Additional Mathematical Tools: Detail

Transcription

1 Additional Mathematical Tools: Detail September 9, 25 The material here is not required, but gives more detail on the additional mathmatical tools: coordinate systems, rotations, the Dirac delta function and the a proof of the Helmholz theorem. Cylindrical coordinates Cylindrical or polar coordinates keep the Cartesian z unchanged, but label points in the xy-plane with a radius, ρ, and an angle ϕ measured counterclockwise looking down the z-axis from the x-axis. The relationship of cylindrical ρ, ϕ, z to Cartesian coordinates x, y, z is therefore with inverse relations x ρ cos ϕ y ρ sin ϕ z z ρ x 2 + y 2 ϕ tan y x z z cos 2 ϕ dϕ dy x ydx x 2 dϕ cos2 ϕ ρ cos ϕ dy ρcos2 ϕ sin ϕdx ρ 2 cos 2 ϕ dϕ cosϕ ρ dy sin ϕ ρ dx We may define unit vectors in the direction of increase of each of these coordinates. The easiest way to develop the relationships we need is to express them in terms of the constant Cartesian basis. Since ˆρ lies in the xy plane, ˆρ xî + yĵ x2 + y 2 ρ cos ϕî + ρ sin ϕĵ ρ 2 cos ϕî + sin ϕĵ and ˆϕ is orthogonal to this, ˆϕ sin ϕî + cos ϕĵ

2 while in the z direction we still have ˆk. For any vector we may expand in either basis, Inverting the relations we also have v v ρˆρ + v ϕ ˆϕ + v z ˆk î cos ϕˆρ sin ϕˆϕ ĵ sin ϕˆρ + cos ϕˆϕ We can find the length, dl, of infinitesimal displacements in two ways. The difficult way is to change coordinates from the Cartesian distance, where we can find the differentials, Substituting, dl 2 dx 2 + dy 2 + dz 2 dx cos ϕdρ ρ sin ϕdϕ dy sin ϕdρ + ρ cos ϕdϕ dz dz dl 2 cos ϕdρ ρ sin ϕdϕ 2 + sin ϕdρ + ρ cos ϕdϕ 2 + dz 2 cos 2 ϕdρ 2 2ρ cos ϕ sin ϕdρdϕ + ρ 2 sin 2 ϕdϕ 2 + sin 2 ϕdρ 2 + 2ρ sin ϕ cos ϕdρdϕ + ρ 2 cos 2 ϕdϕ 2 + dz 2 cos 2 ϕ + sin 2 ϕ dρ 2 + ρ 2 sin 2 ϕ + cos 2 ϕ dϕ 2 + dz 2 dρ 2 + ρ 2 dϕ 2 + dz 2 This is just the length of the infinitesimal displacement vector dl dρˆρ + ρdϕˆϕ + dzˆk The second way to find the expression for dl 2 is to draw a picture. In the xy-plane, we can draw two vectors from the origin that are separated by a small angle dϕ and differ in length by dρ. Then, since these are infinitesimals, we can use the Pythagorean theorem to write down the distance between the ends of the two vectors. The angular separation is by an arc-length ρdϕ and the lengths differ by dρ, so the Pythagorean theorem gives the length of the diagonal as Adding the Cartesian third direction, we have diag 2 dρ 2 + ρ 2 dϕ 2 dl 2 dρ 2 + ρ 2 dϕ 2 + dz 2 We also need derivatives. We can transform using the chain rule, x ρ x ρ + ϕ x ϕ + z x z x ρ ρ sin ϕ ρ ϕ cos ϕ ρ sin ϕ ρ ϕ 2

3 For y, y ρ y ρ + ϕ y ϕ + z y z sin ϕ ρ + cos ϕ ρ ϕ so the gradient in cylindrical coordinates is f îf x + ĵf y f + ˆk z cos ϕˆρ sin ϕˆϕ cos ϕ f ρ sin ϕ f + sin ϕˆρ + cos ϕˆϕ sin ϕ f ρ ϕ ρ + cos ϕ f ρ ϕ ˆρ cos 2 ϕ f ρ sin ϕ cos ϕ f ρ ϕ + sin2 ϕ f ρ + sin ϕ cos ϕ f ρ ϕ +ˆϕ sin ϕ cos ϕ f ρ + sin2 ϕ f + sin ϕ cos ϕf ρ ϕ ρ + cos2 ϕ f f + ˆk ρ ϕ z ˆρ cos 2 ϕ + sin 2 ϕ f ρ + ˆϕ sin 2 ϕ + cos 2 ϕ f f + ˆk ρ ϕ z ˆρ f ρ + ˆϕ f f + ˆk ρ ϕ z f + ˆk z We can compute the divergence and the curl in the same way, writing out the Cartesian expressions and changing coordinates. The process is lengthy. Fortunately, there is a much easier way to do it that depends on the metric. 2 General formulas for gradient, divergence and curl In any coordinates, if we can write the infinitesmal distance between two points as dl 2 f 2 dx 2 + g 2 dx h 2 dx 2 3 then we can think of the distance as a double sum over the displacement vector dx, dx 2, dx 3, where with inverse dl 2 g ij g ij dx i dx j i,j f 2 g 2 h 2 f ḡ ij 2 g 2 h 2 We can even include off-diagonal terms in g ij as long as it is symmetric, i.e., if you flip the matrix across the diagonal, you get the same matrix back. In Euclidean space, there always exist coordinates where g ij is diagonal.let ê i be three orthonormal basis vectors in the x i directions. The formulas below hold for any symmetric metric g ij, not just diagonal ones. 3

