Multidimensional Calculus: Mainly Differential Theory

Transcription

1 9 Multidimensional Calculus: Mainly Differential Theory In the following, we will attempt to quickly develop the basic differential theory of calculus in multidimensional spaces You ve probably already seen much of this theory Hopefully, we will develop a better understing of the material than is usually imparted in the more elementary treatments, see how to extend it to more general spaces coordinate systems By the way, in keeping with the common practice in physics of denoting the position of a moving object by r, I will relent often use this notation instead of x 91 Motion, Curves, Arclength, Acceleration the Christoffel Symbols For all the following, assume we are considering motion in some space of positions S, that {( x 1, x 2,, x N)} is any coordinate system for this space As before, {h 1, h 2,,h N }, {ε 1,ε 2,,ε N } {e 1, e 2,,e N } are the associated scaling factors, tangent vectors normalized tangent vectors We also assume that we have some position-valued function r(t) that traces out a curve C as t varies over some interval (t 0, t 1 ) Since this is a math/physics course, we naturally view r(t) as being the position of an object, say, George the Gerbil, at time t In terms of our coordinate system, we have some coordinate formula for r r(t) ( x 1 (t), x 2 (t),, x N (t) ) for t 0 < t < t 1 Velocity Speed The formula for velocity v at any given time is easily computed using the chain rule: v dr r x i h i e i, 11/24/2013

2 Chapter & Page: 9 2 Multidimensional Calculus: Differential Theory which we may prefer to rewrite as v dr dx h i i e i or v dr The corresponding speed, then, is ds dr [ N [ dx h i N i e dx i h j j e j j1 That is, ds N j1 h i h j dx j e i e j N j1 ε i dx j g i j (91) If the coordinate system is orthogonal, this reduces to ds N ( ) dx h i 2 i (92)! Example 91: Assume that, using polar coordinates in the plane, the position of an object at time t is given by r(t) ( ρ(t), φ(t) ) ( t 2, 2πt ) Then the velocity at time t is dr dρ h ρ e dφ ρ + h φ e φ ( d [ 1 ) ( ) t 2 d e ρ + ρ [2πt 1 ( 2t ) e ρ + t 2( 2π ) e φ 2t e ρ + 2πt 2 e φ the corresponding speed is ds dr (2t) 2 + ( 2πt 2) 2 2t 1 + π 2 t 2? Exercise 91: In the above example exercises, I ve completely forgotten about explicitly stating the general polar coordinate formulas for velocity v speed for a given r(t) (ρ(t), φ(t)) So you should derive state them That is, show that [dρ v dρ e ρ + ρ dφ 2 [ 2 e ds dφ φ + ρ 2? Exercise 92: Let r(t) be the position at time t of George the Gerbil in a three-dimensional Euclidean space, derive the spherical coordinate formulas for velocity speed (Compare your results to the formulas given in problem on page 200 of Arfken, Weber & Harris Ignore their hint for deriving these things!) e φ

3 Motion, Curves, Arclength, Acceleration the Christoffel Symbols Chapter & Page: 9 3 Distance Traveled / Arclength The distance traveled by our object (George) as t goes from t 0 to t 1 (ie, the arclength of the curve traced out by the object) is simply the integral of the speed, arclength t1 t 0 ds t1 t 0 v(t) In general, the coordinate formula for this is t1 arclength N dx h i h i dx j j tt 0 j1 e i e j t1 tt 0 N j1 dx j g i j, which reduces to arclength t1 tt 0 N ( ) dx h i 2 i when the coordinate system is orthogonal, further reduces to t1 arclength N ( ) dx i 2 if the space is Euclidean the coordinate system is Cartesian tt 0! Example 92: Consider the curve C parameterized in polar coordinates by r(t) ( ρ(t), φ(t) ) ( t 2, 2πt ) for 0 < t < 2, as in example 91 on page 9 2 Its arclength is t1 arclength N ( ) dx h i 2 i tt ( ) 2 ( ) 2 dρ dφ hρ + h φ ( 1 d [ ) 2 ( t 2 + ρ d ) 2 [2πt 2t 1 + π 2 t 2 2 ( 1 + π 2 t 2)3 / 2 3π [ (1 + 4π 2 )3 / 2 1 3π 2 version: 11/24/2013

