It is a standard result that any Cartan subalgebra of a (complex) semisimple Lie algebra has the same size, so what’s going on?

The answer is simple: Poincaré isn’t a semisimple Lie algebra. Therefore we have to be careful about how to define the rank. First, let’s see why Poincaré isn’t semisimple.

Definition. If is a complex Lie algebra, then an ideal in is a complex subalgebra of such that, for all and , .

The brackets of a Lie algebra can be thought of as a product of two elements in that algebra. Then, an ideal (as always), is a sort of ‘zero’, making anything it multiplies a member of itself (just like for any ).

Definition. A complex Lie algebra is called simple if , and the only ideals in are and .

Definition. A complex Lie algebra is called semisimple if it’s (isomorphic to) a direct sum of simple Lie algebras.

We can now see why the Poincaré algebra isn’t simple. Translations, namely , form a basis for an ideal: Translations commute, and the commutator of a translation with a rotation is a sum of translations. It is also not semisimple, which I guess can be seen by considering the decomposition of the Lorentz subalgebra, then adding translations which will ‘link’ the two components.

The Cartan can be defined for non-semisimple algebras, and it turns out it is the Cartan of the largest semisimple subalgebra. In the case of Poincaré, the largest semisimple subalgebra is the Lorentz subalgebra. So we can still define the rank to be the size of the Cartan, with this more general definition. I don’t know if the relation between the number of Casimirs and the rank still holds in this case, but at least for the Poincaré algebra it does turn out correctly, since the rank of the Lorentz algebra is 2.

I tried using Yahoo! Pipes to solve the problem, but Pipes somehow messed up the whole feed, even when it wasn’t supposed to do anything.

Anyway, to fix this I wrote a small ‘feed proxy’ that removes the offending HTML code. If you subscribe to it instead of the official feed, you’ll get good looking author names.

The feed proxy address for hep-th is:
http://4by12.com/cgi-bin/arxiv_feed.pl?feed=hep-th

For other feeds, replace ‘hep-th’ by any other arXiv topic. For example, try gr-qc or math-ph.

Disclaimer: This feed proxy is provided as-is. I can’t guarantee uptime, data integrity, or anything else regarding this service.

Enjoy!

I want to give my own derivation, which I hope is clear and straightforward. If you’re already familiar with the transform, you can skip straight to Derivation of the Legendre Transform.

Some Background

Let’s start with the definition. If you have a function f(x), its Legendre transform is a function g(p), where p is thought of as f(x)’s derivative. g(p) is defined as:

This definition is already confusing, because x is considered a function of p. This function is found by solving the following equation for x:

To clarify matters, let’s consider an example. Take , then:

The Legendre transform is then:

Okay, what is it good for? The usual explanation starts with some geometric construction like the one you see on Wikipedia(*). Then, to explain why the transform works, you are shown the differential:

And the conclusion is that g is indeed a function of p, which is f’s derivative. Mathematically speaking, this argument is quite unconvincing, because saying that g is a function of f’(x) has no meaning. g is a function of some real number, and this number has no intrinsic connection to any derivatives.

One possible relation could be that g and f agree if you give the corresponding arguments:

but this is not the case! Some better explanation is obviously needed.

Derivation of the Legendre Transform

We define as before:

And we are looking for a function g(p). The crucial point is this: We require that g’s derivative will correspond to x:

Rigorously, this means:

Where p(x) is the function defined by the first requirement.

Now we have something we can use. We look at the function g(p(x)) and we calculate:

Both requirements were used in the last transition. We now integrate this equation by dx:

Integrating by parts on the right-hand side:

Thus g(p) is defined up to a constant, and choosing C=0 we get the familiar Legendre transform:

Motivation

Finally, I want to comment on this second requirement that x=g’(p), which is how I came up with this derivation. In Thermodynamics, the important parameters of a problem always come in pairs: Pressure and Volume, Temperature and Entropy, Number of particles and the Chemical constant, and so on. They are paired by the First Law, which is conservation of energy:

So here, the internal energy U is a natural function of S,V,N. Its partial derivatives are T,-P, and mu. Of course it can depend on other variables and there are other derivatives, but these are the ones that are easiest to calculate.

The Legendre transform is used to get new forms of energy that are naturally dependent on other parameters. This is done by switching between pairs of variables. For instance, if you know the temperature T instead of the entropy S, you can transform like this:

F is called the ‘Helmholtz Free Energy’, and its natural parameters are T,V,N:

So T, which was a derivative before, is now a parameter. But equally as important, S is now a derivative. This is very important because it allows us to calculate S very easily. If we had a new function F=F(T,V,N) that wouldn’t allow simple calculation of S, it would be useless. This is what makes the Legendre transform so useful.