4 We need one more thing, that looks a lot like the square root of the metric. If the metric is diagonal, it actually is the square root. It s called a triad or a dreibein, f h ij g h f h ij g h The general definition of h ij is that it satisfies g ij h ik h jm δ km k,m h ik h jk k Let the determinant of h ij be h. Then we can write everything we need in terms of h ij and its determinant. I have used advanced techniques to derive each of these within a few lines. The techniques are not beyond your ability, but it would take us too far afield to discuss them in detail now. Gradient: Divergence: Curl: i,j i,j ê i hij x j v h hij v j h x i v h ê i h in ε njk j h km v m i,n,j,k,m Laplacian: 2 f i,j,k h x k [ ] f h h ki hij x j Example: cylindrical or polar coordinates Let s compute these for cylindrical and spherical coordinates. The metric is the matrix g ij ρ 2 4

5 so that h ij ρ with determinant h ρ. The basis vectors are ê i ˆρ, ˆϕ, ˆk Gradient: Divergence: Curl: v h Laplacian: ρ ˆρ i,n,j,k,m j,k,m v i,j ê i hij x j ˆρ ρ + ˆϕ ρ ϕ + ˆk z i,j h hij v j h x i ρv ρ + ρ x i ρ ϕ v ϕ + z v z ê i h in ε njk j h km v m ˆρε jk j h km v m + ρ ρ ϕv z z v ϕ 2 f Example: spherical coordinates j,k,m + ˆϕ z v ρ ρ v z + ˆk i,j,k ρ ρ h x k ρ f ρ ˆϕρε 2jk j h km v m + ρ [ f h h ki hij x j j,k,m ρ ρ ρv ϕ ρ ϕv ρ ] + ρ 2 2 f ϕ f z 2 Now consider spherical coordinates, r, θ, ϕ. The metric is the matrix g ij r 2 r 2 sin 2 θ ˆkε 3jk j h km v m so that with inverse h ij r r sin θ h ij r r sin θ 5

6 and determinant h r 2 sin θ. The basis vectors are ê i ˆr, ˆθ, ˆϕ. Gradient: Divergence: Curl: v i,j r 2 sin θ r 2 r i,j ê i hij x j ˆr h r + ˆθ h 22 θ + ˆϕ h 33 ϕ ˆr r + ˆθ r θ + ˆϕ r sin θ ϕ h hij v j h x i sin θ r 2 v r + r r r 2 v r + r sin θ θ sin θv θ + r ϕ v ϕ θ sin θv θ + r sin θ ϕ v ϕ v ê i h in ε njk h km v m h x j i,n,j,k,m ˆrε r 2 jk h km v m + sin θ x ˆθrε 2jk h km v m + ˆϕr sin θε 3jk h km v m j x j x j j,k,m ˆr r 2 sin θ θ h 33v 3 ϕ h 22v 2 + ˆθr ϕ h v r h 33v 3 + ˆϕr sin θ r h 22v 2 θ h v ˆr r 2 sin θ θ r sin θv ϕ ϕ rv θ + ˆθr ϕ v r r r sin θv ϕ + ˆϕr sin θ r rv θ θ v r ˆr r sin θ θ sin θv ϕ ϕ v θ + ˆθ r sin θ ϕ v r r r rv ϕ + ˆϕ r r rv θ r θ v r Laplacian: 2 f i,j,k r 2 r h x k r 2 f r f h h ki hij x j + r 2 sin θ θ sin θ f + θ 2 f r 2 sin 2 θ ϕ 2 3 Rotations We wish to describe a rotation through an angle θ around and axis in ˆn direction. Positive θ is to correspond to a counterclockwise rotation when looking down at the tip of ˆn. Let the rotation, O, act on an arbitrary vector v to give the rotated vector v. To describe what happens, we decompose v into parts parallel and perpendicular to ˆn. The parallel part had magnitude ˆn v and direction ˆn so we define v ˆn v ˆn 6

