Title: Part IA / Vector Calculus
Tags: Curves Poisson_and_Laplace_Equations Surfaces Tensors Grad_Div_Curl Integral_Theorems

Q: Definition. A vector function is differentiable at t if
A: as \(\delta t \to 0\), we can write \[\mathbf{x}(t + \delta t) - \mathbf{x} = \dot{\mathbf{x}}(t)\delta t + O(\delta t^2)\]

Q: Definition. Regular Parametrisation.
A: If \(\dot{\mathbf{x}}(t)\neq 0\) for any t

Q: Proposition. Leibniz Identities in Vector Calculus.
A: \[\frac{d}{dt}(f\mathbf{g}) = \dot{f}\mathbf{g} + f\dot{\mathbf{g}}\]
\[\frac{d}{dt}(\mathbf{g \cdot h}) = \dot{\mathbf{g}} \cdot \mathbf{h} + \mathbf{g} \cdot \dot{\mathbf{h}}\]
\[\frac{d}{dt}(\mathbf{g \times h}) = \dot{\mathbf{g}} \times \mathbf{h} + \mathbf{g} \times \dot{\mathbf{h}}\]

Q: Definition. Arc Length.
A: Length of the curve between two points.
For two nearby points we see that \(\delta s = |\dot{\mathbf{x}} \delta t| + O(\delta t^2)\) .
So we want to sum over lots of these nearby lengths so we integrate to get \[s = \int_{t_0}^t dt' |\dot{\mathbf{x}}(t')|\]

Q: Claim. Arc length is independent of the choice of parameterisation
A: Pick a different parameterisation \(\tau (t)\), with \(\frac{d \tau}{dt} > 0\). Then \[\frac{d \mathbf{x}}{dt} = \frac{d \mathbf{x}}{d \tau} \frac{d \tau}{dt}\]
We can then compute the arc length using the new parameterisation.
\[s = \int_{t_0}^t dt' |\dot{\mathbf{x}}(t')| = \int_{\tau_0}^{\tau} d \tau' \frac{dt'}{d\tau '} \left |\frac{d \mathbf{x}}{d \tau '}\frac{d\tau '}{dt'} \right | = \int_{\tau_0}^{\tau} d\tau ' \left |\frac{d\mathbf{x}}{d \tau '}\right |\]

Q: Definition. Curvature and Principal Normal.
A: We know that the tangent vector \(\mathbf{t} = \frac{d\mathbf{x}}{ds}\) which has unit length, where s is the arc length. We choose the arc length so that the curvature doesn't depend on our choice of parameter.
Then the curvature is \[\kappa (s) = \left | \frac{d^s \mathbf{x}}{ds^2}\right |\]
This also gives us a unit vector called the principal normal \[\mathbf{n} = \frac{1}{\kappa} \frac{d^2 \mathbf{x}}{ds^2} = \frac{1}{\kappa}\frac{d \mathbf{t}}{ds}\]

Q: Definition. Binormal.
A: \[\bf b = t \times n\]

Q: Definition. Torsion.
A: We know that \(|\bf b| = 1\) so \(\mathbf{b} \cdot \frac{d\mathbf{b}}{ds} = 0\).
Also \(\bf t \cdot b = 0\) which we can differentiate to get \[0 = \frac{d\mathbf{t}}{ds} \cdot \mathbf{b} + \mathbf{t} \cdot \frac{d\mathbf{b}}{ds} = \kappa \mathbf{n \cdot b} + \mathbf{t} \cdot \frac{d\mathbf{b}}{ds}\]
So \(\mathbf{t} \cdot \frac{d\mathbf{b}}{ds} = 0\) so we define \[\frac{d\mathbf{b}}{ds} = - \tau(s) \mathbf{n}\]

Q: Definition. Line Integral. For scalar field \(\phi : \mathbb{R}^n \to \mathbb{R}\) and for vector field \(\mathbf{F}: \mathbb{R}^n \to \mathbb{R}^n\)
A: \[\int_C \phi \; ds = \int_{t_a}^{t_b} \phi(\mathbf{x}(t))|\mathbf{\dot{x}}(t)|dt\]
\[\int_C \mathbf{F}(x) \cdot dx = \int_{t_a}^{t_b} \mathbf{F(x}(t)) \cdot \mathbf{\dot{x}}(t) dt\]
The choice of direction along C is called orientation.
The line integral of a scalar field does not depend on the orientation, the line integral of a vector field does depend on orientation.

