Mathematical Physics Vol 1

Chapter 4. Field theory

126

From here we obtain

∇ R = ∇ q ( x − a ) 2 +( y − b ) 2 +( z − c ) 2 = = ( x − a ) i +( y − b ) j +( z − c ) k p ( x − a ) 2 +( y − b ) 2 +( z − c ) 2 = R R ,

which is the unit vector of vector R = −→ AP .

Problem44 Let P be a point on the ellipse, and A and B be the foci of that ellipse. Show that the angle between AP and the tangent of the ellipse at point P is equal to the angle between BP and the tangent at the same point.

Solution Let R 1 = −→ AP and R 2 = −→ BP be vectors connecting points A and B with point P , respectively, and T be the unit vector of tangent in point P (see Fig. 4.24). From the definition of an ellipse it follows that the sum of the distances from the foci to the point P is a constant p , i.e. R 1 + R 2 = p = const .

As the unit vector of the normal to the ellipse is ∇ ( R 1 + R 2 ) , it follows that [ ∇ ( R 1 + R 2 )] · T = 0 or ∇ R 2 · T = − ∇ R 1 · T . Given that ∇ R 1 and ∇ R 2 are unit vectors of vectors R 1 and R 2 (se Example 43, p. 125), it follows that the cosine of the angle between R 2 and T is equal to the cosine of the angle between R 1 and − T . Consequently, these angles are equal.

Figure 4.24: Ellipse.

R Note that the physical meaning of this result can be found in optics. Namely, the light emitted from point A is reflected by an elliptical mirror and passes through point B and vice versa.

4.6.2 Divergence

Problem45 Find div A for the vector function A = x 2 z i + 2 y 3 z 2 j − xy 2 z k at point (1,1,-1).