University of Alabama Department of Physics and Astronomy. PH 125 / LeClair Spring A Short Math Guide. Cartesian (x, y) Polar (r, θ)

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1 University of Alabama Department of Physics and Astronomy PH 125 / LeClair Spring 2009 A Short Math Guide 1 Definition of coordinates Relationship between 2D cartesian (, y) and polar (r, θ) coordinates. Note that we use the convention that the cartesian unit vectors are ˆ, ŷ, and ẑ, rather than î, ĵ, and ˆk, using the substitutions ˆ î, ŷ ĵ, and ẑ ˆk. Cartesian (, y) Polar (r, θ) θ r y r 2 + y 2 θ tan 1 (y/) r cos θ y r sin θ Common 3-D coordinate systems and their relationships: cartesian (, y, z), cylindrical (R, ϕ, z), and spherical (r, θ, ϕ) systems. Note that the cartesian system is a right-handed one if your fingers point along the ais (ˆ), and your thumb along the y ais (ŷ), the z ais (ẑ) points out of the back of your hand for a right-handed system. z z z (, y, z) (R, ϕ, z) (r, θ, ϕ) θ r y ϕ R y ϕ y Figure 1: Common 3-D coordinate systems. left: Cartesian (, y, z), center: cylindrical (R, ϕ, z), right: spherical (r, θ, ϕ).

2 Table 1: Relationships between coordinates in different 3D systems. Cartesian (, y, z) Cylindrical (R, ϕ, z) Spherical (r, θ, ϕ) R 2 + y 2 R cos ϕ r sin θ cos ϕ ϕ tan 1 ( y ) y R sin ϕ y r sin θ sin ϕ z z z z z r cos θ r 2 + y 2 + z 2 r ( ) + z 2 R r sin θ θ tan 1 2 +y 2 z θ tan 1 R z ϕ ϕ ϕ tan 1 ( y ) ϕ ϕ z r cos θ 2 Relationships between unit vectors Note that these follow directly from the coordinate relationships above plus the normalization condition (i.e., unit vectors must have unit length). Table 2: Relationships between unit vectors in cartesian and polar systems. Cartesian (, y) Polar (r, θ) ˆr cos θ ˆ + sin θ ŷ ˆ sin θ ˆr + cos θ ˆθ ˆθ sin θ ˆ + cos θ ŷ ŷ cos θ ˆr + sin θ ˆθ One can also transform between unit vectors using the following matri equations: ] [ˆr ˆθ [ ] ] ] [ cos θ sin θ [ˆ [ˆ sin θ cos θ sin θ cos θ ŷ ŷ cos θ sin θ ] [ˆr ˆθ Table 3: Relationships between unit vectors in different 3D systems. ] Cartesian (, y, z) Cylindrical (R, ϕ, z) Spherical (r, θ, ϕ) ˆR R ˆ + y R ŷ ˆ cos ϕ ˆR sin ϕ ˆϕ ˆ sin θ cos ϕˆr + cos θ cos ϕ ˆθ sin ϕ ˆϕ ˆϕ y R ˆ + R ŷ ŷ sin ϕ ˆR + cos ϕ ˆϕ ŷ sin θ sin ϕˆr + cos θ sin ϕ ˆθ + cos ϕ ˆϕ ẑ ẑ ẑ ẑ ẑ cos θ ˆr sin θ ˆθ ˆr 1 r ( ˆ + y ŷ + z ẑ) ˆr 1 r R ˆR + z ẑ ˆR sin θ ˆr + cos θ ˆθ ˆθ cos θ cos ϕ ˆ + cos θ sin ϕ ŷ sin θ ẑ ˆθ 1 r z ˆR Rẑ ˆϕ ˆϕ ˆϕ sin ϕ ˆ + cos ϕ ŷ ˆϕ ˆϕ ẑ cos θ ˆr sin θ ˆθ The 3D transformations can also be written as matri equations; for eample, transforming from cartesian to spherical polar coordinates, or cartesian and cylindrical coordinates ˆ sin θ cos ϕ cos θ cos ϕ sin ϕ ˆr ŷ sin θ sin ϕ cos θ sin ϕ cos ϕ ˆθ ẑ cos θ sin ϕ 0 ˆϕ ˆ cos θ sin ϕ 0 ˆR ŷ sin ϕ cos ϕ 0 ˆϕ ẑ ẑ