4 Chapter & Page: 9 4 Multidimensional Calculus: Differential Theory Acceleration the Christoffel Symbols Computing Acceleration The acceleration a of George at each instant of t is, of course, a dv d h i e i [ d dx i h ie i If the coordinate system is basis based, then the scaling factors local basis vectors do not vary, the above simplifies to a dv d 2 x i 2 h ie i This is certainly the case when our coordinate system is Cartesian If, however, the scaling factor /or local basis vectors vary from point to point (as with polar coordinates), then we must use the product rule yielding Recalling that [ d dx i h ie i a dv d2 x i 2 h ie i [ d 2 x i 2 h ie i ε i h i e i, + dxi + dxi we can rewrite the above in the slightly more convenient form a dv [ d 2 x i 2 ε i + dxi Applying the chain rule to the derivatives of the ε i s, we get 1 d[h i e i d[h i e i dε i dε i dxi k1 dx k ε i x k k1 dx k ε i x k So, a dv [ d 2 x i ε 2 i + k1 dx k ε i x k (93) The last acceleration formula is probably the most important new formula of this section Rather than memorizing it, you can always easily rederive it using the product chain rules from calculus However, to use formula (93), we still need to know each ε i/ x k Since this a partial derivative of the vector-valued function ε i, it is, itself, a vector-valued function That is, for 1 I m cheating We ve not verified the chain rule for the differentiation of vector-valued functions, only for scalar position-valued functions The chain rule used here with a vector-valued function can be rigorously justified in a Euclidean space or in a non-euclidean space contained in a Euclidean space (trust me) Justifying it in a more general space is more involved requires clever definitions

5 Motion, Curves, Arclength, Acceleration the Christoffel Symbols Chapter & Page: 9 5 each position p in S, each ε i/ x k is a vector in the tangent vector space at that point p So each ε i/ x k can be expressed in terms of the {ε1,ε 2,,ε N } basis at each point in space This means that, for each point of the space S, there must be corresponding scalars Ŵik 1, Ŵ2 ik, Ŵik N such that ε i x k Ŵik m ε m (94) m1 These Ŵik m s are called the Christoffel symbols (for the coordinate system)2 They are just the components of each ε i/ x k with respect to the {ε 1,ε 2,,ε N } basis at each point Using the Christoffel symbols, formula (93) becomes a dv [ d 2 x i ε 2 i + d 2 x i 2 ε i + With a little relabeling of the indices, we have k1 m1 k1 m1 dx k Ŵm ik ε m dx k Ŵm ik ε m k1 m1 dx k Ŵm ik ε m j1 k1 dx j dx k Ŵi jk ε i, which allows us to rewrite the last formula for acceleration as a dv d2 x i dx + j dx k 2 Equivalently, a dv d2 x i + 2 j1 k1 j1 k1 dx j Ŵi jk dx k Ŵi jk ε i (95) h i e i (95 ) Of course, for the above to be useful, we need to find the values of the Christoffel symbols Computing ε i/ x k the Christoffel Symbols Since the Christoffel symbols are, at each point in S, components of vectors with respect to the basis {ε 1,ε 2,,ε N }, we can use what we learn earlier this term about finding components to find the Ŵ i jk s, provided we can, somehow, compute each ε j x k The simplest case is where the coordinate system is basis based Then the scaling factors local basis vectors do not vary Thus, ε j x k 0, 2 More precisely, they are Christoffel symbols of the second kind version: 11/24/2013

6 Chapter & Page: 9 6 Multidimensional Calculus: Differential Theory equation (94) becomes 0 ε j x k which means that, for every j, k i, Ŵ i jk ε i, Ŵ i jk 0 Remember, this is only if the coordinate system is basis based More generally, if the coordinate system is orthogonal, then (as you showed in 219 on page ), the Ŵ i jk s in ε j x k Ŵ i jk ε i are given by Ŵ i jk ε j x k ε i ε i 2 By the definitions of the scaling factors normalized tangent vectors, this can also be written as Ŵ i jk 1 ε j h 2 i x k ε i 1 [h j e j h i x k e i (96) We should also note that we do have some symmetry in these computations This is because Hence, ε j x k x k r x j x j r x k Ŵ i jk Ŵ i k j, ε k x j a fact that can reduce the number of computations we may have to carry out Of course, computing formula (96) requires that we can compute this dot product If the space is Euclidean you ve determined how to express the ε i s e i s in terms of the orthonormal basis corresponding to a Cartesian system, then computing the Ŵ i jk s by the above equation is relatively straightforward! Example 93: Consider the polar coordinate system {(ρ, φ)} In example 89 on page 8 33 we saw that h ρ 1 h φ ρ,, in terms of the stard basis associated with the Cartesian system {(x, y)}, So, e ρ cos(φ) i + sin(φ) j e φ sin(φ) i + cos(φ) j ε ρ ρ [h ρe ρ ρ ρ [ cos(φ) i + sin(φ) j 0 (97a) 3 Ie, that if a basis {b 1, b 2,, b N } is orthogonal, then v N v i b k v i v b i b i 2

