Four-dimensional reflection groups and electrostatics
Maxim Olshanii, Yuri Styrkas, Dmitry Yampolsky, Vanja Dunjko, Steven G. Jackson
SSciPost Physics Submission
Four-dimensional reflection groups and electrostatics
M. Olshanii , Y. Styrkas , D. Yampolsky , V. Dunjko , S. G. Jackson Department of Physics, University of Massachusetts Boston, Boston, MA 02125, USA Princeton High School, Princeton, NJ 08540, USA* [email protected] 15, 2019
Abstract
The two-element cyclic group consisting of the identity and a sphere inversion can be viewed asa stereographic image of a one-mirror reflection group in 4D. Using this, we identify 19 three-parametric families of finite groups formed by at most four sphere inversions. Exactly as in thesingle sphere case, each member of each of the 19 families generates a solvable electrostaticsproblem of a charge inside a piecewise-spherical cavity with grounded conducting walls. Wepresent a worked example of a member of the D family: a cavity formed by three mutuallyorthogonal planes (i.e. spheres of infinite radius) and a spherical segment at to each of theflat walls. In this case, generating the induced potential requires 191 image charges. Contents D
76 Summary and outlook 10References 11 a r X i v : . [ phy s i c s . c l a ss - ph ] J u l ciPost Physics Submission Every classical electrodynamics textbook (e.g. [1, 2]) will include a problem about the field created bya point charge in the vicinity of a grounded conducting sphere, or in the vicinity of the surface of aspherical cavity in a conducting medium. These problems are solvable using the method of images: thefield created by the conducting medium can be generated by a single image charge. The position of theimage charge is related to the position of the physical charge by a sphere inversion, while the value ofthe charge is linked to the two positions.A conducting half-space is a natural limiting case of problems involving spheres. In the planar case,it is natural to attempt to add additional flat conducting walls in the hope that the generated imageswill form a finite set. This indeed happens when the mutual orientation of the walls coincides withthat of the generating mirrors of a known finite reflection group [3]. To the contrary, little is knownabout the group-theoretical properties of the sphere- and circle inversions. As a result, there have beenno proposals for generalizing the problem of a conducting sphere or cavity to the problem of multiplespherical surfaces. And even if a set of spheres can be arranged in such a way that it produces a finitenumber of image charge positions—all inside the conductor—it is not clear a priori if the correspondingvalues of the image charges can be consistently assigned.In what follows, we will show that if the walls of a conducting cavity are generated by the 4Dstereographic images of the grand hypercircles formed on the intersection of, on the one hand, thegenerating mirrors of a known 4D finite reflection group, and, on the other hand, the hypersphere theprojection originates from, the field induced by a point charge inside the cavity can found using amethod of images. As a particular example, we consider a cavity formed by three mutually orthogonalwalls and a spherical surface at 60° to each of the three walls.
Consider a point charge q at a position p inside an empty spherical cavity of radius R centered ata point O , surrounded by a grounded conductor (Fig. 1). Consider a point p related to the point p through a sphere inversion, with the cavity wall playing the role of the inversion sphere: p = O + (cid:18) R | p − O | (cid:19) ( p − O ) . (1)Let r and r be the distances between a point on the cavity surface and the points p and p ,respectively. For any point on the cavity surface, in can be proven that r | p − O | = r | p − O | . (2)From this property, it follows immediately that if the value of the image charge is assigned as q = − (cid:115) | p − O || p − O | q , (3)then the electrostatic potential created by the physical charge and its image on any point on the wall willvanish: q / r + q / r = ciPost Physics Submission | ⃗ p − ⃗ O | ⃗ p ⃗ p Rr r ⃗ O | ⃗ p − ⃗ O | Figure 1: A charge inside a conducting spherical cavity: notations and definitions. O is the center ofthe cavity. The vector p is the position of the physical charge and r is the distance between a particularpoint on the cavity wall and the physical charge, while p and r are the corresponding quantities forthe image charge. The distances | p − O | and | p − O | govern the assignment of the value of the imagecharge. Imagine an empty cavity surrounded by a grounded conductor. Assume that its walls are formed bysegments of spherical surfaces. The field induced by a point charge placed inside the cavity can beconstructed using the method of images if the following three solvability conditions are satisfied:3 ciPost Physics SubmissionSolvability conditions
I. The set of image charge locations produced via chains—of any length—of sequentialapplication of the inversions from Eq. (1) with respect to any of the spheres involved isfinite;II. The values of the image charges can be unambiguously assigned via a sequentialapplication of the rule (3), where the charge value assigned does not depend on theparticular sequence of inversions that produced its location.III. No image charges are produced inside the cavity.Indeed, consider one of the spherical surfaces defining the cavity. According to the condition (I.),any charge—regardless of whether it is the physical charge or an image charge—will have a counterpartlinked to it, in both directions, by an inversion with respect to the surface of interest. According to thecondition (II.), the two will create a zero potential on the cavity surface. The rest of the charges willform similar pairs, with similar results. Finally, the condition (III.) guarantees that no “ghost” chargesare predicted by the solution obtained, i.e. that inside the cavity there is only the charge that is presentin the statement of the problem.
