domain: phys math

text:
Adjunct Professor S D Agashe
email: eesdaia@ee.iitb.ac.in

A New Approach to 3-Dimensional Fields

Introduction 1.A New Operator

A new approach, which uses the differential operator (x ∂ ∂x + y ∂ ∂y + z ∂ ∂z ), is presented for deriving some results, old and new, in the calculus of 3-dimensional scalar and vector fields, i.e., of functions f : R 3 → R and F : R 3 → R 3 . The approach exploits the following property, which does not seem to have been noticed before, of the differential operator (see Theorem 1 below), namely:

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 3 2 dxdydz = -4πf (0, 0, 0), (1)
where f is a scalar-valued function f : R 3 → R, and its partial derivatives are evaluated at (x, y, z) .

The familiar Laplacian operator ∇ 2 , namely:

∇ 2 = ∂ 2 ∂x 2 + ∂ 2 ∂y 2 + ∂ 2
∂z 2 , has a similar property, namely:

R 3 ( ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 ) (x 2 + y 2 + z 2 ) 1 2 dxdydz = -4πf (0, 0, 0), (2)
The Laplacian result assumes that the functions involved are twice-differentiable whereas the new approach requires them to be only once-differentiable. The approach also exploits some nice properties of the sphericalpolar coordinate system and its relation with the rectangular coordinate system. Incidentally, Gauss [?] exploited these in his Memoir on the "inverse square force law" to prove that the potential function is twice-differentiable. (Most textbooks on Electromagnetism give expressions for the "del" operators in various curvilinear coordinate systems, but use them only when solving problems with spherical or cylindrical symmetry. Who would remember the curvilinear forms of the del operators, anyway?)

Both the results above should be distinguished from the Dirac δ-function idea. In these results, we have operators acting on a function whereas the δ-function multiplies the function. Moreover, the integration is rigorous and not merely symbolic.

In view of the similarity of the two results, one may expect "identities" like the Green identities which involve two functions to hold for the new operator. In fact, they do ; see Corollaries to Theorem 1 and 4 below.

A Synopsis of New Results

A result which is not as famiiliar as the property of the Laplacian is Poisson's Formula [?] , namely:

F (a, b, c) = - 1 4π R 3 ∇(∇ • F ) -∇ × (∇ × F ) r dV. (3)
where

r = [(x -a) 2 + (y -b) 2 + (z -c) 2 ]
1 2 and the numerator of the integrand is evaluated at (x, y, z).

Using the new approach,we derive a better alternative to Poisson's Formula (Theorem 6 below), namely:

F (a, b, c) = - 1 4π R 3 1 r 3 [(∇ • F ) r -r × (∇ × F ) ] dV, (4)
where r denotes the vector from (a, b, c) to (x, y, z),or a relative position vector, and r denotes its length. Further, ∇ • F and ∇ × F are evaluated at (x, y, z).

Using the new approach, we give a new proof of Helmholtz's Theorem [?], namely:

F = ∇φ + ∇×A (5)
where the "scalar potential" φ and "vector potential" A are given by some volume integrals involving F . We also give new proofs of the classical Poisson Theorem and the Divergence Theorem.

Interestingly, results similar to (1) above hold for the operator (x ∂ ∂x ) in one variable, the operator (x ∂ ∂x + y ∂ ∂y ) in two variables, and even the operator (x 1

∂ ∂x1 + x 2 ∂ ∂x2 + x 3 ∂ ∂x3 + x 4 ∂ ∂x4
) in four variables; see Theorem 1(R), Theorem 1(R 2 ) and Theorem 1(R 4 ) below.

More generally, we show (Theorem 2 below) that, if (a, b, c) is in R 3 :

R 3 [(x -a) ∂f ∂x + (y -b) ∂f ∂y + (z -c) ∂f ∂z ] [(x -a) 2 + (y -b) 2 + (z -c) 2 ] 3 2 dxdydz = -4πf (a, b, c), (6)
where the derivatives are evaluated at (x, y, z), and also:

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 3 2 dxdydz = -4πf (a, b, c), (7)
where the derivatives are evaluated at (x + a, y + b, z + c).

An auxiliary useful result (Theorem 3 below) is proved regarding operators like (x ∂ ∂y -y ∂ ∂x ).

Our proofs of these results do not invoke the Dirac δ-function , or "singularity functions", like ∇( 1 r ), and ∇ 2 ( 1 r ). We avoid "δ-function identities". In fact, we do not use any properties of the operator ∇. The operator that we use here, namely, (x ∂ ∂x + y ∂ ∂y + z ∂ ∂z ) , can be expressed as → r •∇, but we use it as a simple "scalar operator", i.e., we use it to transform scalar-valued functions into scalar-valued functions and vector-valued functions into vector-valued functions, and its use seems to be new.

The value of the function 1 r (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) at the "point" (x, y, z) can be seen to be the directional derivative [?], at (x, y, z), of f in the direction of the position vector of the point, and we can write the property of the operator also as: R 3 r•∇ r 3 dV = -4πf . In "Field Physics", it is customary to talk about "space", "field point" and "space point" and then [START_REF] Oliiver | Foundations of Potential Theory[END_REF] can be written as:

f (f ield point) = - 1 4π space r • ∇f (source point) r 3 , (8)
where r denotes the vector "drawn" from the field point to the source point, and r its length or the distance between the field point and the source point. The numerator of the integrand above is the directional derivative of the field (function) at the source point in the direction from the field point to the source point.

We prove an extension of Theorem We conclude with a list of formulas about scalar and vector fields. Many of these seem to be new, and involve integrals that have "time-retardation" or "time-advance". Our initial results are easily seen to apply for time-dependent vector fields with integrals that do not involve any delayed or advanced action. As noted in the next Section, thanks to the use of the spherical-polar coordinate system, only a slight modification suffices for delayed or advanced integrals. We use some of them to derive an interesting result regarding Maxwell's equations with delayed integrals, proved in [?] using the traditional "δ-function" approach.

The paper may also be regarded as a plea for more rigour in classroom and textbook treatments of Electromagnetism that do not use Theory of Distributions, or Differential Forms.

Preliminaries

We start with the defining relations between the rectangular coordinates (x, y, z) and the spherical-polar coordinates (r, θ, φ):

x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ.
Let us denote by R 3 sph the set of spherical-polar coordinate values, i.e., the set { (r, θ, φ) : r ≥ 0, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π}, and by T the transformation from the spherical-polar to rectangular coordinates, so that T is a function on R 3 sph onto R 3 , and T (r, θ, φ) = (x, y, z) where x, y, z are given by the equations above. From these one easily obtains, by differentiation, the operator relation

r ∂ ∂r = x ∂ ∂x + y ∂ ∂y + z ∂ ∂z . (9)
Indeed,

r ∂ ∂r = r ( ∂ ∂x ∂x ∂r + ∂ ∂y ∂y ∂r + ∂ ∂z ∂z ∂r ) = r (sinθ cosφ ∂ ∂x + sinθ sinφ ∂ ∂y + cosθ ∂ ∂z ) = x ∂ ∂x + y ∂ ∂y + z ∂ ∂z .
Similarly,

∂ ∂φ = ∂ ∂x ∂x ∂φ + ∂ ∂y ∂y ∂φ = x ∂ ∂y -y ∂ ∂x .
We note also the following relations involving the Jacobian matrix JT of the transformation T corresponding to the change of coordinates:

      ∂ ∂r ∂ ∂θ ∂ ∂φ       =       sinθcosφ sinθsinφ cosθ rcosθcosφ rcosθsinφ -rsinθ -rsinθsinφ rsinθcosφ 0             ∂ ∂x ∂ ∂y ∂ ∂z       , (10)
the determinant, |JT |, of the matrix being r 2 sinθ, and, by inverting these relations, if sinθ = 0:

      ∂ ∂x ∂ ∂y ∂ ∂z       = 1 sin θ       sin 2 θcosφ sinθcosθcosφ -sinφ sin 2 θsinφ sinθcosθsinφ cosφ sinθcosθ -sin 2 θ 0             ∂ ∂r 1 r ∂ ∂θ 1 r ∂ ∂φ       . ( 11
)
Then, if f is a function on R 3 into R, and g is the function on R 3 sph defined by

g(r, θ, φ) = f (rsinθcosφ, rsinθ sin φ, rcosθ), (12)
we have

∂g ∂r = 1 r (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ), (13)
and

∂g ∂φ = x ∂f ∂y -y ∂f ∂x , (14)
the two sides of these equations being evaluated at corresponding triples (r, θ, φ) and (x, y, z).

Remark 1:

The two functions f and g have different domains, but their values are related. They are usually denoted by a single symbol, and notation like f (x, y, z) and f (r, θ, φ) is used to indicate the two different meanings. We can also write (f • T ) for g, where "•" denotes the composition of two functions. We will say that the function g is associated with the function f . Note that the Jacobian matrix JT and its determinant |JT | are also functions on R 3 sph .