Q: Definition. Line integral around a closed curve C.
A: \[\oint_C \mathbf{F} \cdot \mathbf{x}\]
This quantity is called the circulation of \(\bf F\) around C.

Q: Definition. Gradient of scalar field.
A: \[\nabla \phi = \frac{\partial \phi}{\partial x^i}\mathbf{e}_i\]
Where we are summing over i and the \(\{\mathbf{e}_i\}\) are an orthonormal basis of vectors associated to the coordinates \(x_i\)
We can write \[\phi(\mathbf{x + h}) = \phi(\mathbf{x}) + \mathbf{h} \cdot \nabla \phi + \Theta(|\mathbf{h}|^2)\]

Q: Definition. Directional Derivative.
A: \[D_{\mathbf{n}}\phi = \mathbf{\hat{n}} \cdot \nabla \phi\]
Is maximal at any point when \(\mathbf{\hat{n}}\) lies parallel to \(\nabla \phi(\mathbf{x})\)

Q: Definition. Conservative vector field.
A: \[\mathbf{F} = \nabla \phi\]
for some scalar field \(\phi\) which, in this context, is referred to as a potential

Q: Claim. The line integral around any closed curve vanished if and only if \(\bf F\) is conservative.
A: Consider a conservative field, \(\mathbf{F} = \nabla \phi\). Integrate along a curve C to get \[\int_C \mathbf{F} \cdot d\mathbf{x} = \int_C \nabla \phi \cdot d\mathbf{x} = \int_{t_a}^{t_b} \frac{\partial \phi}{\partial x^i} \frac{\partial dx^i}{dt}dt = \int_{t_a}^{t_b} \frac{d}{dt} \phi(\mathbf{x}(t)dt = \phi(\mathbf{b}) - \phi(\mathbf{a})\]
So it depends only on the end points
Now to prove the converse. First take \(\phi(\mathbf{0}) = 0\), then at any other point \(\bf y\), we define \(\bf x = y\) \[\phi(\mathbf{y}) = \int_{C(\mathbf{y})} \mathbf{F} \cdot d\mathbf{x}\] where \(C(\mathbf{y})\) is a curve that starts at the origin and ends at the point \(\bf y\).
Using the definition of the partial derivative we have \[\frac{\partial \phi}{\partial x^i}(\mathbf{y}) = \lim_{\epsilon \to 0}\frac{1}{\epsilon}\left [ \int_{C(\mathbf{y} + \epsilon \mathbf{e}_i)} \mathbf{F} \cdot d\mathbf{x} - \int_{C(\mathbf{y})} \mathbf{F} \cdot d\mathbf{x}\right]\]
Now we get rid of the closed curve and are basically only left with the integral of the tiny straight line we have moved by.
This is roughly \(F_i \epsilon\), so when we divide by \(\epsilon\) and take the limit we get \[\frac{\partial \phi}{\partial x^i}(\mathbf{y}) = F_i(\mathbf{y})\].
This is our desired result \(\nabla \phi = \mathbf{F}\)

Q: Claim. Given F, how do you know if its conservative?
A: \[\frac{\partial F_i}{\partial x^j} = \frac{\partial^2 \phi}{\partial x^i \partial x^j} = \frac{\partial F_j}{\partial x^i}\]
So we check this holds for all pairs of co-ordinates

Q: Definition. Differential and Exact Differential.
A: Given a function \(\phi(\mathbf{x})\) on \(\mathbb{R}^n\), the differential is\[d\phi = \frac{\partial \phi}{\partial x^i} = \nabla \phi \cdot d\mathbf{x}\]
A differential form is said to be exact if it can be written as \[\mathbf{F} \cdot d\mathbf{x} = d\phi\]
So this is the same as \(\bf F\) being conservative.