It is not immediately clear how to construct a set of spherical surfaces that satisfy the solvabilityconditions outlined in the previous section. Consider, however, a set of spheres each of is a 4Dstereographic projection of a grand hypercircle on a surface of a 4D hypersphere (Fig. 2). A 4Dstereographic projection takes a point (cid:112) (cid:48) on a hypersphere of a center (cid:79) and radius (cid:82) , (cid:112) (cid:48) ≡ (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) p (cid:48) x p (cid:48) y p (cid:48) z p (cid:48) w (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) = (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) sin ( Θ p (cid:48) ) sin ( θ p (cid:48) ) cos ( φ p (cid:48) ) (cid:82) sin ( Θ p (cid:48) ) sin ( θ p (cid:48) ) sin ( φ p (cid:48) ) (cid:82) sin ( Θ p (cid:48) ) cos ( θ p (cid:48) ) (cid:82) cos ( Θ p (cid:48) ) (cid:82) (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) . (4)and converts it to a point (cid:112) on the tangential “horizontal” 4D hyperplane—identified with the physical3D space—that touches the hypersphere at the “South Pole” ( x = , y = , z = , w = − (cid:82) ) : (cid:112) ≡ (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) p x p y p z p w (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) = (cid:169)(cid:173)(cid:173)(cid:173)(cid:173)(cid:171) ( Θ p (cid:48) ) sin ( θ p (cid:48) ) cos ( φ p (cid:48) ) (cid:82) ( Θ p (cid:48) ) sin ( θ p (cid:48) ) sin ( φ p (cid:48) ) (cid:82) ( Θ p (cid:48) ) cos ( θ p (cid:48) ) (cid:82) − (cid:82) (cid:170)(cid:174)(cid:174)(cid:174)(cid:174)(cid:172) . (5)Notice that the 4D coordinate system we are using has the origin at the center of the hypersphere, (cid:79) .The fourth coordinate axis, corresponding to the w coordinate, is normal to the “physical” hyperplane.4 ciPost Physics Submission ℛ 𝒪 O ⃗ n ⃗ p ⃗ p ⃗ p ′ R ⃗ p ′ Figure 2: An artistic rendering of the relationship between a 4D reflection on a hypersphere and a 3Dsphere inversion, through a 4D stereographic projection. The points p (cid:48) and p (cid:48) are on a hypersphereof radius (cid:82) whose center is at (cid:79) . These two points are related by a 4D reflection with respect to ahyperplane through (cid:79) whose unit normal vector is n . The 3D points p and p are the images of p (cid:48) and p (cid:48) under the stereographic projection from the hypersphere to a “horizontal” hyperplane. Thelatter hyperplane is identified with the physical 3D space. It can be shown that p and p are thenrelated via a 3D sphere inversion with respect to a sphere whose center is O and whose radius is R . Thissphere is the stereographic image of the grand hypercircle formed at the intersection of the hyperplanecharacterized by n and the hypersphere.A great hypercircle on our hypersphere is a set of 4D points that satisfy ( (cid:112) (cid:48) ) = (cid:82) (6) (cid:112) (cid:48) · (cid:110) = , (7)where (cid:110) ≡ (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) n x n y n z n w (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) = (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) sin ( Θ n ) sin ( θ n ) cos ( φ n ) sin ( Θ n ) sin ( θ n ) sin ( φ n ) sin ( Θ n ) cos ( θ n ) cos ( Θ n ) (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) . (8)5 ciPost Physics Submission We may interpret (cid:110) as the unit vector normal to a hyperplane that passes through (cid:79) , the center of thehypersphere. The intersection of this hyperplane and the hypersphere is the great hypercircle of interest.Under the stereographic projection (5, 4), the great hypercircle (6-8) transforms to a sphere of radius R = | cot ( Θ n )| (cid:82) (9)centered at (cid:79) ≡ (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) O x O y O z O w (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) = (cid:169)(cid:173)(cid:173)(cid:173)(cid:171) − ( Θ n ) sin ( θ n ) cos ( φ n ) (cid:82) − ( Θ n ) sin ( θ n ) sin ( φ n ) (cid:82) − ( Θ n ) cos ( θ n ) (cid:82) − (cid:82) (cid:170)(cid:174)(cid:174)(cid:174)(cid:172) . (10)Notice that all the spheres produced this way yield