Conference rusure::math

Physicist's reply follows formfeed...

To measure the resistance between grid node A and adjacent node B,
we could attach a known current source between the two nodes and
then measure the potential difference (voltage).  Then we calculate
the resistance using Ohm's law, Resistance = Voltage / Current.

Ohm's law is linear, therefore we may consider this case as the
linear superposition of a current source at node A and a current
sink at node B.

Consider a current source at node A.  Say the current into node A
is 1 Amp and define the zero of electric potential at infinity.
By symmetry, each of the 4 1-Ohm resistors connected to node A
carries 0.25 Amp.  The direction of the current is away from node A.

Now consider the current sink at node B.  Again by symmetry, each
of the 4 resistors connected to node B carries 0.25 Amp.  The
direction of the current is into node B.

Now superpose the two solutions.  The single resistor between nodes
A and B carries 0.25 + 0.25 Amp = 0.50 Amp.  Use Ohm's law to
calculate the potential difference between A and B:  V = IR =
(0.50 Amp)(1 Ohm) = 0.50 Volt.

So what do we have?  We have a total current of 1 Amp into the grid
at node A and a total current of 1 Amp out of the grid at node B.
And the potential between A and B is 0.50 Volt.  Therefore the
equivalent resistance of the entire grid is R = V/I =
(0.50 Volt)/(1 Amp) = 0.50 Ohm.

Now the physicist haughtily says, "Here is a problem that a
mathematician would find very difficult to solve but yields
readily to solution when we apply a little physical understanding."
This is all well and good as long as we don't think too hard about
what's going on at infinity.  The problem is:  when we superpose
our two solutions, does the current in any resistor far, far away
from nodes A and B approach zero in such a way to guarantee that
charge is conserved at every node?  Are we not sweeping under the
rug any problems we may have at infinity?  The question is beyond
*this* physicist's ken.

Re .1:

Not only will you not be able to generate enough electrons to fill an
infinite grid, but the charge will not reach all regions of the grid in
any finite time.  I assume the voltage drop will have a limit, but I
won't swear that it is non-zero.
				/AHM

    re .2:
    
    Seems to me that if you can postulate an infinite grid of resistors
    then you can also assume that electricity is continuous and not
    composed of discrete electrons.
    
    I can prove easily (not mathematical proof) that the voltage drop
    will be non-zero. The infinite grid is just a discrete model of
    of a continuous system, otherwise known as "free-space". If the voltage
    drop were zero, then electromagnetism could not radiate.
                                                   
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Interesting question.  Here are some natural followups:

o	What is the resistance between ANY two nodes (x1,y1) and (x2,y2),
	not just adjacent ones ?

o	Solve the problem in three dimensions.  That is, 1-ohm resistor
	connecting every integer (x,y) with each of the six orthogonally
	adjacent nodes.  What is resistance between any two (x1,y1,z1)
	and (x2,y2,z2) ?

/Eric

    re: .1, "the symmetry" argument.
    
    I found your solution intriguing, but not convincing.  I don't see
    that you can break a non-symmetric situation into two symmetric
    ones, solve, and then just add them back together to get the answer.
    
    Understand that I'm not saying your answer is wrong, just unconvincing.
    
    I also have the following difficulty:
    
    	Given this infinite grid, and a current source at the (0,0)
        point, what VOLTAGE is required between that source and the
    	"edge" to drive 1 amp of current out of the source?  Answer:
    	Infinite!  I'm not sure that Ohms Law, or any other laws,
    	apply to sinks operating 1-unit away from an infinite voltage
    	point!  In some sense you have done +infinite + -infinity =
    	0, but it doesn't, it is undefined.
                                           
    /Bevin

    The limiting argument to support .1 is similar to that used to justify
    contour integration of improper integrals via residue calculus.  I don't
    recall the details since it's been years since I saw the problem in the
    base note.  But you basically show that the currents on the boundry
    of a large but finite grid are perturbed only slightly by moving the
    source of current near the center by one grid point.  Letting the boundry
    grow lets you have any accuracy you desire via the usual epsilon delta
    arguments.

