involves equations which look impenetrable to the layman. But by the time you’ve spent years studying them, they aren’t so terrifying!

The real difficulty in research is psychological. The natural state for a scientist is failure – most ideas simply do not succeed! Developing the resilience, maturity and sheer bloody mindedness to just keep on plugging away is a vital but tough skill.

This letter, written by an experienced academic to her PhD student is a wonderfully candid account of the minefield of academic criticism, both professional and personal. What’s more, it lays bare some important coping strategies – I certainly wish I’d read it before embarking on my PhD.

Above all, this letter is an admission of humanity. As researchers, we face huge challenges in our careers. But the very personal process of responding to them is precisely what makes us better scientists, and perhaps even improves us as people.

Originally posted on The Thesis Whisperer:

This letter was written by an experienced academic at ANU to her PhD student, who had just presented his research to a review panel and was still licking her wounds.

The student sent it to me and I thought it was a great response I asked the academic in question, and the student who received it, if I could publish it. I wish all of us could have such nuanced and thoughtfu feedback during the PhD. I hope you enjoy it.

A letter to…My PhD student after her upgradeWell you did it. You got your upgrade. But from the look on your face I could tell you thought it was a hollow victory. The professors did their job and put the boot in. I remember seeing that look in the mirror after my own viva. Why does a win in academia always have the sting of defeat?

Yeah, it’s a…

View original 1,120 more words

# Mathematica for Physicists

I’ve just finished writing a lecture course for SEPnet, a consortium of leading research universities in the South East of Britain. The course comprises a series of webcasts introducing Mathematica – check it out here!

Although the course starts from the basics, I hope it’ll be useful to researchers at all levels of academia. Rather than focussing on computations, I relay the philosophy of Mathematica. This uncovers some tips, tricks and style maxims which even experienced users might not have encountered.

I ought to particularly thank the Mathematica Summer School for inspiring this project, and demonstrating that Mathematica is so much more than just another programming language. If you’re a theorist who uses computer algebra on a daily basis, I thoroughly recommend you come along to the next edition of the school in September.

# Using Thunderbird Client with Office 365

I’ve just wasted a good half hour trying to migrate my email to an Office365 SMTP server. It seems that QMUL have decided to discontinue their in-house email server, but have not provided sufficient details about the new settings needed for email clients.

So here they are, in case anyone else runs into difficulties.

SMTP server : smtp.office365.com
Port : 587 (not the default)
Encryption : STARTTLS (not SSL/TLS)

I imagine that similar settings should work for other institutions which have moved to an Office365 system.

# T-duality and Isometries of Spacetime

I’ve just been to an excellent seminar on Double Field Theory by its co-creator, Chris Hull. You may know that string theory exhibits a meta-symmetry called T-duality. More precisely, it’s equivalent to put closed strings on circles of radius $R$ and $1/R$.

This is the simplest version of T-duality, when spacetime has no background fields. Now suppose we turn on the Kalb-Ramond field $B$. This is just an excitation of the string which generalizes electromagnetic potential.

This has the effect of making T-duality more complicated. In fact it promotes the $R\to 1/R$ symmetry to $O(d,d;\mathbb{Z})$ where $d$ is the dimension of your torus. Importantly for this to work, we must choose a $B$ field which is constant in the compact directions, otherwise we lose the isometries that gave us T-duality in the first place.

Under this T-duality, the $B$ field and $G$ metric get mixed up. This can have dramatic consequences for the underlying geometry! In particular our new metric may not patch together by diffeomorphisms on our spacetime. Similarly our new Kalb-Ramond field $B$ may not patch together via diffeomorphisms and gauge transformations. We call such strange backgrounds non-geometric.

To express this more succintly, let’s package diffeomorphisms and gauge transformations together under the name generalized diffeomorphisms. We can now say that T-duality does not respect the patching conditions of generalized diffeomorphisms. Put another way, the $O(d,d)$ group does not embed within the group of generalized diffeomorphisms of our spacetime!

This lack of geometry is rather irritating. We physicists tend to like to picture things, and T-duality has just ruined our intuition! But here’s where Double Field Theory comes in. The idea is to double the coordinates of your compact space, so that $O(d,d)$ transformations just become rotations! Now T-duality clearly embeds within generalized diffeomorphisms and geometry has returned.