the following constraint: R = O + (cid:82) , (11)with O being the 3D version of the corresponding 4D vector (cid:79) : O ≡ ( O x , O y , O z ) . (12)Let p (cid:48) , and p (cid:48) , be two points on the hypersphere. Suppose they are related by a reflection viaa 4D mirror defined by the unit normal vector (cid:110) , in other words that p (cid:48) , = p (cid:48) , − ( (cid:110) · p (cid:48) , ) (cid:110) . Then it can be explicitly shown that the stereographic images of p (cid:48) , and p (cid:48) , will be related by asphere inversion (1), with p and p being the 3D versions of the corresponding 4D vectors p , and p , : p ≡ ( p , x , p , y , p , z ) p ≡ ( p , x , p , y , p , z ) . Let us finally consider the 3D spheres forming the cavity of interest. Let us assume that these 3Dspheres stereographically originate from grand hypercircles that are, in turn, produced by the generatingmirrors of a 4D reflection group [4].
Let us place a charge ( q , p ) inside the cavity described above. Let (cid:112) (cid:48) be the stereographic inverseimage of p (so that p is the stereographic image of (cid:112) (cid:48) ). Sequential sphere inversions applied to thisoriginal charge location p will be stereographically connected to the sequential 4D reflections of theposition (cid:112) (cid:48) on the hypersphere. If the mirrors in question generate a finite reflection group, the numberof locations generated by the 4D reflections will be finite. Therefore, so will be the number of positionsproduced by the sequential 3D spherical inversions.6 ciPost Physics Submission If an inversion sphere is a stereographic image of a great hypercircle on a hypersphere of radius (cid:82) , thenany two points p and p related by the inversion obey | p − O || p − O | = F ( p ) F ( p ) , (13)with F ( p ) = p + (cid:82) . (14)Notice that F ( p ) depends neither on the other point p , nor on the center O of the inversion sphere,nor on the radius R of the inversion sphere. The same holds for F ( p ) , upon a p ↔ p substitution.The charge assignment rule (3) becomes q = − (cid:115) F ( p ) F ( p ) q . (15)It is easy to show that this relationship will become q + m = (− ) m (cid:115) F ( p ) F ( p ) q (16)if q + m and q are linked by a chain of m inversions (instead of a single inversion). In particular, anyimage charge q image will be related to the physical charge q physical as q image = (− ) m image (cid:115) F ( p image ) F ( p physical ) q physical , (17)where (− ) m image is the parity of the number of inversions linking the physical charge and the imagecharge in question. Recall also that for reflection groups, the parity of the number of reflections leadingto a particular member of the group is the same for any chain of reflections. Hence, the sphere inversionsstereographically linked to the members of a 4D reflection group will have the same property, and thecharge value assigned using (17) is indeed unique. Consider again (cid:112) (cid:48) , the stereographic inverse image of p , where the latter is the location of the physicalcharge. From the general properties of reflection groups it follows that for the 4D reflections, all thereflection images of (cid:112) (cid:48) will lie outside of the principal chamber —the space delineated by the generatingmirrors. The intersection of the principal chamber with the hypersphere will, under the stereographicprojection, become the 3D cavity of interest. Likewise, the 4D reflection images will become thelocations of the 3D image charges, and none of them will be located inside the cavity of interest. D The 4D reflection group D is generated by three mutually orthogonal mirrors, and a fourth one at 60°to the first three [4]. The total number of group elements is is 192, of which 12 are pure reflections.7 ciPost Physics Submission Once the mutual orientation of the generating mirrors is fixed, we have to choose the orientation of themirrors as a group: in four dimensions, rotations are parametrized by 6 real parameters (of which 3 willgovern the 3D orientation of the resulting 3D cavity, and only the remaining 3 