Q: Definition. Area Integral. How to integrate a scalar function \(\phi : \mathbb{R}^2 \to \mathbb{R} \) over a region \(D \subset \mathbb{R}^2\)
A: \[\int_D \phi(x,y) dA = \int_a^b dy \int_{x_1(y)}^{x_2(y)}dx \phi(x,y)\]

Q: Claim. Area integral when changing from \((x,y)\) plane to \((u,v)\) plane.
A: \[\int_D dx \; dy  \phi(x,y) = \int_{D'} du \; dv |J(u,v)| \phi(u,v)\] where \[J(u,v) = \left | \begin{matrix} \partial x / \partial u & \partial x / \partial v \\ \partial y / \partial u & \partial y / \partial v \end{matrix} \right | \]
Sometimes we write \[J = \frac{\partial(x,y)}{\partial(u,v)}\]

Q: Definition. Volume Integrals. How to do them when changing from Cartesian to spherical polar or cylindrical polar.
A: \[\int_D \phi(x,y,z) dV = \int_{D'} \phi(r, \theta, \phi) r^2 \sin \theta \; dr \; d\theta \; d\phi = \int_{D''} \phi(\rho, \phi, z) \rho\;  d\rho \; d\phi \; dz\]
You get this from the fact that \[dV = |J| \; du\; dv \;dw\] where J is the Jacobian

Q: Definition. Normal Vector of Surface. What is a regular surface?
A: \[\mathbf{n} = \pm \frac{1}{|\nabla F|} \nabla F \]
Points perpendicularly away from the surface at all points
If \(\mathbb{n} \neq 0\) anywhere on the surface, then the surface is said to be regular

Q: Definition. Surface and Boundary in \(\mathbb{R}^3\).
A: A surface can be viewed as the level set of a function \[F(x,y,z) = 0\]
A boundary is a piece wise smooth closed curve, several boundaries means this curve has several disconnected pieces. We can view it as \[\partial V_r = \lim_{\epsilon \to 0} \frac{1}{\epsilon}(V_{r+\epsilon}\setminus V_r)\]

Q: Definition. Bounded and Closed Surface.
A: A surface is bounded if it can be contained within a solid sphere of fixed radius.
A bounded surface with no boundary is said to be closed.

Q: Definition. Orientable Surface.
A: A surface is said to be orientable if there is a consistent choice of unit normal which varies smoothly over the surface.

Q: Definition. Surface integral of a parameterised surface.
A: \[ \int_S \phi (\mathbf{x})dS = \int_D du \; dv \; \left | \frac{\partial \mathbf{x}}{\partial u } \times \frac{\partial \mathbf{x}}{\partial v} \right | \phi (\mathbf{x}(u,v)) \]

Q: Definition. Surface Integral over Vector Field F(x).
A: \[\int_S \mathbf{F(x) \cdot n} dS = \int_D du\;dv\; \left( \frac{\partial \mathbf{x}}{\partial u} \times \frac{\partial \mathbf{x}}{\partial v}\right) \cdot \mathbf{F(x}(u,v))\]
This is called the flux of F through S

Q: Definition. Vector Area Element.
A: \[d\mathbf{S} = \mathbf{n}dS = \left ( \frac{\partial \mathbf{x}}{du} \times \frac{\partial \mathbf{x}}{dv}\right) du\;dv\]
This has magnitude dS and points in the normal direction n.

Q: Definition. Gradient Operator.
A: \[\nabla = \mathbf{e_i}\frac{\partial}{\partial x^i}\]
This is both a vector and an operator.

Q: Definition. Divergence.
A: \[\nabla \cdot \mathbf{F} = \left ( \mathbf{e_i}\frac{\partial}{\partial x^i}\right ) \cdot (\mathbf{e}_jF_j) = \frac{\partial F_i}{\partial x_i}\]
Measures the net flow of F into a point x.

Q: Definition. Curl.
A: \[\nabla \times \mathbf{F} = \left ( \mathbf{e}_i \frac{\partial}{\partial x^i}\right ) \times (\mathbf{e}_jF_j) = \epsilon_{ijk} \frac{\partial F_j}{\partial x_i}\mathbf{e}_k\]
Or written out fully: \[\nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial x_2} - \frac{\partial F_2}{\partial x^3}, \frac{\partial F_1}{\partial x^3} - \frac{\partial F_3}{\partial x^1}, \frac{\partial F_2}{\partial x^1} - \frac{\partial F_1}{\partial x^2}\right ) = \left | \begin{matrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ \frac{\partial}{\partial x^1} & \frac{\partial}{\partial x^2} & \frac{\partial}{\partial x^3} \\ F_1 & F_2 & F_3\end{matrix}\right |\]
Measures the rotation of the vector field.