Remark 2: If the action of a time-dependent source field f (x, y, z, t) is delayed or advanced in time, the associated delayed/advanced function g(r, θ, φ, t) is defined as follows:

g(r, θ, φ, t) = f (rsinθcosφ, rsinθsinφ, rcosθ, t - r v ), ( 15
)
where v is a "speed" parameter; v > 0 for delayed action, and v < 0 for advanced action. We then have:

∂g ∂r = 1 r (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) - 1 v ∂f ∂t , (16)
and so, in equation ( 11) above, we have in the column on the right-hand side

( ∂ ∂r + 1 v ∂ ∂t ) instead of ∂ ∂r .
This simple modification leads to simple changes in the results below. Note that ∂g ∂t = ∂f ∂t . We could, of course, introduce a modified operator:

(x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) -r v ∂f ∂t .
3 The Main Results for 3-Dimensional Scalar Fields

The Basic Result

We are now ready to prove the basic result (2) mentioned in the Introduction.

Theorem 1: If f : R 3 → R has continuous first-order partial derivatives and lim r→∞ f (rsinθcosφ, rsinθsinφ, rcosθ) = 0 , uniformly for all 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π, then

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 3 2 dxdydz = -4πf (0, 0, 0) (17)
Remark 3: The integrand in the integral above is to be evaluated at (x, y, z), and r = x 2 + y 2 + z 2 . The point (x, y, z) is often referred to as the "source" point, and the point where the function value is obtained, here the point (0, 0, 0), as the "field" point. The integrand is undefined at (0, 0, 0).

Remark 4:

The integral is, of course, an "improper" integral, and is to be understood as the limit of a "definite" integral extended over the set {(x, y, z) : 0 < ≤ r ≤ R} as → 0 and R → ∞. We need the since the integrand is not well-defined at (0, 0, 0).

Proof of Theorem 1: If we denote the integrand function above by h, then the "change of variables formula for multiple integrals" [?] tells us that

R 3 h = R 3 sph (h • T )(|JT |).
We use the fact that |JT | (r, θ, φ) = r 2 sinθ, and that ∂g ∂r = x ∂f ∂x + y ∂f ∂y + z ∂f ∂z , so that we have: = -f (0, 0, 0)×2×2π = -4πf (0, 0, 0).

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 )
We have used the notation [g(r, θ, φ)] r=∞ r=0 to denote the difference [g(∞, θ, φ) -g(0, θ, φ)].

In the above sequence of calculations, we have changed a triple integral into a succession of two integrals and changed the order of integration because all the intervals of integration are finite "intervals".

We have our first Green-like identity, which we state as a Corollary, involving two functions f and h:

Corollary: dxdydz = -4πf (0, 0, 0) h(0, 0, 0).

R
Proof: We simply use the distributive property of our operator, namely:

(x ∂ ∂x + y ∂ ∂y + z ∂ ∂z )(f g) = h(x ∂ ∂x + y ∂ ∂y + z ∂ ∂z )f + f (x ∂ ∂x + y ∂ ∂y + z ∂ ∂z )h. Remark 5:
In the computations above we see the advantages of the spherical-polar coordinates over rectangular. The multiple integral can be reduced to iterated integrals. The improperness of the integral can be handled with limits on only one va riable, namely, r. One can see also how the number π makes its appearance in the formula. Of course, unlike x, y, z, there is no symmetry between r, θ, φ.

Remark 6: We could put a multiplier ψ(r), say, with the integrand to enable us to treat modifications of the inverse square law of force such as are involved in the Yukawa potential. We then have:

Theorem 1 : R 3 ψ(r)(x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) dxdydz = 4π(α -β) - R 3 (rψ (r) + 3ψ(r))f (x, y, z) dxdydz,
where α = lim r→∞ r 3 ψ(r)g(r, θ, φ) and β = lim r→0 r 3 ψ(r)g(r, θ, φ) .

Proof: We have: sinθ(r 3 ψ (r) + 3r 2 ψ(r))g(r, θ, φ) drdθdφ

= 4π(α -β) - R 3 sph (rψ (r) + 3ψ(r))g(r, θ, φ)r 2 sinθ drdθdφ = 4π(α -β) - R 3 (rψ (r) + 3ψ(r))f (x, y, z) dxdydz, .
In one of the steps above we have used "integration by parts" with respect to r. If we choose ψ(r) = 1 r 3 , we will obtain Theorem 1 as a special case, if lim r→∞ g(r, θ, φ) = 0. If we choose ψ(r) = 1 r , we obtain, if lim r→∞ r 2 g(r, θ, φ) = 0:

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 1 2 dxdydz = - R 3 2 r f (x, y, z) dxdydz.
3.2 Basic Results for R, R 2 , and R 3

We first have a result similar to Theorem 1 in one variable, namely:

Theorem 1(R): +∞ -∞ 1 |x| (x df dx )dx = -2 f (0).
We next have a similar result in two variables, namely:

Theorem 1(R 2 ): R 2 (x ∂f ∂x + y ∂f ∂y ) r 2 dxdy = -2πf (0, 0).
Proof: The usual rectangular-to-polar transformation has, for the Jacobian determinant JT , the value r, and so we need only r 2 in the denominator of the integrand. The rest of the calculations proceed as in the proof of Theorem 1. Note that we have a multiplier -2π for f (0, 0).

To prove a similar result for four variables, we need an unusual coordinate transformation which, however, has the desired features of the usual rectangular-to-spherical transformation in two and three variables. We note the following simple fact:

( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) 2 = ( x 1 2 + x 2 2 ) 2 + ( x 3 2 + x 4 2 ) 2 ,
which suggests the transformation:

x 1 = rcosθcosψ 1 , x 2 = rcosθsinψ 1 , x 3 = rsinθcosψ 2 , x 4 = rsinθsinψ 2 , with r = √ x 1 2 + x 2 2 + x 3 2 + x 4 2 , 0 ≤ r, 0 ≤ θ ≤ π 2 , 0 ≤ ψ 1 ≤ 2π, 0 ≤ ψ 2 ≤ 2π, and |JT | = -r 3 sinθcosθ. Also, choose x 2 1 + x 2 2 = r cosθ, x 2 3 + x 2 4
= r sinθ, so that θ, ψ 1 , ψ 2 can be appropriately determined. We then have:

Theorem 1(R 4 ): R 4 (x 1 ∂f ∂x1 + x 2 ∂f ∂x2 + x 3 ∂f ∂x3 + x 4 ∂f ∂x4 ) r 4 dx 1 dx 2 dx 3 dx 4 = 2π 2 f (0, 0, 0, 0).
Proof:Note that we have r 4 in the denominator and the multiplier of f (0, 0, 0, 0) is 2π 2 . We use the fact that r ∂g ∂r = (x

1 ∂f ∂x1 + x 2 ∂f ∂x2 + x 3 ∂f ∂x3 + x 4 ∂f ∂x4 ).

Extensions of the Basic Result

Next, using the slightly modified expression for ∂g ∂r noted above for delayed/advanced action, we immediately have:

Theorem 1 with delayed/advanced Action:

R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 3 2 dxdydz = -4πf (0, 0, 0, t) + 1 v R 3 ∂f ∂t (x, y, z, t -r v ) (x 2 + y 2 + z 2 ) dxdydz, (18)
= -4πf (0, 0, 0, t)

+ 1 v ∂ ∂t R 3 f (x, y, z, t -r v ) (x 2 + y 2 + z 2 ) dxdydz. (19)
We can easily prove a generalization of Theorem 1 to obtain two slightly different formulas for the value of f at points other than the origin.

Theorem 2: Under the same assumptions as those of Theorem 1, if (a, b, c) R 3 ,then

(i) R 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (x 2 + y 2 + z 2 ) 3 2 dxdydz = -4πf (a, b, c), (20)
where the partial derivatives are evaluated at (x + a, y + b, z + c), and

(ii) R 3 [(x -a) ∂f ∂x + (y -b) ∂f ∂y + (z -c) ∂f ∂z ] [(x -a) 2 + (y -b) 2 + (z -c) 2 ] 3 2 dxdydz = -4πf (a, b, c), (21)
where the partial derivatives are evaluated at (x, y, z).

Note that in ( ) above, a slightly different operator, namely,

[(x-a) ∂ ∂x +(y-b) ∂ ∂y +(z-c) ∂ ∂z ],
appears. The form of the integral in ( ) is convenient for interpretation and computation, whereas the form in ( ) is useful for derivations where the integral needs to be differentiated with respect to a, b, c which appear as parameters.

Proof of (i) :

Define a related function f by

f (x, y, z) = f (x + a, y + b, z + c) so that f (0, 0, 0) = f (a, b, c).
Applying Theorem 1 to f , we get:

R 3 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) dxdydz = -4πf (a, b, c), (22)
where r = (x 2 + y 2 + z 2 ) and the partial derivatives are evaluated at (x, y, z). Further, ∂f ∂x (x, y, z) = ∂f ∂x (x + a, y + b, z + c), keeping in view the definition of f . Similar relations hold for the other two partial derivatives.

Remark 7: In his calculation of the derivative of a potential function, Gauss used this idea of a "translation". He showed that the potential of a "mass distribution" at any point has the same value as the potential at the origin -or any chosen reference point -of a suitably translated distribution.

