# Vectors

## The basic ideaEdit

A vector is a mathematical concept that has both magnitude and direction. Detailed explanation of vectors may be found at the Wikibooks module Linear Algebra/Vectors in Space. In physics, vectors are used to describe things happening in space by giving a series of quantities which relate to the problem's coordinate system.

A vector is often expressed as a series of numbers. For example, in the two-dimensional space of real numbers, the notation (1, 1) represents a vector that is pointed 45 degrees from the x-axis towards the y-axis with a magnitude of ${\displaystyle {\sqrt {2}}}$.

Commonly in physics, we use position vectors to describe where something is in the space we are considering, or how its position is changing at that moment in time. Position vectors are written as summations of scalars multiplied by unit vectors. For example:

${\displaystyle a{\hat {i}}+b{\hat {j}}+c{\hat {k}}}$

where a, b and c are scalars and ${\displaystyle {\hat {i}},{\hat {j}}}$ and ${\displaystyle {\hat {k}}}$ are unit vectors of the Cartesian (René Descartes) coordinate system. A unit vector is a special vector which has magnitude 1 and points along one of the coordinate frame's axes. This is better illustrated by a diagram.

A vector itself is typically indicated by either an arrow: ${\displaystyle {\vec {v}}}$, or just by boldface type: v, so the vector above as a complete equation would be denoted as:

${\displaystyle {\vec {v}}=a{\hat {i}}+b{\hat {j}}+c{\hat {k}}}$

The magnitude of a vector is computed by ${\displaystyle |{\vec {v}}|={\sqrt {\sum _{i}(x_{i}^{2})}}}$. For example, in two-dimensional space, this equation reduces to:

${\displaystyle |{\vec {v}}|={\sqrt {x^{2}+y^{2}}}}$.

For three-dimensional space, this equation becomes:

${\displaystyle |{\vec {v}}|={\sqrt {x^{2}+y^{2}+z^{2}}}}$.

# ExercisesEdit

Find the magnitude of the following vectors. Answers below.

${\displaystyle {\vec {v}}=(4,3)}$

${\displaystyle |{\vec {v}}|={\sqrt {4^{2}+3^{2}}}=5}$

${\displaystyle {\vec {v}}=(5,3)}$

${\displaystyle |{\vec {v}}|={\sqrt {5^{2}+3^{2}}}={\sqrt {34}}}$

${\displaystyle {\vec {v}}=(1,0)}$

${\displaystyle |{\vec {v}}|={\sqrt {1^{2}+0^{2}}}=1}$

${\displaystyle {\vec {v}}=(4,4)}$

${\displaystyle |{\vec {v}}|={\sqrt {4^{2}+4^{2}}}={\sqrt {32}}}$

${\displaystyle {\vec {v}}=(5,0,0)}$

${\displaystyle |{\vec {v}}|={\sqrt {5^{2}+0^{2}+0^{2}}}=5}$

## Using vectors in physicsEdit

Many problems, particularly in mechanics, involve the use of two- or three-dimensional space to describe where objects are and what they are doing. Vectors can be used to condense this information into a precise and easily understandable form that is easy to manipulate with mathematics.

Position - or where something is, can be shown using a position vector. Position vectors measure how far something is from the origin of the reference frame and in what direction, and are usually, though not always, given the symbol ${\displaystyle {\vec {r}}}$. It is usually good practice to use ${\displaystyle {\vec {r}}}$ for position vectors when describing your solution to a problem as most physicists use this notation.

Velocity is defined as the rate of change of position with respect to time. You may be used to writing velocity, v, as a scalar because it was assumed in your solution that v referred to speed in the direction of travel. However, if we take the strict definition and apply it to the position vector - which we have already established is the proper way of representing position - we get:

${\displaystyle {\frac {d{\vec {r}}}{dt}}={\frac {da}{dt}}{\hat {i}}+{\frac {db}{dt}}{\hat {j}}+{\frac {dc}{dt}}{\hat {k}}}$

However, we note that the unit vectors are merely notation rather than terms themselves and are in fact not differentated, only the scalars which represent the vector's components in each direction differentiate.

Assuming that each component is not a constant and thus has a non-zero derivative, we get:

${\displaystyle {\vec {v}}=a'{\hat {i}}+b'{\hat {j}}+c'{\hat {k}}}$ where a', b' and c' are simply the first derivatives with respect to time of each original position vector component.

Here it is clear that velocity is also a vector. In the real world this means that each component of the velocity vector indicates how quickly each component of the position vector is changing - that is, how fast the object is moving in each direction.