Q: Definition. Irrotational. A vector field \(\mathbf F\) is irrotational if [...]
A: \(\nabla \times \mathbf{F} = 0\)
Back Extra: This is equivalent to \(\mathbf F\) being conservative if the field is defined everywhere on \(\mathbb{R}^3\)

Q: Definition. A vector field \(\mathbf{F}\) is called divergence free or solenoidal if [...]
A: \(\nabla \cdot \mathbf{F} = 0\)
Back Extra: Any divergence free field can be written as the curl of something else \[\nabla \cdot \mathbf{F} = 0 \iff \mathbf{F} = \nabla \times \mathbf{A}\]

Q: Definition. The Laplacian is a second order differential operator defined by [...]
A: \[\nabla^2 = \nabla \cdot \nabla = \frac{\partial^2}{\partial x^i \partial x^i}\]

Q: Definition. Orthogonal Curvilinear Coordinates.
A: \[\frac{\partial \mathbf{x}}{\partial u} \cdot \left ( \frac{\partial \mathbf{x}}{\partial v} \times \frac{\partial \mathbf{x}}{dw}\right ) \neq 0\]
\(\frac{\partial \mathbf{x}}{\partial u}\) is the tangent vector to the lines defined by v,w = constant. So we need the three tangent vectors to be mutually orthogonal.

Q: Theorem. Gauss' Divergence Theorem. For a smooth vector field \(\mathbf{F(x)}\) over \(\mathbb{R}^3\) \[\int_V \nabla \cdot \mathbf{F} \; dV = \int_S \mathbf{F} \cdot d\mathbf{S}\]
where V is a bounded region whose boundary \(\partial V = S\) is a piece wise smooth closed surface. The integral on the RHS is taken with normal pointing outward.
A: The intuition is that we can divide a volume up into a bunch of tiny cubes, then the flux of \(\mathbf{F}\) through the those sides that lie in the (y,z) plane is given by \[ [F_x(x + \delta x, y, z) - F_x(x,y,z)]\delta y \delta z \approx  \frac{\partial F_x}{\partial x}\delta x \delta y \delta z\] where the minus sign is because the outward normal points in opposite directions on opposite sides of a cube. We repeat this over z and y directions also and since the LHS terms cancel on interior walls, we are left with \[\int_S \mathbf{F} \cdot d\mathbf{S} = \int_V \nabla \cdot \mathbf{F} \; dV\]

Q: Claim. For \(S = \partial V\), we have \[\int_V \nabla \phi \; dV = \int_S \phi \; d\mathbf{S}\]
A: Consider the divergence theorem with \(\mathbf{F} = \phi \mathbf{a}\) where \(\mathbf{a}\) is a constant vector.
\[\int_V \nabla \cdot (\phi \mathbf{a})dV = \int_S(\phi \mathbf{a}) \cdot d\mathbf{S} \implies \mathbf{a} \cdot \left (\int_V \nabla \phi dV - \int_S \phi d\mathbf{S} \right ) = 0\]True for any constant vecotor, so the expression in the brackets must be equal to 0.

Q: Proposition. Conservation Laws. We think about the energy contained in a hot gas. Since energy is conserved there is a corresponding continuity equation \[\frac{\partial \epsilon}{\partial t} + \nabla \cdot \mathbf{J} = 0\] where \(\epsilon(\mathbf{x},t)\) is the energy density of the gas and \(\bf J\) is the heat current current which tells us how energy is transported from one region of space to another.
Then we use \[\epsilon(\mathbf{x},t) = cT(\mathbf{x},t)\] with \(c_V\) the specific heat capacity. Next we use the fact that heat flow is due to temperature differences between different parts of the system. \[\mathbf{J} = -\kappa \nabla T\] where \(\kappa\) is the thermal conductivity and the minus sign ensures that heat flows from hot to cold. This is Fick's Law. Combining these equation we get [...]
A: \[\frac{\partial T}{\partial t} = D \nabla^2 T\]which is the heat equation and tells us how the temperature of a system evolves.

Q: Theorem. Green's Theorem in the Plane. Let \(P(x,y)\) and \(Q(x,y)\) be smooth functions on \(\mathbb{R}^2\). Then \[\int_A \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right )dA = \oint_C Pdx + Q dy \] where A is a bounded region in the plane and \(C = \partial A\) is a piecewise smooth, non-intersecting closed curve which is traversed anti-clockwise.
A: Equivalent to the 2D divergence theorem. Let \(\mathbf{F} = (Q,-P)\) be a vector field in \(\mathbb{R}^2\). We then have \[\int_A \nabla \cdot \mathbf{F}dA = \int_A \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right )dA\] If \(\mathbf{x}(s) = (x(s),y(s))\) is the parameter used curve C, then the tangent vector is \(\mathbf{t}(s) = (x'(s), y'(s))\) and the normal vector \(\mathbf{n} = (y'(s), -x'(s))\) obeys \(\mathbf{n \cdot t} = 0\).
We then have \(\mathbf{F \cdot n} = Q\frac{dy}{ds} + P \frac{dx}{ds}\) and so the integral around C is \[\int_C \mathbf{F \cdot n}ds = \int_C Pdx + Qdy\]
By the divergence theorem the LHS of both eqauctions are equal. Green's theorem in the plane is the statement that the RHS are equal.