7 Motion, Curves, Arclength, Acceleration the Christoffel Symbols Chapter & Page: 9 7 ε ρ φ [h ρe ρ φ φ [ cos(φ) i + sin(φ) j sin(φ) i + cos(φ) j (97b) Just plugging into formula (96), above, we get Ŵρρ ρ 1 [h ρ e ρ e ρ 1 h ρ ρ 1 0 e ρ 0, Ŵρρ φ 1 [h ρ e ρ e φ 1 h φ ρ ρ 0 e φ 0, Ŵ ρ ρφ 1 [h ρ e ρ e ρ 1 [ [ sin(φ) i + cos(φ) j cos(φ) i + sin(φ) j 0 h ρ φ 1 Ŵ φ ρφ 1 [h ρ e ρ e φ 1 [ [ 1 sin(φ) i + cos(φ) j sin(φ) i + cos(φ) j h φ φ ρ ρ Thus (using (94)), ε ρ ρ Ŵρ ρρ ε ρ + Ŵ φ ρρ ε φ 0ε ρ + 0ε φ 0 ε ρ φ Ŵρ ρφ ε ρ + Ŵ φ ρφ ε φ 0ε ρ + 1 ρ ε φ 1 ρ ε φ In terms of the normalized tangent vectors, ε ρ ρ 0 ε ρ φ e φ Keep in mind that the important quantities are the partial derivatives of the ε j s If we can compute these, as we did in equation set (97), above, then the only value of the Christoffel symbols is to help us express those partial derivatives in terms of the coordinate system s tangent vectors ε 1, ε 2, (instead of i, j, ) With some coordinate systems, though, you can just look at your initial formulas for these partial derivatives, without using Christoffel symbols, write out the formulas in terms of the ε j s Certainly, for example, we did not really need to find the Christoffel symbols to derive from formulas (97) ε ρ ρ 0 ε ρ φ 1 ρ ε φ e φ? Exercise 93: a: Verify that Continue the work in the last example In particular: Ŵ ρ φρ 0, Ŵφ φρ 1 ρ, Ŵ ρ φφ ρ Ŵφ φφ 0 version: 11/24/2013

8 Chapter & Page: 9 8 Multidimensional Calculus: Differential Theory b: Verify that ε φ ρ 1 ρ ε φ e φ ε φ φ ρε ρ ρe ρ two ways: first, by inspecting these partial derivatives computed in terms of i j, then using the Christoffel symbols c: Write out the full formula for acceleration in polar coordinates Corresponding to a function of position r(t) (ρ(t),φ(t)), you should get something like [ ( ) d a 2 2 [ ρ dφ ρ e 2 ρ + ρ d2 φ + 2 dρ dφ e 2 φ d: Compute the acceleration at time t when, in polar coordinates, as in example 91 r(t) ( ρ(t), φ(t) ) ( t 2, 2πt ),? Exercise 94 (spherical coordinates): Let {(r, θ, φ)} be the spherical coordinate system described at the start of section 83, page 8 9 Recall that you found the associated scaling factors tangent vectors (in terms of i, j k ) in exercise 821 on page 8 34 a: Find the nine partial derivatives ε r r, ε r θ, ε r φ, ε θ r, ε φ φ (You might not need the Christoffel symbols for this) b: How many Christoffel symbols are there? Find a few c: Let r(t) (r(t),θ(t),φ(t)) be the position at time t of George the Gerbil in a threedimensional Euclidean space, derive the spherical coordinate formulas for his acceleration (Compare your results to the formula given in problem on page 200 of Arfken, Weber & Harris Again, ignore their hint for deriving this!)? Exercise 95: Once again, consider the circular paraboloid S we discussed in examples 83, 85, (pages 8 16, ) Recall that the coordinate system corresponding to polar coordinates, {(ρ,φ)} is orthogonal that ε ρ cos(φ) i + sin(φ) j + 2ρk, ε φ ρ sin(φ) i + ρ cos(φ) j, h ρ 1 + 4ρ 2, h φ ρ Now, compute the Christoffel symbols, write out the full formula for acceleration for an object at position r(t) (ρ(t), φ(t))

9 Scalar Vector Fields Chapter & Page: 9 9 Final Comments on Christoffel Symbols Acceleration Using the above, we can find formulas for acceleration ( Christoffel symbols) in Euclidean spaces in noneuclidean subpaces contained in Euclidean spaces (eg, curves spheres in Euclidean three-space), at least when the coordinates on the subspace are a subset of the coordinates of the coordinates of the larger Euclidean space To compute the Christoffel symbols when we do not have a convenient Cartesian system lying around, we will need to further develop the metric tensor We will do that in the next chapter (If you want, you can glance at the resulting formula in theorem 112 on page 11 16) 92 Scalar Vector Fields Basic Definitions Concepts In the following, all points/positions refer to points in some N-dimensional space S A scalar field Ψ (on S ) is a scalar-valued function of position, Some examples: Ψ (x) some scalar value corresponding to position x in S T(x) temperature at position x, Φ(x) the gravitational potential at x due to the surrounding masses, x 2 (x) the second coordinate of x with respect to some given coordinate system r(x) distance between x some fixed point O dist(x, O) The coordinate formula for a scalar field Ψ with respect to a given coordinate system {(x 1, x 2,, x N )} is just the formula ψ(x 1, x 2,,x N ) for computing the value of Ψ (x) from the coordinates for x So Ψ (x) ψ(x 1, x 2,,x N ) with x (x 1, x 2,, x N ) For example, if our space is a plane we are using Cartesian coordinates {(x, y)} with O as the origin, then the coordinate formula for r(x) dist(x, O) is x 2 + y 2 If, instead, we are using polar coordinates {(ρ,φ)} with O as the origin, then the coordinate formula for r(x) dist(x, O) is ρ A vector field (on S ) is a vector-valued function of position, V(x) some vector corresponding to position x in S version: 11/24/2013