3 The remaining transformations can also be written as matri equations... which we leave as an eercise to the reader! 3 Rotation of coordinate systems We will only worry about 2D cartesian (, y) systems. First, take a normal (, y) system, and pick a point P (, y). A clockwise rotation about the origin by θ results in coordinates after rotation P (, y): P θ Or, [ ] [ ] [ ] [ ] cos θ sin θ cos θ sin θ y P + sin θ cos θ y + sin θ cos θ cos θ y sin θ y sin θ + y cos θ On the other hand, if make a counterclockwise rotation of θ about the origin (equivalently, a clockwise rotation of θ), the new coordinates after rotation are P θ Or, [ ] [ ] [ ] [ ] cos θ + sin θ cos θ + sin θ y P sin θ cos θ y sin θ cos θ cos θ + y sin θ y sin θ + y cos θ 4 Line, surface, and volume elements in different coordinate systems. Cartesian d l ˆ d + ŷ dy + ẑ dz d S dy dz ˆ + d dz ŷ + d dy ẑ dv d dy dz

4 Cylindrical d l ˆR dr + ˆϕ R dϕ + ẑ dz d S R dϕ dz ˆR + dr dz ˆϕ + R dr dϕ ẑ dv R dr dϕ dz Spherical d l ˆr dr + ˆθ r sin θ dθ + ˆϕ r dϕ d S r 2 sin θ dθ dϕˆr + r sin θ dr dϕ ˆθ + r dr dθ ˆϕ dv r 2 sin θ dr dθ dϕ 5 Scalar products Say you have two vectors, a and b: a a ˆ + a y ŷ + a z ẑ b b ˆ + b y ŷ + b z ẑ The dot or scalar product of these two vectors results in a scalar, or just a number. We find this number by multiplying like components: n a b a b + a y b y + a z b z a i b i i1 where the latter sum runs over the number of dimensions in your problem (so from 1 to 3 for a normal 3-dimensional problem). The magnitude one of these vectors is simply related to the dot product: a a a a 2 + a 2 y + a 2 z This gives us a geometric interpretation of the dot prodcut: it is the length of a square of side a. We can also write the scalar product in terms of vector magnitudes, and the angle between a and b: a b a b cos θ

5 Put another way, given two vectors, the angle between them can be found readily: a b θ cos 1 a b Of course, this implies that if a and b are orthogonal (at right angles), their dot product is zero: if a b, then a b 0 Moreover, two vectors are orthogonal (perpendicular) if and only if their dot product is zero, and they have non-zero length, providing a simple way to test for orthogonality. A few other properties are tabulated below, as well as the scalar product between unit vectors in different coordinate systems. Table 4: Algebraic properties of the scalar product formula relationship a b b a commutative a ( b + c) a b + a c distributive a (r b + c) r( a b) + r( a c) bilinear (c 1 a) (c 2 b) (c 1 c 2 )( a b) multiplication by scalars if a b, then a b 0 orthogonality Table 5: Scalar products of unit vectors Cartesian Spherical Cylindrical ˆ ŷ ẑ ˆr ˆθ ˆϕ ˆR ˆϕ ẑ ˆ sin θ cos ϕ cos θ cos ϕ - sin ϕ cos ϕ - sin ϕ 0 ŷ sin θ sin ϕ cos θ sin ϕ cos ϕ sin ϕ cos ϕ 0 ẑ cos θ - sin θ Vector products The cross or vector product between two polar vectors results in a pseudovector, also known as an aial vector. i An easy way to remember how to calculate the cross product of these two vectors, i Pseudovectors act just like real vectors, ecept they gain a sign change under improper rotation. See for eample, the Wikipedia page Pseudovector. An improper rotation is an inversion followed by a normal (proper) rotation, just what we are doing when we switch between right- and left-handed coordinate systems. A proper rotation has no inversion step, just rotation.