Having said that, we now see that we can define other transforms by changing this requirement. We still get functions that can be thought of as ‘functions of the derivative of f’. For example, require:

Where a,b are some arbitrary parameters. Then we get a new transform:

Perhaps this method can generate some other useful transforms.

(*) I just saw that Wikipedia has something that is related to my explanation under ‘Another definition’, although it goes through a different route and still uses the geometric requirements.

I only had a couple of days to do this, and I figured the easiest thing would be to create a relativistic version of the Bohr model. This is a toy model that predicts the energy levels of a non-relativistic hydrogen atom.

The first thing I guessed was that the basic assumption from the Bohr model regarding the angular momentum still holds. That is:

where is some integer and is hbar.

Also, the electron should still travel in a circular orbit, so that:

where is the orbit radius and is the linear momentum. It is easy to show that this is still correct relativistically (*).

Next we have the energy, which for a free particle now obeys the dispersion relation:

Our electron is in the electric field created by the nucleus, so this should be (**) :

Where is the electron charge and is its rest mass.

Let’s find the possible orbit radii. In the classical Bohr model this is usually done by calculating the force acting on the particle. But here we already wrote down this nice expression for the energy so we might as well use it… To do that, we will venture one last guess.

We assume that, given angular momentum , the system chooses the radius where it has minimal energy.

You can verify for example that when you make this assumption in the non-relativistic version (instead of using forces and such), you still get the same result.

To get the orbit radii we then need to solve:

which comes to:

Plugging this into the expression for the energy we get the atom’s energy levels:

And there you have it. I have reason to believe this model works, at least partially, because I used it on measurements of large-Z atoms and it fit the data splendidly (while the non-relativistic model failed miserably). In addition, taking recovers the non-relativistic energies, as expected.

Having said that, you may have noticed a problem in the last expression. For large enough Z, it is imaginary, and energy isn’t imaginary. How large a Z? Well if you’re a physicist may have also noticed that the coefficient is the fine structure constant:

So for the ground state , the largest possible Z is about 137. Above that, according to the model, the system is unstable.

You might be thinking that a more plausible explanation is that I made some mistake in my series of, uhm, guesses. That could be, but a quick look in the periodic table shows that it goes all the way up to… 118.

Well! Could this little toy model tell us something deep about the stability of atoms? Apparently it can. I googled for ‘stability of large z atoms’ and came up with this review, where the instability is discussed. In page 10 it says:

… we are on the ‘primitive’ side.

(*) This can be verified by taking and making a change of coordinates:

Place the electron on the intersection of the X axis and the circular orbit to obtain the answer.

(**) This guess can be justified by writing down the Lagrangian for a relativistic charged particle (c=1):

And converting to a Hamiltonian. Taking the Hamiltonian to be the energy, one arrives at this formula.

Suppose you’re sitting in a car heading east at 100 km/h, while another car passes you by heading west at 100 km/h. According to special relativity, if you compare your clock with the clock of the passenger in the other car (let’s call her Alice), you will notice that her clock runs slower than yours.

The closer your relative speed is to the speed of light, the more noticeable the difference will be. For example, if your relative speed is 90% the speed of light, you will see Alice’s clock running about 2.3 times slower than yours. Of course, it’s not just the clock that will run slower — everything will run slower, including, for example, biological processes. Alice will age 2.3 times slower than you.

But what makes you so special? Nothing, really. Because according to Alice, it’s you whose aging slowly. So whose right? Well, the short answer is that you’re both right, because time isn’t an absolute thing — it depends on who’s measuring it.

This line of reasoning leads to the famous ‘twin paradox’: Two twins, Alice and Bob, are born on earth (if you must know then yes, it’s the same Alice). At birth, Alice is put aboard a spaceship that can reach extremely high speeds and sent off into space for a long trip. She reaches the edge of the galaxy, makes a U-turn, and returns to earth. This round-trip takes her 40 years in earth-time.

Meanwhile, Bob remains on earth. When Alice gets back, Bob is 40 years old. Because Alice traveled at high speed, Bob saw her aging very slowly, so when she gets back she is much younger than Bob, say 20 years old. But according to Alice, it’s Bob who aged slowly, so actually Alice should be older than Bob.

