MATH1011 - Lecture 21

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MATH1011 - Lecture 20
Skeleton Slides

Divergence

• In ${\mathbb{R}}^2$, the divergence of $\mathbf{F}=(f_1,f_2)$ is:
  $\text{div }\mathbf{F}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}$

• In ${\mathbb{R}}^3$, the divergence of $\mathbf{F}=(f_1,f_2,f_3)$ is:
  $\text{div }\mathbf{F}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}$

• Another notation for the divergence is $\text{div }\mathbf{F}=\nabla \cdot \mathbf{F}$

  • Like a “dot product” of the operator $\nabla =\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\right)$ and the vector $\mathbf{F}=(f_1,f_2,f_3)$
  • $\nabla$ is not a vector in the conventional sense

• Divergence of a vector field is a local measure of its “outgoingness”

  • A point at which the flux is outgoing has positive divergence, and is often called a source of the field
  • A point at which the flux is directed inward has negative divergence, and is often called a sink of the field.

• Divergence is a scalar value

“Is more stuff coming out of this point than going in?”

Curl

• In ${\mathbb{R}}^3$, the curl of a vector field $F=(f_1,f_2,f_3)$ is the vector field:
  $\nabla \times \mathbf{F}=\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z},\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x},\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)$
• $\nabla \times \mathbf{F}$ is itself a vector field in ${\mathbb{R}}^3$
  • Thought of like the “cross-product” of the two vectors
  • Curl is a vector in ${\mathbb{R}}^3$
• Divergence extends to any dimension, curl is only for ${\mathbb{R}}^3$
  • Curl is a scalar in ${\mathbb{R}}^2$

Curl and Divergence

• $\nabla \cdot (\nabla \times \mathbf{A})=0$ for any vector field $\mathbf{A}$ of ${\mathbb{R}}^3$
• Let $\mathbf{F}$ be a vector field defined on the whole of ${\mathbb{R}}^3$
  • If the divergence $\nabla \cdot \mathbf{F}=0$, then $\mathbf{F}=\nabla \times \mathbf{A}$ for some vector field $\mathbf{A}$, which we call the vector potential of $\mathbf{F}$

Work Done By Force

• If a vector $\mathbf{F}$ describes a force, the circulation yields the work done by the vector field to move a particle along this path
  $W={\int }_C\mathbf{F}\cdot \mathbf{dr}={\int }_a^b\mathbf{F}(\mathbf{r}(t))\cdot \dot{\mathbf{r}}(t)dt$

The path taken from point $A$ to $B$ does not matter, you will get the same calculated work done whether you used a direct route or a detour

If $\mathbf{F}$ is conservative:

• $W=\Psi (a)-\Psi (b)$ where $a$ and $b$ are the two end points of the path

Potential

• Vector field sometimes indirectly given as gradient of a scalar function:
  $\mathbf{F}=\nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})\text{for some scalar function}f(x,y)$
• $\mathbf{F}$ is called a gradient field and $f$ is called the potential of $\mathbf{F}$
  • The above equation can be rewritten for ${\mathbb{R}}^3$ as well
  • e.g. $\mathbf{F}=\nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})\text{for some scalar function}f(x,y,z)$

