Download Energy in Resistor Networks

Energy in Resistor Networks
Tim Kunisky
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Motivation
We know from empirical physical laws that a resistor with resistance R and a current I moving
across it will, over a fixed period of time, dissipate an amount of energy proportional to the power
dissipated, P = IV = I 2 R. So, it is reasonable to expect that for a given voltage across a resistor
network, current will flow, roughly speaking, in such a way as to minimize the total power dissipated
P
P
P = e I(e)2 R(e), by analogy to the various other physical principles
by the entire network,
of least action. While Ohm’s law gives us an expression for the current that a voltage generates
locally at each edge, this interpretation will characterize the current globally as a minimizer of the
appropriate “energy” quantity.
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Defining Energy
Let G be a finite network with vertex set V and edge set E, equipped with conductances for the
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edges, given as a function c : E → R+ , and resistances r : E → R+ , given by r = c . Recall that a
voltage is just an arbitrary assignment of real values to the vertices, or a function v : V → R. For a
given voltage, the current it induces across each edge is given by Ohm’s law,
i(hx, yi) = c(x, y) v(x) − v(y) .
Note that the current depends on the orientation of the edge hx, yi; in particular, the current is
antisymmetric: i(hx, yi) = −i(hy, xi). So, let us take E to be the set of oriented edges, where we
choose some positive orientation e = hx, yi for each edge e, and then define −e = hy, xi. Denote by
E+ the representative set of positively oriented edges, containing the (arbitrary) positive orientation
of each unoriented edge.
Then, the conductance is a symmetric function on E, satisfying c(−e) = c(e), while the current
is antisymmetric, satisfying i(−e) = −i(e). The set of all antisymmetric functions α : E → R clearly
forms a finite-dimensional real vector space. This structure, however, does not yet enable us to
describe a quadratic function of current such as power, to which end we note that this space can be
equipped with the usual real inner product,
(α, β) =
X
α(e)β(e) =
e∈E+
1 X
α(e)β(e),
2 e∈E
where the factor of 12 is used when summing over all oriented edges because the dimension of the
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2 (E). The inner product then gives us a
space is |E+ | = 2 |E|. Let us call this inner product space `−
natural definition of energy/power.
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df
2 (E) is E(θ) = (θ, θr ).
Definition. The energy of θ ∈ `−
Then, we have for a current flow i that
E(i) = (i, ir ) =
X
i(e)2 r (e)
e∈E+
which is precisely the “total power dissipated” described earlier. There is a rather more elegant
2 (E) by
formulation of this definition: define another inner product on `−
df
(α, β)r = (α, βr ) = (αr , β)
and let k • kr be the corresponding norm. Then, the above definition is simply
E(θ) = kθk2r .
Here we are using r as a positive-definite symmetric form on the space of antisymmetric functions;
indeed, it is easy to check that in the “standard” basis of characteristic functions, r is just a diagonal
matrix whose entry for the basis element corresponding to edge e is r (e).
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The Subspace of Current Flows
2 (E) which arise by Ohm’s law as curNow, we will attempt to characterize those elements of `−
rents induced by voltages. The voltages, being simply the functions f : V → R, also form a finitedimensional real vector space, and giving it the inner product
(f , g) =
X
f (v)g(v)
v∈V
we obtain another inner product space, `2 (V ). Since increasing a voltage function by any constant
yields the same current, the numerical value of this inner product is physically inconsequential, but
as we will see shortly, this inner product is very convenient for certain calculations.
2 (E), defined by
Note that Ohm’s law gives a linear mapping O : `2 (V ) → `−
(Of )(hx, yi) = c(hx, yi)(f (x) − f (y)).
2 (E) as flows (we will later relate them to the usual graph-theoretic
We will refer to the elements of `−
notion of a flow), and to the subspace I = img(O) as the current flows. The map O decomposes into
2 (E) given by (df )(hx, yi) = f (x)−f (y),
the composition of two maps: first, the map d : `2 (V ) → `−
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and second the map c : `− (E) → `− (E) given by pointwise multiplication with the conductances.
Then, we have O = cd, and in particular Ohm’s law becomes the simple statement i = cdv, or
dv = ir . Thus for the energy of a current flow, this gives
E(i) = (i, ir ) = (i, dv),
which is analogous to the physical law P = IV . To work with this expression, it is useful to identify
an adjoint operator of d.
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2 (E) → `2 (V ) given by
Lemma 3.1. The operator d∗ : `−
df
(d∗ θ)(v) =
X
θ(hv, wi)
w∼v
2 (E), we have (d∗ θ, f ) = (θ, df ).
is adjoint to d; that is, for any f ∈ `2 (V ) and θ ∈ `−
Proof. Expanding the definition of the inner product in `2 (V ) and switching the sum to be over
edges, we obtain
(d∗ θ, f ) =
X
f (v)
θ(hv, wi)
w∼v
v∈V
=
X
X
(f (v)θ(hv, wi) + f (w)θ(hw, vi))
hv,wi∈E+
=
X
(f (v) − f (w))θ(hv, wi)
hv,wi∈E+
=
X
(df )(e)θ(e)
e∈E+
= (θ, df ).