Q: Theorem. Stoke's Theorem. Let S be a smooth surface in \(\mathbb{R}^3\) with boundary \(C = \partial S\) a piecewise smooth curve. Stokes' theorem states that, for any smooth vector field \(\mathbf{F(x)}\), we have [...]
A: \[\int_S \nabla \times \mathbf{F} \cdot d\mathbf{S} = \int_C \mathbf{F} \cdot d\mathbf{x}\]
Back Extra: The orientations of \(\mathbf{S} \text{ and } \mathbf{C}\) should be compatible. Basically if \(\bf n\) is normal to \(\bf S\) and \(\bf t\) is tangent to \(\bf C\), then \(\mathbf{t \times n}\) should point out of S.

Q: Corollary. Irrotational Implies Conservative.
A: By Stokes' theorem, if \(\nabla \times \mathbf{F} = 0\), then \[\oint_C \mathbf{F} \cdot d\mathbf{x} = 0\] around any closed curve C. We have showed that any conservative field can be written as \(\mathbf{F} = \nabla \phi\) for some scalar field \(\phi\).

Q: Definition. Gauss' Law for Gravity. Gauss' law tells us that \[[...]\]
A: \int_S \mathbf{g} \cdot d\mathbf{S} = -4 \pi GM
Back Extra: We can then see that \[\int_S \mathbf{g} \cdot d\mathbf{S} = \int_S g(r)dS = 4\pi r^2 g(r)\] which by Gauss' law tells us that \[\mathbf{g}(r) = -\frac{GM}{r^2}\mathbf{\hat{r}}\]

Q: Proposition. Gauss' Laws rewritten. If we use Gauss' divergence theorem combined with the laws we get new equations.
For gravity: \[[...] \implies \nabla \cdot \mathbf{g} = -4\pi G \rho(\mathbf{x})\]For electric fields we do the same thing to get \[[...]\]
A: \int_V \nabla \cdot \mathbf{g} dV = \int_S \mathbf{g} \cdot d\mathbf{S} = 4\pi GM = -4\pi G \int_V \rho(\mathbf{x}) dV; \nabla \cdot \mathbf{E} = \frac{\rho_e (\mathbf{x})}{\epsilon_0}

Q: Definition. Poisson Equations. For gravitational and electric fields we use the fact that \(\nabla \cdot \mathbf{g} = -4 \pi G \rho(\mathbf{x})\) and \(\nabla \cdot \mathbf{E} = \frac{\rho_e(\mathbf{x})}{\epsilon_0}\)  along with the fact that they are both conservative fields to get \[[...]\]
A: \nabla^2\Phi = 4\pi G \rho(\mathbf{x}) \;\;\;\;\nabla^2 \phi = -\frac{\rho_e(\mathbf{x})}{\epsilon_0}

Q: Definition. Laplace Equation. When the 'source' \(\rho(\mathbf{x})\) vanishes in the Poisson equations they reduce to \[[...]\]
A: \nabla^2 \Phi = 0

Q: Definition. Isotropic Solutions. For the Laplacian if we have symmetry in r in spherical coordinates or cylindrical coordinates, we get the following complementary solutions. First in spherical.\[[...]\]Now in cylindrical \[[...]\]
A: \nabla^2 \psi = 0 \implies \frac{d^2 \psi}{dr^2} + \frac{2}{r}\frac{d\psi}{dr} = \frac{1}{r^2}\frac{d}{dr}\left ( r^2 \frac{d\psi}{dr}\right ) \implies \psi(r) = \frac{A}{r} + B; \nabla^2 \psi = 0 \implies \frac{d^2 \psi}{dr^2} + \frac{1}{r}\frac{d\psi}{dr} = \frac{1}{r}\frac{d}{dr}\left ( r \frac{d\psi}{dr}\right ) \implies \psi(r) = A\log r + B
Back Extra: Both diverge at r=0.