will lead to nontrivialdifferences in the cavity wall radii and mutual alignments).As an example, we choose the following set of the generating mirror normals: ( , , , )( , , , )( , , , ) (18) (cid:18) , , , (cid:19) . (19)The corresponding angles (8) are then given by Θ = π ; θ = π ; φ = Θ = π ; θ = π ; φ = π Θ = π ; θ = φ = any (20) Θ = π ; θ = arg (cid:16) + i √ √ (cid:17) ; φ = π . (21)Under a stereographic projection (4, 5), the first three great hypercircles transform to 3D spheres ofinfinite radius (see (9)), i.e. to 3D planes. These planes will cross the origin. The fourth hypercirclebecomes a sphere. The resulting cavity is given by the following set of inequalities: x > , y > , z > , (22)and (cid:18) x + R (cid:19) + (cid:18) y + R (cid:19) + (cid:18) z + R (cid:19) < R , , (23)with R = (cid:82) , (24)where (cid:82) is the radius of the hypersphere the stereographic projection originates from (see (12), (10),and (9) for the formulae for the center and the radius of the ball (23)). Notice that since the stereographicprojection is a conformal transformation, the angles between the resulting 3D surfaces are the same asthe angles between the 4D hyperplanes (18, 19) that generate them. In particular, the spherical segmentof the cavity surface crosses each of the three planar segments at an angle of 60°.An electrostatic potential created by a point charge, placed anywhere inside the cavity of a shape(22), (23) with grounded conducting walls can be found using the method of images (Fig. 3 (b)). Theimage charge locations will be given by sequential applications of the reflections about the planar cavityboundaries and sphere inversions (1). The number of image charges, all situated outside the cavity, is191. The charge values can be unambiguously assigned using the rule (17, 14) (or the original rule(3)). For the reflections, when an inversion sphere degenerates into a plane, the rules (17, 14) and (3)become q image = (− ) m q physical and q = − q , respectively, with m still being the total number of theinversions and reflections linking q image and q physical .8 ciPost Physics Submission (a) (b) Figure 3: A solvable electrostatic problem associated with the four-dimensional reflection group D (asubgroup of the full symmetry group of the tesseract). (a) Shown here is a cubic segment of space,centered at the origin, of a linear size 0 . R . The tetrahedron-like shape in the middle of the subfigurerepresents the shape of an electrostatic cavity—with grounded conducting walls—solvable using themethod of images. A sample charge (red) is placed at the grand diagonal, at a distance 0 . R awayfrom the spherical segment of the surface of the cavity. (Solvability persists for any position of thecharge inside the cavity.) There are 191 image charges, but only a subset of them are visible (white)in the subfigure. See text for the charge values. The total number of charges, corresponding to theone physical charge supplemented by all the image charges, is 192, which is also the total number ofgroup elements—both reflections and rotations—in the reflection group D . Together with the sample(physical) charge, the image charges generate a potential inside the cavity that vanishes at the cavitywalls. (b) The surfaces of zero potential for the electrostatic potential induced by the sample charge ofthe previous subfigure; shown here is a cubic segment of space, centered at the origin, of a linear size0 . R . A part of this zero-potential surface coincides with the surface of the cavity. The view of thatpart is partially obscured by the eight additional spheres (additional to the four surfaces forming thecavity) on which the potential vanishes as well. The potential also vanishes on the continuations of thesurface fragments forming the cavity of interest. In total, there are 12 spherical or planar surfaces onwhich the potential is guaranteed to vanish—the same number as the number of mirror reflections inthe D reflection group. 