Friday, June 21, 2013

Vacuum Cleaner Suction Revisited

A few months ago I wrote about how to calculate the suction of a vacuum cleaner. The article was prompted by something I read in Wikipedia about how typical vacuum cleaners would generate up to 20 kPa of suction. This seemed like an outrageous amount, and so I wrote an article showing that it should only be about 2.5 kPa.

It turns out to be one of my most well-read articles; it is eighth on my all-time hit list, although it only went up last January. And it has attracted a some comments, mostly appreciative. But a recent poster has pointed out to me that reports of high pressure are not limited to Wikipedia, and that in fact he has noted several manufacturers who claim similar ultra-high (by my reckoning) pressures in their catalogues.

So what gives? I know how to calculate the pressure of a centrifugal blower, and I think I have some idea of the dimensions and rpm's involved. So what gives? Do they install ultra-high speed DC motors, like in a router? Or do they have multi-stage blower assemblies? It just seems way out of line. And I'm not sure quite what you'd actually do with 20 kPa of suction. It sounds just a little bit dangerous to be waving around your living room.

Anyone care to weigh in on this question?

Tuesday, June 11, 2013

How I Got Maxwell-Boltzmann for Electron Spin

When we left off the other day I was telling you about some funny spreadsheet calculations. I had just figured out how to count electron spin states, after being stalled on the question for about three months. It turns out there's a kind of modified Pascal's Triangle that tells you how they add up. You know with coins, it's heads or tails, so if you want to know the chance of flipping four heads and two tails, you go to the sixth row of Pascal's Triangle:

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

and you can see that there are 15 ways out of 64 that you can flip 4H2T on six tosses. Another way of saying it is if you take six random steps alternately left and right, your chances are 15/64 of being one two steps to the right of your starting point. It's the one-dimensional Drunkard's Walk.

It's exactly the same with electrons. You have six electrons, and they each "have to" be either spin-up or spin-down. Since they each have spin-1/2, you should get a probability of 15/4 to find them with a total spin of +1.

But that's only part of the story. There are actually three different ways they can end up with z-axis spin of +1. They can have a total spin of 3, 2, or 1. We write these cases in the (l,m) representation as (3,1), (2,1) and (1,1). It turns out there is one way to get the first state, five ways to get the second state, and nine ways to get the third state. Fifteen pathways altogether.

And there's a modified Pascal's Triangle that gives you those parameters. It looks like this:

1 1
1 2
1 3 2
1 4 5
1 5 9 5


The terms of this triangle are actually the differences of the terms in Pascal's triangle, going halfway across each row. So if you add up successive terms, you recover Pascal's Triangle. You can see that the sixth row gives you:

1   1 + 5 = 6        6 + 9 = 15     15 + 5 = 20

I already said there were 15 out of 64 ways to get z-axis spin of +1: here you get the breakdown. One way is with total spin 3, five ways with total spin 2, and nine ways with total spin 1.

So the z-axis spins are distributed according to Pascal's triangle, which is becomes a Gaussian for large numbers. But the y-axis spins must have the same distribution, as must the x-axis spins. But we already know (or should know) what you get when you have Gaussian distributions in three dimensions independently. It's the Maxwell-Boltzmann distribution, the distribution of gas molecule velocities in thermal equilibrium.

I should have seen this coming, but I actually didn't. I did some very intricate spreadsheet calculations with the modified Pascal's Triangle to see what the distribution looked like, and I was dismayed to find that it looked a lot like Maxwell-Boltzmann for quite a ways, but diverged markedly at the high end. This graph shows the ratio of the two curves:

You can see I've added up 100 electrons to get a maximum total spin of +/-50. The RMS average spin is something like 5 in this case, so I get out there quite a few standard deviations before I start to diverge much from Maxwell-Boltzmann. This bothered me quite a lot.

But then I thought about it. I "knew" that Pascal's Triangle gave you a really good Gaussian, but had I ever plotted it to check out just how good? I threw it into the spreadsheet and what do you know? I got exactly the same curve. It was really close about halfway out and then fell off quite rapidly.

And it makes physical sense. The Binomial Distribution really is different from the Gaussian, especially near the ends. It's too much to expect them to line up all the way, and the extent to which they do track each other is really quite excellent. The graph here shows the ratio of the value of the two curves; if you consider that the Gaussian starts with a value of unity, by the time they start to diverge the Gaussian has a value of only 10^-8.