6 c a b, is to take the determinant of the following matri: ˆ ŷ ẑ a a y a z b b y b z Or, eplicitly: ˆ ŷ ẑ c a a y a z b b y b z a y b y a z b z ˆ + a z b z a b ŷ + a b a y b y ẑ (a yb z a z b y ) ˆ+(a z b a b z ) ŷ+(a b y a y b ) ẑ One key to remember is that the unit vectors in a orthogonal coordinate system follow a cyclical permutation of yz. Break the order, and you pick up a minus sign (as that corresponds to an improper rotation, or mirror operation): ˆ ŷ ẑ ŷ ˆ ŷ ẑ ˆ ẑ ŷ ẑ ˆ ŷ ˆ ẑ The magnitude of the cross product is c a b a b sin θ where θ is the smallest angle between a and b. Geometrically, the cross product can be viewed as the area of a parallelogram with sides defined by a and b or the (signed) volume of a parallelepiped defined by the three vectors given. The pseudovector c resulting from a cross product of (polar) vectors a and b is perpendicular to the plane formed by a and b, with a direction given by the right-hand rule: a b ab sin θ ˆn where ˆn is a unit vector perpendicular to the plane containing a and b. Note of course that if a and b are collinear (i.e., the angle between them is either 0 or 180 ), the cross product is zero. We can use the cross product to define a unit normal vector to a plane formed by vectors a and b:

7 ˆn a b a b a b a b sin θ â ˆb sin θ Check this by verifying that the z ais is perpendicular to the y plane. Right-hand rule 1. Point the fingers of your right hand along the direction of a. 2. Point your thumb in the direction of b. 3. The vector c a b points out from the back of your hand. The cross product is also anticommutative, distributive over addition, and has numerous other algebraic properties listed in the table below. We also list decompositions of various higher-order products. Table 6: Algebraic properties of the vector product formula relationship a b ( b ) a anticommutative a b + c a b + ( a c) distributive over addition (r a) b a (r b) r( a b) compatible with scalar multiplication a ( b c) + b ( c a) + c ( a b) 0 not associative; obeys Jacobi identity a ( b c) b( a b) c( a b) triple vector product epansion ( a b) c c ( a b) a( b c) + b( a c) triple vector product epansion a ( b c) b ( c a) c ( a b) triple scalar product epansion a b 2 + a b 2 a 2 b 2 relation between cross and dot product if a b a c then b c iff a b a c lack of cancellation Note that the parentheses may be omitted without causing ambiguity, since the dot product cannot be evaluated first. If it were, it would leave the cross product of a vector and a scalar, which is not defined One higher order product has a particularly nice matri representation: a b c a a y a z b b y b z (a b y c z a b z c y ) + (a y b z c a y b c z ) + (a z b c y a z b y c ) c c y c z

8 7 Vector Derivatives 7.1 Gradient In cartesian coordinates, given a scalar function ϕ(, y, z): ϕ ϕ ϕ ˆ + y ŷ + ϕ z ẑ (1) 7.2 Divergence In cartesian coordinates, given a vector function F(, y, z): F F + F y y + F z z (2) 7.3 Curl In cartesian coordinates, given a vector function F(, y, z): ˆ ŷ ẑ F y z F F y F z 8 Eample: Solving systems of linear equations Say we have three equations and three unknowns, and we are left with the pesky problem of solving them. There are many ways to do this, we will illustrate two of them. Take, for eample, three equations that result from applying Kirchhoff s rules to a particular multiple loop dc circuit: I 1 I 2 I 3 0 R 1 I 1 + R 3 I 3 V 1 I 2 R 3 I 3 V 2 The first way we can proceed is by substituting the first equation into the second: V 1 R 1 I 1 + R 3 I 3 R 1 (I 2 + I 3 ) + R 3 I 3 R 1 I 2 + (R 1 + R 3 ) I 3 V 1 R 1 I 2 + (R 1 + R 3 ) I 3

