In each case the uniform density of the n-simplex is 1 and its mass is 1/n!. r2 = X2 + Y2 + Z2.
The 3D field produced by a 0-simplex at the origin is 1/r. The potential field transforms with Euclidean similarities but not affine.
For the 1-simplex {(x, 0, 0) | 0≤x≤1} is already slightly complex:
p1(X, Y, Z) =
∫[x=0:x=1] dx/√((X-x)2 + Y2 + Z2) at the point (X, Y, Z).
∫[x=0:x=1] dx/√((X-x)2 + Y2 + Z2)
= ∫[x=0:x=1] dx/√(x2 − 2Xx + r2)
= (log(2√(r2+(−2X)x+x2) + (−2X) + 2x))[x=0:x=1]
= log(2√(r2+(−2X)+1) + (−2X) + 2) − log(2√(r2) + (−2X))
= log(2√(r2 − 2X + 1) − 2X + 2) − log(2r − 2X)
= log((2√(r2 − 2X + 1) − 2X + 2)/(2r − 2X))
= log((√(r2 − 2X + 1) − X + 1)/(r − X)).
See p1 here.
To explore simplification we consider
pd(X, Y, Z) = p(X, Y, Z)+p(−X, Y, Z) = p((X+1)/2, Y/2, Z/2).
This is the potential for {(x, 0, 0) | −1≤x≤1} with mass = 2.
See (pq x y z) = (let ((f (lambda (x) (- (sqrt (+ (sq x) (sq y) (sq z))) x)))) (log (/ (f (- x 1)) (f (+ x 1)))))).
We choose the triangle with vertices at (0, 0, 0), (1, 0, 0), (x', y', 0).
Any triangle is, for some x' and y', similar to this.
The potential at (X, Y, Z) is:
p2(x', y')(X, Y, Z) = ∫[λ=0, λ=1]
p1((X−λx')/(1−λ), (Y−λy')/(1−λ), Z/(1−λ))