Path Integral

Path integrals (line integrals), as the name suggests, return the area underneath a parametric curve, as opposed to a simple interval (in the case of normal integrals).

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Definition §

Path Integral (Scalar Function) ^def-scalar

Let $f:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be a continuous scalar function and let $C$ be a parametric curve given by the $C^1$ parametric equation $\vec{c}(t)=(x(t),y(t),z(t))$ for $t\in [a,b]$. Then the path integral:

$$\begin{array}{rl}{\int }_{C}f\text{d}s& ={\int }_{C}f(x,y,z)\text{d}s\\ & ={\int }_a^{b}f(x(t),y(t),z(t))\frac{\text{d}s}{\text{d}t}\text{d}t\end{array}$$

$$\underset{\text{Path Integral}}{\underbrace{{\int }_{C}f\text{d}s}}=\underset{\text{Standard Integral}}{\underbrace{{\int }_a^{b}f(\vec{c}(t))\Vert {\vec{c}}^{{}^{\prime }}(t)\Vert \text{d}t}}$$

where $s$ is the arclength of the path.

• Any parametrisation of the curve $C$ implicitly orients it, by the increasing values of $t$.
• ${\int }_{C}1\text{d}s=s$. So we can get the arclength of $C$ using integrals.
• The path integral is also independent of the orientation of $C$.

Path Integral (Vector Field) ^def-vector

Let $\vec{F}$ be a continuous vector field and let $C$ be a parametric curve given by the $C^1$ parametric equation $\vec{c}(t)=(x(t),y(t),z(t))$ for $t\in [a,b]$. Then the path integral is given by:

$${\int }_C\vec{F}\cdot \text{d}\vec{c}=\underset{\text{Scalar Path Integral}}{\underbrace{{\int }_C\vec{F}\cdot \hat{T}\text{d}s}}=\underset{\text{Standard Integral}}{\underbrace{{\int }_a^b\vec{F}(\vec{c}(t))\cdot {\vec{c}}^{{}^{\prime }}(t)\text{d}t}}$$

• $\cdot$ is the dot product
• $\hat{T}$ is the unit tangent vector

If we have $\vec{F}:{\mathbb{R}}^3\to {\mathbb{R}}^3$ represented as $\vec{F}(x,y,z)=(u(x,y,z),v(x,y,z),w(x,y,z))$. Then the path integral can be represented as:

$${\int }_C\vec{F}\cdot \text{d}\vec{c}={\int }_a^b\frac{\text{d}x}{\text{d}t}[u(\vec{c}(t))]+\frac{\text{d}y}{\text{d}t}[v(\vec{c}(t))]+\frac{\text{d}z}{\text{d}t}[w(\vec{c}(t))]\text{d}t$$

In Differential 1-Form:

$${\int }_{C}u\text{d}x+v\text{d}y+w\text{d}z$$

Notation for closed curves (loops)

Note that some sources (me included) use $\oint$ to refer to closed loops. That is, if $f$ is being integrated over a closed curve $C$, we write:

$${\oint }_{C}f\text{d}s$$

This should only be used if we know that $C$ is closed and that we are working with real numbers (because the same notation is used for contour integrals)

Physical Interpretation §

For Scalar Functions §

In most cases the path integral is useful for any function $f$ where the units are “some property” per length. For example:

• If $f$ is the mass per length of some cable $C$, then ${\int }_{C}f\text{d}s$ returns the total mass of the cable.

#todo connect with physics 2 (advanced).

For Vector Fields §

The most straightforward way to visualise path integrals is to imagine the vector field as the current of a river, which pushes with different direction and strength at different points. Then, the path integral is just how much force the river exerts on a particle moving a certain path:

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• When ${\int }_C>0$, the river (vector field) does work on the particle (it aids movement)
• When ${\int }_C<0$, the river (vector field) does work against the particle (it hinder movement)
• When ${\int }_C=0$, the river neither aids nor hinders movement, and does no work.

Hence, the path integral is extremely useful in finding the work done on an object as it moves through a Force Field.

For example, the work done on a charged particle as it moves through an electric field is given by the path integral:

#todo

Connection To Standard Integrals §

The standard integral of a function $f(x)$ that operates over a one-dimensional interval $[a,b]$ is given by:

$$I={\int }_a^{b}f(x)\text{d}x$$

This integral is actually a degenerate case of the line/path integral.

If we create a scalar parametric equation $c(t)=t$ from $t\in [a,b]$, then this corresponds exactly with the line created by the interval $x\in [a,b]$. Then the path integral is the exact same as the standard integral!

$$\begin{array}{rl}{\int }_{C}f(x)\text{d}s& ={\int }_a^{b}f(\vec{c}(t))\cdot \Vert \frac{\text{d}\vec{c}}{\text{d}t}\Vert \text{d}t\\ & ={\int }_a^{b}f(t)\cdot 1\text{d}t\\ & ={\int }_a^{b}f(t)\text{d}t=I\end{array}\text{\hspace{1em}(dtdc=1)}$$

Another way to think about it is that standard integrals are path integrals where we take the parametric curve to be a straight line.

Theorems §

T1: Path Integrals over scalar functions are independent of the path's orientation

For any suitable scalar function $f$, and for the parametric path $C$, we have:

$${\int }_{C}f\text{d}s={\int }_{-C}f\text{d}s$$

where $-C$ refers to $C$ being oriented the opposite way.

T2: Path Integrals over vector fields depend on the path's orientation

For any suitable vector field $\vec{F}$, and for the parametric path $C$, we have:

$${\int }_C\vec{F}\text{d}s=-{\int }_{-C}\vec{F}\text{d}s$$

where $-C$ refers to $C$ being oriented the opposite way.

Examples §

E1: Wire Mass

Consider a thin cable represented by the parametric curve:

$$\vec{c}(t)=(\cos (t),\sin (t),t)$$

having mass per unit length given by:

$$\mu (x,y,z)=xy+z$$

The total mass $m$ can be given by the path integral: ${\int }_C\mu (x,y,z)\text{d}s$:

$${\int }_C\mu \text{d}s={\int }_0^{2\pi }\mu (\vec{c}(t))\Vert \frac{\text{d}\vec{c}}{\text{d}t}\Vert \text{d}t$$

$$\frac{\text{d}\vec{c}}{\text{d}t}=(-\sin (t),\cos (t),1)\Vert \frac{\text{d}\vec{c}}{\text{d}t}\Vert =\sqrt{{\cos }^2(t)+{\sin }^2(t)+1}=\sqrt{2}$$

$$\begin{array}{rl}m& ={\int }_0^{2\pi }(\cos (t)\sin (t)+t)\sqrt{2}\text{d}t\\ & =\sqrt{2}{\int }_0^{2\pi }\frac{1}{2}\sin (2t)+t\text{d}t=\sqrt{2}{\left[-\frac{1}{4}\cos (2t)+\frac{t^2}{2}\right]}_0^{2\pi }\\ & =\sqrt{2}(-\frac{1}{4}+2{\pi }^2+\frac{1}{4})=2\sqrt{2}{\pi }^2\end{array}$$