Proof of (ii) :

We give a short "technical" proof of the second result."Translate" the origin, or change the variables from (x, y, z) to (x , y , z ) where

x = x + a, y = y + b, z = z + c, so that R 3 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) dxdydz = R 3 ((x -a) ∂f ∂x + (y -b) ∂f ∂y + (z -c) ∂f ∂z ) [ (x -a) 2 + (y -b) 2 + (z -c) 2 ] 3 dx dy dz (23) = -4πf (a, b, c), (24)
where the partial derivatives are evaluated at (x , y , z ). The only problem with this approach is that both the integrals above are improper because their integrands are undefined at two different points, one at (0, 0, 0) and the other at (a, b, c). Finally, we observe that x , y , z are only "dummy variables" so that they can be changed to x, y, z and get the desired result.

We now give a "better" proof, using a different spherical-polar to rectangular coordinate transformation that we will use further on. The transformation, denoted by T a,b,c is defined by:

x = a + r sinθ cosφ, y = b + r sinθ sinφ, z = c + r cosθ, so that r = (x -a) 2 + (y -b) 2 + (z -c) 2 .
We define the associated function g by

g(r, θ, φ) = f (a + rsinθcosφ, b + rsinθsinφ, c + rcosθ) so that ∂g ∂r = ( x -a r ) ∂f ∂x + ( y -b r ) ∂f ∂y + ( z -c r ) ∂f ∂z . ( 25
)
We now compute the integral as in the proof of Theorem 1, noting that r → 0 yields x = a, y = b, z = c.

The Laplacian result says that a function is determined by its second-order partial derivatives, whereas our result says that it is determined by its first-order derivatives. In the partial differential equation (pde) view, the Laplacian reult says that if the pde ∇ 2 f = h has a solution, then we can obtain the solution through a simple volume integration, whereas our result says that if the pde system ∂f ∂x = u, ∂f ∂y = v, ∂f ∂z = w has a solution, then the solution can be obtained through a simple volume integration (see Theorem 5 below).

Employing the terminology of "source point" and "field point", we can write the classical result as:

-4πf (f ield point) = space 1 r ∇ 2 f (source point)
and our result as:

-4πf (f ield point) = space 1 r 3 r • ∇f (source point)
where r denotes the vector from the field point to the source point, and r denotes its length.

A Basic Result for related Operators

We state a result for the operator (x ∂ ∂y -y ∂ ∂x ) that follows from the fact

∂g ∂φ = x ∂f ∂y -y ∂f ∂x , (26)
Result: Under the conditions of Theorem 1,

R 3 1 r 3 (x ∂f ∂y -y ∂f ∂x ) dxdydz = 0. ( 27
)
Proof: The integral is equal to

R 3 sph 1 r 3 ∂g ∂φ r 2 drdθdφ = 1 r ( 2π 0 ∂g ∂φ dφ) drdθ = 1 r [g(r, θ, 2π) -g(r, θ, 0)] drdθ = 0.
Remark 9: One can obtain two more results like the one above, by appealing to "symmetry", namely:

R 3 1 r 3 (y ∂f ∂z -z ∂f ∂y ) dxdydz = 0, (28) R 3 1 r 3 (z ∂f ∂x -x ∂f ∂z ) dxdydz = 0. ( 29
)
But what do we mean by "symmetry" here since we are not talking "physics"? We could, once again, talk about a change of variables, this time a permutation of the variables, from x, y, z to, say, z, y, x, i.e. , new variables x , y , z such that x = z, y = y, z = x, an associated function f , use the result proved above to get

R 3 1 r 3 (z ∂f ∂y -y ∂f ∂z
) dx dy dz = 0 and finally,appeal to the "dummy variables" idea to get

R 3 1 r 3 (z ∂f ∂y -y ∂f ∂z ) dxdydz = 0.
Again, a better approach would be to use yet another spherical-polar to rectangular coordinate transformation, namely:

x = rcosθ, y = rsinθ sin φ, z = rsinθcosφ
and then proceed as in the proof above.

We collect these 3 results together as a Theorem:

Theorem 3: Under the assumption that f has continuous partial derivatives,

R 3 1 r 3 (x ∂f ∂y -y ∂f ∂x ) dxdydz = 0, (30)
R 3 1 r 3 (y ∂f ∂z -z ∂f ∂y ) dxdydz = 0, (31)
R 3 1 r 3 (z ∂f ∂x -x ∂f ∂z ) dxdydz = 0. ( 32
)
Remark 10: These three results show that the three first-order partial derivatives of f are not totally "independent" of one another, even though second-order partial derivatives may not exist. Of course, we do assume that the first-order derivatives are continuous. Also, like Theorem 2, we will have two versions of these results. We can also combine the 3 equations into a single vector equation:

R 3 r × ∇f r 3 = (0, 0, 0) = 0.

A Basic Result for Bounded Regions

Theorem 1 above involved a volume integral extended over whole space. We have a Theorem below that involves a volume integral extended over a bounded region bounded by a "surface" and a surface integral. We use a new definition of a surface in spherical-polar coordinates.

Let S be a positive real-valued function, bounded away from 0, of two variables θ and φ, with 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π,i.e., S(θ, φ) > for some > 0; we mean by the surface S the set

S sph = {(r, θ, φ) : r = S(θ, φ), 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π} in spherical coordinates,
and the set S rect = {(rsinθcosφ, rsinθsinφ, rcosθ) : r = S(θ, φ), 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π} in rectangular coordinates. By the region V bounded by the surface S we mean the set

V sph = {(r, θ, φ) : 0 ≤ r ≤ S(θ, φ), 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π} in spherical coordinates,
and the set

V rect = {(rsinθcosφ, rsinθsinφ, rcosθ) : 0 ≤ r ≤ S(θ, φ), 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π} in rectangular coordinates.
Note that the origin is an interior point of V and that each ray from the origin meets the surface in only one point.

Theorem 4: Let S be a surface such that S has continuous first-order partial derivatives ∂S ∂θ and ∂S ∂φ , let V be the region bounded by S,and let f : R 3 →R be a function with continuous first-order partial derivatives in V . Then:

f (0, 0, 0) = - 1 4π Vrect 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) dV + 1 4π θ=π,φ=2π θ=0,φ=0 f S (θ, φ)sinθ dθdφ. ( 33
)
where we denote the value of the function at a surface point, i.e., f (S(θ, φ)sinθcosφ, S(θ, φ)sinθsinφ, S(θ, φ)cosθ) by f S (θ, φ).

Proof: Let g(r, θ, φ) denote the function associated with f (x, y, z). Then, we have: f S (θ, φ)sinθ dθdφ -4πf (0, 0, 0).

Vrect 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) dV = V sph 1 r 3 (r ∂g ∂r )r 2 sinθ drdθdφ = V sph ∂g ∂r sinθ drdθdφ = θ=π,φ=2π θ=0,φ=0 ( r=S(θ,φ
Corollary: With two functions, we have a Green-like identity:

f (0, 0, 0)h(0, 0, 0) = - 1 4π [ 1 r 3 h(x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) + 1 r 3 f (x ∂h ∂x + y ∂h ∂y + z ∂h ∂z )] dV + 1 4π f S (θ, φ)h S (θ, φ)sinθ dθdφ.
Remark 11: The theorem can be interpreted as follows. Given a region bounded by a surface, the value of a once-differentiable scalar function (field) at an interior point is uniquely determined by the values of the function on the surface and the values of the partial derivatives in the region. It also gives a "formula" for determining the value at an interior point. It could be regarded as a generalization of "the fundamental theorem of the differential calculus of one variable" to 3 variables or 3 dimensions. Note that the 2-dimensional integral above is not the usual surface integral.

Remark 12: With an appropriate definition of "vector element of surface area" dS, we can write the 2-dimensional integral on the right hand side above as a "surface integral":

1 4π S ( f r r 3 ) • dS. ( 34
)
The vector element dS is also written as dS n where dS is the "magnitude" of the surface element and n is the unit normal vector. However, the "surface integral" is harder to visualize and calculate.

Proof of Remark 12: S is not a spherical surface in general. So we have to define what we could mean by the magnitude dS of the surface element and the unit normal n to it. It turns out to be easier to define the vector surface element dS.

Consider a "quadrilateral" ABCD, with the 4 corners determined by 4 pairs of θ, φ values. Thus, let A be the point (S(θ, φ), θ, φ); B: (S(θ, φ+δφ), θ, φ+δφ); C:(S(θ+δθ, φ+δφ), θ+δθ, φ+δφ);D : (S(θ+δθ, φ), θ+δθ, φ).

The r coordinates are given by the values of the function defining the surface.