# Vectors in MechanicsEdit

Vector notation is ubiquitous in the modern literature on solid mechanics, fluid mechanics, biomechanics, nonlinear finite elements and a host of other subjects in mechanics. A student has to be familiar with the notation in order to be able to read the literature. In this section we introduce the notation that is used, common operations in vector algebra, and some ideas from vector calculus.

## VectorsEdit

A vector is an object that has certain properties. What are these properties? We usually say that these properties are:

• a vector has a magnitude (or length)
• a vector has a direction.

To make the definition of the vector object more precise we may also say that vectors are objects that satisfy the properties of a vector space.

The standard notation for a vector is lower case bold type (for example ${\displaystyle \mathbf {a} \,}$).

In Figure 1(a) you can see a vector ${\displaystyle \mathbf {a} }$ in red. This vector can be represented in component form with respect to the basis (${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2}\,}$) as

${\displaystyle \mathbf {a} =a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}\,}$

where ${\displaystyle \mathbf {e} _{1}\,}$ and ${\displaystyle \mathbf {e} _{2}\,}$ are orthonormal unit vectors. Orthonormal means they are at right angles to each other (orthogonal) and are unit vectors. Recall that unit vectors are vectors of length 1. These vectors are also called basis vectors.

You could also represent the same vector ${\displaystyle \mathbf {a} \,}$ in terms of another set of basis vectors (${\displaystyle \mathbf {g} _{1},\mathbf {g} _{2}\,}$) as shown in Figure 1(b). In that case, the components of the vector are ${\displaystyle (b_{1},b_{2})\,}$ and we can write

${\displaystyle \mathbf {a} =b_{1}\mathbf {g} _{1}+b_{2}\mathbf {g} _{2}\,~.}$

Note that the basis vectors ${\displaystyle \mathbf {g} _{1}\,}$ and ${\displaystyle \mathbf {g} _{2}\,}$ do not necessarily have to be unit vectors. All we need is that they be linearly independent, that is, it should not be possible for us to represent one solely in terms of the others.

In three dimensions, using an orthonormal basis, we can write the vector ${\displaystyle \mathbf {a} \,}$ as

${\displaystyle \mathbf {a} =a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+a_{2}\mathbf {e} _{3}\,}$

where ${\displaystyle \mathbf {e} _{3}\,}$ is perpendicular to both ${\displaystyle \mathbf {e} _{1}\,}$ and ${\displaystyle \mathbf {e} _{2}\,}$. This is the usual basis in which we express arbitrary vectors.

## Vector Algebra OperationsEdit

Some vector operations are shown in Figure 2.

If ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$ are vectors, then the sum ${\displaystyle \mathbf {c} =\mathbf {a} +\mathbf {b} \,}$ is also a vector (see Figure 2(a)).

The two vectors can also be subtracted from one another to give another vector ${\displaystyle \mathbf {d} =\mathbf {a} -\mathbf {b} \,}$.

### Multiplication by a scalarEdit

Multiplication of a vector ${\displaystyle \mathbf {b} \,}$ by a scalar ${\displaystyle \lambda \,}$ has the effect of stretching or shrinking the vector (see Figure 2(b)).

You can form a unit vector ${\displaystyle {\hat {\mathbf {b} }}\,}$ that is parallel to ${\displaystyle \mathbf {b} \,}$ by dividing by the length of the vector ${\displaystyle |\mathbf {b} |\,}$. Thus,

${\displaystyle {\hat {\mathbf {b} }}={\frac {\mathbf {b} }{|\mathbf {b} |}}~.}$

### Scalar product of two vectorsEdit

The scalar product or inner product or dot product of two vectors is defined as

${\displaystyle \mathbf {a} \cdot \mathbf {b} =|\mathbf {a} ||\mathbf {b} |\cos(\theta )}$

where ${\displaystyle \theta \,}$ is the angle between the two vectors (see Figure 2(b)).

If ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$ are perpendicular to each other, ${\displaystyle \theta =\pi /2\,}$ and ${\displaystyle \cos(\theta )=0\,}$. Therefore, ${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }=0}$.

The dot product therefore has the geometric interpretation as the length of the projection of ${\displaystyle \mathbf {a} \,}$ onto the unit vector ${\displaystyle {\hat {\mathbf {b} }}\,}$ when the two vectors are placed so that they start from the same point (tail-to-tail).