Anyhow, a couple of weeks ago I left off with the question of how you make an (l,m) state of (1/2,1/2) out of three electrons. Your available states are :


And if you add up those three states, you get the (3/2, 1/2). That's total spin 3/2, z-axis spin 1/2. So what combination gives you total spin 1/2, z-axis spin 1/2? It's a question that has some funny implications and I want to come back to it.

Thursday, June 6, 2013

Gaussian:Pascal's Triange <->Maxwell-Boltzmann:???

I think I've got something very cool here. I was working on the problem of total spin states of multiple electron systems, and a guy on stackexchange posted a function showing the distribution of the states. I'm not exactly smart enough to read his math, but I can pick out his numbers, and I figured out how to generate them out of a kind of modified Pascal's Triangle.

I'm not going to explain Pascal's Triangle here, but there's something about it everyone should know. If you take any row, the higher the better, it starts to look a lot like a Gaussian distribution. There are all kinds of reasons why this makes sense. Now, this new function I'm working with, this modified Pascal's triangle, supposedly describes the number of ways to generate a particular spin state from a random collection of electrons. So I'm it like Pascal's triangle? Is the distribution somehow Gaussian?

Well, in three dimensions you shouldn't really expect a Gaussian distribution. For example, in an ideal gas, the x-, y-, and z-velocities of the molecules are all individually described as Gaussians. But the total distribution described radially is then a Gaussian multiplied by r-squared. that how electron spins are distributed? I think it is.

It's not all that easy to run the numbers, but I've put it in a spreadsheet and I think it works out. The triangle goes like this:

     1  1
   1 2
  1 3 2
 1 4 5
1 5 9 5
etc. (You should be able to see how the numbers get generated.)

The way it works, for example, is if you have three electrons you look in row 3, and there is one way of generating the spin-3/2 states (four altogether) and two ways of generating each of the two spin-1/2 states. So there are eight states, counting duplicates.

The question then becomes: if you assume in a random collection of electrons that each state is equally probable, and you plot the resultant spins as  vectors in three dimensions, are the tips of those vectors distributed in space as a Maxwell-Boltzmann function? I think it turns out they are.


Tuesday, June 4, 2013

Using Three Electrons to Make L = 1/2, M = 1/2

I told you in my last post about how I was trying to line up the spin states of three electrons with the (l,m) descriptions of spin. The real problem was when you had a z-axis spin of 1/2, how to tell whether your total spin was 3/2 or 1/2.

I figured out that you can basically create either of these states only by combining states with two electrons up and one down, which I am going to label A, B, and C:

A = duu
B = udu
C = uud

The one obvious way to create a z-axis spin of 1/2 was to add the three states: A+B+C. But what is the total spin of this system (the l-number)? Is it 3/2 or 1/2?

I'm pretty sure it's 3/2. Here is my reasoning. To get a spin of 1/2, you would want to start by combining two electrons into a single state, and then adding a third electron. Something like this:

{ud -du}*u = udu - duu  = B - C    (Equation 1)

To get a total spin of 3/2, on the other hand, you would want to take a spin-1 combination and add an electron pointing down the other way:

{uu}*d =  uud   =  A                   (Equation 2)

But in fact, neither of these systems is right. The problem is that electrons don't really have separate identities. You just can't say the first electron is doing this and the second electron is doing that. You have to come up with a description whereby it doesn't matter which electron is which. That's not always so easy.

But in the case of Equation 2, we can actually do it. We just have to consider all permutation of whether the first, second or third electron is the one  pointing down. It looks like this:

uud + udu + duu  

But this is just the combination A + B + C (actually in reverse order this time) which I said was my likely candidate for the (l,m) state of (3/2, 1/2).

Can we do the same thing then for the (1/2, 1/2) state? We wrote an expression in Equation 1 for combining a spin-up electron with a pair in the singlet state. (It's a fascinating story as to why the difference ud-du gives you the singlet state! but that's one for another day.) By analogy, all we need to do is symmetrize it by combining all possible permutations of A, B and C. Generalizing  Equation 1, we get:

{B-C} + {C-A} + {A-B} =  .......0  ????

Horrifyingly, just when we've got a nice symmetric expression, we find it adds up!