10 Chapter & Page: 9 10 Multidimensional Calculus: Differential Theory Some examples: V(x) wind velocity at x, F(x) force of gravity on some object at position x ε k x x k rate of change in position as the k th coordinate varies At each point x, V(x) will be a vector in the tangent space at that point Often, though, we will be dealing with a subspace (say, a curve or a sphere in a three-dimensional space), in which case V(x) may or may not be in the tangent space of that subspace, depending on how V is generated For example, V may be the velocity of some object moving on a curve, in which case V(x) is tangent to the curve at each point x On the other h, V(x) may be a vector field perpendicular to a sphere in Euclidean three-space, in which case V(x) will not be in the tangent space of the sphere at any x in the sphere Because V(x) is in the tangent space of our overall space at x, it will have components with respect to both of our associated bases It can be expressed in terms of the ε i s, V(x) V i (x 1, x 2,, x N )ε i where x (x 1, x 2,, x N ), it can be expressed in terms of the e i s, V(x) v i (x 1, x 2,, x N ) e i where x (x 1, x 2,, x N ) These formulas for V(x) should be refered to as coordinate/component formulas for V In practice, it is more common for a vector field to be expressed in terms of the normalized tangent vectors (the e k s ) than the ε k s Still, there are occasions where using the other representation is convenient Since ε i h i e i, these components are related by v i h i V i Note that these component functions are denoted with superscripts That is stard when dealing with the slightly more general theory we are developing here In practice, though, expect a lot of people to use subscripts On the other h, there is no real convention on how to notationally distinguish components with respect to the ε i s from components with respect to the e i s My use here of V i v i will not be consistently followed not even by me If the coordinate system is orthogonal, then {e 1,,e N } is orthonormal at each point,, so, the corresponding components of V are given by v i V e i? Exercise 96: Show that, if the coordinate system is orthogonal, then the components of vector field V with respect to the ε k s are given by V i V ε i (h i ) 2

11 Scalar Vector Fields Chapter & Page: 9 11? Exercise 97: Using the product rule, show that V x j k1 [ V k x j + V i Ŵi k j ε k What would this be in terms of the unit tangents e 1,, e N? (By the way, the expression inside the brackets in the above equation is known as a covariant derivative of V ) Some Warnings 1 Many authors use the terms scalar vector when they mean scalar field vector field 2 People often use the same symbols to denote both a scalar/vector field its coordinate formula This is not particularly bad if there is only one coordinate system being used ( everyone knows which system that is), but it can be quite confusing when we have multiple coordinate systems Change of Coordinates the Coordinate Formulas For us, this should be simple: Suppose ψ ( x 1, x 2,,x N) is the coordinate formula for some scalar field Ψ using some coordinate system {(x 1, x 2,, x N )} To obtain the coordinate formula for Ψ using a different coordinate system {(x 1, x 2,, x N )}, we simply replace each x k in the formula for with the k th change of coordinates formula, ψ ( x 1, x 2,,x N) x k x k ( x 1, x 2,,x N ) If we have a coordinate formula for vector field V, V(x) V i (x 1, x 2,, x N )ε i or V(x) v i (x 1, x 2,, x N ) e i where x (x 1, x 2,, x N ), then, in addition to converting the formulas to corresponding formulas in term of the x k, we need to express V in terms of the local bases associated with {(x 1, x 2,,x N )}, V(x) V i (x 1, x 2,, x N )ε i or V(x) v i (x 1, x 2,, x N ) e i version: 11/24/2013

12 Chapter & Page: 9 12 Multidimensional Calculus: Differential Theory where x (x 1, x 2,, x N ) For this, we can use what we learned earlier about change of basis In particular, if our coordinate systems are orthogonal (so that the sets of associated normalized tangent vectors are orthonormal at every point), then where v 1 v 2 V B M v 2 B B V B v N v N B {e 1, e 2,,e N }, B { e 1, e 2,,e N } [M B B i j e i e j Further details will probably be assigned as exercises v 1 A Few Words About Continuity Differentiability Because our goal is to develop a substantial amount of applicable multidimensional calculus in a relatively short time, we are glossing over some of the mathematical fine points Still, a few words on continuity differentiability are in order, since there will be some issues involving the continuity differentiability of scalar vector fields Continuity is easy to rigorously define: We say that a scalar field Ψ a vector field V are continuous at a point p if only if lim Ψ (x) Ψ ( p) lim V(x) V( p) x p x p We then say that Ψ V are continuous in some region if they are continuous at each point in the region Differentiability is a bit more difficult to rigorously define without going into more mathematical details than appropriate for this course Here is a working definition for us: Ψ is k-times differentiable (at a point or in a region) if only if, using any reasonable coordinate system {(x 1, x 2,,x N )}, the coordinate formula ψ(x 1, x 2,, x N ) for Ψ, as well as all partial derivatives of ψ up to order k, exist are continuous (at the point or in the region) 4 A similar working definition works for vector fields If the order of differentiability k is not mentioned, you should normally assume k 1 That is, Ψ is differentiable means Ψ is 1-time differentiable If, however, we say Ψ is suitably differentiable, then assume k is the order of the highest order derivative relevant to whatever we are discussing 4 Actually, mathematicians call this k-times continuous differentiability or smoothness