The scalar product leads to a scalar quantity and can also be written in component form (with respect to a given basis) as

${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=\sum _{i=1..3}a_{i}b_{i}~.}$

If the vector is ${\displaystyle n}$ dimensional, the dot product is written as

${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }=\sum _{i=1..n}a_{i}b_{i}~.}$

Using the Einstein summation convention, we can also write the scalar product as

${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }=a_{i}b_{i}~.}$

Also notice that the following also hold for the scalar product

1. ${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }={\mathbf {b} }\cdot {\mathbf {a} }}$ (commutative law).
2. ${\displaystyle {\mathbf {a} }\cdot {(\mathbf {b} +\mathbf {c} )}={\mathbf {a} }\cdot {\mathbf {b} }+{\mathbf {a} }\cdot {\mathbf {c} }}$ (distributive law).

### Vector product of two vectorsEdit

The vector product (or cross product) of two vectors ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$ is another vector ${\displaystyle \mathbf {c} \,}$ defined as

${\displaystyle \mathbf {c} ={\mathbf {a} }\times {\mathbf {b} }=|\mathbf {a} ||\mathbf {b} |\sin(\theta ){\hat {\mathbf {c} }}}$

where ${\displaystyle \theta \,}$ is the angle between ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$, and ${\displaystyle {\hat {\mathbf {c} }}\,}$ is a unit vector perpendicular to the plane containing ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$ in the right-handed sense (see Figure 3 for a geometric interpretation)

In terms of the orthonormal basis ${\displaystyle (\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})\,}$, the cross product can be written in the form of a determinant

${\displaystyle {\mathbf {a} }\times {\mathbf {b} }={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{vmatrix}}~.}$

In index notation, the cross product can be written as

${\displaystyle {\mathbf {a} }\times {\mathbf {b} }\equiv e_{ijk}a_{j}b_{k}~.}$

where ${\displaystyle e_{ijk}}$ is the Levi-Civita symbol (also called the permutation symbol, alternating tensor).

## Identities from Vector AlgebraEdit

Some useful vector identities are given below.

1. ${\displaystyle {\mathbf {a} }\times {\mathbf {b} }=-{\mathbf {b} }\times {\mathbf {a} }}$.
2. ${\displaystyle {\mathbf {a} }\times {\mathbf {b} +\mathbf {c} }={\mathbf {a} }\times {\mathbf {b} }+{\mathbf {a} }\times {\mathbf {c} }}$.
3. ${\displaystyle {\mathbf {a} }\times {({\mathbf {b} }\times {\mathbf {c} })}=\mathbf {b} ({\mathbf {a} }\cdot {\mathbf {c} })-\mathbf {c} ({\mathbf {a} }\cdot {\mathbf {b} })}$ ~.
4. ${\displaystyle {({\mathbf {a} }\times {\mathbf {b} })}\times {\mathbf {c} }=\mathbf {b} ({\mathbf {a} }\cdot {\mathbf {c} })-\mathbf {a} ({\mathbf {b} }\cdot {\mathbf {c} })}$ ~.
5. ${\displaystyle {\mathbf {a} }\times {\mathbf {a} }=\mathbf {0} }$~.
6. ${\displaystyle {\mathbf {a} }\cdot {({\mathbf {a} }\times {\mathbf {b} })}={\mathbf {b} }\cdot {({\mathbf {a} }\times {\mathbf {b} })}=\mathbf {0} }$~.
7. ${\displaystyle {({\mathbf {a} }\times {\mathbf {b} })}\cdot {\mathbf {c} }={\mathbf {a} }\cdot {({\mathbf {b} }\times {\mathbf {c} })}}$~.

## Vector CalculusEdit

So far we have dealt with constant vectors. It also helps if the vectors are allowed to vary in space. Then we can define derivatives and integrals and deal with vector fields. Some basic ideas of vector calculus are discussed below.

## Derivative of a vector valued functionEdit

Let ${\displaystyle \mathbf {a} (x)\,}$ be a vector function that can be represented as

${\displaystyle \mathbf {a} (x)=a_{1}(x)\mathbf {e} _{1}+a_{2}(x)\mathbf {e} _{2}+a_{3}(x)\mathbf {e} _{3}\,}$

where ${\displaystyle x\,}$ is a scalar.

