Mathematical Physics Vol 1

4.1 Scalar field

85

Theorem5 Let f ( M )= f ( x , y , z ) be a scalar function, the first partial derivatives of which are continuous functions. Then a vectorgrad f exists, whose magnitude and direction are independent of the coordinate system. If grad f at point M is not equal to zero, then it has the direction of maximum increase of function f at point M .

We will prove this theorem later.

Theorem6 Let the gradient of the function u = f ( x , y , z ) at point M be different from zero. It is then orhtogonal 3 to each line ℓ , which passes through M , and lies in the equiscalar surface f = const .

Proof

Observe line ℓ , passing through point M and lying in surface f = const . (Fig. 4.6). As the value of the function does not change when the point moves along the line ℓ (because it lies in f = const . ), then d f d s = 0 . As, on the other hand, the deriva tive of the function along the line ℓ is (4.23)

Figure 4.6: Gradient of function f .

d f d s = e · grad f = | grad f |· cos ( grad f , e )= 0 ,

D e f =

then, assuming that grad f̸ = 0, we obtain cos ϕ = 0 (Fig. 4.5). Thus, if e is the unit vector of the tangent line ℓ , it follows that the gradient is orthogonal to the equiscalar surface.

The derivative of the function f in the direction of the tangent t ( t is the unit vector of the tangent to the curved line C at point M ) is, according to (4.23)

d f d s

= t · grad f =

D t f = k ·

(4.25)

d x

k =

∂ f ∂ x ·

∂ f ∂ y ·

∂ f ∂ z ·

d r d s ·

d y d s ·

d z d s ·

i +

j +

i +

j +

grad f .

d s ·

From here we obtain

d f = d r · grad f . (4.26) 3 By "orthogonal to the line at point M " it is assumed that it is orthogonal to the tangent plane, passing through M .