GRADIENT OF A SCALAR

I. Definitions

I.1. Operator part: ∂

	We define the vector partial: ∂ as follows:
	∂ =  ( ∂/∂x, ∂/∂y, ∂/∂z) = (∂x, ∂y, ∂z)

I.2. Gradient

	We define the  Gradient   of a scalar as
	follows:
	Grad (φ) = ∇(φ) = (∂φ/∂x, ∂φ/∂y,  ∂φ/∂z)
	the result is a vector.

II. Applications

II.1. The gradient of a vector ?

	Grad (R) = ∇(R) = ∂ (R) = (∂xR, ∂yR, ∂zR)
	= (∂x(Rx, Ry, Rz), ∂y(Rx, Ry, Rz), ∂z(Rx, Ry, Rz))

II.2. The gradient of a scalar product

	If A = (Ax, Ay)
	and B = (Bx, By), we have :
	A.B = AxBx + AyBy	
	Then:
	∇ (A.B) =( ∂x(AxBx + 
	AyBy), ∂y (AxBx + AyBy)) 
	= ( ∂x(AxBx) + ∂x(AyBy), ∂y(AxBx) + ∂yAyBy)
	= (Ax∂xBx + Bx∂xAx + 
	Ay∂xBy + By∂xAy, Ax∂yBx + Bx∂y Ax + Ay∂yBy + By∂yAy)	
	= (Ax∂xBx + Ay∂xBy + 
	Bx∂xAx + By∂xAy,  Ax∂yBx + Ay∂yBy +  Bx∂yAx + By∂yAy)	
	=(A . ∂xB + B . ∂xA, A . ∂yB + B . ∂yA)	
	= A . ∂B + B . ∂A = A . ∇B + B . ∇A
	Let's writte:

	∇ (A.B) = A . ∂B + B . ∂A = A . ∇B + B . ∇A

	We can write then:
	Aj∂iBj + Bj∂iAj
	with j = x, y related to the components of the vectors: A and B; 
	and i = x, y related to the components of the divergence.
	That is the tensor notation.

	The result is exactly the same for a vector with three(3) dimensions.
	Let's then write:

	∇ (A.B) = Aj∂iBj + Bj∂iAj

	with j = x, y, z related to the components of the vectors: A and B; 
	and i = x, y, z related to the components of the divergence.

II.3. The gradient: ∇ (1/|r - r'|)

	
	Let's write:
	r (x, y , z) and r'(x',y', z') the components of 
	the vectors r and r' in  the Cartesian coordinates.

	The x component of (r - r') / |r - r'|3 
	is: 
	(r - r')x = (x - x) / ((x - x')2 + (y - y')2 + (z - z')2)3/2

	We have:
	∂/∂x (1/|r - r'|) =
	∂∂x (1 / ((x - x')2 + (y - y')2 + (z - z')2)1/2

	- (1/2) 2 ( x-x') / ((x - x')2 + (y - y')2 + (z - z')2)3/2

	- ( x - x') / ((x - x')2 + (y - y')2 + (z - z')2)3/2

	- ( x - x') /  (|r - r'|)3

	The same calculation leads to:	
	(r - r')y= - ( y - y') / (|r - r'|)3
	And 	
	(r - r')z= - ( z - z')/ (|r - r'|)3

	Then:
	∂/∂x =  1/|r - r'| = - ( x - x') / |r - r'|3
	∂/∂y =  1/|r - r'| = - ( y - y') / |r - r'|3
	∂/∂z =  1/|r - r'| = - ( z - z') / |r - r'|3
	Hence:
	Grad(1/|r - r'|) = - (r - r') / |r - r'|)3
	Or:
	∇ (1/|r - r'|) = - (r - r') / |r - r'|)3