This section is divided into a number of subsections, links to which are:

Vector products

Tensor products

Triple products

Wedge products

WeRotors

Although physical phase space can extend to thousands of dimensions, our minds are incapable of thinking even in five dimensions—we have no ability to visualize such things. It is impossible to do physics in high dimensions without having the right tools and symbols with which to work. Cross product is an important tool applied only in three dimensional vector space. In Euclidean space, a cross product is related to areas of parallelograms and volumes of parallelepipeds. Thus, it opens the door of high dimensional math and physics, including tensors and wedge products.

In 1881, Josiah Willard Gibbs (1839--1903), and independently Oliver Heaviside (1850--1925), introduced the notation for both the dot product and the cross product using a period (a • b) and an "×" (a × b), respectively, to denote them.

In 1877, to emphasize the fact that the result of a dot product is a scalar while the result of a cross product is a vector, William Kingdon Clifford (1845--1879) coined the alternative names scalar product and vector product for the two operations. These alternative names are still widely used in the literature.

 

Vector or Cross product

It is well-known that a vector space can be equipped with a product operation (besides vector addition) only for dimensions 1 and 2. The cross product is a successful attempt to implement the product in a three-dimensional vector space, but without the commutative property of multiplication. On the other hand, the outer product assigns a matrix to two vectors of arbitrary size.

Cross-product of two vectors together with these vectors form so called right-handed system.

The special basis vectors i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) form a right-handed system in ℝ³. Thus, if the thumb of your right hand points along the x-axis and your index finger points along the y-axis, your middle finger should point along the z-axis. When all three vectors lie in a plane, then we say that the vectors are coplanar. In this case, the system is neither right-handed nor left-handed.

Besides algebraic definition of the cross product, it is possible to give a geometrical one.

We note that the length of the cross product, ∥ a × b ∥ = ∥ a ∥ ∥ b ∥ sin(θ) , given by the formula \eqref{EqCross.2}, is the area of the parallelogram determined by a and b.

We plot with Mathematica cross product of two vectors using DynamicModule command so you can see the value of cross product, which is the area of parallelogram (in blue for the acute angle, pink for the obtuse angle) formed by two given vectors. You can manipulate the plot in Mathematica by dragging the arrowheads.

Example 1: Cross product of two vectors

Despite that the cross product is identified uniquely by two entries, there is something very peculiar about the vector u × v. If u and v are orthogonal unit vectors, then the vectors u, v, u × v form a right-handed coordinate system. But if R ∈ ℝ^3×3 is the linear transformation (whch is 3×3 matrix with real entries) that mirrors vectors in the u, v plane, then { R u, R v, R u×v } = { u, v, −u×v } forms a left-handed coordinate system. In general, the cross product obeys the following identity under matrix transformation:

where M is a 3×3 matrix and M^−T is the transpose of the inverse matrix.

Thus, cross product u×v is not really transformed as a vector. This anomaly should alert us to the fact that cross product is not really a true vector. In fact, cross product is transformed more like a tensor than a vector.

ar1 = Graphics[{Blue, Arrowheads[0.05], Thickness[0.01], Arrow[{{0, 0}, {1, 0.2}}]}]; ar2 = Graphics[{Blue, Arrowheads[0.05], Thickness[0.01], Arrow[{{0, 0}, {0.5, 1.5}}]}]; ar1m = Graphics[{Red, Arrowheads[0.05], Thickness[0.01], Arrow[{{0, 0}, {-1, -0.2}}]}]; ar2m = Graphics[{Red, Arrowheads[0.05], Thickness[0.01], Arrow[{{0, 0}, {-0.5, -1.5}}]}]; ar = Graphics[{Blue, Arrowheads[0.07], Thickness[0.02], Arrow[{{0, 0}, {-0.3, 1.2}}]}]; arr = Graphics[{Red, Arrowheads[0.07], Thickness[0.02], Arrow[{{0, 0}, {-0.3, 1.2}}]}]; txt1 = Graphics[{Black, Text[Style["a", 20, Bold], {1, 0.15}]}]; txt2 = Graphics[{Black, Text[Style["b", 20, Bold], {0.4, 1.5}]}]; txt1m = Graphics[{Red, Text[Style["(-a)", 20, Bold], {-0.9, -0.3}]}]; txt2m = Graphics[{Red, Text[Style["(-b)", 20, Bold], {-0.3, -1.4}]}]; tm = Graphics[{Red, Text[Style["(-a)\[Cross](-b)", 20, Bold], {-0.2, 1.3}]}]; t = Graphics[{Blue, Text[Style["a\[Cross]b", 20, Bold], {-0.45, 1.1}]}]; Show[ar1, ar2, ar1m, ar2m, txt1, txt2, txt1m, txt2m, ar, arr, t, tm]

Note when BOTH vectors change sign, the cross product is unchanged. However, this is not true when only one vector changes sign in cross product.