Q: Definition. Dirichlet and Neumann Conditions. These are boundary conditions on \(\partial V\) when solving Poisson or Laplace equations in some bounded region V.
Dirichlet condition: [...]
Neumann condition: [...]
A: We fix \(\phi(\mathbf{x}) = f(\mathbf{x})\) for some specific f on \(\partial V\); We fix \(\mathbf{n} \cdot \nabla \psi(\mathbf{x}) = g(\mathbf{x})\) for some specific g on the boundary, where \(\mathbf{n}\) is the outwardly pointing normal of \(\partial V\)
Back Extra: Sometimes Neumann condition written as \(\partial \psi / \partial \mathbf{n} := \mathbf{n} \cdot \nabla \psi\).

Q: Claim. There is a unique solution to the Poisson equation on a bounded region V, with either Dirichlet or Neumann coundary conditions specified on each boundary \(\partial V\).(For Neumann boundary conditions everywhere, the solution is only unique up to a constant.)
A: Let \(\psi_1(\mathbf{x})\) and \(\psi_2(\mathbf{x})\) both satisfy the Poisson equation with the specified boundary conditions. Then \(\psi = \psi_1 - \psi_2\) obeys \(\nabla^2 \psi = 0\) and either \(\psi = 0\) or \(\mathbf{n} \cdot \nabla \psi - 0\) on \(\partial V\). Then consider \[\int_V \nabla \cdot (\psi \nabla \psi) dV = \int_V (\psi \nabla \cdot \nabla \psi + \psi \nabla^2 \psi)dV = \int_V |\nabla \psi|^2 dV\]But by the divergence theorem, we have \[\int_V \nabla \cdot (\psi \nabla \psi)dV = \int_{\partial V} \psi \nabla \psi d\mathbf{S} = \int_{\partial V} \psi(\mathbf{n} \cdot \partial \psi) dS = 0\]where either Dirichlet or Neumann boundary conditions set the boundary term to zero. The term in the first integral will only be 0 if \(\nabla \psi = 0\) everywhere in V, so \(\psi\) must be constant. If Dirichlet boundary conditions are imposed anywhere, then that constant must be 0.

Q: Definition. Green's Identities. They are like a 3d version of integration by parts.
1st Identity: [...]
2nd Idenitity: [...]
A: \(\int_V \phi \nabla^2 \psi dV = - \int_V \nabla \phi \cdot \nabla \psi dV + \int_S \phi \nabla \psi \cdot d\mathbf{S}\); \(\int_V (\phi \nabla^2 \psi - \psi \nabla^2 \phi)dV = \int_S (\phi \nabla \psi - \psi \nabla \phi)\cdot d\mathbf{S}\)

Q: Definition. Harmonic Functions.
A: Solutions to Laplace equation \(\nabla \psi^2 = 0\)

Q: Claim. Mean Value Property. Suppose that \(\psi\) is harmonic in a region V that includes the solid sphere with boundary \(S_r : |\mathbf{x - a}| = R\). Then the value of \(\psi\) at \(\bf a\), the centre of the sphere, is given by \(\psi(\mathbf{a}) = \overline{\psi}(R)\) where \[\overline{\psi}(R) = \frac{1}{4\pi R^2} \int_{S_R} \psi(\mathbf{x})dS\]
A: In spherical coordinates centred on a, the area element is \(dS = r^2 \sin \theta d\theta d\phi\), so \[\overline{\psi}(r) = \frac{1}{4\pi} \int d\phi \int d\theta \sin \theta \psi(r, \theta, \phi)\]and \[\frac{d \overline{\psi}(R)}{dr} = \frac{1}{4\pi} \int d\phi \int d\theta \sin \theta \frac{\partial \psi(R)}{\partial r} = \frac{1}{4\pi R^2} \int_{S_R} \frac{\partial \psi(R)}{\partial r} dS = \frac{1}{4\pi R^2} \int_{S_R} \nabla \psi \cdot d\mathbf{S} = \int_{Ball} \nabla^2 \psi dV = 0\]But clearly \(\overline{\psi}(R) \to \psi(\mathbf{a})\) as \(R \to 0\) so we must have \(\overline{\psi}(R) = \psi(\mathbf{a})\) for all R

Q: Claim. A harmonic function can have neither a maximum not minimum in the interior of a region V. Any maximum or minimum must lie on the boundary \(\partial V\)
A: If \(\psi\) has a local maximum at \(\mathbf{a}\) in V, then there exists an \(\epsilon\) such that \(\psi(\mathbf{x}) < \psi(\mathbf{a})\) for all \(|\mathbf{x-a}| < \epsilon\). But we know from the mean value property that \(\overline{\psi}(R) = \psi(\mathbf{a})\) and this contradicts the assumption for any \(0 < R < \epsilon\).