All this complexity got me thinking about an easier problem – what do we mean by an isometry in a theory with background fields? In vacuum isometries are defined as diffeomorphisms which preserve the metric. Infinitesimally these are generated by Killing vector fields, defined to obey the equation

$\displaystyle \mathcal{L}_K g=0$

Now suppose you add in background fields, in the form of an energy-momentum tensor $T$. If we want a Killing vector $K$ to generate an overall symmetry then we’d better have

$\displaystyle \mathcal{L}_K T=0$

In fact this equation follows from the last one through Einstein’s equations. If your metric solves gravity with background fields, then any isometry of the metric automatically preserves the energy momentum tensor. This is known as the matter collineation theorem.

But hang on, the energy momentum tensor doesn’t capture all the dynamics of a background field. Working with a Kalb-Ramond field for instance, it’s the potential $B$ which is the important quantity. So if we want our Killing vector field to be a symmetry of the full system we must also have

$\displaystyle \mathcal{L}_K B=0$

at least up to a gauge transformation of $B$. Visually if we have a magnetic field pointing upwards everywhere then our symmetry diffeomorphism had better not twist it round!

So from a physical perspective, we should really view background fields as an integral part of spacetime geometry. It’s then natural to combine fields with the metric to create a generalized metric. A cute observation perhaps, but it’s not immediately useful!

Here’s where T-duality joins the party. The extended objects of string theory (and their low energy descriptions in supergravity) possess duality symmetries which exchange pieces of the generalized metric. So in a stringy world it’s simplest to work with the generalized metric as a whole.

And that brings us full circle. Double Field Theory exactly manifests the duality symmetries of the generalized metric! Not only is this mathematically helpful, it’s also an important conceptual step on the road to unification via strings. If that road exists.

# Integrating Differentials in Thermodynamics

“My bad!”, he said.

I’ve just realised I made a mistake when teaching my statistical physics course last term. Fortunately it was a minor and careless maths mistake, rather than any lack of physics clarity. But it’s time to set the record straight!

Often in thermodynamics you will derive equations in differential form. For example, you might be given some equations of state and asked to derive the entropy of a system using the first law

$\displaystyle dE = TdS - pdV$

My error pertained to exactly such a situation. My students had derived the equation

$\displaystyle dS = (V/E)^{1/2}dE+(E/V)^{1/2}dV$

and were asked to integrate this up to find $S$. Naively you simply integrate each separately and add the answers. But of course this is wrong! Or more precisely this is only correct if you get the limits of integration exactly right.

Let’s return to my cryptic comment about limits of integration later, and for now I’ll recap the correct way to go about the problem. There are four steps.

1. Rewrite it as a system of partial DEs

This is easy – we just have

$\displaystyle \partial S/\partial E = (V/E)^{1/2} \textrm{ and } \partial S / \partial V = (E/V)^{1/2}$

2. Integrate w.r.t. E adding an integration function $g(V)$

Again we do what it says on the tin, namely

$\displaystyle S(E,V) = 2 (EV)^{1/2} + g(V)$

3. Substitute in the $\partial S/\partial V$ equation to derive an ODE for $g$

We get $dG/dV = 0$ in this case, easy as.

4. Solve this ODE and write down the full answer

Immediately we know that $g$ is just a constant function, so we can write

$\displaystyle S(E,V) = 2 (EV)^{1/2} + \textrm{const}$

Contrast this with the wrong answer from naively integrating up and adding each term. This would have produced $4(EV)^{1/2}$, a factor of $2$ out!

So what of my mysterious comment about limits above. Well, because $dS$ is an exact differential, we know that we can integrate it over any path and will get the same answer. This path independence is an important reason that the entropy is a genuine physical quantity, whereas there’s no absolute notion of heat. In particular we can find $S$ by integrating along the $x'$ axis to $x' = x$ then in the $y'$ direction from $(x',y')=(x,0)$ to $(x',y')=(x,y)$.

Mathematically this looks like

$\displaystyle S(E,V) = \int_{(0,0)}^{(E,V)} dS = \int_{(0,0)}^{(E,0)}(V'/E')^{1/2}dE' + \int_{(E,0)}^{(E,V)}(E'/V')^{1/2}dV'$

The first integral now gives $0$ since $V=0$ along the $E$ axis. The second term gives the correct answer $S(E,V) = 2(EV)^{1/2}$ as required.

In case you want a third way to solve this problem correctly, check out this webpage which proves another means of integrating differentials correctly!

So there you have it – your health warning for today. Take care when integrating your differentials!

