Mathematical Physics Vol 1

Chapter 4. Field theory

104

Thus, the potential U satisfies the Laplace equation

∂ 2 U ∂ x 2

∂ 2 U ∂ y 2

∂ 2 U ∂ z 2

∆ U =

= 0 .

(4.99)

+

+

Functions that satisfy the Laplace equation are called harmonic functions . Thus, Newton’s potential is a harmonic function. Plane task. Logarithmic potential

Consider now the force in a plane by which some point O attracts a material point M .

Suppose this force of attraction is given by the expression F = − 2 c r r r (4.100) where r = | r | is the magnitude of the position vector r .

Figure 4.12: Force of attraction - logarithmic potential.

Once again, the question is whether there exists a potential U for the force defined in such a manner. Similar to the previous case, we start from:

∂ U ∂ x ∂ U ∂ y

2 c r 2 2 c r 2

X =

x ,

= − = −

(4.101)

Y =

y .

From here we obtain

U = − 2 c Z

x r 2

dx + f ( y ) .

Further, given that

r 2 = x 2 + y 2 , it follows that 2 x d x = 2 r d r ,

and we obtain the potential U

U = − 2 c ln r + f ( y ) .

Let us now find the partial derivative by y : ∂ U ∂ y = ∂ ∂ y

d f d y

( − 2 c ln r )+

=

∂ ∂ r

∂ r ∂ y

d f d y

y r 2

d f d y

= − 2 c

( ln r )

= − 2 c

(4.102)

+

+

.

Thus, according to (4.101) (condition for Y ), we can conclude that d f here also that this constant is equal to 0, and we thus obtain the potential

dy = const . We shall assume

U = − 2 c ln r .

(4.103)