7 The part of v perpendicular to ˆn has magnitude ˆn v v sin α where α is the angle between the two vectors. However, the vector ˆn v is perpendicular to both ˆn and v. We need a vector with this magnitude which is perpendicular to ˆn but lies in the ˆnv-plane. We can get it by taking another cross product with ˆn v. Using the BAC-CAB rule, so we can write ˆn ˆn v ˆn ˆn v v ˆn ˆn [v ˆn ˆn v] v ˆn ˆn v ˆn ˆn v The two vectors on the right are perpendicular to one another and lie in the ˆnv-plane. perpendicular component of v to be We define the v ˆn ˆn v v ˆn ˆn v Now we can easily write the rotated vector, v, because the parallel part is unchanged while the part lying in the ˆnv-plane is a simple 2-dimensional rotation. We therefore have Adding these, v v v v cos θ + ˆn v sin θ v v + v cos θ + ˆn v sin θ ˆn v ˆn + v ˆn ˆn v cos θ + ˆn v sin θ ˆn v ˆn cos θ + v cos θ + ˆn v sin θ To find the rotation matrix, O, we can look at the three Cartesian unit vectors, î, ĵ, ˆk. Since î,,, the matrix product Oî gives the first column of Oî, Oĵ gives the second column, and Oˆk the third. The first column is therefore Oî ˆn î ˆn cos θ + î cos θ + ˆn î sin θ n x î + n y ĵ + n z ˆk n x cos θ + î cos θ + n z ĵ n y ˆk sin θ Similarly, n 2 x cos θ + cos θ î + n y n x cos θ + n z sin θ ĵ + n zn x cos θ n y sin θ ˆk v v + v cos θ + ˆn v sin θ ˆn v ˆn + v ˆn ˆn v cos θ + ˆn v sin θ Oĵ ˆn ĵ ˆn cos θ + ĵ cos θ + ˆn ĵ sin θ n x n y cos θ î + n2 y cos θ ĵ + n yn z cos θ ˆk + ĵ cos θ + n x sin θˆk n z sin θî n x n y cos θ n z sin θ î + n 2 y cos θ + cos θ ĵ + n y n z cos θ + n x sin θ ˆk Oˆk ˆn k ˆn cos θ + ˆk cos θ + ˆn ˆk sin θ n z n x î + n y ĵ + n z ˆk cos θ + ˆk cos θ + n y sin θî n x sin θĵ n x n z cos θ + n y sin θ î + n yn z cos θ n x sin θ ĵ + n 2 z cos θ + cos θ ˆk 7

8 As a matrix, [O] is therefore, n 2 x cos θ + cos θ n x n y cos θ n z sin θ n x n z cos θ + n y sin θ [O] ij n y n x cos θ + n z sin θ n 2 y cos θ + cos θ n y n z cos θ n x sin θ n z n x cos θ n y sin θ n y n z cos θ + n x sin θ n 2 z cos θ + cos θ Example: Consider a rotation of 2π 3 2 about the direction ˆn 3,,. Substituting the vector components with cos 2π 3 2, sin 2π The first column given above is then Oî ˆn î ˆn cos θ + î cos θ + 3 ˆn cos θ + î cos θ + ˆn î ˆn î sin θ sin θ Oî n 2 x cos θ + cos θ î + n y n x cos θ + n z sin θ ĵ + n zn x cos θ n y sin θ ˆk + î ĵ ˆk ĵ so we recover the result we found easily from the symmetry argument. 4 Dirac delta function The Dirac delta function actually a distribution is defined as the limit of a sequence of functions, such that δ x x lim n h n x δ x x for x x f x δ x x dx f x for any test function f x By setting f x, it follows from the integral expression that δ x x dx and also that the integral of δ x x over any interval not containing the point x will give zero. It is useful to think of an analogy with the Kronecker delta, v i f x δ ij v j i f x δ x x dx 8

9 where the discrete sum in the first expression is replaced by a continuous sum over x in the second. One such limit is a sequence of Gaussians, h n x n exp [ 2 ] 2ϕ n2 x x 2 where we choose each Gaussian to have unit integral. In the limit as n, the exponential suppresses the n in the amplitude, except when x x. At this point the limit distribution diverges. We can prove the basic integral property by considering small n. f x δ x x dx lim n n 2ϕ f x exp [ 2 ] n2 x x 2 dx For any point x x, we know that the limit goes to zero, so we can restrict the range of integration to a small neighborhood of x, f x δ x x dx lim n x+ε ˆ n f x exp 2ϕ x ε [ 2 n2 x x 2 ] dx with the becoming exact in the limit. We can set ε n 2 and choose n large enough that we may approximate f x by the first terms of its Taylor series, f x f x + f x x x Since x x < n 2 the integral for large n is approximately f x δ x x dx lim n f x lim n x+ε ˆ n f x + f x x x exp 2ϕ x ε +f x lim n x+ε ˆ n exp 2ϕ x ε < f x + f x lim n [ 2 n2 x x 2 ] dx x+ε ˆ n x x exp 2ϕ x ε x+ε ˆ n 2ϕ n 2 exp x ε [ 2 n2 x x 2 ] dx [ 2 n2 x x 2 ] dx [ 2 n2 x x 2 ] dx f x + f x lim n x+ε ˆ n n 2 exp [ 2 ] 2ϕ n2 x x 2 dx f x where we use the fact that the Gaussian is normalized and in the finiteness of the second integral. We can evaluate integrals of expressions containing derivatives of a Dirac delta function as well, using integration by parts. Remembering that d df fg dx dx g + f dg dx x ε 9