Example 2: Cross product under transformations

We are interested in more general formula than Eq.(3). Suppose that a linear transformation is defined by a 3×3 matrix

where p₁, p₂, and p₃ are row vectors in matrix P. This matrix (as any 3×3 matrix) defines a linear transformation in ℝ^3×1 by multiplication from the left. Then a natural question arises whether it is possible to express a transformed cross product in terms of transformed cross product v × u, similar to Eq.(3). For example, if P is a diagonal matrix, then This expression for diagonal matrix tells us that there is no simple formula for vector product (Pv) × u in terms of v × u because vector product is a bilinear transformation ℝ³ × ℝ³ ↦ ℝ³. On the other hand, if we fix vector u, then cross product becomes a linear transformation A square matrix A = [T] of this linear transformation depends on vector u: This singular skew-symmetric matrix has one real eigenvalue λ₁ = 0 and two pure imaginary eigenvalues (j² = −1) The null space of T is the line through the origin spanned by vector u and its image is the plane through the origin orthogonal to u.

One of the main motivations to use cross product comes from classical mechanics: A torque τ about a point due to a force F acting at another point at distance r is expressed through cross product: τ = r × F.

The cross product is convenient to define with the Levi-Civita symbol that was invented in the late 1800's by the Italian mathematician Tullio Levi-Civita (1873--1941), a student of Gregorio Ricci:

The Levi-Civita symbol ε_ijk is a tensor of rank three and is anti-symmetric on each pair of indexes. The determinant of a matrix A with elements 𝑎_ij can be written in term of ε_ijk upon dropping sign of summation using the Einstein summation rule. The Levi-Civita symbol possesses the following properties

• The Levi-Civita tensor has 3×3×3 = 27 components;
• 3 × (6+1) = 21 components are equal to 0;
• 3 components are equal to 1 = ε₁₂₃ = ε₂₃₁ = ε₃₂₁;
• 3 components are equal to −1 = ε₃₂₁ = ε₂₁₃ = ε₁₃₂.

Cross product in index notation using Levi-Civita symbol is Given two space vectors, a and b, we can find a third space vector c, called the cross product of a and b, and denoted by c = a × b. The magnitude of c is defined by |c| = |a| |b| sin(θ), where θ is the angle between a and b.

The direction of c is given by the right-hand rule: If a is turned to b (note the order in which a and b appear here) through the angle between a and b, a (right-handed) screw that is perpendicular to a and b will advance in the direction of a × b. This definition implies that

This property is described by saying that the cross product is antisymmetric. The definition also implies that where • denotes dot product. That is, a × b is perpendicular to both a and b. The vector product has the following properties: Using these properties, we can write the vector product of two vectors in terms of their components. We are interested in a more general result valid in other coordinate systems as well. So, rather than using x, y, and z as subscripts for unit vectors, we use the numbers 1, 2, and 3. In that case, our results can also be used for spherical and cylindrical coordinates which we shall discuss shortly. Using the antisymmetry property of cross product, we have Also, if we assume that \( \hat{\bf e}_1 , \ \hat{\bf e}_2 , \ \hat{\bf e}_3 \) form a so-called right-handed set, i.e., if then we obtain Proof of Theorem 3: Example 3: Mathematica verifies some cross product properties The property of cross product, a × (b × c) = b(c • a) − c(a • b), discussed in the previous example, can be used to prove the identity involving the gradient operator ∇: Here f = f(x) = (f₁, f₂, f₃) is a 3-dimensional vector function depending on three variables, x ∈ ℝ³, and \( \displaystyle \nabla = \left( \frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z} \right) \) is the gradient operator. It is common to use a shortcut: ∇ = ∂_xi + ∂_yj + ∂_zk = (∂_x, ∂_y, ∂_z). Then Proof of gradient identity:

Actually, there does not exist a cross product vector in space with more than 3 dimensions. The fact that the cross product of 3 dimensions vectors gives an object which also has 3 dimensions is just pure coincidence.