    Calculating the 'transfer function' from one pair of grid points
    to any other is an intriguing problem.  One way to approach it may be
    by invoking Tellegen's theorem, and a variational principle that
    states that the currents will distribute themselves thru the grid of
    resistors in such a way as to minimize the total dissipated power.

    From this point of view, you could even make a probabalistic argument out
    of it. I've heard of an MAA book on random walks and electric networks
    that may be relevant - has anyone seen this book?

    - Jim

    My intuition tells me that with a single current source connected
    between the center and infinity, there will be equipotential lines
    surrounding the source. The first equipotential is easy, it is the
    four closest nodes. Connecting those (graphically, not electrically)
    forms a diamond shape. It seems reasonable to me that the next
    equipotential will be the next larger diamond, parallel to the first,
    and so on. 
    
    Now to determine the resistance between the source and a particular
    equipotential, electrically connect those nodes on the diamond and
    ground it. The parallel combination can then be found relatively
    easily.
    
    If we start with the equipotential closest to the source and work
    outwards, I get the following series for the resistance.
    
    	1/4, 1/12, 1/20, 1/28, 1/36, ...
    
    thus, the resistance between the source and the "n"th equipotential
    is:
    
    	       n
       1     -----      1
      --- *  \      ________
       4     /       2*i - 1
    	     -----
             i = 1      
                                                            
    So the voltage difference between any two points would be twice
    this (by the same argument as the original problem)
    
    However, I admit that this is just an intuitive argument and I have
    no formal "proof".
    
    I'm ready to be torn to shreds.
    
                                                   
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                 /
     

    > I found your solution intriguing, but not convincing.  I don't see
    > that you can break a non-symmetric situation into two symmetric
    > ones, solve, and then just add them back together to get the answer.
    
I don't have a problem with linear superposition.  It's just 
decomposing the solution to a linear equation into the sum of
two other solutions.  You can do this trick to construct the 
field from an electric dipole.  You superpose the (easily solved
for) field from a unit positive charge with the (translated) field
from a unit negative charge.  The fields from the point charges are
radially symmetric about the charges and the dipole field is anti-
symmetric about the plane equidistant from the two charges.
    
    	> Given this infinite grid, and a current source at the (0,0)
        > point what VOLTAGE is required between that source and the
    	> "edge" to drive 1 amp of current out of the source?  Answer:
    	> Infinite!  I'm not sure that Ohms Law, or any other laws,
    	> apply to sinks operating 1-unit away from an infinite voltage
    	> point!  In some sense you have done +infinite + -infinity =
    	> 0, but it doesn't, it is undefined.
                                           
I'm not happy about "sweeping infinities under the rug" either! :-)
You always have problems with infinities near point charges, current
sources, and infinitesimally thin current carring wires.  The point
is, I guess, that here's a technique that gives the correct answer
to an otherwise difficult problem.  Physicists make arguments like
this all the time.  Something similar happens in quantum electro-
dynamics (QED).  Some of the answers you calculate are infinite.
But if you muck about a bit with other infinities you can get 
answers that have been experimentally verified to one part in 
billions (this "mucking about" is called "renormalization").
QED is the most accurately verified theory in the history of physics.

Re .3:

Perhaps I misunderstood .1's reference to conservation of charge.  And
I know I'm not used to reasoning with constant current sources.

Will the power output by the constant current source vary with time?
				/AHM/THX

    re .9:
    
    An ideal current source is defined as being able to provide the
    specified current regardless of the resulting voltage potential
    across it. 
    
    Likewise, an ideal voltage source will maintain the specified potential
    across it regardless of the amount of current necessary to maintain
    that potential.
    
    Thus, if the current source is maintaining a constant current, and
    the load is not changing with time, then the power supplied will
    not vary.
    