That’s the twin paradox, but it isn’t really a paradox. The problem is that in order to make this round-trip, Alice has to accelerate — she can’t go away and back again with constant speed. Therefore, the problem is no longer symmetrical: Alice is accelerating, Bob isn’t. To calculate exactly what happens in this case you need to use general relativity, but the answer is that Bob is right and Alice is wrong: Alice will have aged more slowly than Bob.

Closed Universe

I’ve been wondering, what happens to the twin paradox if the universe is closed? What I mean by ‘closed’ is that if you travel in a straight line, you end up back where you started. So the universe looks like a closed loop, only in three dimensions. (*)

In such a universe, the Alice twin can leave earth, travel with constant speed, and return to earth by simply going straight (**). Meanwhile, the Bob twin remains on earth like before. No one is accelerating, so the usual explanation of ‘Alice is accelerating and that’s cheating’ doesn’t apply here. Apparently, the situation is symmetrical: Bob thinks Alice is moving and aging slowly, Alice thinks Bob is. Who is right?

Well the answer is rather interesting. In special relativity we are used to saying that there are no ‘preferred frames of reference’ — that all inertial observers are the same, and that there are no absolute speeds, only relative ones. It turns out that in a closed universe this is no longer the case.

Consider the following experiment. Bob, who is on non-moving earth, simultaneously fires two photons at opposite directions. The photons travel around the universe and return to Bob, both at exactly the same instant.

Now let Alice perform this experiment on her spaceship. She sends out two photons, traveling at the speed of light. Let’s look at this experiment from Bob’s frame of reference, where the photons are still traveling at the speed of light (***). Bob sees the photons going around the universe, but meanwhile Alice is moving. So one photon will reach Alice before the other. Alice must observe the same thing (although they will not agree on the timing).

The same experiment, conducted by Bob or by Alice, produces different results. This is because Bob’s frame is special: It has zero speed, an absolute speed. Thus the symmetry between Bob and Alice is broken. Further calculations are required, but it turns out that Bob is correct: Alice ages less than Bob.

If you want more details, I found an interesting paper that discusses this paradox, and also develops the Lorentz transformations for such a universe.

(*) The mathematical term for this is actually ‘compact’, but that’s because mathematicians enjoy complicating things.

(**) You may be wondering how it’s possible to go in a circle without accelerating. You can try looking at it this way: In circular motion with constant speed, the vector of acceleration points inward toward the center of the circle. If you live in a 2-dimensional world, you can measure this vector because it points in a direction you can move in. For instance, you can hang a ball from a piece of string and watch the string stretch.

But if you live in a closed, 1-dimensional world, and you’re going in a circle, where is the acceleration vector pointing? It can’t point inward, because for you there’s no ‘inward’. In your 1D world there’s only front and back. As a consequence, you can’t measure this acceleration experimentally. And what you can’t measure experimentally doesn’t exist.

By the way, in terms of differential geometry, this is what geodesic curvature is all about: It’s the part of the acceleration that’s due to the shape of the manifold.

(***) This is one of the premises of special relativity — that light travels at the same speed for all observers. If you’re unconvinced it’s still true in a closed universe, consider the following. The metric of a cylinder is the same as that of a plane — a cylinder isn’t curved. Similarly, the metric of a closed, 1-dimensional universe is the same as that of an open one:

The constancy of light speed follows from the metric, and from assuming that the correct transformations (boosts) preserve the metric.

In my first implementation of the solution I used a simple method to calculate the derivative:

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This is just the average of the right and left derivatives. I wanted to improve the accuracy of the derivative calculation because of various reasons, so I turned to Savitsky-Golay filters, which were also described in the previous post. Using this method you get a digital filter that you apply to your data. If you’re not familiar with digital filters, you can think of a filter as an array of numbers which are applied to your function f as follows:

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(I’m probably mixing up some of the signs here, but the intention is clear I’m sure). To calculate a derivative using S-G, you first apply their filter to your data, and then divide by the spatial resolution dx — the distance between adjacent points on the grid. It is worth noting that there is a highly efficient way of doing this calculation using FFT. For more details, see Numerical Recipes.

I implemented S-G and tested it using a simple sin(x) function, and it worked flawlessly. But after I inserted it into my main program, the simulation started spewing out strange results. After some debugging I discovered something very strange: Using S-G, the derivative of a constant function wasn’t zero, and in one case reached . How odd! Debugging further, I found that non-zero derivatives only happened when the initial function f had very large (and constant) values, on the order of , and dx was very small, about .