We next calculate the first-order approximations to the vectors AB and AD. In rectangular coordinates these are:

AB = ( ∂S ∂φ sinθ cos φ -S(θ, φ)sinθsinφ, ∂S ∂φ sinθsinφ + S(θ, φ)sinθcosφ, ∂S ∂φ cosθ)dφ AD = ( ∂S ∂θ sinθcosφ + S(θ, φ)cosθcosφ, ∂S ∂θ sinθsinφ + S(θ, φ)cosθsinφ, ∂S ∂θ cosθ -sinθ)dθ.
We then define the vector surface element dS as dS = AD × AB which turns out to be, in rectangular cordinates:

dθdφ(α, β, γ)
where, writing S in place of S(θ, φ) for easy readability:

α = Ssinφ ∂S ∂φ -Ssinθcosθcosφ ∂S ∂θ + S 2 sin 2 θcosφ, β = -Scosφ ∂S ∂φ -Ssinθcosθsinφ ∂S ∂θ + S 2 sin 2 θsinφ, γ = Ssin 2 θ ∂S ∂θ + S 2 sinθcosθ,
and if r denotes the position vector of the point A, we have:

r • dS = (S(θ, φ)) 3 sinθdθdφ,
hence the desired result.

Like Theorem 2, we can have a "translated" version of Theorem 4. We can write it in coordinate-free form as:

f (f ield point) = -1 4π V ( 1 r 3 r • ∇f (source point)) + 1 4π S ( f (surf ace point) r r 3 • dS)
where r denotes the vector from field point to source point and r denotes its length.

Theorem 4 with delayed/advanced Action:

f (0, 0, 0, t) = - 1 4π Vrect 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z )dV + 1 4π 1 v ∂ ∂t Vrect f (x, y, z, t -r v ) (x 2 + y 2 + z 2 ) dxdydz + 1 4π θ=π,φ=2π θ=0,φ=0
f S (θ, φ)sinθ dθdφ.

(35)

An Existence Theorem for a PDE System

We next have a "converse" of Theorem 2 which amounts to a solution of the simplest 3-variable partial differential equation problem:

find a function w(x, y, z) such that

∂w ∂x = f (x, y, z), ∂w ∂y = g(x, y, z), ∂w ∂z = h(x, y, z),
where f, g, h are given functions that satisfy:

∂f ∂y = ∂g ∂x , ∂g ∂z = ∂h ∂y , ∂h ∂x = ∂f ∂z .
Theorem 5: Under the assumption that the functions f, g, h have continuous partial derivatives and that ∂f ∂y = ∂g ∂x , ∂g ∂z = ∂h ∂y , ∂h ∂x = ∂f ∂z , and that f, g, h → 0 as r → ∞, if w is given by:

w(a, b, c) = - 1 4π R 3 1 r 3 [xf (x + a, y + b, z + c) + yg(x + a, y + b, z + c) + zh(x + a, y + b, z + c)] dxdydz, (36)
where r = x 2 + y 2 + z 2 , then we have:

∂w ∂a = f, ∂w ∂b = g, ∂w ∂c = h.
Proof: It is tempting to bring the derivative under the integral sign, but the inegrand is not defined at one point, namely,(0, 0, 0). So, we use the modified spherical-polar rectangular transformation T a,b,c and write:

f (r, θ, φ) = f (a + rsinθcosφ, b + rsinθ cos φ, c + rcosθ), ḡ(r, θ, φ) = g(a + rsinθcosφ, b + rsinθ cos φ, c + rcosθ), h(r, θ, φ) = h(a + rsinθcosφ, b + rsinθ cos φ, c + rcosθ), so that w(a, b, c) = - 1 4π R 3 sph 1 r 3 [rsinθcosφ f + rsinθsinφḡ + rcosθ h]r 2 sinθ drdθdφ = - 1 4π R 3 sph [sinθcosφ f + sinθsinφḡ + cosθ h]sinθ drdθdφ.
Note that the functions f , ḡ, h in the integrand above contain a, b, c as parameters and are differentiable with respect to them, so that using the rule of "differentiating under the integral sign" [?], we get:

∂w ∂a = -1 4π R 3 sph [sinθcosφ ∂ f ∂1 + sinθsinφ ∂ ḡ ∂1 + cosθ ∂ h ∂1 ]sinθ drdθdφ where ∂ f ∂1 , ∂ ḡ ∂1 , ∂ h ∂1
denote the partial derivatives of f , ḡ, h with respect to the first "component". But by the assumption above

∂ ḡ ∂1 = ∂ f ∂2 , ∂ h ∂1 = ∂ f ∂3
, so we have:

∂w ∂a = - 1 4π R 3 sph [sinθcosφ ∂ f ∂1 + sinθsinφ ∂ f ∂2 + cosθ ∂ f ∂3 ]sinθ drdθdφ = - 1 4π R 3 sph ( ∂ f ∂r )sinθdrdθdφ = - 1 4π [-4π f (0, 0, 0)] = f (a, b, c).
Remark 13: Note how use of spherical-polar coordinates has allowed differentiation under the integral sign with impunity, which would not be possible if we had ( [(x -a) 2 + (y -b) 2 + (z -c) 2 ]) 3 in the denominator of the integrand. Gauss uses spherical-polar coordinates in his paper on the "inverse square law of force" to calculate derivative of the potential function in his proof of Poisson's equation. Green somehow does not use spherical-polar cordinates.

Remark 14: In the "recovery" formula given by Theorem 1, one does not require conditions of equality of the second-order mixed partial derivatives. Indeed, we did not require even the existence of the second-order derivatives. However, in proving the existence theorem on the solution of the pde problem, we have invoked the equality of the second-order mixed partial derivatives. Perhaps, with a suitable modification of our argument, one may be able to dispense with that requirement.

A New Alternative to the Poisson Formula

We now turn to 3-dimensional vector fields, i.e., functions F : R 3 → R 3 . We can immediately extend the results for scalar fields to vector fields by considering a 3-dimensional vector-valued function F as a set of 3 scalar valued functions F x , F y , F z and have a recovery formula,using Theorem 1:

F x (0, 0, 0) = - 1 4π R 3 1 r 3 (x ∂F x ∂x + y ∂F x ∂y + z ∂F x ∂z ), (37)
F y (0, 0, 0) = - 1 4π R 3 1 r 3 (x ∂F y ∂x + y ∂F y ∂y + z ∂F y ∂z ), (38)
F z (0, 0, 0) = - 1 4π R 3 1 r 3 (x ∂F z ∂x + y ∂F z ∂y + z ∂F z ∂z ). ( 39
)
Writing F as a vector:

F = F x i + F y j + F z k,
we have

F (0, 0, 0) = -1 4π R 3 1 r 3 (x ∂F ∂x + y ∂F ∂y + z ∂F ∂z ).
This recovery involves nine partial derivatives but only three combinations of these appear in the recovery formula. However, using Theorem 3 and similar results,putting a multiplier -1 4π , we have:

- 1 4π R 3 1 r 3 (x ∂F y ∂y -y ∂F y ∂x )dxdydz = 0, - 1 4π R 3 1 r 3 (x ∂F z ∂z -z ∂F z ∂x )dxdydz = 0.
"Adding" the left-hand sides of these two equations to the right-hand side of the equation above for F x (0, 0, 0) and rearranging terms, we get:

F x (0, 0, 0) = -1 4π R 3 1 r 3 [x( ∂Fx ∂x + ∂Fy ∂y + ∂Fz ∂z ) -y( ∂Fy ∂x -∂Fx ∂y ) + z( ∂Fx ∂z -∂Fz ∂x )].
We can obtain similar expressions for F y (0, 0, 0) and F z (0, 0, 0).

We recognize that ( ∂Fx ∂x + ∂Fy ∂y + ∂Fz ∂z ) is the divergence of F , i.e., ∇ • F . We also see that y(

∂Fy ∂x -∂Fx ∂y ) - z( ∂Fx ∂z -∂Fz ∂x ) is the x-component of r × (∇ × F ) because r × (∇ × F ) =       i j k x y z ( ∂Fz ∂y - ∂Fy ∂z ) ( ∂Fx ∂z -∂Fz ∂x ) ( ∂Fy ∂x -∂Fx ∂y )       .
Here, r is (x i + y j + z k). Consideration of the other components of F and using the notation of Theorem 2, we obtain the following

Theorem 6: F = -1 4π R 3 1 r 3 [(∇ • F ) r -r × (∇ × F )] dV.
This theorem appears to be a new alternative to Poisson's formula stated in the Introduction, namely:

F = -1 4π R 3 ∇(∇•F )-∇×(∇×F ) r
dV .

Theorem 6 (Delayed/advanced version of Theorem 6):

F = -1 4π R 3 1 r 3 [(∇ • F ) r -r × (∇ × F )] + 1 4π 1 v ∂ ∂t R 3 1 r 2 F .
We can see immediately that by applying Theorem 4 to the components of F and then combining the results using the above technique, we will have the following result which gives F in terms of its integral over a bounded region and a surface integral .