13 The Classic Gradient, Divergence Curl Chapter & Page: The Classic Gradient, Divergence Curl Basic Definitions in Euclidean Space For expediency, we will first define the classical differential operators for scalar vector fields in a Euclidean space E using the del operator Moreover, we will assume that our coordinate system {(x 1, x 2,,x N )} is Cartesian The del operator is the vector differential operator given in our Cartesian system by N k1 x k e k k1 e k x k k1 x x k x k For any sufficiently differentiable scalar field Ψ vector field F with corresponding coordinate formulas we define Ψ (x) ψ ( x 1, x 2,, x N) F(x) F ( k x 1, x 2,,x N) e k, k1 the gradient of Ψ grad(ψ ) Ψ, the divergence of F div(f) F, the curl of F curl(f) F by the coordinate formulas grad(ψ) Ψ div(f) F curl(f) F k1 k1 e k x k ψ ( x 1, x 2,,x N) k1 e k x k [ N x k e k F ( j x 1, x 2,,x N) e j k1 j1 [ N x k e k F ( j x 1, x 2,,x N) e j j1 k1 x k e k, k1 F k x k, e 1 e 2 e 3 x 1 x 2 x 3 F 1 F 2 F 3 [ F 3 x [ F2 F 3 e 2 x 3 1 x [ F1 F 2 e 1 x x F1 e 1 x 2 3 Both grad(ψ ) div(f) can be defined assuming our space is of any dimension However, curl(f) requires that our space be three-dimensional (in which case, we usually use the traditional {(x, y, z)} {i, j, k} notation) version: 11/24/2013

14 Chapter & Page: 9 14 Multidimensional Calculus: Differential Theory Observe that grad(ψ) curl(f) are vector fields, while div(f) is a scalar field One thing not obvious from the above definitions is why anyone would be interested in these things There are good reasons arising from both basic mathematics from physics The geometric/physical significance of grad(ψ) will be discussed in a little bit The importance of the divergence curl of a vector field will be discussed later, in conjunction with some classic integral theorems In anticipation of that discussion, let me introduce some terminology that you may encounter in homework: F is solenoidal F is divergence free F 0 F is irrotational F is curl free F 0 Invariance Under Change of Cartesian Coordinates Because these formulas are given in terms of one Cartesian coordinate system, there is the issue of what these formulas become when we switch to any other Cartesian coordinate system (Hopefully, these formulas are Cartesian coordinate system independent ) To examine this issue, let {(x 1, x 2,, x N )} be another Cartesian coordinate system corresponding to an orthonormal basis {e 1, e 2,,e N } Remember, in problem J 1 of Homework Hout VII you showed (among other things) that x e j j x k e k Using the chain rule from elementary calculus the above, we get k1 k1 x k e k k1 j1 x j x k x j e k x j1 j k1 x j x k e k j1 x j e j This shows that the basic formula k1 x k e k is invariant under any change of Cartesian coordinates Consequently, the formulas given for Ψ F F are valid using any Cartesian coordinate system, yield the same vector or scalar fields Product Chain Rules Using the Cartesian formulas elementary calculus, we can derive or verify a number of product rules chain rules that might simplify calculations later on In particular, if Ψ Φ are suitably differentiable scalar fields, F G are suitably differentiable vector fields, then (ΦΨ ) Ψ Φ + Φ Ψ, (98a) (ΦF) ( Φ) F + Φ F, (98b) (ΦF) ( Φ) F + Φ( F) (98c)

15 The General Gradient Operator Chapter & Page: 9 15 (F G) ( F) G F ( G) (98d) These are multidimensional versions of the product rule (note the in (98d) due to the antisymmetry in the cross product) You can easily derive or verify any of them, yourself? Exercise 98 a: Verify equation (98a) b: Pick any one of the other equations in set (98), verify it Now let f be a real-valued function on (so f (a real number) a real number ) Then, for each position x, we can plug the real number Φ(x) into f, getting a new scalar field You can then easily verify the chain rule? Exercise 99: Verify equation (99) f (Φ(x)) [ f (Φ(x)) [ f (Φ(x)) Φ(x) (99) Another chain rule involving the gradient will be derived in the next section 94 The General Gradient Operator Geometric Significance of Ψ Consider a function f of the form f (t) Ψ(x(t)) where Ψ (x) is some scalar field on a Euclidean space E x(t) is any (sufficiently smooth) position-valued function (with t being in some interval (α,β) ) Both Ψ x(t) will have coordinate formulas x(t) ( x 1 (t), x 2 (t),, x N (t) ) for α < t < β Ψ (x) ψ ( x 1, x 2,, x N) where x ( x 1, x 2,, x N) Taking the composition of the two, f (t) Ψ (x(t)) ψ ( x 1 (t), x 2 (t),, x N (t) ) for α < t < β, gives us an ordinary real-valued function f such as you studied in Calc III Its derivative can be found by the classical chain rule developed in that course Computing this derivative, we see version: 11/24/2013