# Bibliographies and The arXiv

I’m currently writing up my first paper! The hope is that my collaborators and I will release the paper in the next couple of months. When we do, it’ll go on the arXiv – a publically accessible preprint server.

This open-access policy is adopted pretty much universally throughout mathematics and theoretical physics. I think it’s extremely good for science to be freely accessible to all. There’s still a place for journals, allowing research to be ranked by quality and rigorously peer reviewed. But the arXiv is vital in maintaining the pace of research, particularly in hot topic areas.

Every piece of work on the arXiv gets its own unique identifier. I rather like codes, so I tend to remember these numbers for my favourite papers. Just typing the number into Google search immediately takes you to the relevant document.

My current paper draft is peppered with arXiv numbers referring to important papers we need to cite. When we come to making a bibliography I’ll need to convert these into a standard form. Technically this involves making a BibTex file, and referring to it in my typesetting program.

I thought this would take ages, but it turns out that there’s an online Easter Egg solving the problem in a flash. Inspire HEP is a database of physics papers, providing all the metadata you could ever need including ready formatted BibTex. And it even has a feature which automatically generates a bibliography for you – check it out!

If you’re writing up your first paper and this tip helped you out, do drop me a line in the comments! And to the curators of arXiv and Inspire HEP – a huge thank you from me and the whole physics community.

# Counting Microstates with Lateral Thinking

I’ve spent some time this morning stuck on the following undergraduate problem

Consider a quantum system with $N$ distinguishable particles, each of which can have energy $E=n\epsilon$. Show that the number of microstates consistent with a macrostate of energy $E$ is given by the binomial coefficient

$\displaystyle \left(\frac{E/\epsilon + N - 1}{N - 1}\right)$

This is a classic example of a problem which needs some lateral thinking! It’s pretty trivial when you get the right idea, but I think it’s not entirely obvious at first. I’ll share with you my (embarrassingly slow) reasoning – let me know whether you agree with my philosophy in the comments.

To start with, I tried some examples – attempting to arrange $3$ particles in $5$ energy levels for instance. My approach was to work out how I could partition $5$ as a sum of smaller numbers, then evaluate the number of possible configurations consistent with this.

More concretely, I had a configuration where the particles had energies $\{0,1,4\}$. There are $3 \times 2=6$ such arrangements, because the particles can be distinguished (think of them as different coloured balls, if you like).

That’s all very well, but generalizing this idea is tricky. The problem is that the total energy constraint $\sum E_i = E$ makes it hard to enumerate all possible configurations. So I sat, stumped, for a good few minutes.

But thankfully, my failure contained a vital clue. My difficulties lay with that irritating total energy constraint. What if I could remove it from the problem altogether?

To do this requires a bit of lateral thinking. We’ve been trying to fit particles into energy levels. But you can turn the problem around. Equivalently we can try to distribute the $E/\epsilon$ units of energy among the particles. This effectively trivializes the troublesome constraint.

We’re not quite out of the woods yet. We need to work out how to distribute the energy blocks into the particle buckets. Here a second piece of lateral thinking helps us out. Rather than throwing the energy into buckets, we can think of partitioning it into sections. It’s just like being at the supermarket till – different customers (particles) separate their shopping (energy) with plastic dividers.

So how many ways can we divvy up the shopping on the conveyor belt? Well, there are $N$ customers so we’ll need $N-1$ dividers. We’ve also got $E/\epsilon$ items ready to be bought. This means that you have

$\displaystyle (E/\epsilon + N - 1)!$

possible arrangements of the items and dividers. But hang on, every unit of energy looks exactly the same. It’s as if every customer has bought exactly the same product! And clearly it doesn’t matter if you exchange the dividers – the overall partition is unchanged. Taking this into account, the correct number of microstates is

$\displaystyle \frac{(E/\epsilon + N - 1)!}{(E/\epsilon)!(N - 1)!}$

This is exactly the binomial coefficient in the question above!

Although this problem was pretty simple, there are two important morals. First, always examine a problem from every angle. Second, never completely discard your failed attempts. Chances are they hold vital clues which will point you in the right direction!

## Musings Mathematical and Otherwise

The Thesis Whisperer

Just like the horse whisperer - but with more pages

Gauging Thoughts

Symmetry is beauty

Of Particular Significance

Conversations About Science with Theoretical Physicist Matt Strassler

Gowers's Weblog

Mathematics related discussions

4 gravitons

The trials and tribulations of four gravitons and a postdoc

Eventually Almost Everywhere

A blog about probability by Dominic Yeo