Proof of Theorem 4:

ill2 = {Opacity[1, Black], PointSize[Large], Point[p], Arrowheads[Medium], Thick, Black, Arrow[{p, u}], Red, Arrow[{p, v}], Green, Arrow[{p, w}]}; paraP1 = Labeled[Graphics3D[{Opacity[0.2], Parallelepiped[p, {u, v, w}], ill2}, Axes -> False, Boxed -> False, Epilog -> {Text["u", {.15, .22}], Red, Text["v", {.3, .37}], Text["w", {.1, .5}]}], "Parallelepiped", LabelStyle -> FontFamily -> "Times"]

Since the base of the parallelepiped is the parallelogram determined by the vectors u and v, we find is volume, which cross product u × v, as the product of the area and the height, which is projection of ∥ w ∥ on the perpendicular line:

p = {0, 0, 0}; {u, v, w} = {{1, 0, 0}, {1, 1, 0}, {0, 1, 1}}; \[Theta] = VectorAngle[w, Cross[u, v]]; arc = ParametricPlot3D[ 1/4 (1/Sqrt[2] (Cos[t]) w + (Sin[t]) Cross[u, v]), {t, 0, Pi/2}, PlotStyle -> Directive[Thick, Black], Boxed -> False, Axes -> False]; ill3 = {Opacity[1, Black], PointSize[Large], Point[p], Arrowheads[Medium], Thick, Black, Arrow[{p, u}], Red, Arrow[{p, v}], Green, Arrow[{p, w}], Blue, Arrow[{p, Cross[u, v]}]}; paraP2 = Labeled[Show[ Graphics3D[{Opacity[.156, Hue[.148/3]], Parallelepiped[p, {u, v, w}], ill3}, Axes -> False, Boxed -> False, Epilog -> {Text["\[Theta]", {.070, .35}], Text["u", {.15, .22}], Red, Text["v", {.3, .37}], Text["w", {.18, .7}], Blue, Text["u\[Cross]v", {.07, .5}]}], arc, BoxRatios -> Automatic], "Figure 9: Parallelepiped", LabelStyle -> FontFamily -> "Times"]

Therefore, we get the volume of the parallelepiped as

where middle dot stands for multiplication of real numbers and • denoted the dot product. Also This expression is known as the box product.

Example 4: Area of parallelogram
Example 5: Volume of parallelepiped Example 6: right-handed system vs left-handed Proof of Theorem 6:

Coordinates are “functions” that specify points of a space. The smallest number of these functions necessary to specify a point P is called the dimension of that space. There are three widely used orthogonal coordinate systems in a three-dimensional vector space ℝ³: Cartesian (x(P), y(P), z(P)), cylindrical (r(P), θ(P), z(P)), and spherical (ρ(P), θ(P), φ(P)). The latter φ(P) is called the azimuth or the azimuthal angle of P, while θ(P) is called its polar angle.

 

A cylindrical coordinate system in ℝ³ is a marriage of polar coordinates on the plane ℝ² and regular applicate direction from the Cartesian one. A cylindrical coordinates (r ≥ 0, 0 ≤ θ < 2π, −∞ < z < ∞) of a point in ℝ³ are defined by the equations:

ii = Graphics[{Arrowheads[0.07], Thickness[0.007], Arrow[{{0.4, 0.4}, {-0.5, -0.5}}]}]; jj = Graphics[{Arrowheads[0.07], Thickness[0.007], Arrow[{{-0.6, 0}, {1, 0}}]}]; kk = Graphics[{Arrowheads[0.07], Thickness[0.007], Arrow[{{0, -0.1}, {0, 1}}]}]; ti = Graphics[Text[Style["x", Bold, FontSize -> 22], {-0.47, -0.3}]]; tj = Graphics[Text[Style["y", Bold, FontSize -> 22], {0.9, 0.12}]]; tk = Graphics[ Text[Style["z", Bold, FontSize -> 22], {0.1, 0.9}]]; circle = Graphics[{Blue, Thick, Circle[{0, 0}, {0.5, 0.25}]}]; circle2 = Graphics[{Blue, Thick, Circle[{0, 0.8}, {0.5, 0.25}]}]; line1 = Graphics[{Blue, Thick, Line[{{-0.22, -0.22}, {-0.22, 0.58}}]}]; line2 = Graphics[{Blue, Thick, Line[{{0.22, -0.22}, {0.22, 0.58}}]}]; line3 = Graphics[{Blue, Thick, Dashed, Line[{{0.22, -0.22}, {0, 0}}]}]; line4 = Graphics[{Blue, Thick, Dashed, Line[{{0.22, 0.58}, {0, 0.8}}]}]; tr = Graphics[ Text[Style["r", Bold, FontSize -> 22], {0.16, -0.09}]]; tt = Graphics[ Text[Style["\[Theta]", Bold, FontSize -> 22], {0., -0.18}]]; tz = Graphics[ Text[Style["(r,\[Theta],z)", Bold, FontSize -> 22], {0.4, 0.58}]]; disk = Graphics[{Purple, Disk[{0.22, 0.58}, 0.025]}]; ar = Graphics[{Blue,Arrowheads[0.07], Thickness[0.01], Arrow[{{0, 0}, {0.21, 0.56}}]}]; Labeled[ arr = Graphics[{Arrowheads[0.05], Thickness[0.01], Arrow[{{0.15, -0.238}, {0.2, -0.227}}]}]; Labeled[Show[ii, jj, kk, ti, tj, tk, circle, circle2, line1, line2, line3, line4, tr, tt, tz, disk, ar,arr], "Figure10: Cylindrical coordinates"]

Variable r is the radius of the cylinder passing through point P(r, θ, z) or the radial distance distance from the applicate (z-axis); θ is called azimuthal angle and it is measured from the abscissa (x-axis) in the xy-plane, and z is the same as in the Cartesian system. A vector v (with endpoint P) in cylindrical coordinates can be written as where \( \displaystyle \hat{\bf r}, \ \hat{\theta}, \ \hat{\bf z} \) are nit vectors in r, θ, and z directions, respectively. They are pointed in direction of increasing of corresponding coordinates. Note that the radial unit vector \( \displaystyle \hat{\bf r} (\theta ) \) depends on angle θ.

The unit vectors in the cylindrical coordinate system are functions of position. It is convenient to express them in terms of unit vectors of the rectangular coordinate system that are not themselves functions of position.

or Here \( \displaystyle \hat{\bf x} = {\bf i}, \ \hat{\bf y} = {\bf j}, \ \hat{\bf z} = {\bf k} \) are unit vectors in Cartesian coordinates. The inverse formulas are The position vector r of a point in terms of unit vectors is

Cross products of the utit vectors in cylindrical coordinates show that they are orthogonal

Cross product of two vectors written in cylindrical coordinates is expressed in Cartesian coordinates as In cylindrical coordinates, we have where \( \displaystyle \hat{\bf r}_1 = \hat{\bf r} (\theta_1 ) \) is the unit radial vector for point (r₁ , θ₁) and \( \displaystyle \hat{\bf r}_2 = \hat{\bf r} (\theta_2 ) \) is the unit radial vector towards point (r₂ , θ₂). These radial vectors are applied at two different points, so their cross product has no natural (simple) way to be specified.

The gradient operator in cylindrical coordinates is given by

Example 7: Cross product in cylindrical coordinates

 

A spherical coordinate system is a coordinate system for three-dimensional space ℝ³, where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin; its zenith angle (or inclination angle or polar angle) measured from an applicat direction; and the azimuthal angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the fixed axis, measured from another fixed reference direction on that plane. The spherical coordinates (ρ, θ, ϕ) are related to the Cartesian coordinates (x, y, z) by where the radial, azimuth, and zenith angle coordinates are taken as ρ, θ, and ϕ, respectively. As usual, the location of a point (x, y, z) is specified by the distance ρ of the point from the origin, the angle ϕ between the position vector and the z-axis, the polar angle measured down from the north pole, and the azimuthal angle θ from the x-axis to the projection of the position vector onto the xy plane, analogous to longitude in earth measuring coordinates: While zenith angle is take from the closed interval 0 ≤ ϕ ≤ π, the azimuth angle belongs to semiclosed interval 0 ≤ θ < 2π.