    In the solution to the problem, it should probably be stated that
    the system is in steady-state, that is, the current source was
    connected to the grid an infinite time ago.
    
                                                   
                  /
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                 /
     
    P.S. Is anybody verifying my intuitive solution to the resistance
    between any two points?

>    P.S. Is anybody verifying my intuitive solution to the resistance
>    between any two points?

Yes.  Give me a bit of time to work my way through the replies; this has
turned out to be more complicated than I thought !

re .1:

When you break up the current source into two current sources, I assume
you mean the other end of the current source to be attached to the
'zero of electric potential' you mention later.

I think I agree with this solution although I find the result a bit
surprising.  It means that the equivalent resistance between two
adjacent nodes if I remove the resistor between the adjacent nodes is
also 1 ohm (by resistors in parallel).

I notice that the argument depends only on symmetry which is also interesting.

re .1, .2, .3:

As to the questions on charge conservation, I think that the intent was
to treat the resistors as ideal.  Every node in the network (albeit an
infinite network) is thus assumed to respond instantaneously to any change
in driving current or voltage.  To simulate the effect of finite propagation
delay time, the grid might need ideal capacitors which is a different problem.
I didn't intend it to be that complicated.

re .4:

The first one is hard.

To the second question for adjacent nodes: I think it must be 1/3 ohm by
an analogous symmetry argument to that in .1 (3 dimensions, current divides
evenly 6 ways, etc.).  The general answer for this second question also
seems hard.

re .5:

I don't understand how you derive the voltage needed between the source and
the edge to drive 1 amp of current as infinite.  I think it can be finite.

re .6:

What is Tellegen's theorem ?  Sounds intriguing.

re .7:

In what follows, I'm talking about a single current source.

I'm not sure I buy the idea that all the nodes on the diamond lines are
equipotential.  It's plausible that { (x,y), (x,-y), (-x,y), (-x,-y) } are
at the same potential, but I suspect that it's not true for all (x,y)
such that abs(x) + abs(y) = c.  My analogy would be
to the falloff in current density in a sheet conductor where total flux
across any circle's diameter whose center is the point current source.
Current density would be radially outward and the equipotentials are the
diameters.  I would therefore guess that the points that are approximately
equipotential would satisfy x^2 + y^2 = k for the same k.  This is
admittedly a fuzzy argument.

Re: .7,.12
    With considerable equation juggling, I convinced myself that the
    'diamond' argument gives incorrect results.  The argument in .12
    is very satisfying.

    So what *is* the right answer?

    re .12,.13:
    
    as author of .7, I think that .12 is correct. but as before, I have
    no proof.
                                                   
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re .8:

I agree with the correctness of the superposition argument.

> I'm not happy about "sweeping infinities under the rug" either! :-)
> You always have problems with infinities near point charges, current
> sources, and infinitesimally thin current carring wires.  The point
>is, I guess, that here's a technique that gives the correct answer
>to an otherwise difficult problem.  Physicists make arguments like
> this all the time.  Something similar happens in quantum electro-
>dynamics (QED).  Some of the answers you calculate are infinite.
>But if you muck about a bit with other infinities you can get 
>answers that have been experimentally verified to one part in 
>billions (this "mucking about" is called "renormalization").
>QED is the most accurately verified theory in the history of physics.

I thin your solution is correct and I like the technique.

But I don't think this problem as originally stated involves currents or
voltages going to infinity.  Even if this were treated as a network of
elements with energy storage, I think that infinities wouldn't appear.

For example, consider the same grid with each resistor replaced with a
shunt capacitor&resistor.  This is (sort of) like a transmission line.
When a voltage step is applied to a transmission line, the currents and
voltages along the line don't go infinite.  What happens is that the step
propagates down the line.

It wouldn't matter if the transmission line were infinite; the currents and
voltages established at each point in space and time are determined by what
the electrons see locally (that is, looking into the next segment of line
that they are moving into).

Your statements about QED are probably true; QED is over my head.  I'm still
struggling with classical electromagnetism.