We then have that img(d) = ker(d∗ )⊥ , because for any θ, we have (θ, dv) = (d∗ θ, v), which means
that θ is orthogonal to dv for all v if and only if d∗ θ = 0. So, we have an orthogonal vector space
2 (E) = img(d) ⊕ ker(d∗ ).
decomposition, `−
Remark. In claiming that this decomposition holds, we are using implicitly that the graph is finite (if
2 (E) were not finite-dimensional, it would hold if and only if the subspaces in question were closed
`−
under the metric topology; in the finite-dimensional case it follows just by choosing orthonormal
bases for the components separately and noting that their union forms an orthonormal basis for
the entire space).
So, the subspace I is the image of ker(d∗ )⊥ under (pointwise multiplication by) c. Alternatively,
2 (E) are perpendicular if
note that under the weighted inner product (•, •)r , two elements α, β ∈ `−
and only if α and cβ are perpendicular under the ordinary unweighted inner product (•, •). Hence,
if I is the image of ker(d∗ )⊥ under c, then it is also the subspace orthogonal to ker(d∗ ) under the
weighted inner product, which we write as I = ker(d∗ )⊥r . Our first main result is essentially a series
of easy corollaries of this statement.
2 (E), there is a unique current flow i for which
Theorem 3.2 (Thomson’s Principle). For any θ ∈ `−
∗
∗
∗
∗
d i = d θ, and moreover for any η with d η = d θ, we have E(i) ≤ E(θ), with equality if and only
if η = i.
Proof. By the orthogonal decomposition, any θ ∈ `2 (E o ) is uniquely expressible as θ = i + (θ − i)
for some i ∈ I for which θ − i ∈ ker(d∗ ). But, this last condition is the same as d∗ (θ − i) = 0, or
d∗ θ = d∗ i. Hence, there is a unique current flow i for which d∗ i = d∗ θ.
If we have any η for which d∗ η = d∗ θ = d∗ i, then we have η = i + (η − i) is the orthogonal
decomposition of η, and thus by the Pythagorean theorem kηk2r = kik2r + kη − ik2r ≥ kik2r , with
equality if and only if kη − ik2r = 0, which is equivalent to η = i. Since E(θ) = kθk2r , the theorem as
stated follows immediately.
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Hence, for fixed values of the operator d∗ , the current flows are precisely those that minimize
energy; thus, we have obtained our minimization principle, now it remains to interpret the operator
d∗ , whose “level sets” give us the sets over which the current flows are minimal in energy.
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Flows and d∗
2 (E) to standard graph-theoretic flows between
We begin by relating the elements of the space `−
vertex sets.
2 (E) is a flow from A to Z, or an A-Z flow, if the
Definition. For disjoint sets A, Z ⊂ V , we say θ ∈ `−
following conditions hold:
1. d∗ θ(a) ≥ 0 for a ∈ A.
2. d∗ θ(z) ≤ 0 for z ∈ Z.
3. d∗ θ(x) = 0 for x ∉ A ∪ Z.
The usual interpretation is that an A-Z flow represents some material entering the network at vertices of A (the sources), flowing through the network with no loss, and exiting the network at vertices
of Z (the sinks). The value of d∗ θ(x) then represents the net amount of material entering the network at x. Note that d∗ gives us a reformulation of Kirchoff’s node law: if v is a voltage harmonic
off A ∪ Z and i is the induced current, then d∗ i is zero off A ∪ Z.
It is easy to verify that the above definition is equivalent to any other reasonable definition of an
object representing material flowing through a network. Indeed, our first deduction will be that a
flow between sets defined as such necessarily satisfies the intuitive property that the total amount
of material entering the flow is the same as the total amount exiting.
Lemma 4.1. Define
In(θ) =
X
d∗ θ(a),
a∈A
Out(θ) =
X
d∗ θ(z).
z∈Z
Then, In(θ) = −Out(θ).
Proof. Using that d∗ θ is zero off of A and Z and that A and Z are disjoint, we have
In(θ) + Out(θ) =
X
x∈A∪Z
d∗ θ(x) =
X
d∗ θ(x) = (d∗ θ, 1) = (θ, d1) = (θ, 0) = 0
x∈V
by adjointness, where we write the constants 1 or 0 for the corresponding constant maps.
Definition. The strength of a flow θ is the non-negative quantity Str(θ) = In(θ) = −Out(θ).
Recall that for current flows, the energy was given by E(i) = (i, dv). When the voltage v is constant
on the sets A and Z, we can simplify expressions of this sort significantly in terms of the strength.
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Lemma 4.2. If f ∈ `2 (V ) is identically equal to some α on A and some ζ on Z, then for any A-Z
flow θ, we have
(θ, df ) = (α − ζ) · Str(θ).