		az = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, -0.2}, {0, 1.2}}]}]; ax = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{0.4, 0}, {-0.65, -0.65}}]}]; ay = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.4, 0}, {1.2, 0}}]}]; line = Graphics[{Blue, Thickness[0.01], Line[{{0, 0}, {0.6, 0.86}}]}]; line1 = Graphics[{Black, Dashed, Line[{{0.6, 0.86}, {0.6, -0.6}}]}]; line2 = Graphics[{Black, Line[{{0, 0}, {0.6, -0.6}}]}]; circle1 = Graphics[{Red, Thick, Circle[{0, 0}, 0.8, {Pi/2, 0.97}]}]; circle2 = Graphics[{Red, Thick, Circle[{0, 0}, 0.5, {-Pi/4, -3*Pi/4}]}]; ar1 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0.36, 0.72}, {0.42, 0.67}}]}]; ar2 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0.3, -0.395}, {0.35, -0.355}}]}]; disk = Graphics[Disk[{0.6, 0.86}, 0.03]]; tx = Graphics[{Black, Text[Style["x", 20], {-0.6, -0.45}]}]; ty = Graphics[{Black, Text[Style["y", 20], {1.1, 0.15}]}]; tz = Graphics[{Black, Text[Style["z", 20], {0.15, 1.1}]}]; tf = Graphics[{Black, Text[Style[\[Phi], 20], {0.18, 0.6}]}]; tt = Graphics[{Black, Text[Style[\[Theta], 20], {0, -0.4}]}]; tp = Graphics[{Black, Text[Style["(x,y,z)", 20], {0.6, 1.0}]}]; tr = Graphics[{Black, Text[Style[\[Rho], 20], {0.5, 0.57}]}]; Labeled[Show[ax, ay, az, line, line1, line2, circle1, circle2, ar1, ar2, disk, tx, ty, tz, tf, tt, tp, tr], "Spherical Coordinates"]
Spherical coordinates.		Mathematica code

Unit vectors in spherical coordinates are

They can be expressed through Cartesian unit vectors as follows: and A position vector in spherical coordinates is described in terms of the radial parameters ρ, which depends on the zenith (or polar) angle ϕ and the azimuthal angle θ as follows:

The cross product of unit vectors in spherical coordinates is given by the rule (which shows that they are orthogonal):

However, these formulas are of little help for determinantion of cross product in spherical coordinates because unit vectors depend on location of the given two vectors. Therefore, the rules above are not directly applicable for cross product.

Suppose that two vectors are given in Cartesian and spherical coordinates

The radial distance ρ_ab of their cross product a × b is known to be where α is the angle between these two given vectors. Its value can be determined from their dot product: Using expressions of Cartesian components via spherical coordinates, we find With the aid of trigonometric identity cosα cosβ + sinα sinβ = cos(α − β), we slightly simplify the expression which resembles the formula This allows us determine the components of cross product: where ε takes care of signs of cosines of angles ϕ_a and ϕ_b.

The gradient in spherical coordinates is

Example 8: Cross product spherical coordinates

Mathematica has three multiplication commands for vectors: the dot (or inner) and outer products (for arbitrary vectors), and the cross product (for three dimensional vectors).

The cross product can be done on two vectors. It is important to note that the cross product is an operation that is only functional in three dimensions. The operation can be computed using the Cross[vector 1, vector 2] operation or by generating a cross product operator between two vectors by pressing [Esc] cross [Esc]. ([Esc] refers to the escape button)

Applications in Physics

The simplest physical model of a vector product is the moment M of a force F about a point 0 (the origin), defined as where r is the vector joining the origin to the initial point of F. The positive direction of the vector M depends on the choice of the coordinate system, like the direction of the angular velocity vector due to the action of the moment M.

An electric charge e in an electric field E experiences a force

while an electric charge e moving with velocity v in a magnetic field H experiences a force where c is the velocity of light.9 Hence, the total force experienced by an electric charge e moving with velocity v in an electromagnetic field with electric field E and magnetic field H is given by If the charge moves a distance dr in time dt, the work done by the electromagnetic field is This work goes into changing the kinetic energy U of the moving charge. The magnetic field does no work because the force F_m due to the magnetic field is perpendicular to v at every instant. Hence the work done by the electromagnetic field is due to the electric field alone and is done at the rate The magnetic field changes the direction of the charge's velocity, but not the magnitude of the velocity. Subsection: Exercises

 

References