10 we may write d dx fδ x x df dx δ x x + f d dx δ x x If we integrate this over all x, d dx fδ x x df dx δ x x + [fδ x x ] df dx x + f d dx δ x x f d dx δ x x Since the delta function vanishes at ± as does f x, the left side vanishes and we have f d dx δ x x df dx x In higher dimensions, we simply multiply Dirac delta functions δ 3 x x δ x x δ y y δ z z This allows us to write a complete expression for charge densities even when they are idealized. For example, an ideal point charge at the origin has charge density ρ x qδ 3 x while a uniformly charged, infinitesimally thick ring of total charge Q and radius R, lying in the xy plane and centered at the origin, has charge density in polar coordinates, since σ x Q δ z δ ρ R 2πR 2π ˆ σ x ρdρdϕdz 4. Example: spherical shell ˆ2π Q ρdρ dz dϕδ z δ ρ R 2πR Q 2πR 2π ρdρ dzδ z δ ρ R Q 2πR 2π ρdρδ ρ R Q 2πR 2πR Q Write a charge density using the Dirac delta function, δ x, and/or the unit step function { x a > Θ x a x a <

11 Example : Spherical shell, radius R, total uniform charge Q: ρ x ρ r, θ, ϕ Aδ r R Fix the constant by computing the total charge, by integrating the density over all space, ˆ Q ρd 3 x ˆ A δ r R r 2 sin θdrdθdϕ A δ r R r 2 dr ˆπ A δ r R r 2 dr 2 2π A4πR 2 ˆ2π sin θdθ dϕ so A Q 4πR 2 and we have ρ x Q δ r R 4πR2 5 Helmholtz theorem Suppose we have a vector field, v, in some region or all space, where we know the values of v on the entire boundary or at infinity, and know its divergence and its curl, v and v, everywhere. Then we may find v. To begin, separate v into a divergence-free part and a curl-free part, with a possible remainder We can do this if we can find u satisfying and v satisfying Then s must satisfy v u + w + s u v u u S w v w w S v u + w + s s v + s

12 and and the boundary condition. This gives v u + w + s s v + s s s s S v S We set about solving these three sets of equations. First, for s, we take a second curl s s 2 s 2 s Therefore, s satisfies the Laplace equation with boundary conditions: 2 s s S v S We will show in our study of electrostatics that this problem has a unique solution. Therefore, s is determined. As a special case, s vanishes if v S. Next, we seek u satisfying u v u u S This time, we may find a potential. Consider the line integral ˆ u dl C ˆ u dl C 2 u dl along any curves C, C 2 from a fixed point P to some point x. The difference of these gives a closed loop integral, and by Stokes theorem gives ˆ ˆ u dl u dl C C 2 C C 2 ˆn u d 2 x ˆx C ;P ˆx C 2;P S since the curl of u vanishes. This means that we do not need to specify a curve when we integrate to x; the value is unique and defines a function, ˆx f x u dl 2 P u dl u dl

13 and by the fundamental theorem of calculus, u f The negative sign is a convention. Rewriting the conditions for u in terms of f, the vanishing curl is now automatic and f satisfies the Poisson equation with vanishing derivative on the boundary: This system again has a unique solution. Finally, consider w: We claim there is a vector field A such that 2 f v f S w v w w S w A and show that A exists. Indeed, if one does there are many because we may set A A + h and we still have w A. Now, either A or A must satisfy, The first equation becomes A v A A S A 2 A v where the divergence of A, A is some unknown function. This means that the divergence of A is given by A A + 2 h so if we can solve 2 h A then A. This is once again the Poisson equation, and therefore has a solution, so we now have A 2 A v Letting J v, A now satisfies the Poisson equation 2 A v 2 A J with A S as boundary condition. Combining these results, we have written our original vector v as v u + w + s f + A + s 3

14 where: 2 s s S v S and 2 f v f S and 2 A v A S This is the Helmholz theorem. 4