Theorem 7:

F = - 1 4π R 3 [(∇ • F ) r -r × (∇ × F )] r 3 dV + 1 4π θ=π,φ=2π θ=0,φ=0 F S (θ, φ)sinθ dθdφ = - 1 4π R 3 [(∇ • F ) r -r × (∇ × F )] r 3 dV + 1 4π S ( r • dS r 3 )F dV.
Remark 15: It is indeed surprising that whereas the expresions for F x , F y , F z separately involved 3 partial derivatives multiplied by x, y, z, the vector F , i.e., the 3 scalars put together, has an expression that involves 4 different "disjoint" combinations of the partial derivatives multiplied by x, y, z, and these happen to be the combinations occurring in ∇ • F and ∇ × F . But it is even more surprising that we can have combinations different from the usual ones, obtained by permuting the variables x, y, z, but not the three components F x , F y , F z . Thus, one may check that the following 4 combinations will work: ) k by ∇ ⊗ F ("pseudo-curl"), and the vector (y i + z j + x k) by r, we will have:

(
F = -1 4π 1 r [(∇ F )r -r × (∇ ⊗ F )] dV.
The point is that the usual combinations of partial derivatives in divergence and curl are not the only ones that will work. Remark 16: Incidentally, it may be stated that we were led to our new formula above by starting with Poisson's formula and then using two little-known formulas below, namely:

R 3 1 r [∇(∇ • F )] = R 3 1 r 3 [(∇ • F ) r] (40) R 3 1 r [∇ × (∇ × F )] = R 3 1 r 3 [ r × (∇ × F )]. ( 41
)
This led us, in turn, to our formula (6) by writing the expression for F x (0, 0, 0) in our formula above and expanding out the expressions for ∇ • F and ∇ × F .

The above two little-known formulas were obtained by application of the following new formulas which are proved in the next Section:

R 3 1 r [∇f ] = R 3 1 r 3 [(f ) r] (42) R 3 1 r [∇ × A] = R 3 1 r 3 [ r × A)]. ( 43
)

Removing a Derivative occurring inside an Integral

Our main result in Section 3 can be regarded as enabling us to "integrate out" completely differential expressions occurring inside an integral. We now prove a number of results which only remove a derivative occurring inside an integral, without succeeding in integrating out completely. Our first result removes a ∇ operator occurring inside an integral. Note that we have a 1 r multiplier on the operator side and a 1 r 3 multiplier on the other side.

Theorem 8 : Assuming f is differentiable, we have:

R 3 1 r [∇f ] = R 3 1 r 3 [(f ) r]
. Proof: On considering the x-, y-, and z-components of the two sides of the result to be proved, we see that the Theorem is equivalent to the following three lemmas. Lemma 1:

R 3 1 r ∂f ∂x = R 3 1 r 3 (xf ). Lemma 2: R 3 1 r ∂f ∂y = R 3 1 r 3 (yf ). Lemma 3: R 3 1 r ∂f ∂z = R 3 1 r 3 (zf ).

Proof of Lemma 1:

Denoting the associated function by g, we have:

R 3 1 r ∂f ∂x = R 3 sph 1 r 1 sinθ [sin 2 θcosφ ∂g ∂r + sinθcosθcosφ 1 r ∂g ∂θ -sinφ 1 r ∂g ∂φ ]r 2 sinθ drdθφ = R 3 sph r[sin 2 θcosφ ∂g ∂r + sinθcosθcosφ 1 r ∂g ∂θ -sinφ 1 r ∂g ∂φ ] drdθdφ
We consider the 3 terms in the last integral above separately. First, we have: where we have used integration by parts with respect to r and that rg(r, θ, φ) → 0 as r → ∞.

For the second term we have: Here, we have used integration by parts with respect to θ.

For the third term we have:

- integrating by parts again with respect to φ this time.

Putting together these 3 results we have:

Proof of Lemma 2:

We have:

R 3 1 r ∂f ∂y = R 3 sph 1 r 1 sinθ [sin 2 θsinφ ∂g ∂r + sinθcosθsinφ 1 r ∂g ∂θ + cosφ 1 r ∂g ∂φ ]r 2 sinθ = R 3 sph r[sin 2 θsinφ ∂g ∂r + sinθcosθsinφ 1 r ∂g ∂θ + cosφ 1 r ∂g ∂φ ].
Once again, we use integration by parts, with respect to r on the first term in the integral above, with respect to θ on the second term, and with respect to φ on the third term. Further manipulations lead to the right hand side.

Proof of Lemma 3:

Similar operations show the equality of the two sides.

Remark 17: By choosing ∂f ∂x , ∂f ∂y , ∂f ∂z in place of f respectively in the three Lemmas above and adding, we get the following result involving the Laplacian operator:

R 3 1 r ( ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 ) = R 3 1 r 3 (x ∂f ∂x + y ∂f ∂y + z ∂f ∂z ) (44)
so that using our basic result, we get:

R 3 1 r ( ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 ) = -4πf (0, 0, 0). ( 45
)
Our next rseult enables us to remove a ∇× operator occurring inside an integral. Note again that we have a 1 r multiplier on the operator side and a 1 r 3 multiplier on the other side.

Theorem 9: Assuming the vector function A is differentiable:

R 3 1 r [∇ × A] = R 3 1 r 3 [ r × A)].
Proof: Consider the three components of the two sides and apply the 3 Lemmas.

Finally, as one may expect, we have a result which removes the ∇• operator. Once again, the multiplier on the operator side is 1 r on the operator side and 1 r 3 on the other side.

Theorem 10: Assuming the vector function A is differentiable:

R 3 1 r [∇ • A] = R 3 1 r 3 [ r • A)].
These three theorems have perhaps appeared as Exercises in some textbooks or have been used when carrying out some derivations. But it is useful to highlight them as we have done. Also, there are variations obtained by changing the multiplier on the operator side to 1 r 2 , and even dropping it altogether. They can be proved by changing the multipliers in the three Lemmas.

Moving Differentiation into inside of an Integral :

We now illustrate how the operation of differentiation with respect to a rectangular coordinate applied to an integral is equivalent to an integral in spherical-polar coordinates involving an integrand which has a derivative. Commonly, this is known as interchanging the order of integration and differentiation, or carrying out a differentiation into the inside of an integral [?]. In textbooks on Electromagnetism this procedure is used to move the "del" operators into the inside of an integral for later manipulations.

Helmholtz's Theorem

We now state and prove a theorem, which gives one aspect of Helmholtz's Theorem, requiring weaker assumptions. This aspect of Helmholtz's Theorem is an existence theorem which shows the existence of a vector field having a prescribed divergence and curl, subject to the condition that the prescribed curl has zero divergence. The other aspect is a decompostiion theorem which states that any continuously differentiable vector field can be decomposed into two "components", one of which is the gradient of a scalar field and the other is the curl of a vector field, and that these "generating" fields can be obtained from the original vector field.

Theorem 11: Helmholtz Theorem, Existence If f is a given continuously differentiable scalar function and A is a given continuously differentiable vector function, and that ∇ • A = 0, then the function W defined by:

W (a, b, c) = -1 4π R 3 1 r 3 [f r -r × A], satisfies ∇ • W = f, and ∇ × W = A.
Proof: For the components of W we have:

W x (a, b, c) = - 1 4π R 3 1 r 3 [xf (x + a, y + b, z + c) -yA z (x + a, y + b, z + c) + zA y (x + a, y + b, z + c)], W y (a, b, c) = - 1 4π R 3 1 r 3 [yf (x + a, y + b, z + c) -zA x (x + a, y + b, z + c) + xA z (x + a, y + b, z + c)], W z (a, b, c) = - 1 4π R 3 1 r 3 [zf (x + a, y + b, z + c) -xA y (x + a, y + b, z + c) + yA x (x + a, y + b, z + c)]. 20
Carrying out differentiation under the integral sign, we get:

∂W x ∂a (a, b, c) = - 1 4π R 3 1 r 3 [x ∂f (x + a, y + b, z + c) ∂a -y ∂A z (x + a, y + b, z + c) ∂a + z ∂A y (x + a, y + b, z + c) ∂a ], ∂W y ∂b (a, b, c) = - 1 4π R 3 1 r 3 [y ∂f (x + a, y + b, z + c) ∂b -z ∂A x (x + a, y + b, z + c) ∂b + x ∂A z (x + a, y + b, z + c) ∂b ], ∂W z ∂c (a, b, c) = - 1 4π R 3 1 r 3 [z ∂f (x + a, y + b, z + c) ∂c -x ∂A y (x + a, y + b, z + c) ∂c + y ∂A x (x + a, y + b, z + c) ∂c ].
Noting that the partial derivatives of the functions with respect to a have the same value as the derivatives with respect to x, similarly for b and y, and for c and z, adding the three partial derivatives above, and using Theorem 2 and Theorem 3 , we obtain:

∇ • W = f.
Next, we compute (∇ × W ) x :

(∇ × W ) x = ∂W z ∂b - ∂W y ∂c = - 1 4π R 3 1 r 3 [ (z ∂f ∂b -x ∂A y ∂b + y ∂A x ∂b ) -(y ∂f ∂c -z ∂A x ∂c + x ∂A z ∂c )] = - 1 4π R 3 1 r 3 [ (z ∂f ∂y -x ∂A y ∂y + y ∂A x ∂y ) -(y ∂f ∂z -z ∂A x ∂z + x ∂A z ∂z ) ] = - 1 4π R 3 1 r 3 [ (x ∂A x ∂x + y ∂A x ∂y + z ∂A x ∂z ) + (z ∂f ∂y -y ∂f ∂z ) -x( ∂A x ∂x + ∂A y ∂y + ∂A z ∂z ) ] = A x
because of the assumption that ∇ • A = ( ∂Ax ∂x + ∂Ay ∂y + ∂Az ∂z ) = 0 and using Theorem 2 and Theorem 3 , once again.