16 Chapter & Page: 9 16 Multidimensional Calculus: Differential Theory that, at least if the coordinate system is Cartesian, d f d [Ψ (x(t)) d [ ( ψ x 1 (t), x 2 (t),, x N (t) ) But remember dx 1 + dx dx N x 1 x 2 x N [ x e x e [ dx x N e 1 N e 1 + dx2 e dx N e N [ N [ N x k e dx k k e k k1 Ψ k1 k1 x k e k dx k1 dx k e k So the above boils down to d dx [Ψ (x(t)) Ψ (910) This is another multidimensional chain rule Keep in mind that the Ψ in this equation is being evaluated at the position corresponding to whatever value of t is relevant To be a little more precise, we should write this last equation as d [Ψ (x(t)) [ [ Ψ x(t0 ) t0 dx t0, but most of us are too lazy Some consequences of equation (910) can be seen if we rewrite this dot product as d [Ψ (x(t)) Ψ dx dx cos(θ) where θ angle between Ψ Clearly, then: d 1 As a function of θ, the above is maximum when θ 0 That is, Ψ (x(t)) is largest when Ψ dx / are lined up With a little thought, you ll realize this means that, at each point x 0 in space, Ψ evaluated at x 0 points in the direction in which Ψ (x) is increasing most rapidly as x moves through position x 0 2 On the other h, if S is a curve or surface 5 on which Ψ is constant, if x(t) traces out a curve in S (with x (t) being nonzero), then Ψ (x(t)) is a constant function of t Hence Ψ dx cos(θ) d Ψ(x(t)) 0, which means that θ π / 2 (or Ψ 0 ) In other words, at each point on this surface, Ψ must be perpendicular to this curve or surface S 5 or hyper-surface if E has dimension greater than three

17 The General Gradient Operator Chapter & Page: 9 17 General Definition Formula for the Gradient Look, again, at equation (910) It gives a coordinate-free description of the gradient; namely, that Ψ is the vector field making that equation true Being coordinate-free, this is actually a better description for the gradient than the Cartesian formula given in the previous section It can even apply when our space is not Euclidean So let us drop our old definition of the gradient,, instead, define the gradient of a scalar field Ψ to be the vector field denoted by either grad(ψ ) or Ψ such that d dr [Ψ (r(t)) [ Ψ (r(t)) (911) whenever r is a differentiable position-valued function Note that this definition is free of coordinates, does not require the space to be Euclidean Since Ψ is a vector field, it has a coordinate/component formula in terms of the local basis at each point, Ψ ( Ψ ) i e i (912) For convenience, we ve used ( Ψ ) i to denote the i th component of Ψ with respect to the local basis Keep in mind that each ( Ψ ) i is a scalar field, may vary with position just as Ψ each e k varies with position To find the coordinate formulas for these components, let us see what the left right sides of equation (911) (defining the gradient) become when using an arbitrary position-valued function with its coordinate formula, r(t) ( x 1 (t), x 2 (t),, x N (t) ) First look at what we get when using this with the coordinate formula for Ψ in the left side of equation (911) Applying the chain rule from elementary calculus, d [Ψ (r(t)) d [ ( ψ x 1 (t), x 2 (t),, x N (t) ) x i (913) On the other h, using the coordinate/component formulas for Ψ dr / in the right side of (911) yields: [ [ Ψ (r(t)) dr N ( Ψ ) i e i r dx j x j j1 [ N ( Ψ ) i e i dx h j j Assuming the coordinate system is orthogonal, the local basis {e 1,,e 2 } is orthonormal, the last set of equations reduces to j1 e j [ Ψ (r(t)) dr ( Ψ ) i dx h j j Thus, using equations (913) (914) any choice of (914) r(t) ( x 1 (t), x 2 (t),, x N (t) ), version: 11/24/2013

18 Chapter & Page: 9 18 Multidimensional Calculus: Differential Theory we can now rewrite as This clearly means that 6 d dr [Ψ (r(t)) [ Ψ (r(t)) x i ( Ψ ) i dx h i i x i ( Ψ ) i h i for i 1, 2,, N Hence ( Ψ ) i 1 h i x i for i 1, 2,, N, the coordinate/component formula for the gradient when using any orthogonal coordinate system is Ψ 1 h i x i e i (915)! Example 94: In polar coordinates for the Euclidean plane, {(ρ,φ)}, Ψ Recalling that h ρ 1 h φ ρ, we see that 1 h i x i e i 1 h ρ ρ e ρ + 1 h φ φ e φ Ψ ρ e ρ + 1 ρ φ e φ In particular, if the polar coordinate formula for Ψ is then, for x (ρ,φ), ψ(ρ,φ) ρ 2 sin(φ) Ψ (x) ρ e ρ + 1 ρ φ e φ 2ρ sin(φ) e ρ + 1 [ ρ 2 cos(φ) e φ 2ρ sin(φ) e ρ + ρ cos(φ) e φ ρ? Exercise 910: Find the formula for the gradient in spherical coordinates ( ) 6 If this isn t clear, first fix your position x 0 x0 1, x2 0,, x 0 N choose r(t) ( ) x 1 (t), x 2 (t),, x N (t) ( t, x 2 0,, x N 0 ) Then equation (914) reduces to { 1 if i 1 0 if i 1 x 1 ( Ψ )1 h 1

