Mathematical Physics Vol 1

Chapter 4. Field theory

168

Exercise 104 Let F be a conservative force, that is, F = − ∇ φ . Suppose a particle of constant mass m moves in its field. If A and B are any two points in that field, prove that

1 2

1 2

mv 2

mv 2 B

φ ( A )+

A = φ ( B )+

where v A and v B are velocities of the particle in points A and B , respectively.

Solution According to Newton’s second law

d 2 r d t 2

d v d t

F = m a = m

= m

.

Given that

d d t

m v 2 ,

d v d t ·

m 2 ·

d d t

1 2

( v ) 2 =

F · v = m

v =

by integration from point A to point B we obtain B Z A F · d r = B Z A F · d r d t d t = B Z A F · v d t = B Z A d d t

m v 2 d t =

1 2

d

m v 2 =

B

B Z A

1 2

m 2

1 2

1 2

v 2

mv 2

mv 2

(4.202)

B −

=

=

A .

A

As F = − ∇ φ it follows that B Z A

B Z A

A Z B

∇ φ · d r =

d φ = φ ( A ) − φ ( B ) .

F · d r = −

(4.203)

Comparing relations (4.202) and (4.203) we obtain

1 2

1 2

mv 2

mv 2

φ ( A ) − φ ( B )=

B −

A ⇒

1 2

1 2

mv 2

mv 2

φ ( A )+

B + φ ( B ) .

A =

Note that φ ( A ) is also called the potential energy at point A , and 1 2 mv 2 A is the kinetic energy of the particle at point A . The result shows that the total energy, i.e. the sum of the kinetic and potential energy, at point A is equal to the total energy at point B . The law of conservation of energy, in this form, applies only to the fields of conservative forces.

Exercise 105 Let a φ = 2 xyz 2 be a scalar function, F = xy i − z j + x 2 k a vector field, and c a curve in