Mathematical Physics Vol 1

Chapter 4. Field theory

106

d Z = − d F · cos β .

(4.107)

Further, from the relation z = r cos β + R cos θ (see Fig. 4.14), we obtain

z − Rcos θ r

cos β =

(4.108)

,

and thus finally the projection of the elementary force d F on the z – axis

k 2 m 1 ρ d S R 2 + z 2 − 2 zR cos θ · 1 ρ dS ( z − R cos θ ) ( R 2 + z 2 − 2 zR cos θ ) 3 / 2 . k 2 m

z − R cos θ r

d Z = −

=

(4.109)

= −

The total force can then be determined by integrating the above expression over the whole sphere S Z = − Z S k 2 m 1 ρ ( z − R · cos θ ) ( R 2 + z 2 − 2 Rz cos θ ) 3 / 2 d S . (4.110) Let us now express the elementary surface d S in spherical coordinates d S = R 2 sin θ d ϕ d θ , (4.111)

and consequently obtain the expression

2 Z

π 0 Z

2 π

( R cos θ − z ) · sin θ d ϕ d θ ( R 2 + z 2 − 2 Rz cos θ ) 3 / 2 .

Z = k 2 m

1 ρ R

(4.112)

0

Further, given that r 2 = R 2 + z 2 − 2 Rz cos θ (see eq. (4.105)), where R , as the radius, and z , as the distance between the fixed point M and the center of the sphere, are fixed distances, we obtain by differentiating that r d r = Rz sin θ d θ , that is

r 2 + z 2 − R 2 2 z ,

R cos θ − z = −

(4.113)

so that

z + R | z − R |

r 2

Z π 0

1 2 Rz 2 Z 1 2 Rz 2

z 2 − R 2

( R cos θ − z ) sin θ d θ ( R 2 + z 2 − 2 Rz cos θ ) 3 / 2

1 +

d r =

(4.114)

= −

r

z + R

z 2 − R 2

r −

= −

.

| z − R |

Let us now analyze this result. As θ = 0 ⇒ r = | z − R | , two cases should be distinguished. First , when the point is outside the sphere then z > R , so that | z − R | = z − R , from where it follows that Z = − k 2 m 1 4 π R 2 ρ z 2 = − k 2 m 1 m 2 z 2 . (4.115) Here we assumed that the volume of a sphere of unit thickness V = 4 π R 2 · d = 4 π R 2 , where d is the thickness, which is equal to 1, so that the mass is m 2 = ρ V = 4 πρ R 2 . And the second case, when the point is inside the sphere. In that case z < R , so that | z − R | = R − z , from where it follows that Z = 0.