19 General Formulas for the Divergence Curl Chapter & Page: General Formulas for the Divergence Curl Later, we will discover the geometric significance of the divergence the curl of a vector field Then, using those, we can both redefine divergence curl in a coordinate-free manner, obtain more general formulas for these This will come from our development of the divergence (or Gauss s) theorem the Stokes theorem In the meantime, I will simply tell you what those formulas are, we will use them, as needed, with the understing that, eventually, we will see how to derive them As usual, let {( x 1, x 2,, x N)} be an orthogonal coordinate system for our space S with {h 1, h 2,,h N }, {ε 1,ε 2,,ε N } {e 1, e 2,,e N } being the associated scaling factors, tangent vectors normalized tangent vectors at each point Assume F is a vector field with component/coordinate formula F(x) F ( i x 1, x 2,,x N) e i where x ( x 1, x 2,,x N) The corresponding formulas for the divergence of F depend somewhat on the dimension N of the space, though the pattern will become obvious If N 2, then F 1 { } [ F 1 [ h h 1 h 2 x F 2 h x 2 1 (916a) If N 3, then F 1 h 1 h 2 h 3 { } [ F 1 [ h x 1 2 h 3 + F 2 [ h x 2 1 h 3 + F 3 h x 3 1 h 2 (916b) If N 4, then F 1 h 1 h 2 h 3 h 4 { x 1 [ F 1 h 2 h 3 h 4 + x 2 [ F 2 h 1 h 3 h 4 + x 3 [ F 3 h 1 h 2 h 4 + x 4 [ F 4 h 1 h 2 h 3 } (916c) And so on The curl still only makes sense if the dimension N is three Using Stokes theorem, we will version: 11/24/2013

20 Chapter & Page: 9 20 Multidimensional Calculus: Differential Theory be able to show that F h 1 e 1 h 2 e 2 h 3 e 3 1 h 1 h 2 h 3 x 1 x 2 x 3 h 1 F 1 h 2 F 2 h 3 F 3 1 ( [ h3 F 3 [ h2 F 2 ) h 2 h 3 x 2 x 3 1 h 1 h h 1 h 2 e 1 ( [ h3 F 3 [ h1 F 1 ) x 1 x 3 ( [ h2 F 2 [ h1 F 1 ) x 1 x 2 e 2 e 3 (917)? Exercise 911: Verify that the above formulas reduce to the Cartesian formulas originally given when the coordinate system is Cartesian? Exercise 912: Using the above, verify that the two-dimensional formula for divergence using polar coordinates {(r, φ)} is F 1 { [ ρ F ρ + } Fφ, ρ ρ φ? Exercise 913: Verify that the the three-dimensional formulas for divergence curl are, using cylindrical coordinates {(ρ, φ, z)}, F 1 { [ ρ F ρ + } Fφ + F z, ρ ρ φ z F [ 1 F z ρ φ Fφ e ρ + z [ F ρ z F z e φ + 1 [ ρ ρ ρ ( ρ F φ ) Fρ φ e z? Exercise 914: Find the formulas for the divergence curl in spherical coordinates? Exercise 915: Let S 3 be a sphere of radius 3 with the two-dimensional spherical coordinates system {(φ, θ)} described in exercise 815 on page 8 27 Find the corresponding coordinate formula for the divergence of a vector field on S 3 (Use the above along with the results derived in exercise 815) It should be noted that you can also derive the formulas in the above exercises by taking the Cartesian coordinate formulas for these entities carefully converting these formulas to polar/cylindrical/spherical coordinates

21 Repeated Applications of the Del Operator Chapter & Page: Repeated Applications of the Del Operator General Observations For any sufficiently differentiable scalar field ψ vector field F, we can compute ( ψ), ( ψ), ( F), ( F) ( F) I ve included parentheses to emphasize that we are taking a del operation on an object resulting from a previous del operation However, except for the last expression, the parentheses are not really necessary are not usually written? Exercise 916 a: Which of the above ends up being i: a scalar field? ii: a vector field? b: Give an example of an expression involving repeated use of that is nonsense Some of these formulas simplify greatly, with the verifications being rather straightforward (at least when our space is Euclidean) Consider, for example, ψ If we are in a Euclidean space, then we can easily compute this using a stard Cartesian coordinate system: [ ψ x i + y j + z k i j k x y z x y z ( ) 2 ψ y z 2 ψ i z y ( ) 2 ψ x z 2 ψ j z x ( ) 2 ψ x y 2 ψ k y x But remember, the order in which we compute two partial derivatives of a sufficiently differentiable function is irrelevant So 2 ψ y z 2 ψ z y, 2 ψ x z 2 ψ z x 2 ψ x y 2 ψ y x, the above computations reduce to In fact, this equation holds more generally ψ 0 version: 11/24/2013