Re .9:

> Will the power output by the constant current source vary with time?

For the problem stated in .0, I don't think so.  The equivalent impedance
of the infinite grid is a simple resistance.  Since the application of
a constant current source to a linear time-invariant resistance is a
constant voltage, the product (that is, V*I) is also constant.

If the network had energy storage elements, then the answer would be
a yes.

re .10:
    
Yes to everything.
     
re .13:

>   So what *is* the right answer?

Sorry; to which question ?

The general question posed by Mr. Osman regarding the resistance between
any two nodes is unresolved.  If I can track down Mr. Roth's reference to
Telllegen's theorem, I will work that one.

The answer to the original question (that I saw) is that given in .1.  The
solution I saw did not have the worked out details but just an answer.

re .14:
    
>   As author of .7, I think that .12 is correct. but as before, I have.
>   no proof.

Well, neither do I.  The answer isn't going to be the same as sampling the
continuous solution at the grid point coodinates because the infinite
grid voltage distribution the result of solving an infinite set of discrete
difference equations and the continuous solution is the result of solving
a single differerential equation.  But I'd wager that they'll look similar.

I'll see if I can write down the recursive equations for the voltage
V(x,y) for each point on the grid this weekend and try to solve it.

Re .15:

>If the network had energy storage elements, then the answer would be
>a yes.

And since a real-world network of resistors has capacitance and inductance,
the answer would be "yes" in the real world.  But I agree that .0 discusses
an ideal, not the real world.
				/AHM/THX

    how about a word from an amateur in math but fairly experienced
    in tronics. 
     I think the answer is   .5>r>.4999...

program grid(output);

var v,vp: array [ 1..101, 1..101 ] of real;
    i, j, k:integer;

begin

for j := 1 to 101 do
 for k := 1 to 101 do
  v[j,k] := 0.0;

v[51,51] := 1.0;

for i := 1 to 200 do
 begin
 for j := 2 to 100 do
  for k := 2 to 100 do
   vp[j,k] := (v[j-1,k]+v[j+1,k]+v[j,k-1]+v[j,k+1])/4.0;
 for j := 2 to 100 do
  for k := 2 to 100 do
   v[j,k] := vp[j,k];
 v[51,51] := 1.0
 end;

 i := 0;

 for j := 2 to 100 do
  for k := 2 to 100 do
   begin
   i := i + 1;
   if i = 3 then
    begin
    i := 0;
    writeln( j:3,',',k:3,':',v[j,k] )
    end
   else write( j:3,',',k:3,':',v[j,k] )
   end
end.

1. Out with the old!

	In addressing the problem, the first thing to do is to define it
properly.   Firstly, let's forget about times of propagation, etc, etc.   I'll
assume the standard Kirchoff model for circuits.   This is already hard
enough for us to deal with.   The real difficulty with "infinity" here is the 
following:

	Imagine that I have no sources, and no sinks, then I can still define
a non-zero solution!   For instance, I can say that from each point (x,y)
1 Amp flows to (x-1,y).   I define the voltage at (x,y) to be x Volra.
This satisfies:

	Kirchoff current law:
	Total current outflow at any point is zero.

	Kirchoff voltage law:
	The potential differences round any cycle of points is zero.


	Why does this matter?   Well, normally for a finite circuit we can
specify the input and output current at sinks and sources, and this will
*define* the currents & potential differences throughout the system.   In
the infinite case, upon any given solution, I can *superpose* the anomalous
flow discussed above, to get a different solution.

	For instance, if I specify that at the point (1,0) I have 1 Amp
entering the system, and at (0,0) 1 Amp leaves the system.   Previous
replies to this note suggest that by symmetry, 1/2 Amp passes in the edge
(1,0)->(0,0).   Thus the Voltage across the edge is 1/2 Volt.   Since the 
current through the entire circuit is 1 Amp, we know that the total resistance
is 1/2 Ohm.   So ran the argument.

