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Defining energy

Initial remarks

Let a surface S be defined by the parametrization

Φ ( θ , t ) = r ( t ) cos ( θ ) , r ( t ) sin ( θ ) , t : 0 θ 2 π , 0 t h .

S is a surface of revolution with radius r ( t ) at each height t [ 0 , h ] . Suppose that V ( θ , t ) = ( a ( θ , t ) , b ( θ , t ) , c ( θ , t ) ) is a vector field tangent to S , so that for any θ 0 , t 0 , V ( θ 0 , t 0 ) is tangent to S at Φ ( θ 0 , t 0 ) . (We are particularly interested in vector fields with unit length, but for now we consider the general case.) We would like to define the notion of the “kinetic energy" of V . Energy should measure how much the vector field changes over the surface; thus, our definition should resemble

S | D V | 2 d Area .

Clarifying this expression is simple in one dimension (if S is a curve), but trickier in two: we must consider the change of V (a three-dimensional function) in two coordinate directions, and account for the curvature of the surface on which it lies. To achieve this, we take | D V | 2 to be the sum of the squares of the derivatives of V , taken in perpendicular directions, with respect to arc length. If V is a function of θ and t , the θ -directional derivative of V can be expressed as V θ θ s , where s = r ( t ) θ is the horizontal arc length of S . Similarly, the t -directional derivative of V can be expressed as V t t σ , where σ = 0 t 1 + r ' ( τ ) 2 d τ .

d Area is given by the square root of the determinant of the gram matrix, | Gram | . By [link] , the Gram matrix is

Φ θ · Φ θ Φ θ · Φ t Φ t · Φ θ Φ t · Φ t .

For our given Φ ( θ , t ) ,

d Area = | Gram | = r ( t ) 1 + r ' ( t ) 2 .

With these considerations in mind, we arrive at the following expression:

E ( V ) = 0 1 0 2 π 1 + r ' ( t ) 2 r ( t ) a θ 2 + b θ 2 + c θ 2 + r ( t ) 1 + r ' ( t ) 2 a t 2 + b t 2 + c t 2 d θ d t .

Some special cases

Perhaps the simplest surface of rotation is the unit cylinder with unit radius, parametrized by Φ ( θ , t ) = cos ( θ ) , sin ( θ ) , t : 0 θ 2 π , 0 t 1 . On this cylinder, the expression for energy simply becomes

E ( V ) =

0 1 0 2 π a θ 2 + b θ 2 + c θ 2 + a t 2 + b t 2 + c t 2 d θ d t .

It can be difficult to define a vector field of the form ( a ( θ , t ) , b ( θ , t ) , c ( θ , t ) ) while ensuring that the vector field remains tangent and of unit length. To this end, we can represent any unit length tangent vector field V at Φ ( θ , t ) on the cylinder by the angle ϕ ( θ , t ) that V ( θ , t ) makes with the horizontal tangent vector ( - sin ( θ ) , cos ( θ ) , 0 ) . Note that if V is continuous on S , ϕ must be continuous in both variables and periodic (up to multiples of 2 π ) in θ ; henceforth, we will assume that all ϕ are continuous and (strictly) periodic. We then have the relation V ( ϕ ) = V ˜ ( θ , t ) = ( - sin ( θ ) cos ( ϕ ( θ , t ) ) , cos ( θ ) cos ( ϕ ( θ , t ) ) , sin ( ϕ ( θ , t ) ) ) . The expression for energy on the unit cylinder becomes

E ( V ˜ ) = 0 1 0 2 π cos ( ϕ ( θ , t ) ) 2 + ϕ θ θ , t 2 + ϕ t θ , t 2 d θ d t .

Examples

It is helpful to consider some specific examples of vector fields on surfaces and to calculate their energies.

Example 1. Let S be the unit cylinder, and define V ( θ , t ) = ( 0 , 0 , 1 ) (or equivalently, in our alternate notation, ϕ ( θ , t ) = π 2 ). V is the unit vector field pointing directly “up" at every point on the surface. Since every partial derivative of V (and of ϕ ) is 0, it is easy to check, using either expression for energy in the previous section, that E ( V ) = 0 . See the following figure.

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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