Finding Potentials

• Read Conservative Vector Fields below first

1. Check that $\nabla \times \mathbf{F}=0$
2. $\frac{\partial \Psi }{\partial x}=f_1$ → $\Psi =\int f_{1}dx+g(y,z)$
   • $g(y,z)$ is unknown and represents the integral constant
3. Insert $\Psi =\int f_{1}dx+g(y,z)$ into 2nd equation:
   • $f_2=\frac{\partial }{\partial y}(\int f_{1}dx)+\frac{\partial g}{\partial y}(y,z)$
     • Since the curl is 0, the terms with $x$ cancel out
   • Thus we obtain: $g(y,z)=\hat{g}(y,z)+h(z)$
     • $\hat{g}(y,z)$ is known but $h(z)$ is unknown
   • This all becomes: $\Psi =\int f_{1}dx+\hat{g}(y,z)+h(z)$
4. Insert $\Psi =\int f_{1}dx+\hat{g}(y,z)+h(z)$ into 3rd equation:
   $f_3=\frac{\partial }{\partial z}\left(\int f_{1}dx\right)+\frac{\partial \hat{g}}{\partial z}(y,z)+\frac{dh}{dz}(z)$
   • Rearrange: $h^{\prime }(z)=f_3-\frac{\partial }{\partial z}\left(\int f_{1}dx\right)-\frac{\partial \hat{g}}{\partial z}(y,z)$
   • Since the curl is 0, the terms with $x$ and $y$ cancel out
   • $h(z)=\hat{h}(z)+C$
     • $\hat{h}(z)$ is known but $C$ is unknown
   • Finally:
     $\Psi =\int f_{1}dx+\hat{g}(y,z)+\hat{h}(z)+C$

• We can choose $C$, so usually we should set $C$ equal to $0$
• Full example of this process in lecture recording

Conservative Vector Fields

A vector field is called conservative if the circulation ${\int }_C\mathbf{F}\cdot \mathbf{dr}$ from a starting point $S$ to an end point $E$ depends on the locations of $S$ and $E$ but not on the particular path $C$ taken between them

Suppose $\mathbf{F}=\nabla \psi$ for some scalar function $\psi (x,y,z)$

• Let $C$ be a path from $S$ to $E$ parametrised by:
  $\mathbf{r}(t)=(x(t),y(t),z(t))\text{for}t\in [a,b]$
• Hence:
  ${\int }_C\mathbf{F}\cdot \mathbf{dr}={\int }_a^b\nabla \psi (\mathbf{r}(t))\cdot \dot{\mathbf{r}}(t)dt=\psi (E)-\psi (S)$
• $\therefore$ circulation depends only on $S$ and $E$
  • See slides for full proof/expansion

The Fundamental Theorem for Line Integrals

Let $\mathbf{F}$ be a vector field in an open subset $U$ of ${\mathbb{R}}^3$ or ${\mathbb{R}}^2$

• If $\mathbf{F}=\nabla \Psi$ or some scalar function $\Phi$, then $\mathbf{F}$ is conservative in $U$
  $\text{for any}C\text{in}U:{\int }_C\mathbf{F}\cdot \mathbf{dr}=\psi (E)-\psi (S)$
• If $\mathbf{F}$ is conservative in $U$, then $\mathbf{F}=\nabla \Psi$ for some scalar function $\Psi$
  • Both statements imply each other

If $C$ is a closed curve and $\mathbf{F}$ is conservative, then ${\oint }_C\mathbf{F}\cdot \mathbf{dr}=0$This follows immediately from the fact that $S=E$ for a closed curve

Criterion for Conservative Fields

• Let $\mathbf{F}$ be a conservative vector field in an open subset $U$ of ${\mathbb{R}}^3$, then: $\nabla \times \mathbf{F}=0$
• The converse is not always true, another condition must be met:
  • The domain of $\mathbf{F}$ must also be simply-connected
• A curve is simple if it does not intersect itself.

An open subset $U$ of ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$ is called simply-connected if every closed simple curve in $U$ can be continuously contracted to a single point without leaving $U$. An easy way to check this is if there is a hole within the region bounded by $C$, as if there is a hole, the hole would be crossed as it contracts to a single point making it not simply-connected.

This applies to ${\mathbb{R}}^2$ as well:

• If vector field $\mathbf{F}$ is conservative in open subset $U$ of ${\mathbb{R}}^2$:
  $\nabla \times \mathbf{F}=0$
• If $U$ is simply-connected and $\nabla \times \mathbf{F}=0$, $\mathbf{F}$ is a conservative vector field

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MATH1011 - Lecture 22
Annotated Slides