9 Now our three equations look like this: I 1 I 2 I 3 0 R 1 I 2 + (R 1 + R 3 ) I 3 V 1 I 2 R 3 I 3 V 2 The last two equations now contain only I 1 and I 2, so we can solve the third equation for I 2... I 2 I 3R 3 V 2... and plug it in to the second one: V 1 R 1 I 2 + (R 1 + R 3 ) I 3 R 1 ( I3 R 3 V 2 V 1 + V 2R 1 I 3 I 3 ( R 1 + R 3 + R 1R 3 V 1 + V 2R 1 R 1 + R 3 + R 1R 3 V 1 + V 2 R 1 R 1 + R 3 + R 1 R 3 ) I 3 0 ) + (R 1 + R 3 ) I 3 Now that you know I 3, you can plug it in the epression for I 2 above. What is the second way to solve this? We can start with our original equations, but in a different order: I 1 I 2 I 3 0 I 2 R 3 I 3 V 2 R 1 I 1 + R 3 I 3 V 1 The trick we want to use is formally known as Gaussian elimination, but it just involves adding these three equations together in different ways to eliminate terms. First, take the first equation above, multiply it by R 1, and add it to the third:

10 [ R 1 I 1 + R 1 I 2 + R 1 I 3 ] 0 + R 1 I 1 + R 3 I 3 V 1 R 1 I 2 + (R 1 + R 3 ) I 3 V 1 Now take the second equation, multiply it by R 1 /, and add it to the new equation above: R 1 [ I 2 R 3 I 3 ] R 1 [ V 2 ] + R 1 I 2 + (R 1 + R 3 ) I 3 V 1 R1 R 3 + R 1 + R 3 I 3 R 1 V 2 + V 1 Now the resulting equation has only I 3 in it. Solve this for I 3, and proceed as above. Optional: There is one more way to solve this set of equations using matrices and Cramer s rule, ii if you are familiar with this technique. If you are not familiar with matrices, you can skip to the net problem - you are not required or necessarily epected to know how to do this. First, write the three equations in matri form: R 1 0 R 3 I 1 V 1 0 R 3 I 2 V I 3 0 ai V The matri a times the column vector I gives the column vector V, and we can use the determinant of the matri a with Cramer s rule to find the currents. For each current, we construct a new matri, which is the same as the matri a ecept that the the corresponding column is replaced the column vector V. Thus, for I 1, we replace column 1 in a with V, and for I 2, we replace column 2 in a with V. We find the current then by taking the new matri, calculating its determinant, and dividing that by the determinant of a. Below, we have highlighted the columns in a which have been replaced to make this more clear: ii See Cramer s rule in the Wikipedia to see how this works.

11 V 1 0 R 3 V 2 R I 1 det a I 2 R 1 V 1 R 3 0 V 2 R det a I 3 R 1 0 V 1 0 V det a Now we need to calculate the determinant of each new matri, and divide that by the determinant of a. iii First, the determinant of a. det a R 1 R 1 R R 3 (R 1 + R 3 + R 1 R 3 ) We can now find the currents readily from the determinants of the modified matrices above and that of a we just found: I 1 V 1 V 1 R V 2 R 3 0 (R 1 + R 3 + R 1 R 3 ) V 1 ( + R 3 ) V 2 R 3 R 1 + R 3 + R 1 R 3 I 2 R 1V 2 0 V 1 R R 3 V 2 (R 1 + R 3 + R 1 R 3 ) R 3V 1 V 2 (R 1 + R 3 ) R 1 + R 3 + R 1 R 3 I 3 0 R 1V V 1 (R 1 + R 3 + R 1 R 3 ) R 1 V 2 + V 1 R 1 + R 3 + R 1 R 3 These are the same results you would get by continuing on with either of the two previous methods. Both numerically and symbolically, we can see from the above that I 1 I 2 +I 3 : I 2 + I 3 R 3V 1 V 2 (R 1 + R 3 ) + R 1 V 2 + V 1 R 1 + R 3 + R 1 R 3 V 1 ( + ) + V 2 (R 1 R 1 R 3 ) R 1 + R 3 + R 1 R 3 V 1 ( + ) V 2 R 3 R 1 + R 3 + R 1 R 3 I 1 iii Again, the Wikipedia entry for determinant is quite instructive.

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