Remark 18: Note that in the proof above , we have virtually proved the following two results for the two parts of W ,

W 1 = 1 4π R 3 1 r 3 f r and W 2 = 1 4π R 3 1 r 3 r × A : ∇ • W 1 = f, ∇ × A = 0,
and

∇ • W 2 = 0, ∇ × W 2 = A.
To prove the decomposition aspect of the Helmholtz Theorem, in view of our new Poisson formula, it suffices to prove the following two results:

∇ f r = 1 r 3 f r (46) ∇ × A r = 1 r 3 ( r × A). (47)
We sketch proofs of these results under the assumption that f and A are differentiable. (Many other results are listed in Section 7.2 and their proofs can be given along similar lines.)

Result 1: ∇ f r = 1 r 3 f r. Proof: We show the equality of the i-components of the two sides. Denoting the integral on left-hand side by V , we have:

V (a, b, c) = R 3 sph f r .
Carrying out differentiation into the inside of the integral, we have:

∂V ∂a = 1 r ∂f ∂a .
Using Lemma 1 in Section 5, the result follows immediately. Result 2: ∇ × B r = 1 r 3 ( r × B). Proof: Again, we show the equality of the i-components of the two sides. Denoting the integral on the left-hand side by F , we have:

F (a, b, c) = R 3 sph (Bxi+Byj+Bzk) r
.

So, the i-component of the left-hand side of ∇ × F is:

1 r ( ∂Bz ∂b -∂By ∂c ). Using Lemmas 2 and 3 in Section 5, the result follows . We now state: Theorem 12: Helmholtz Theorem, Decomposition If F is a given continuously differentiable vector function, there exists a scalar function V and a vector function A such that

F = ∇V + ∇ × A, where V = -1 4π ∇•F r , and
A = 1 4π ∇×F r .
Proof: This follows from the two results above and our formula

F = -1 4π R 3 1 r 3 [(∇ • F ) r -r × (∇ × F )] dV.
The function V is usually called the scalar potential function and A the vector potential function generating F . The above two expressions are the ones commonly given. But using our Theorems 9 and 10, the same functions are also given by the following expressions which do not involve any ∇, that is to say, differentiation operation.

V = -1 4π 1 r 3 ( r • F ), A = 1 4π 1 r 3 ( r × F ).
Interestingly, there is a close similarity between these expressions and the following one:

F = 1 r 2 [( r • F ) r + r × ( r × F )]
. (Such a "decomposition" or "representation" of any arbitrary vector in terms of another arbitrary vector follows immediately from the "vector algebra" identity:

u × ( v × w) = ( u • w) v -( u • v) w.)
Perhaps, there is some deeper connection here !

A New Proof of Poisson's Theorem

We now prove Poisson's Theorem and the Divergence Theorem, employing spherical-polar coordinates, which will further illustrate the "power" of these cordinates used along with rectangular coordinates.

We first prove Poisson's Theorem, namely:

Poisson's Theorem:

∇ • ( R 3 1 r 3 f r) = -4πf .
Proof: We first explicate how the integral on the left hand side is to be understood. By it, we mean a function F : R 3 → R 3 defined by:

F (a, b, c) = R 3 (x-a)f (x,y,z) i+(y-b)f (x,y,z) j+(z-c)f (x,y,z) k [(x-a) 2 +(y-b) 2 +(z-c) 2 ] 3 2
dxdydz which, we have shown, can equally well be given by:

F (a, b, c) = R 3 xf (x+a,y+b,z+c) i+yf (x+a,y+b,z+c) j+zf (x+a,y+b,z+c) k [x 2 +y 2 +z 2 ] 3 2 dxdydz.
Note that in the integrals above, a, b, c are to be regarded as parameters. In the first integral, we cannot bring in diffrentiation under the integral sign, whereas in the second one we can. Denote the three components of F (a, b, c) by F x (a, b, c), F y (a, b, c), F z (a, b, c). Considering the first component and its value from the second integral, we have:

F x (a, b, c) = R 3 xf (x+a,y+b,z+c) [x 2 +y 2 +z 2 ] 3 2

dxdydz.

We now calculate the partial derivetives of F x as in Theorem 5, except that we have only one function f instead of 3 functions f, g, h. So, we get:

∂Fx ∂a = R 3 sph sinθcosφ ∂f ∂1 sinθ drdθdφ.
Similarly, we get:

∂F y ∂b = R 3 sph sinθsinφ ∂f ∂2 sinθ drdθdφ ∂F z ∂c = R 3 sph cosθ ∂f ∂3 sinθ drdθdφ
Adding the three derivatives,

∇ • F = R 3 sph [sinθcosφ ∂f ∂1 + sinθsinφ ∂f ∂2 + cosθ ∂f ∂3 ]sinθ drdθdφ = R 3 sph ∂ f ∂r sinθ drdθdφ = -4πf (a, b, c).
Remark 19: The Poisson Theorem is more commonly stated as ∇ 2 V = ρ where V denotes the potential function corresponding to the source density function ρ. The proofs given in most textbooks use the Divergence Theorem and "suffer" from the defect that they assume that the potential function is twicedifferentiable. Gauss's proof was probably the first to show that the potential function is twice-differentiable, though under the assumption that the density function ρ is once-differentiable. Our proof also makes this assumption. Interestingly, Kellogg[?] proves the result without making this assumption, but assumes what is known as a Holder condition.

A New Proof of the Divergence Theorem

Here is our proof of the classical Divergence Theorem: If a region V is enclosed by a surface S and F : R 3 → R 3 , then

V ∇ • F = S F • dS. Proof:
We use the notation of Theorem 4. As in the classical proof of the Theorem, we consider the three terms on either side of the equation separately. In the rectangular coordinates proof, as is noted by some authors, it is required that the surface is such that it is "raised"on its projections on each of the three coordinate planes. We do not require this because we have used a different definition of a surface. We first consider the integral of the component ∂Fx ∂x of the divergence and the corresponding component on the right hand side. To avoid long expressions which might obscure the argument, we will use letter abbreviations for the various integrals. We use the expression for the surface element dS as calculated earlier. We denote the functions associated with F x , F y , F z by G x , G y , G z respectively. We will show that

Vrect ∂F x ∂x = R sph G x [Ssinφ ∂S ∂φ -Ssinθcosθcosφ ∂S ∂θ + S 2 sin 2 θcosφ] dθdφ.
We have:

Vrect ∂F x ∂x = V sph 1 sinθ (sin 2 θcosφ ∂G x ∂r + sinθcosθcosφ 1 r ∂G x ∂θ -sinφ 1 r ∂G x ∂φ ]r 2 sinθ drdθdφ = V sph [sin 2 θcosφ(r 2 ∂G x ∂r ) + rcosφ(sinθcosθ ∂G x ∂θ ) -r(sinφ ∂G x ∂φ )] drdθdφ.
Let us use the abbreviations

I 1 = V sph sin 2 θcosφ(r 2 ∂G x ∂r ) drdθdφ, I 2 = V sph rcosφ(sinθcosθ ∂G x ∂θ ) drdθdφ I 3 = V sph -r(sinφ ∂G x ∂φ ) drdθdφ,
so that Vrect ∂Fx ∂x = I 1 + I 2 + I 3 . Then:

I 1 = θ=π θ=0 φ=2π φ=0 sin 2 θcosφ[r 2 G x (r, θ, φ)] r=S(θ,φ) r=0 ] dθdφ - V sin 2 θcosφ[2rG x ]drdθdφ (48) = θ=π θ=0 φ=2π φ=0 sin 2 θcosφS(θ, φ) 2 G x (S(θ, φ), θ, φ) dθdφ -2 V sph r[sin 2 θcosφG x ] drdθdφ. (49)
We have used integration by parts with respect to r first. Note that this is justified although the upper limit of the integral with respect to r is not a constant but S(θ, φ) which depends on θ and φ which are involved in the other two integrations. We cannot integrate by parts with respect to θ in I 2 and with respect to φ in I 3 because this will involve interchanging the order of integration with respect to r. (However, we can interchange the order if integration with respect to θ and φ because their limits are constant.) So we use a "trick". Let

Now, because -2sin 2 θcosφ -(cos 2 θ -sin 2 θ)cosφ + cosφ = 0,adding (39),( 42),(44) we have:

Vrect ∂F x ∂x dV = θ=π θ=0 φ=2π φ=0 G x (S 2 sin 2 θcosφ -Ssinθcosθcosφ ∂S ∂θ + Ssinφ ∂S ∂φ ) dθdφ,
and the last term is the same as the first term in the expansion of S F • dS.

It remains to patiently verify the equality of the remaining two terms on the two sides -or appeal to "symmetry".

A List of Some Results

We state a number of useful results without proof. They can be easily proved in the same way as those proved above, namely, by using the rectangular to spherical-polar transformation and and carrying out integration by parts, with respect to r,θ, or φ, appropriately. The results are stated for delayed/advanced integrals. All the integrals are over R 3 .