Q: Clain. The Poisson equation \(\nabla^2 \psi(\mathbf{x}) = -\rho(\mathbf{x})\) has the integral solution \[\psi(\mathbf{x}) = \frac{1}{4\pi} \int_{V'} \frac{\rho(\mathbf{x}')}{|\mathbf{x - x'}|} dV'\]where the integral is over a region V' parametrised by \(\mathbf{x'}\)
A: \[\nabla^2 \psi = \frac{1}{4\pi} \int_{V'} \rho(\mathbf{x}') \nabla^2 \left(\frac{1}{|\mathbf{x-x'}|} \right ) dV'\]since the \(\mathbf{x'}\) is just some arbitrary point.
We then have the result \[\nabla^2 \frac{1}{|\mathbf{x - x'}|} = -4\pi \delta^3 (\mathbf{x - x'})\]The result follows from the definition of the Dirac delta function.

Q: Definition. Tensor.
A: A tensor is somcething whose components transforms like a tensor. \[T'_{i_1\dotsi_p} = R_{i_1j_1}\cdots R_{i_pj_p} T_{j_1\dots j_p}\]Where in this case we are using a tensor of rank p/ a p-tensor.

Q: Proposition. Tensors as Maps. A tensor of rank p is a multilinear map that takes p vectors, \(\bf a,b, \dots ,c\) and spits out a number in \(\mathbb{R}\), \[[...]}\]where multilinear means that T is linear in each of the entries individually.
A: T(\mathbf{a,b,\dots,c}) = T_{i_1i_2\dots i_p}a_{i_1}b_{i_2} \dots c_{i_p
Back Extra: T is independent of the choice of bases, and we can equally well write \[T(\mathbf{a,b,\dots,c}) = T'_{i_1i_2\dots i_p}a'_{i_1}b'_{i_2} \dots c'_{i_p}= (R_{i_1j_1}R_{i_2j_2} \dots R_{i_pj_p}T_{i_1i_2\dots i_p})(R_{i_1k_1}a_{k_1})(R_{i_2k_2}b_{k_2})\dots (R_{i_pk_p} c_{k_p}) = T_{j_1j_2 \dots j_p} a_{j_1}b_{j_2}\dots c_{j_p}\]

Q: Proposition. Tensors as maps between vectors. A rank p tensor can be viewed as taking (p-1) vectors and spitting out a single vector. \[[...]\]
A: a_i = T_{ij_1 \dots j_{p-1} } b_{j_1} \dots c_{j_{p-1} }
Back Extra: For example with a rank 2 tensor, \[\mathbf{u} = T\mathbf{v} \implies u_i = T_{ij}v_j\]

Q: Definition. Tensor Product. Is S is a tensor of rank p and T a tensor of rank q, then the tensor product \(S \otimes T\) is a rensor of rank p+q defined by, \[[...]\]
A: (S \otimes T)_{i_1 \dots i_p j_1 \dots j_q} =S_{i_1 i_2 \dots i_p} T_{j_1 \dots j_q}

Q: Definition. Contraction. Given a tensor T of rank p, we can construct a new tensor S of rank p-2 by contracting on two indices using \(\delta_{ij}\), \[[...]\]
A: S_{k_1 \dots k_{p-2} } = \delta_{ij} T_{ijk_1 \dots k_{p-2} }
Back Extra: For a rank 2 tensor, the contraction is what we call the trace.

Q: Definition. Quotient Rule. [...]
A: If for every tensor \(u_{j_1 \dots j_p}\), the output \(v_{i_1 \dots i_p}\) is a tensor, then \(T_{i_1 \dots i_p j_1 \dots j_q}\) are the components of a tensor.
Back Extra: It's sufficient to restrict attention to tensors u formed from the tensor product of vectors \(u_{j_1 \dots j_q} = c_{j_1} \dots d_{j_q}\). Then by assumption, \(v_{i_1 \dots i_p} = T_{i_1 \dots i_p j_1 \dots j_q}\) is a tensor. If we then contract with p further vectors \(\bf a, \dots b\) then \(v_{i_1 \dots i_p}a_{i_1} \dots b_{i_p} = T_{i_1 \dots i_p j_1 \dots j_q}a_{i_1} \dots b_{i_p} c_{j_1} \dots d_{j_q}\) is necessarily a scalar. This is then enough to ensure the correct transformation rule for the components \(T_{i_1 \dots i_p j_1 \dots j_q}\).