But there's one possible way out. I've taken all permuations of {B-C} and added them up (actually I've taken only the even permutations, as defined in Group Theory, but don't worry about that). Who said we had to add them? Why not subtract them? Well, there's a problem subtracting three things from each other, but I'm going to borrow some ideas from the Theory of Equations that I wrote about last summer. We're going to use the three cube roots of unity, which I will call {1, w, and w^2}. Here is the combination I will try:

{B-C} + w{C-A} + w^2{A-B}

It's like I've taken three vectors in the complex plane separated by 120 degrees and applied them to my spin states. It turns out you can reduce this expression with algebra and it comes to:

{B-C} + w{C-A} + w^2{A-B} = -i{A + wB + w^2C}

Interesting results. The phase of a quantum state is of course arbitrary, so this is the same as:

{A} + w{B} + w^2{C}    =  (?)  (1/2, 1/2) ???

That's the question. Have we written a description of the (1/2, 1/2) state in terms of three electrons? I think we probably have, but we'll consider the implications later. Because there are some very important parity conditions that have to be observed, and it's not at all clear that we've done so.

Counting Spin States for Three Electrons

I told you recently that I don't know how to count spin states when I have four or more electrons. I thought at least I could count up to three but it turns out I was wrong. I have a problem counting the spin states with three electrons.

I said when there were four electrons I could specify the total spin state by listing eight numbers, corresponding to the complex amplitudes of each electron being up or down. My problem was that there were actually nine spin states you could make with four electrons...five of spin-2, three of spin-1, and one of spin-0. I posted this problem on to see if anyone could help me out.

I got a very good answer from one Lagerbeer who unfortunately had a totally different way of counting. He didn't have eight states and he didn't have nine states. He had sixteen states, corresponding to every possible ordered combination of four electrons, listed as being either up or down:


And of course this is right. The way I was counting, if I had n electrons I had to specify 2n parameters. That was wrong. You need to specify 2^n parameters, the amplitudes to be in each of those possible compound states.

In a way that's a little better. I had a real problem specifying nine independent states with only eight numbers. With sixteen numbers at least I have too many parameters, which in a way is better. But there's a new wrinkle. I thought I was OK up to three electrons, because I had six paramters to describe six possible spin states spanning l=(3/2, 1/2) and m=(-3/2...+3/2). It turns out I was wrong. I don't have six parameters, I have eight, corresponding to 2^3. So I can't even count up to three.

But this clarifies the problem tremendously. I know how to work the up/down basis for two electrons...that is, I know how to relate the up/down states to the (l,m) states. I even wrote a couple of blog posts about it. For two electrons it works like this (where the bracketed terms are (l,m):

uu          <---->   (1,1)
ud+du    <---->   (1,0)
ud-du     <---->   (0,0)
dd           <---->   (1,-1)

The l number is of course total spin and the m number is z-axis spin. And I haven't bothered to normalize the mixed states. But you get the idea.

Can I do it for three electrons? Let's list all eight states. For the first and last states, it's obvious what they line up with in terms of (l,m). But for the others it's not so obvious....

1.   uuu    <---->  (3/2, 3/2)
2.   duu
3.   udu     <---->    ?(3/2, 1/2)?
4.   uud                      ?(1/2, 1/2)?
5.   ddu
6.   dud     <---->    ?(3/2, -1/2)?      
7.   ddu                     ?(1/2, -1/2)?
8.   ddd     <---->  (3/2, -3/2)

The problem boils down to this: how do you combine the 2, 3 and 4 states to make the two different m=1/2 states?

Well, by analogy with the two-electron system we can add those three states together: {1} + {2} + {3}. (Again, I'm not bothering to normalize). That definitely gives us a state with z-axis spin of 1/2. But what is the total spin? Is it 1/2 or 3/2? That's not so obvious.

And with two electrons, we got our two different states by taking sums and differences. Well, we've taken the sum, but how do you take the difference of three states? 

I have some ideas on this but I think I'll save that for my next post.

Sunday, June 2, 2013

Still can't figure out how to count spin

I still can't figure out how to count spin states, so I posted this question on

I can't figure out how many different spin states I can create with a four-electron system. I think I can create a spin-zero state, three spin-one states, and five spin-two states. That gives me nine possible states altogether.

My problem is that I can specify a maximum of eight (complex) numbers to completely describe the spin states of the four electrons. But the nine spin-states I can seemingly create correspond to the spherical harmonic functions and I know for sure that they are linearly independent. It seems very wrong.

This discrepancy in the counting doesn't appear until you get to four electrons. It gets worse as you add more electrons.

Does anyone else have a problem with this?