9 ciPost Physics Submission We showed that the solution of a classic electrostatic problem—finding the field induced by a pointcharge placed inside a spherical cavity with grounded conducting walls—can be generalized to anycavity whose walls are represented by 4D stereographic projections of grand hypercircles formed by theintersections of a 4D hypersphere and the mirrors of any known finite 4D reflection group. We used thegroup D as a worked example.The scope of problems solvable by this method is as follows: • Any finite reflection group can be used, including those that are reducible and those that arelower-dimensional but embedded in 4D [4]; see Table 1. A • — • — • — •( B = C ) • — • — • — • D • — • •/\• F • — • — • — • H • — • — • — • A × A • — • — • •( B = C ) × A • — • — • • H × A • — • — • • I ( m ) × I ( m ) • m — • • m — • I ( m ) × A × A • m — • • • A × A × A × A • • • • A • — • — •( B = C ) • — • — • H • — • — • I ( m ) × A • m — • • A × A × A • • • I ( m ) • m — • A × A • • A • . . Table 1: Here m , m , and m are any integers greater than or equal to 3. The Coxeter diagrams aboveencode the angles between the generating mirrors of the corresponding reflection group. In a Coxeterdiagram, each vertex corresponds to a mirror. The angle between two mirrors is π / π / π / n if the vertices are joined and the edge is marked by n . Thefinal entry in the table above corresponds to the classic problem with one spherical cavity—the veryinspiration for this paper. 10 ciPost Physics Submission • For each of the reflection groups listed above, there will be a three-parametric family of orientationsof the mirrors in the 4D space, leading to a three-parametric family of 3D sphere radii andpositions. The remaining three parameters of the six-parametric family of the 4D rotations willcontrol the trivial 3D rotation of the resulting 3D cavity.A 2D generalization of our scheme (a 3D stereographic projection from a sphere to a plane) isworth considering. There, one obtains a family of solvable problems involving cylindrical conductingcavities and charged wires. The problem of assigning the values of charges is expected to disappear (asone trivially gets a set of sign-alternating charges). However, this property alone only guarantees thatthe electrostatic potential will be a non-zero constant on each of the cylindrical segments; it is still inprinciple possible that the potential will not be given by the same constant on each of the cylinders.However, we expect that a relationship analogous to (13, 14) can be proven in the 2D case as well. If so,then it will be possible to prove that the potential in fact has the same value at every point on the cavitysurface.Finally, the property (13, 14)—that so far seems completely accidental—may prove to be aconsequence of a deeper connection. One may conjecture that a map exists between the solutionsof the Poisson equation on the hypersphere and the solutions of the Poisson equation on a planestereographically connected to that hypersphere. In this case, the solutions for the class of problemswe considered will become images of the hyperspherical solutions obtained via pure reflections . Thecharges on the hypersphere will have the same magnitude and an alternating sign. The relationshipbetween a hyperspherical charge q (cid:48) and the 3D charge q will be then given by q = const × (cid:112) F ( p ) q (cid:48) ,where p is the location of the charge q , and the function F is given by (14). The proof of this conjecturemay involve (a) the fact that the 3D sphere-, and, potentially, 4D hypersphere inversions convert solutionsof the Poisson equation to solutions of the Poisson equation and (b) the fact that a 4D stereographicprojection can be reinterpreted as a 4D hypersphere inversion. References [1] L. D. Landau and L. P. Pitaevskii,
Electrodynamics of Continuous Media: v. 8 (Course of TheoreticalPhysics) , Pergamon, New York, 2nd edn., doi:10.1016/B978-0-08-030275-1.50025-4 (1984).[2] J. D. Jackson,
Classical Electrodynamics , Wiley, Hoboken, NJ, 3rd edn. (2009).[3] J. Vanderlinde,
Classical Electromagnetic Theory , Springer, Dordrecht, The Netherlands,doi:10.1007/1-4020-2700-1 (2004).[4] J. Humphreys,