Q: Definition. Invariant and Isotropic Tensors. A tensor is said to be invariant under a given rotation R if \[[...]}\]
A: T'_{i_1 \dots i_n} = R_{i_1 j_1} \dots R_{i_n j_n} T_{j_1 \dots j_n} = T_{i_1 \dots i_n
Back Extra: A tensor that is invariant under all rotations R is said to be isotropic.

Q: Definition. Symmetric or Antisymmetric Tensors. A tensor is said to be symmetric or anti-symetric in the indices i and j if it obeys \[[...]\]If a tensor is (anti)-symmetric in one coordinate system then it is (anti)-symmetric in any coordinate system. \[[...]}\]
A: T_{ijp \dots q} = \pm T_{jip\dots q}; T'_{ijp \dots q} = R_{ik}R_{jl}R_{pr} \dots R_{qs} T_{klr\dots s} = \pm R_{ik}R_{jl}R_{pr}\dots R_{qs} T_{lkr \dots s} = \pm T'_{jip\dots q

Q: Claim. The only non zero isotropic tensors in \(\mathbb{R}^3\) of rank 1,2 or 3 are \(T_{ij} = \delta_{ij}\) and \(T_{ijk} = \epsilon_{ijk}\) up to a multiplicative constant.
A: For rank 1, we need \[T_i' = R_{ij}T_j \;\;\;\;\text{with } R_{ij} = \left( \begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right )\] Requiring \(T'_i = T_i\) gives \(T_1 = T_2 = 0\). Clearly a similar argument, using a different R, also gives \(T_3 = 0\)
For a rank 2 tensor consider the transformation \[T'_{ij} = R_{ik}R_{jl}T_{kl} \text{ with } R_{ij} = \left( \begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{matrix} \right )\]This gives \(T'_{13} = T_{23}\), \(T'_{23} = - T_{13}\) so if \(T'{ij} = T_{ij}\), we must have \(T_{13} = T_{23} = 0\). Meanwhile \(T'_{11} = T_{22}\). Similar arguments tell you that all off-diagonals are 0 and the diagonal elements are equal. So \(T_{ij} = \delta_{ij}\).
For rank 3, use both of these previous matrices to get the result that \(T_{ijk} = \beta \epsilon_{ijk}\). Do it!

Q: Proposition. Derivatives of Tensors. We can differentiate a rank p tensor q times to get a rank p+q tensor by \[[...]\]
A: X_{i_1 \dots i_1 j_1 \dots j_p} = \frac{\partial}{\partial x_{i_1} } \dots \frac{\partial }{\partial x_{i_q} } T_{j_1 \dots j_p}(\mathbf{x})
Back Extra: We can verify that this is a tensor, in a new basis \(x'_i = R_{ij}x_j\) and so \[\frac{\partial x_i'}{\partial x_j} = R_{ij} \implies \frac{\partial}{\partial x_i'} \frac{\partial}{\partial x_j} = R_{ij} \frac{\partial}{\partial x_j}\]We can form a rank 2 tensor field if we start with a vector field \(\bf F(x)\). By splitting up the rank 2 tensor field into anti-symmetric, trace and symmetric traceless pieces we get the following.
Anti-symmetric: \(A_{ij}(\mathbf{x}) = \epsilon_{ijk} B_k(\mathbf{x}), B_k = \frac{1}{2}\epsilon_{ijk} \frac{\partial F_i}{\partial F_j} = -\frac{1}{2} (\nabla \times \mathbf{F})_k\)
Trace: \(Q = \frac{\partial F_i}{\partial x_i} = \nabla \cdot \mathbf{F}\)
Symmetric, traceless: \(P_{ij}(\mathbf{x}) = \frac{1}{2}\left ( \frac{\partial F_i}{\partial x_j} + \frac{\partial F_j}{\partial x_i}\right ) - \frac{1}{3} \nabla \cdot \mathbf{F}\)