Removing Derivative in the Integrand

Here,r = (x 2 + y 2 + z 2 ) 1 2 and f and ∂f ∂x are evaluated at (x, y, z).

1) ψ(r) ∂f ∂x = - ψ (r) r xf + 1 v ∂ ∂t 1 r ψ(r)xf, (54)
provided [ψ(r)r 2 g(r, θ, φ)] is finite as r → 0 and r → ∞.

provided [ ψ(r)r 3 g(r, θ, φ) ] is finite as r → 0 and r → ∞.

3) ψ(r)y ∂f ∂x = -ψ (r)

r xyf + 1 v ∂ ∂t 1 r ψ(r)xyf, (56)
provided [ ψ(r)r 3 g(r, θ, φ) ] is finite as r → 0 and r → ∞.

Gradient, Divergence and Curl of Integrals

Here, r is the relative position vector and r its length. If the left-hand sides of the equations below are to be evaluated at (a, b, c), then in the integrals on their right-hand sides, either (i) r = (x 2 + y 2 + z 2 ) 1 2 and the functions f and F are evaluated at(x + a, y + b, z + c), or (ii) r = [(x -a) 2 + (y -b) 2 + (z -c) 2 ] 1 2 and the functions ar evaluated at(x, y, z).

Gradient of a Scalar Integral

1)

∇ ψ(r)f = ψ(r)(∇f ) = -ψ (r)

r f r + 1 v ∂ ∂t 1 r ψ(r)f r, (57)
provided [ ψ(r)r 2 g(r, θ, φ) ] is finite as r → 0 and r → ∞.

2)

∇ ψ(r)( r • F ) = ψ(r)(∇ • F ) r.

(58) Also,

3) ∇ ψ(r)( r

• F ) = 4 3 π[ψ(r)r 3 F ] ∞ 0 -ψ(r) F - 1 r ψ (r)( r • F ) r + 1 v ∂ ∂t ψ(r) 1 r ( r • F ) r, (59)
provided [ ψ(r)r 3 G(r, θ, φ) ] is finite as r → 0 and r → ∞.

Divergence of a Vector Integral

Here, again, r is the relative position vector and r its length. provided [ ψ(r)r 2 g(r, θ, φ) ] is finite as r → 0 r → ∞.

3) ∇ • ψ(r)( r × F ) = 0. (62)

Curl of a Vector Integral

Here, too, r is the relative position vector and r its length.

1)

∇ × ψ(r)f r = 0. (63)

2)

∇ × ψ(r) F = ψ(r)(∇ × F ) = - 1 r ψ (r)( r × F ) + 1 v ∂ ∂t 1 r ψ(r)( r × F ), (64)
provided [ ψ(r)r 2 g(r, θ, φ) ] is finite as r → 0 and r → ∞.

3) (66)

∇ × ψ(r)( r × F ) = ψ(r)(∇ • F ) r -4π[ ψ(r)r 3 F ] ∞ 0 -ψ (r)r F + 3 ψ(r) F - 1 v ∂ ∂t ψ(r)[ 1 r ( r • F ) r -r F ]. (65
8 Application to Derivation of Heras's Result :

We now prove the interesting result proved by Heras[?], using our approach, which does not invoke the δ-function nor "expressions" like ∇( 1 r ) involving differentiation with respect to the rectangular coordinates x, y, z .We are using our notation and all integrals involve delayed/advanced action. Our relative position vector r points from the field point to the source point whereas his vector R points from the source point to the field point, so that r = -R. Incidentally, Heras does not specify some limit conditions as r → 0 and r → ∞. Heras's Result: Suppose ρ is a scalar field and J a vector field satisfying the equation ("continuity equation"):

∇ • J + ∂ρ ∂t = 0. ( 67
)
Let a vector field F be defined by:

F = - 1 4π [ 1 r 3 ρ r - 1 v 1 r 2 ( ∂ρ ∂t ) r - 1 v 2 1 r ( ∂J ∂t ) ] (68)
and let G be the vector field defined by:

G = - 1 4π [ 1 r 3 (J × r) - 1 v 1 r 2 ( ∂J ∂t ) × r] (69)
Then: (76)

∇ • F = ρ, (70)
∇ • G = 0, (71)
∇ × F + ∂G ∂t = 0, (72)
By using these, the continuity equation, and the expression for ∂F ∂t , one can verify that the desired equation holds.

(r 3 ψ 3

 33 (r) + 3r 2 ψ(r))g(r, θ, φ) dr] dθdφ rθφ sinθ(r 3 ψ (r) + 3r 2 ψ(r))g(r, θ, φ) drdθdφ = (α -β)

  S(θ, φ), θ, φ) -g(0, θ, φ)]sinθ dθdφ =

(cos 2 θ 3 sph-

 23 -sin 2 θ)g dθ)dφdr = R cosφ(cos 2 θ -sin 2 θ)g.

3 sph sin 2 3 1r 3 [ 3 sph 1 r 3

 3233313 θcosφg. On the right hand side of Lemma 1 we have: R (xf ) which equals: R (rsinθcosφ)r 2 sinθg which equals the expression on the left hand side of Lemma 1 above.

I 4 (((

 4 θ, φ) = r=S(θ,φ) r=0 (rcosφsinθcosθG x ) dr.Note I 4 (0, φ) = 0 and I 4 (π, φ) = 0. Using Leibnitz's rule, we get: cos 2 θ -sin 2 θ)cosφG x + rsinθcosθcosφ ∂G x ∂θ ] dr.Integrating the above between θ = 0 and θ = π, cos 2 θ -sin 2 θ)cosφG x + rsinθcosθcosφ ∂G x ∂θ ]dr) dθ, cos 2 θ -sin 2 θ)cosφG x + rsinθcosθcosφ ∂G x ∂θ ]dr) dθdφ, 2 θ -sin 2 θ)cosφG x drdθdφ. rcosφG x + rsinφ ∂G x ∂φ) dr and so integrating between φ = 0 and φ = 2π, we get: rcosφG x ) drdθdφ.

1 )

 1 ∇ • ψ(r)f r = ψ(r)( r • ∇f ) = 4π[ψ(r)r 3 g(r, θ, φ)] ∞ 0 -3 ψ(r)f -rψ (r)ψ(r) F = ψ(r)(∇ • F ) = -1 r ψ (r)( r • F )

  )( r • F ) r + ψ (r)r F + 2 ψ(r) F F ) r -r F ].

1 v 2 ∂G∂t = 0 :

 120 We use our results above to calculate the left-hand sides of the equations above. Proof of ∇ • F = ρ : equation, we obtain the desired result.Proof of ∇ • G = 0: Looking at the form of G, ∇ • G = 0 is immediate.Proof of ∇ × F + ∇× of the first two terms in F is 0, looking at their form. For the third term we have: and the expression for ∂G ∂t , one can verify that the desired equation holds. Proof of ∇ × G -∂F ∂t = J:


Bhargav Teja Nallapu

Nicolas P Rougier

Dynamics of reward based decision making a computational study

Keywords: Decision making, neural dynamics, basal ganglia, optimal behavior

We consider a biologically plausible model of the basal ganglia that is able to learn a probabilistic two armed bandit task using reinforcement learning. This model is able to choose the best option and to reach optimal performances after only a few trials. However, we show in this study that the influence of exogenous factors such as stimuli salience and/or timing seems to prevail over optimal decision making, hence questioning the very definition of action-selection. What are the ecological conditions for optimal action selection ?

Introduction

Basal ganglia are known to be involved in decision making and action selection based on reinforcement learning and a number of models have been designed to give account on such action selection [START_REF] Redgrave | The basal ganglia: a vertebrate solution to the selection problem?[END_REF][START_REF] Gurney | A computational model of action selection in the basal ganglia. I. A new functional anatomy[END_REF][START_REF] Gurney | A computational model of action selection in the basal ganglia. II. Analysis and simulation of behaviour[END_REF]. We have been studying a specific computational model of the basal ganglia that has been introduced in [START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF] and replicated in [START_REF] Topalidou | re] interaction between cognitive and motor corticobasal ganglia loops during decision making: a computational study[END_REF]. This model has been used to explain, to some extent, decision making in primates on a two armed bandit task. One of the questions we attempt to address in this study is to what extent the physical properties of the stimulus such as the visual salience or other characteristics affect the decision and lead to a suboptimal choice. For example (and quite obviously), a stimulus is very likely to be selected, if it is presented before the other stimuli and this selection will be made irrespectively of the potential reward associated with this stimulus. Moreover, there may be other factors such as stimulus salience or population size that may also disrupt the optimal performance. This led us to do a systematic study of the influence of such exogenous factors to understand what are the ecological conditions for optimal decision making.