To continue debugging, I dropped the fancy FFT convolution and switched to straightforward calculation — the one that’s shown in the last equation. This finally revealed the problem: When multiplying each value of the function f by the factor and summing, some of the least significant bits are lost, because the numbers don’t all have the same exponent. So even though the calculation is accurate, the limitation of the computer’s double precision causes a loss of data. When the convolution is done, you’re left with a small value that’s not zero. But then to get the derivative you divide by dx, a very small number, and this shoots that small error through the roof. The smaller dx, the larger the derivative of the constant function! So once again, decreasing dx actually caused a larger error.

What helped me solve this problem was noticing that the S-G coefficients are anti-symmetric, i.e. . Specifically, anti-symmetric coefficients have the same exponent. Therefore I changed the summing order to sum pairs of anti-symmetric factors:

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This solved the problem: Constant functions now had a zero derivative, because each term in the sum was exactly zero, even in double precision. And the happy consequence was that this fix also solved the weird simulation results I started with.

PDE Primer

In case you’re not familiar with the terminology, I’ll first explain what a PDE is. A simple equation contains one or more unknowns which represent numbers. For example: . An ordinary differential equation (ODE) is similar, except the unknown is a function rather than a number. Such an equation involves derivatives of the function. Here is an example:

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The solution of this particular equation is , where can be any number. Finally, a partial differential equation (PDE) involves a function that has two or more parameters, and includes partial derivatives of this function. For example, the following equation describes waves propagating through a medium:

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PDEs are very important in physics. In fact, many of the basic laws of nature are described as PDEs. Examples include Maxwell’s equations, Shcrodinger’s equation, and Einstein’s field equations.

On to the simulation!

Pitfall 1: Exploding Waves

So, I built my model for the problem, derived the equations, and was ready to solve them. By ‘solving’ I mean that I start out with the known function at time t=0, and I want to find out what that function is at a later time. My function initially looked like this:

Some thousands of time-steps later, it evolved into this:

So far so good, but then it completely exploded:

Going back a bit in time, I was able to trace the beginning of this explosion:

And zooming in on the ‘wavy’ part:

It looked as though waves were forming on my function, and then ‘exploding’.

Inherent Instabilities

I was certain I had a bug, but I couldn’t find it. While debugging, at one point I decreased the spatial resolution — using less points per unit of space to describe the function… and the problem was gone! So, decreasing the accuracy of my solution actually solved the instability… That was very weird.

Mentioning this to a Ph.D student at the lab, he said this problem sounded familiar to him. And as it turns out, this is a universal problem with PDEs: If the time step is too large compared with the spatial resolution, the amplitude of small waves with short wavelengths quickly increases with time until they dominate the solution. This is due to the way numerical derivatives are calculated. The difficulty here is that the time step needs to be incredibly small, making calculation unfeasible. For some equation, the situation is even worse, as they are unstable for any time step, no matter how small.

For simple PDEs, it is very easy to see this effect by taking the function f to be a wave, and watching what happens to the amplitude over time. You can see a derivation of this result here. For a more in-depth discussion, Numerical Recipes is your friend. This method of analyzing equations is called von Neumann stability analysis.

In Comes Lax

Okay, so I found out not alone, but what can be done to solve this problem? The first thing I tried was to calculate the derivative more accurately. There is a method called Savitzky-Golay, where you fit a polynomial to your function at each point, and calculate the polynomial’s derivative at that point. The brilliant thing is that this whole operation (fit + derive) can be done using a single convolution, which costs a meager O(n log n) of processing time.

So I implemented S-G, only to discover it doesn’t solve the problem. More on that in a future post.

As it turns out, there is an incredibly simple solution due to Lax, which says the following. When advancing the function value to the next time step, you do something like this for each position:

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The Lax method says that the f[j] at the right-hand side should be replaced by an average of it’s neighboring cells:

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And that’s it! This replacement causes a numerical diffusion that ‘sedates’ the unruly waves, causing them to decay instead of explode. The time step used in the simulation still needs to be below some value, but now it decreases linearly with the spatial distance dx, which is much better than before. So Lax saved the day — and that was the end of my first pitfall. This is getting to be quite a long post, so I’ll describe the second problem in another post. Cheers!

a. be moving to the left
b. be moving to the right
c. not be moving

What a great question! To clarify the problem, let’s say we don’t just puncture the right wall, but remove it altogether. So at t=0 we have a vessel with a left wall but without a right wall, and with vacuum inside. What happens next is pretty straightforward: gas molecules are hitting the left wall from the outside of the vessel, thus exerting a force to the right. No molecules are hitting the vessel from the right however, so we have a net force to the right. Hence at t=0 the vessel accelerates to the right.