	However, suppose that we superimpose the anomalous current discussed
above.   Then there is current of 3/2 Amps through the critical edge.   Thus
the Voltage across the edge is 3/2 Volts.   Thus the total resistance is
3/2 Ohm.    And we could clearly give a proof for *any* resistance!   Something
is terribly wrong here.

	In summary, given some sink/source specification, the two Kirchoff 
laws are the *only* constraints that the circuit has to satisfy.   Any 
symmetry arguments *presuppose* that there is a unique solution to the
Kirchoff laws.   In an infinite network, this is just not valid.


2. In with the new!

	Given that the Kirchoff theory only makes sense for finite networks,
I think that we need to interpret the question as: given a sequence of finite
networks N_n, which approximate to a finite grid as n goes to infinity, what
is the limiting behaviour of r_n, the resistance across adjacent edges.

	The question is, what networks N_n should we take?

	Candidate (a):  Square array of points, n by n.
	What about other finite "flat" arrays, (eg: candidate (a'): all points
of Euclidean distance =< n from the origin)?   There is a pretty argument that
tells us the we can ignore such other arrays.   r_n is monotonically
decreasing as n increases.   Any candidate a' array will be bounded between
two square arrays (A_n & N_root(n), roughly).   So the large-n fate of a & a'
are just the same.

	However, candidate (a) is not trivial to analyse, so let's deal first
with:

	Candidate (b):   Toroidal array of points, n by n.
	The resistance of such an array is easy to calculate, using a cute
theorem (proof later).

	Theorem 1: Suppose that a connected network is edge-transitive (for any
pair of edges, e, e', there is a graph automorphism of the network, which
maps e onto e'.)   Then the total resistance between two adjacent vertices
is (v-1)/m, where v is the number of vertices, and m the number of edges in
the network.

	This theorem gives us r_n = (n^2-1)/2(n^2), which clearly tends 
downwards to 1/2, as n increases, as the resistance of the torus.

	[If we choose a projective plane instead of a torus, then the
resistance will be the same, since once again the graph is edge-transitive.
A Klein bottle will not be edge-transitive, but (using theorem 2 below) it
is easy to see that the resistance will tend to 1/2, whichever way of the
two ways the Klein bottle is oriented.]

	But, I think that candidate a is closer to my intuition about
what we are trying to approximate.   We can get part of the way to our
solution by using the candidate b result.

	Say that we are measuring the resistance across two points linked by
an edge e.   If n is odd, then by symmetry, we can cut the torus into a tube 
along a line, l, parallel to e, since (by symmetry) no current will flow across
l.

	[A M�bius strip will also have the same limiting resistance, 1/2 Ohm,
by a symmetry argument starting with a projective plane.]

	However, we are not entitled to use symmetry cut the tube into a plane 
array, so we need to use either sneakiness or more theory, in order to solve 
candidate a.

	I believe that a somewhat long-winded argument based on a comparison
of the plane and the tube, and using the excellent Theorem 2, will do the
job, but it's too tedious and too ugly to write or read.   If some one has an 
elegant idea, that would be much more satisfactory.


Appendix.

Theorem 2:	Given an edge ab denote by N(s,a,b,t) the number of spanning
trees of G in which the (unique) path from s to t contains a and b, in this
order.   Define N(s,b,a,t) analogously, and write N for the total number of
spanning trees.   Finally, let w(a,b) = (N(s,a,b,t)-N(s,b,a,t))/N.

	Distribute currents in the edges of G by sending a current of size 
w(a,b) from a to b for every edge ab.   Then there is a total current size 1 
from s to t.

Note: (1) a spanning tree in a graph with v vertices is just a subgraph which
contains all the vertices, and which is a tree.

(2) This Theorem immediately implies Theorem 1, since every edge appears in
the same number of spanning trees, X.   N = mX/(v-1) but:
	N(s,s,t,t) = X &
	N(s,t,s,t) = 0, 
whence the result.

(3) Proof of the Theorem is "straightforward".