Then the derivative of ${\displaystyle \mathbf {a} (x)\,}$ with respect to ${\displaystyle x\,}$ is

${\displaystyle {\cfrac {d\mathbf {a} (x)}{dx}}=\lim _{\Delta x\rightarrow 0}{\cfrac {\mathbf {a} (x+\Delta x)-\mathbf {a} (x)}{\Delta x}}={\cfrac {da_{1}(x)}{dx}}\mathbf {e} _{1}+{\cfrac {da_{2}(x)}{dx}}\mathbf {e} _{2}+{\cfrac {da_{3}(x)}{dx}}\mathbf {e} _{3}~.}$

Note: In the above equation, the unit vectors ${\displaystyle \mathbf {e} _{i}}$ (i=1,2,3) are assumed constant.
If ${\displaystyle \mathbf {a} (x)\,}$ and ${\displaystyle \mathbf {b} (x)\,}$ are two vector functions, then from the chain rule we get

{\displaystyle {\begin{aligned}{\cfrac {d({\mathbf {a} }\cdot {\mathbf {b} })}{x}}&={\mathbf {a} }\cdot {\cfrac {d\mathbf {b} }{dx}}+{\cfrac {d\mathbf {a} }{dx}}\cdot {\mathbf {b} }\\{\cfrac {d({\mathbf {a} }\times {\mathbf {b} })}{dx}}&={\mathbf {a} }\times {\cfrac {d\mathbf {b} }{dx}}+{\cfrac {d\mathbf {a} }{dx}}\times {\mathbf {b} }\\{\cfrac {d[{\mathbf {a} }\cdot {({\mathbf {b} }\times {\mathbf {c} })}]}{dt}}&={\cfrac {d\mathbf {a} }{dt}}\cdot {({\mathbf {b} }\times {\mathbf {c} })}+{\mathbf {a} }\cdot {\left({\cfrac {d\mathbf {b} }{dt}}\times {\mathbf {c} }\right)}+{\mathbf {a} }\cdot {\left({\mathbf {b} }\times {\cfrac {d\mathbf {c} }{dt}}\right)}\end{aligned}}}

## Scalar and vector fieldsEdit

Let ${\displaystyle \mathbf {x} \,}$ be the position vector of any point in space. Suppose that there is a scalar function (${\displaystyle g\,}$) that assigns a value to each point in space. Then

${\displaystyle g=g(\mathbf {x} )\,}$

represents a scalar field. An example of a scalar field is the temperature. See Figure4(a).

If there is a vector function (${\displaystyle \mathbf {a} \,}$) that assigns a vector to each point in space, then

${\displaystyle \mathbf {a} =\mathbf {a} (\mathbf {x} )\,}$

represents a vector field. An example is the displacement field. See Figure 4(b).

## Gradient of a scalar fieldEdit

Let ${\displaystyle \varphi (\mathbf {x} )\,}$ be a scalar function. Assume that the partial derivatives of the function are continuous in some region of space. If the point ${\displaystyle \mathbf {x} \,}$ has coordinates (${\displaystyle x_{1},x_{2},x_{3}\,}$) with respect to the basis (${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\,}$), the gradient of ${\displaystyle \varphi \,}$ is defined as

${\displaystyle {\boldsymbol {\nabla }}{\varphi }={\frac {\partial \varphi }{\partial x_{1}}}~\mathbf {e} _{1}+{\frac {\partial \varphi }{\partial x_{2}}}~\mathbf {e} _{2}+{\frac {\partial \varphi }{\partial x_{3}}}~\mathbf {e} _{3}~.}$

In index notation,

${\displaystyle {\boldsymbol {\nabla }}{\varphi }\equiv \varphi _{,i}~\mathbf {e} _{i}~.}$

The gradient is obviously a vector and has a direction. We can think of the gradient at a point being the vector perpendicular to the level contour at that point.

It is often useful to think of the symbol ${\displaystyle {\boldsymbol {\nabla }}{}}$ as an operator of the form

${\displaystyle {\boldsymbol {\nabla }}{}={\frac {\partial }{\partial x_{1}}}~\mathbf {e} _{1}+{\frac {\partial }{\partial x_{2}}}~\mathbf {e} _{2}+{\frac {\partial }{\partial x_{3}}}~\mathbf {e} _{3}~.}$

## Divergence of a vector fieldEdit

If we form a scalar product of a vector field ${\displaystyle \mathbf {u} (\mathbf {x} )\,}$ with the ${\displaystyle {\boldsymbol {\nabla }}{}}$ operator, we get a scalar quantity called the divergence of the vector field. Thus,

${\displaystyle {\boldsymbol {\nabla }}\cdot {\mathbf {u} }={\frac {\partial u_{1}}{\partial x_{1}}}+{\frac {\partial u_{2}}{\partial x_{2}}}+{\frac {\partial u_{3}}{\partial x_{3}}}~.}$

In index notation,

${\displaystyle {\boldsymbol {\nabla }}\cdot {\mathbf {u} }\equiv u_{i,i}~.}$

If ${\displaystyle {\boldsymbol {\nabla }}\cdot {\mathbf {u} }=0}$, then ${\displaystyle \mathbf {u} \,}$ is called a divergence-free field.