The acceleration continues as the gas fills the vessel, until the first molecules that entered the vessel begin hitting the left wall — this time from the inside. The process of filling the vessel takes a finite amount of time, and therefore at the end of it the vessel is moving to the right with velocity V.

But that’s not the end of the story. Suppose the vessel is now completely filled with gas from the outside, so we have the same particle density with the same velocity distribution hitting the left wall from the inside as from the outside, and the vessel as a whole is moving to the right. To understand what happens next, let’s look at matters from the vessel’s point of view.

So we’re sitting on the vessel, completely still, and watching particles hitting our wall from the left and from the right. Whenever a particle hits the wall, we see it completely reverse its velocity and momentum. The vessel absorbs the difference in momentum and accelerates. But from our point of view, the particles to our left are slower than the ones to our right! That’s because, in the original frame of reference, the vessel is moving to the right — i.e. toward the particles on the right, and away from the particles on the left. As a result, the particles on the left are depositing less momentum with each hit on the wall — in other words, they are exerting a weaker force than those particles on the right.

We’ve shown that F<F’, so we have a net force to the left — slowing the vessel down. The slowing down will continue until V=0, at which time F=F’ and the process is complete. So the answer is that at the end of the process, the vessel isn’t moving, but is positioned to the right of its original position.

Oh, and One Last Thing . If you haven’t already, go feast your eyes on the iPhone. The revolution has come!

I’ll give my view on this in a minute, but please take it with a grain of salt, me being a mere undergrad and all

Most problems nature presents us with are too complicated for us to solve. If we are ever to solve anything, we must make our models as simple as possible: Take into account only those aspects of the system that are absolutely essential to describe the phenomena we are interested in. Otherwise, the models will lie beyond that tiny region of mathematics that we know how to handle. Deciding which parts to keep and which to throw out is a guessing game, and sometimes we get it wrong. One of the clues that we got it wrong is the appearance of a singularity: A region of the problem where the model breaks down, and fails to provide a solution.

There are several examples. One is the ultraviolet catastrophe in blackbody radiation, which provided evidence that classical physics was insufficient, and which also showed that quantization might be a good way of looking at things. Another example, I believe, is from hydrodynamics, where viscosity (internal fluid friction) is sometimes introduced to get rid of singularities.

So what does this have to do with quantum gravity? Well, our classical description of gravity, general relativity, contains singularities. One such singularity occurs at the beginning of time — the big bang. Einstein’s equations allow us to calculate what happened right after the big bang, but they don’t tell us what went on prior to it. They sort of break down.

Judging from history, this suggests that general relativity is not a complete description of gravity–that something is missing. And betting on quantum gravity as that something seems to me a pretty good bet, seeing how we were able to solve similar problems we had with the other forces using this method.

I’d like to thank prof. Oded Agam from the HU, whose discussion of singularities in physics during a recent colloquium led me to this post

Here’s a nice question from the list:

If the mass of one teaspoon of water could be converted entirely into energy in the form of heat, what volume of water, initially at room temperature, could it bring to a boil? (litres).

Let’s see now. First, how much energy would we get by converting a teaspoon of water into heat? By far the largest contributor would be the rest mass of the water. The mass/energy conversion ratio is given by . A teaspoon contains a few grams of water, say 4 grams. So the amount of heat we get from it is:

Where J stands for Joule. 1 Joule is about 0.25 kilo-calories. Next, how much energy do we need to invest to boil 1 litre of water starting at room temperature? First, we need to heat the water from about 20 degress celsius to 100 degrees. Raising the temperature of 1 gram of water by 1 degree celsius takes about 4 Joules. So to heat 1 litre:

(K means Kelvin, which is for our purposes the same as degrees celsius)

But that’s just half the story. Once we’re at 100 degrees, transforming the water from liquid to gas requires an additional investment of energy, known as latent heat. This has nothing to do with temperature — the temperature doesn’t change at all during the process of vaporization itself. For water at 100 deg celsius, the latent heat for vaporization is about 2 kJ/gr. So to vaporize 1 litre of water we need:

Clearly the latent heat is the important factor here. So, the energy required to boil a litre of water is . Hence, the energy in our teaspoon can be used to boil this many litres:

Or about 150 million litres. To put that number in perspective, let’s measure the water volume in olympic swimming pools. i.e. how many olympic pools can a single teaspoon of water evaporate? According to this site, an olympic pool holds

of water. Therefore, a teaspoon-worth of water can vaporize about 60 olympic swimming pools!