It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.3K
- Chat 501
- Study Groups 21
- Petri Nets 9
- Epidemiology 4
- Leaf Modeling 2
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- Baez ACT 2019: Online Course 339
- Baez ACT 2019: Lectures 79
- Baez ACT 2019: Exercises 149
- Baez ACT 2019: Chat 50
- UCR ACT Seminar 4
- General 69
- Azimuth Code Project 110
- Statistical methods 4
- Drafts 9
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 148
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 714

Options

This is a thread to discuss any topics related to this amazing book:

- John Baez and Jacob Biamonte,
*Notes on Quantum Techniques for Stochastic Mechanics*.

Re: Section 15, Dirichlet operators and electrical circuits.

Gloss: This shows how a Hamiltonian framework can be applied to a network of resistors. Let the nodes in the circuit be x1,...,xn. Between xi and xi is a resistor, with conductance (= reciprocal of resistance) cij. Form the symmetric matrix H which has cij in each non-diagonal entry, i.e., Hij = Hji = cij, and let Hii be minus the sum of all the other values in the ith row (or ith column, same thing here). By construction, H is both self-adjoint and infinitesimal stochastic. Such a matrix is called a Dirichlet operator. Because H is self-adjoint, it is a valid quantum mechanical operator, and because it is infinitesmal stochastic, it is a valid stochastic mechanical operator. So it is in an overlapping territory between the two theories.

Let V be a vector in R^n, representing the voltage at each of the points x1,...,xn. The book shows that <V, H(V)> equals the power consumed by circuit!

Here I will add a few points, to this charming topic that has been introduced.

(1) H(V) itself has a physical interpretation, which is the vector I of *currents* that is induced by the voltage vector across the resistor network. The signs are oriented so Ij is the net flow of current *into* the node xj.

It just requires a simple calculation to show this. To get some practice, let's do it for an example with three nodes x1, x2, x3, where there is a 1 ohm resistor connecting each pair of points.

Then H = ((-2, 1, 1), (1, -2, 1), (1, 1, -2)).

Let the voltage vector be (a,b,c).

Then:

H(a,b,c) = (-2a + b + c, a - 2b + c, a + b - 2c) = (b-a + c-a, a-b + c-b, a-c + b-c) = (inflow into a, inflow into b, inflow into c) = current I

(2) As a conclusion, it follows that <V, H(V)> = <V, I> = power consumed by the network.

(3) To give a full physical interpretation of the Hamiltonian dynamic on this network of resistors, we should connect a capacitor between each node and a ground point. Give them all the same unit capacitance. This is what will make the current H(V)(j) into node j actually produce a rising voltage at V(j) -- this condition is needed to fulfill the equation dV / dt = H(V).

(4) As was pointed out in the text, any constant voltage vector K = (a,a,a,...) will satisfy H(K) = 0. Interpretation: There is no current when all voltages are equal.

(5) Intuitively, we see that any given initial state of the network, given by a voltage vector V, the network will asymptotically charge/discharge to an equilibrium state where, for each connected component of the graph, all voltages are the same.

For each connected component C, the final voltage is easily calculated, by dividing the total charge in C in the network by the number of nodes in C:

FinalVoltage = Sum(V(j)) / n = mean(V(C)), for all j in C.

(6) An eigenvector is a voltage vector V that heads in a "straight line" towards the equilibrium vector. A non-eigenvector does some "turning" as it heads towards the equilibrium vector.

## Comments

Just saw the blog discussion regarding the book. I will port my comments over there. Regards.

`Just saw the blog discussion regarding the book. I will port my comments over there. Regards.`

Thanks! You're right, I'd prefer to talk about this over on the blog, since that's where people interested in this course have been congregating all along. We've got a few dozen very smart people involved there.

I don't see your comment above on the blog yet, so let me say:

If you have comments about an individual chapter of the book, it's best to go here and then click on the things that say 'Also available on Azimuth'. The chapter numbers of the book don't perfectly match those here, because we reshuffled things a bit when writing the book, moving the chapter on Perron-Frobenius theory up some. But apart from that, they're pretty darn similar.

For example, the chapter on circuits made of resistors is being discussed here on the blog. That would be an ideal place to put your above comment.

Hey, this is a great idea - this problem has been bugging me for a long time, as you'll see in my comment on the blog. But let's talk about it there!

`Thanks! You're right, I'd prefer to talk about this over on the blog, since that's where people interested in this course have been congregating all along. We've got a few dozen very smart people involved there. I don't see your comment above on the blog yet, so let me say: If you have comments about an individual chapter of the book, it's best to go [here](http://math.ucr.edu/home/baez/networks/) and then click on the things that say 'Also available on Azimuth'. The chapter numbers of the book don't perfectly match those here, because we reshuffled things a bit when writing the book, moving the chapter on Perron-Frobenius theory up some. But apart from that, they're pretty darn similar. For example, the chapter on circuits made of resistors is being discussed [here](http://johncarlosbaez.wordpress.com/2011/11/04/network-theory-part-16/) on the blog. That would be an ideal place to put your above comment. > To give a full physical interpretation of the Hamiltonian dynamic on this network of resistors, we should connect a capacitor between each node and a ground point. Give them all the same unit capacitance. This is what will make the current H(V)(j) into node j actually produce a rising voltage at V(j) – this condition is needed to fulfill the equation dV / dt = H(V). Hey, this is a great idea - this problem has been bugging me for a long time, as you'll see in my comment [on the blog](http://johncarlosbaez.wordpress.com/2011/11/04/network-theory-part-16/#comment-9694). But let's talk about it there!`

Thanks -- I ported my comments to the blog, and modularized them into a few separate posts.

`Thanks -- I ported my comments to the blog, and modularized them into a few separate posts.`