Proof. By adjointness,
(θ, df ) = (d∗ θ, f ) = α
X
a∈A
d∗ θ(a) + ζ
X
d∗ θ(z) = α · In(θ) + ζ · Out(θ) = (α − ζ) · Str(θ).
z∈Z
In particular, this means that if a voltage equal to some p on A and equal to zero on Z (and harmonic
off A ∪ Z), then the energy of the induced current is
E(i) = (i, dv) = p · Str(i)
If we now choose p so that i is a unit current flow, that is so that Str(i) = 1, then we will obtain
simply E(i) = p. On the other hand, we know by the “effective version” of Ohm’s law that when
A = {a}, then the effective resistance R(a → Z) satisfies p = Str(i) · R(a → Z), and hence for a unit
current flow i, we find p = R(a → Z), so R(a → Z) = E(i) for a unit current flow i induced by a
constant voltage. This allows us to generalize the definition of effective resistance to arbitrary sets
in a natural way.
Definition. The effective resistance between disjoint subsets A, Z ⊂ V is R(A → Z) = E(i) for i the
unique unit current flow induced by a voltage that is constant on A and zero on Z.
And, from Thomson’s Principle we immediately obtain a simple characterization of this resistance
as a minimization, that does not make a distinction between current flows and general flows at all.
Lemma 4.3.
R(A → Z) = min E(θ) : θ is a unit flow from A to Z .
Proof. Let θ0 be the choice of θ that minimizes this. Since the property of θ0 being a flow from A
to Z is determined by d∗ θ0 , we obtain by Thomson’s Principle a unique current flow i = dv with
d∗ i = d∗ θ0 , and thus in particular i is a current flow from A to Z. Additionally, the property of θ0
having strength 1 is also determined by d∗ θ0 , so i is also a unit flow. We have E(i) ≤ E(θ0 ), so
since θ0 minimizes energy among unit flows from A to Z and i is another unit flow from A to Z, it
must be that θ0 = i.
Thus, we know that the minimizer is a unit current flow i. On the other hand, we know that
R(A → Z) is the energy of a unit current flow, specifically the unit current flow j induced by a
constant voltage w. So, it suffices to show that E(i) ≥ E(j). Consider the energy of i − j:
E(i − j) = (i − j, i − j)r
= (i, i)r − 2(i, j)r + (j, j)r
= E(i) + E(j) − 2(i, j)r
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Since we have dw = jr , if we have that w is some fixed p on A and zero on Z then by the previous
lemma, the last term is
(i, j)r = (i, jr ) = (i, dw) = p · Str(i) = p = E(j).
Thus, we have E(i − j) = E(i) − E(j), and hence E(i) = E(j) + E(i − j) ≥ E(j), as desired.
Remark. Note that the property (i, j)r = E(j) is the same as the projection of i onto j being j
itself. Hence, any i − j a current flow of strength zero that is orthogonal to j. On the other hand, by
the previous lemma every current flow of strength zero is orthogonal to j = dw. Thus, the content
of this lemma (stripped of the terminology of resistance) is that the unit current flows are just an
orthogonal translation of the current flows of strength zero, by the unique unit current flow j that
arises from a constant voltage, which hence is minimal in energy among unit current flows.
By the same argument, we obtain that for any fixed strength s, if Xs ⊂ I is the set of current
flows from A to Z of strength s and js is the unique such current flow induced by a constant voltage,
then Xs = js + X0 , and moreover js = sj1 is orthogonal to the subspace X0 .
An interesting connection that this interpretation brings up is with the max-flow problem: there,
we impose strict bounds on the flow across each edge, and attempt to find the flow of maximum
strength (usually taking A and Z to be singletons) that does not violate these bounds. Here, we
instead have a “smoothed” version of these strict upper bounds given by the energy norm, and if
we fix some strength s then we can characterize the flow of least energy as arising from a constant
voltage by the above result. Hence, solving the smoothed max flow problem amounts to finding the
greatest constant voltage whose current does not exceed a fixed energy bound. This in turn can be
studied directly using that E(i) = (i, dv) = (cdv, dv), and using that the harmonicity requirement
on v can be significantly simplified when v is assumed to be constant on A and Z.
We now easily obtain the second main theorem.
Theorem 4.4 (Rayleigh’s Monotonicity Principle). Let G be a finite connected graph with two conductance assignments c and c 0 so that c ≤ c 0 on each edge. Then, the effective resistances between
arbitrary disjoint subsets A, Z ⊂ V satisfy
Rc (A → Z) ≥ Rc 0 (A → Z).
Proof. We know by the lemma that there is a unit flow θ from A to Z for which Rc (A → Z) = Ec (θ).
Then, it suffices to observe that Rc 0 (A → Z) ≤ Ec 0 (θ) ≤ Ec (θ), where the first inequality follows by
the characterization of effective resistance from the lemma, and the second from the expression
Ec (α) =
1 X
1 X α(e)2
r (e)α(e)2 =
2 e∈E
2 e∈E c(e)
and hence replacing the c(e) with larger quantities (in our case, the c 0 (e) for each edge) will only
cause the energy to decrease.
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