Methods

Task

The task that has been used to demonstrate action selection in the model is a probabilistic learning task that is described in [START_REF] Pasquereau | Shaping of Motor Responses by Incentive Values through the Basal Ganglia[END_REF]. Four target shapes are associated with different reward probabilities (see figure 1). A trial is a time period in which any two of the four possible shapes are presented at two random positions (out of the four possible positions -up, right, down and left). The model is allowed to settle for the first 500ms of the trial and then two random cues are presented. By the end of trial period, a choice is made and the reward is given according to the reward probability associated with the chosen shape. A trial is

Reward probabilities

Fig. 1. The two armed bandit task as described in [START_REF] Pasquereau | Shaping of Motor Responses by Incentive Values through the Basal Ganglia[END_REF][START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF].

considered to be successful if a decision is made by the model, irrespective of the reward received. In a single independent trial, the cognitive decision (shape of the cue) and motor decision (position of the cue) are independent of each other. At any decision-making level of the model, each of the four cue shapes and each of the four motor movement directions is represented by one unit (neuron) each. Thus in a given trial, when two cue shapes are presented at two different positions, two cognitive, two motor and two associative (in cortex and striatum, see figure 2) neurons are activated. The task is run for a session, a number of trials, while at the end of each trial, the model learns the reward associated to its selection. (see Learning).

Model

In [START_REF] Leblois | Competition between feedback loops underlies normal and pathological dynamics in the basal ganglia[END_REF], authors demonstrated an action selection mechanism in the cortico-basal ganglia loops based on a competition between the positive feedback, direct pathway through the striatum and the negative feedback, hyperdirect pathway through the subthalamic nucleus. In [START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF], authors investigated further how multiple level action selection could be performed by the basal ganglia, and the model has been extended in a manner consistent with known anatomy and electrophysiology of the basal ganglia in the monkey (see figure 2). This model allows a bidirectional information flow between loops such that during early trials, a direction can be selected randomly, irrespective of the cue positions. However, after repeated trials, the model is able to consistently make the cognitive decision before the motor decision in each trial (see figure 3) and most frequently the motor decision, biased by the cognitive decision, towards the position of the more rewarding cue shape.

Learning . Learning has been derived from a simple actor-critic algorithm [12] that shapes the gain between the cognitive cortex and the cognitive striatum. According to the amount of reward received at the end of each trial (0 or 1), the model learns to estimate the value of chosen stimulus and then the cognitive pathway is biased in favor of the stimulus with the highest value.

Neuronal dynamics . This model uses the same, simple neuronal rate model as in [START_REF] Leblois | Competition between feedback loops underlies normal and pathological dynamics in the basal ganglia[END_REF] and [START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF] to focus on the network dynamics. Within each structure, each neuron (in each channel of any loop) is modeled as a single rate coded neuron with the equation:

τ dm dt = -m + I s + I Ext -T (1)
decay time constant of the synaptic input τ , negative values of activation, m and threshold of the neuron T are set to respective constant values as per the model in [START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF]. I Ext is an external input representing the sensory visual salience of the cue, which is unchanged throughout the process of learning.

Results

In all the following cases, we consider 4 stimuli A, B, C, D respectively associated with reward probability of 1.0, 0.66, 0.33 and 0.0. Learning is performed over 120 trials until the model reaches a performance of 0.90, meaning it chooses the best stimulus 90% of the time. We then stopped learning in the model and simulated a scenario where one stimulus is presented first and the other follows after a certain delay. Another scenario involved presenting one stimulus with more salience than the other. In both the scenarios, the intent was to emphasize the advantage (earlier presentation or higher salience) to the lesser rewarding stimulus and see if that leads the model to a suboptimal decision. We presented the model with various scenarios involving different delays and saliences. (see figure 4). ∆V is expressed as the relative ratio between the less salient and the most salient stimuli (∆V = (VA -VB)/VB. ∆t is expressed as the delay separating the two stimuli onsets (∆t = tA -tB).

Influence of delay

We first tested the influence of a small delay (between 0ms and 60ms) between the presentation of the two stimuli. The worst stimulus, that is the one associated with the lesser probability of reward, is presented first and after the delay, the second (better rewarding) stimulus is presented. We have been testing systematically all combinations of stimuli (A/B, A/C, A/D, B/C, B/D, C/D) and averaged the mean performance over 25 trials (see figure 5). As expected, the performance decreased with the increase in delay and the crossing (i.e. performance is random) happens around 35ms for all combinations but the last one (C/D) that happens very early, around 20ms. This specific case can be explained by the poor estimation of the value of C and D during learning because those stimuli are almost never chosen (most of the time, they are presented with a better stimuli).

Influence of salience

We tested the influence of salience by presenting simultaneously the two stimuli but the worst stimulus, i.e, the one associated with a lesser probability of reward, has been made virtually stronger than the other. The model, having learned the rewarding probabilities, is expected to select the higher rewarding stimulus irrespective of the salience. However, the increased salience of the lesser rewarding stimulus affects the model and leads it to take a bad decision (see figure 6). As the salience of the lesser rewarding stimulus increases, a consistent decrease in the performance of model is observed. Interestingly, the threshold percentage of salience difference after which the performance of the model decreases, is a characteristic of the difference in the reward probabilities of both the stimuli presented. Quite visibly (figure 6), it takes higher salience difference for a lesser rewarding stimulus to be chosen against the best rewarding one (in this case, A) whereas a lesser increase in salience seems to be sufficient to compromise the decisions involving lesser rewarding stimuli, like B.

In various neuropsychological studies on humans, like in [START_REF] Mormann | Relative visual saliency differences induce sizable bias in consumer choice[END_REF] and [START_REF] Pooresmaeili | The effect of visual salience on memorybased choices[END_REF], it has been emphasized that the visual saliency of stimuli influences the choices over the learned preferences and visual working memory. Interestingly, in [START_REF] Mormann | Relative visual saliency differences induce sizable bias in consumer choice[END_REF] where at an exposure time of 1500ms, which is quite similar to that of the model discussed here, the influence of visual saliency was particularly evident when there were no strong preferences among the options. This observation is supported by the early performance decline of the model discussed here, when presented with two closely rewarding stimuli (Stimuli C/D in figure 6). Change in performance with relative salience Fig. 6. Decrease in performance of the model when the lesser rewarding stimulus is presented with stronger salience than the higher rewarding stimulus.

Joint influence of delay and salience

We further tested the model and studied the joint influence of delay and salience by make the worst stimulus to be presented earlier and with a stronger salience. As shown in figure 7, this dramatically degrades the performance and the domain of optimal performance is even more restricted compared to original results.

Conclusion

These early results tend to question the very notion of optimal action selection as defined in a number of theoretical works. The action may be considered optimal provided the two options are presented simultaneously and with an equivalent representation. In the reinforcement learning paradigm, such consideration does not hold much relevance. However, from a more behavioral and embodied perspective, we think this is an important dimension to consider because an animal is scarcely confronted by a set of perfectly equivalent options (but their associated value). One may come first or one may just appear more "obvious" (i.e. more salient). In such a case, the inner dynamics of the model may lead to a suboptimal choice as it is the case using the model from [START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF]. Although we did not perform the study presented here on primates yet, the results from the model assert the need for a closer look at the way we perceive decision making paradigms.

The question is to know to what extent some dedicated brain mechanisms are able to cope with these problems. For example, concerning the time delay, a stop signal, as it has been reported in [START_REF] Schmidt | Canceling actions involves a race between basal ganglia pathways[END_REF], may represent a potential mechanism to be able to solve the problem for small delays (< 200ms).

For the salience difference however, and to the best of our knowledge, there is no such dedicated mechanism. In [START_REF] Mogami | Reward association affects neuronal responses to visual stimuli in macaque te and perirhinal cortices[END_REF], a stimulus-reward association study on macaque monkeys, spike recordings showed significant reward dependence in their responses to the visual cues. In [START_REF] Chelazzi | Rewards teach visual selective attention[END_REF], rewards were shown to teach visual selective attention maximizing the positive outcomes. However both the studies do not identify the underlying mechanisms that caused the observations on the effect of salience. This study hence suggests that measuring experimentally performance using different salience levels could bring useful insights into decision making.

Fig. 2 .

 2 Fig. 2. Architecture of the basal ganglia model which is organised around three parallel loops: cognitive, associative and motor. Only the direct and hyperdirect pathways have been modelled.

Fig. 3 .

 3 Fig.3. Time course of a decision in the motor cortex (blue curves) and cognitive cortex (red curve) before learning. At trial start (t=500ms), there is a first bifurcation between stimuli that are actually presented and those who are not. The second bifurcation around t=750ms is the actual decision of the model.

Fig. 4 .

 4 Fig.[START_REF] Guthrie | Interaction between cognitive and motor cortico-basal ganglia loops during decision making: a computational study[END_REF]. Two stimuli A & B can differ in salience (∆V ) and/or in timing (∆t). ∆V is expressed as the relative ratio between the less salient and the most salient stimuli (∆V = (VA -VB)/VB. ∆t is expressed as the delay separating the two stimuli onsets (∆t = tA -tB).

  in performance with delay in presentation 5. Performance of the model as a function of the delay between the worst and the best stimuli. All combinations have been tested and mean performance has been averaged over 25 trials.

Fig. 7 .

 7 Fig.7. Joint influence of delay and salience on the performance of the model. The two effects appears to linearly sum up and the domain of optimal performance is even more restricted.