22 Chapter & Page: 9 22 Multidimensional Calculus: Differential Theory? Exercise 917: Use the more general formulas for the gradient curl to verify that ψ 0 even when the space is noneuclidean (but still three-dimensional)? Exercise 918: Verify that F 0 a: when the space is Euclidean b: when the space is arbitrary One other identity that can be easily verified by simply computing things out in Cartesian coordinates is ( F) ( F) F where F actually denotes the Laplacian of F, which is discussed (briefly) at the end of the next subsection The Laplacian Because many problems in physics involve conservative, divergence-free vector fields (to be discussed later), the expression ψ arises fairly often in applications Because of this, a shorth has been universally adopted of using either 2 or for, ψ 2 ψ ψ This operator, whether denoted or 2, is called the Laplacian In Cartesian coordinates, 2 ψ ψ [ N k1 x k e k k1 ( ) x k x k k1 2 ψ x k 2 Using the material developed a few pages ago, you can easily verify that, in polar coordinates, {(ρ,φ)}, 2 ψ 1 [ ρ ψ, ρ ρ ρ ρ 2 φ 2 in spherical coordinates, {(r, θ, φ)}, 2 ψ 1 [ r 2 r 2 r r + 1 r 2 sin(θ) θ [ sin(θ) θ + 1 r 2 sin 2 (θ) 2 ψ φ 2? Exercise 919: Using the material developed a few pages ago, derive the above polar spherical coordinate formulas for the Laplacian Of course, using the even more general formulas for divergence gradient given in section 95, you can derive more general formulas for the Laplacian for arbitrary two- threedimensional spaces

23 Repeated Applications of the Del Operator Chapter & Page: 9 23? Exercise 920: Let ψ be a scalar field on a two-dimensional space with orthogonal coordinates system associated scaling factors {(x 1, x 2 )} {h 1, h 2 } Show that 2 ψ 1 { [ h2 h 1 h 2 x 1 h 1 x 1 + [ } h1 x 2 h 2 x 2? Exercise 921: Let ψ be a scalar field on a three-dimensional space with orthogonal coordinates system associated scaling factors {(x 1, x 2, x 3 )} {h 1, h 2, h 3 } Show that 2 ψ 1 h 1 h 2 h 3 { [ h2 h 3 x 1 h 1 x 1 + x 2 [ h1 h 3 h 2 x 2 + x 3 [ h1 h 2 h 3 } x 3 It turns out that Laplace s equation 2 ψ 0 Poisson s equation 2 ψ some known nonzero function are two of the fundamental partial differential equations of physics Their solutions describe both potentials of conservative vector fields the steady state behavior of various phenomenon We ll discuss one occasionally useful approach to solving these equations in a moment, more general methods next term The Laplacian also appears in many other fundamental equations of physics This includes the wave equation 2 ψ c 2 2 ψ 0, t 2 the heat equation κ 2 ψ 0, t the Schrödinger wave equation i h t h 2 2m 2 ψ + Vψ As a side note, it s worth mentioning that the Laplacian of a vector field can also be defined If the vector field F F k e k has sufficiently differentiable components (the F k s ), then, using Cartesian coordinates, k1 2 F ( 2 F k) e k k1 version: 11/24/2013

24 Chapter & Page: 9 24 Multidimensional Calculus: Differential Theory 97 Multidimensional Differential Operators under Radial Symmetry Many problems in physics ( mathematics engineering) involve radially symmetric scalar or vector fields in an N-dimensional Euclidean space So let us restrict ourselves to an N- dimensional Euclidean space E to consider such things Let us even assume that we ve chosen some point to be the origin O Remember, in such cases we can define the corresponding position vector for each point r(x) Ox Along with this position vector, we have the corresponding radial distance (distance from the origin) r r(x) Ox, the unit radial vector at each point r r(x) r(x) r(x) A scalar field Ψ a vector field F on E are said to be radially symmetric if ( only if) there are functions ψ f on (0, ) such that (x) ψ(r) F(x) f (r) r(x) Arfken, Weber Harris somewhat discuss these functions their gradients, divergences, curls, Laplacians (mainly for the case with N 3 ) However, you can better develop ( may already have done so) this material yourself in the homework (problems O, P Y in of Homework Hout VIII) In particular, you will show (or have shown) that, Lemma 91 If Ψ is a radially symmetric scalar field on an N-dimensional Euclidean space, Ψ (x) ψ(r) where r Ox, then 2 Ψ ψ (r) + N 1 ψ (r) r An immediate consequence is that Laplace s equation reduces to the simple ( easily solved) ordinary differential equation when we can assume radial symmetry ψ (r) + N 1 ψ (r) 0 r