The physical significance of the divergence of a vector field is the rate at which some density exits a given region of space. In the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or out of the region.

## Curl of a vector fieldEdit

The curl of a vector field ${\displaystyle \mathbf {u} (\mathbf {x} )\,}$ is a vector defined as

${\displaystyle {\boldsymbol {\nabla }}\times {\mathbf {u} }=\det {\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\{\frac {\partial }{\partial x_{1}}}&{\frac {\partial }{\partial x_{2}}}&{\frac {\partial }{\partial x_{3}}}\\u_{1}&u_{2}&u_{3}\\\end{vmatrix}}}$

The physical significance of the curl of a vector field is the amount of rotation or angular momentum of the contents of a region of space.

## Laplacian of a scalar or vector fieldEdit

The Laplacian of a scalar field ${\displaystyle \varphi (\mathbf {x} )\,}$ is a scalar defined as

${\displaystyle \nabla ^{2}{\varphi }:={\boldsymbol {\nabla }}\cdot {{\boldsymbol {\nabla }}{\varphi }}={\frac {\partial ^{2}\varphi }{\partial x_{1}}}+{\frac {\partial ^{2}\varphi }{\partial x_{2}}}+{\frac {\partial ^{2}\varphi }{\partial x_{3}}}~.}$

The Laplacian of a vector field ${\displaystyle \mathbf {u} (\mathbf {x} )\,}$ is a vector defined as

${\displaystyle \nabla ^{2}{\mathbf {u} }:=(\nabla ^{2}{u_{1}})\mathbf {e} _{1}+(\nabla ^{2}{u_{2}})\mathbf {e} _{2}+(\nabla ^{2}{u_{3}})\mathbf {e} _{3}~.}$

## Identities in vector calculusEdit

Some frequently used identities from vector calculus are listed below.

1. ${\displaystyle {\boldsymbol {\nabla }}\cdot {(\mathbf {a} +\mathbf {b} )}={\boldsymbol {\nabla }}\cdot {\mathbf {a} }+{\boldsymbol {\nabla }}\cdot {\mathbf {b} }}$~.
2. ${\displaystyle {\boldsymbol {\nabla }}\times {(\mathbf {a} +\mathbf {b} )}={\boldsymbol {\nabla }}\times {\mathbf {a} }+{\boldsymbol {\nabla }}\times {\mathbf {b} }}$~.
3. ${\displaystyle {\boldsymbol {\nabla }}\cdot {(\varphi \mathbf {a} )}={({\boldsymbol {\nabla }}{\varphi })}\cdot {\mathbf {a} }+\varphi ({\boldsymbol {\nabla }}\cdot {\mathbf {a} })}$~.
4. ${\displaystyle {\boldsymbol {\nabla }}\times {(\varphi \mathbf {a} )}={({\boldsymbol {\nabla }}{\varphi })}\times {\mathbf {a} }+\varphi ({\boldsymbol {\nabla }}\times {\mathbf {a} })}$~.
5. ${\displaystyle {\boldsymbol {\nabla }}\cdot {({\mathbf {a} }\times {\mathbf {b} })}={\mathbf {b} }\cdot {({\boldsymbol {\nabla }}\times {\mathbf {a} })}-{\mathbf {a} }\cdot {({\boldsymbol {\nabla }}\times {\mathbf {b} })}}$~.

## Green-Gauss Divergence TheoremEdit

Let ${\displaystyle \mathbf {u} (\mathbf {x} )\,}$ be a continuous and differentiable vector field on a body ${\displaystyle \Omega \,}$ with boundary ${\displaystyle \Gamma \,}$. The divergence theorem states that

${\displaystyle {\int _{\Omega }{\boldsymbol {\nabla }}\cdot {\mathbf {u} }~dV=\int _{\Gamma }{\mathbf {n} }\cdot {\mathbf {u} }~dA}}$

where ${\displaystyle \mathbf {n} \,}$ is the outward unit normal to the surface (see Figure 5).

In index notation,

${\displaystyle \int _{\Omega }u_{i,i}~dV=\int _{\Gamma }n_{i}u_{i}~dA}$