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Lecture 13: More on Gradient; the Operator 'Del' 13. 1. Examples Of

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Lecture 13: More on Gradient; the Operator 'Del' 13. 1. Examples Of Lecture 13: More on Gradient; the Operator `Del' 13. 1. Examples of the Gradient in Physical Laws Gravitational force due to Earth: Consider the potential energy of a particle of mass m, a height z above the Earth's surface V = mgz. Then the force due to gravity can be written as F = V = mg e . −∇ − 3 Gravitational attraction: Now consider the gravitational force on a mass m at r due to a 2 mass m0 at the origin i.e. Newton's law of Gravitation: F = (Gmm0=r )^r. We can write this as − F = V −∇ where the potential energy V = Gmm0=r (see p.51 for how to calculate (1=r)). − r In these two examples we see that force acts down the potential energy gradient. Diffusion: In Physics 2 you may have encountered the idea of diffusion: for example in a gas the molecular motion effectively smoothes out the density. This can be described by the current of particles j(r) being proportional to the gradient of the density (Fick's Law) j(r) = D n(r) − r i.e. the diffusion current flows down the concentration gradient. 13. 2. Examples on gradient Last lecture some examples using `xyz' notation were given. Here we do some exercises with suffix notation (in the lecture we will repeat using `xyz'). As usual suffix notation is most convenient for proving more complicated identities. 2 2 2 2 1. Let φ(r) = r = x1 + x2 + x3, then @ @ @ 2 2 2 φ(r) = e1 + e2 + e3 (x1 + x2 + x3) r @x1 @x2 @x3 = 2x1 e1 + 2x2 e2 + 2x3 e3 = 2r In suffix notation 2 @ r = ei (xjxj) = ei (δijxj + xjδij) = ei 2xi = 2r r @xi In the above we have used the important fact @xi = δij @xj The level surfaces of r2 are spheres centred on the origin, and the gradient of r2 at r points radially outward with magnitude 2r. 49 2. Let φ = a r where a is a constant vector. · @ (a r) = ei (ajxj) = ei ajδij = a r · @xi This is not surprising, since the level surfaces a r = c are planes orthogonal to a. · 3. Let φ(r) = r = x2 + x2 + x2 = (x x )1=2 p 1 2 3 j j @ 1=2 r = ei (xjxj) r @xi 1 1=2 @ = ei (xjxj)− (xkxk) (chain rule) 2 @xi 1 = e 2 x i 2r i 1 = r =r ^ r The gradient of the length of the position vector is the unit vector pointing radially outwards from the origin. It is normal to the level surfaces which are spheres centered on the origin. 13. 3. Identities for gradients If φ(r) and (r) are real scalar fields, then: 1. Distributive law φ(r) + (r) = φ(r) + (r) r r r Proof: @ φ(r) + (r) = ei φ(r) + (r) = φ(r) + (r) r @xi r r 2. Product rule φ(r) (r) = (r) φ(r) + φ(r) (r) r r r Proof: @ φ(r) (r) = ei φ(r) (r) r @xi @φ(r) @ (r) = ei (r) + φ(r) @xi @xi = (r) φ(r) + φ(r) (r) r r 3. Chain rule: If F (φ(r)) is a scalar field, then 50 @F (φ) F (φ(r)) = φ(r) r @φ r Proof: @ F (φ(r)) = ei F (φ(r)) r @xi @F (φ) @φ(r) @F (φ) = ei = φ(r) @φ @xi @φ r Example of Chain rule: If φ(r) = r @F (r) @F (r) 1 F (r) = r = r r @r r @r r where we used a result from 13.2. If F (φ(r)) = φ(r)n = rn we find that n n 1 1 n 2 (r ) = (n r − ) r = (n r − ) r r r In particular 1 r = r r −r3 13. 4. Transformation of the gradient Here we prove the claim that the gradient actually is a vector (so far we assumed it was!). Let the point P have coordinates xi in the ei basis and the same point P have coordinates x0 in the e 0 basis i.e. we consider the vector transformation law xi x0 = λ xj. i i ! i ij φ(r) is a scalar if it depends only on the physical point P and not on the coordinates xi or xi0 used to specify P . The value of φ at P is invariant under a change of basis λ (but the function may look different). φ(x1; x2; x3) φ0(x0 ; x0 ; x0 ) = φ(x1; x2; x3) ! 1 2 3 Now consider φ in the new (primed) basis, its components are r @ φ0(x10 ; x20 ; x30 ) @xi0 Using the chain rule, we obtain @ @xj @ φ0(x10 ; x20 ; x30 ) = φ(x1; x2; x3) : @xi0 @xi0 @xj Since xj = λkj xk0 (inverse vector transformation law) @x j @xk0 = λkj = λkj δik = λij : @xi0 @xi0 51 Hence @ @ φ(x10 ; x20 ; x30 ) = λij φ(x1; x2; x3) : @xi0 @xj which shows that the components of φ respect the vector transformation law. Thus φ(r) transforms as a vector field as claimed.r r 13. 5. The Operator `Del' We can think of the vector operator (confusingly pronounced \del") acting on the scalar field φ(r) to produce the vector fieldr φ(r). r @ @ @ @ In Cartesians: = ei = e1 + e2 + e3 r @xi @x1 @x2 @x3 We call an `operator' since it operates on something to its right. It is a vector operator since it hasr vector transformation properties. (More precisely it is a linear differential vector operator!) We have seen how acts on a scalar field to produce a vector field. We can make products of the vector operatorr with other vector quantities to produce new operators and fields in the same way as we couldr make scalar and vector products of two vectors. For example, recall that the directional derivative of φ in direction ^s was given by ^s φ. Generally, we can interpret A as a scalar operator: · r · r @ A = Ai · r @xi i.e. A acts on a scalar field to its right to produce another scalar field · r @φ(r) @φ(r) @φ(r) @φ(r) (A ) φ(r) = Ai = A1 + A2 + A3 · r @xi @x1 @x2 @x3 Actually we can also act with this operator on a vector field to get another vector field. @ @ (A ) V (r) = Ai V (r) = Ai Vj(r) ej · r @xi @xi = e (A ) V1(r) + e (A ) V2(r) + e (A ) V3(r) 1 · r 2 · r 3 · r The alternative expression A V (r) is undefined because V (r) doesn't make sense. · r r N.B. Great care is required with the order in products since, in general, products involving operators are not commutative. For example A = A r · 6 · r A is a scalar differential operator whereas · r @A A = i gives a scalar field called the divergence ofA r · @xi 52.
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