Solution to Laplace’s Equation

This post deals with the familiar (to the physics student) Laplace’s equation. I am solving this equation in the context of physics, instead of a pure mathematical perspective. This problem is considered most extensively in the context of electrostatics. This equation is usually considered in the spherical polar coordinate system. A lot of finer details are also considered in a mathematical physics course where the topic of spherical harmonics is discussed. Assuming spherical-polar coordinates, Laplace’s equation is

\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\bigg\{r^{2}\frac{\partial \psi}{\partial r}\bigg\}+\frac{1}{r^{2}\sin{\theta}}\frac{\partial}{\partial \theta}\bigg\{\sin{\theta}\frac{\partial \psi}{\partial \theta}\bigg\}+\frac{1}{r^{2}\sin{\theta}}\frac{\partial^{2}\psi}{\partial \phi^{2}}=0, (1)

Suppose that the function \psi\rightarrow \psi(r,\theta,\phi). Furthermore, by the method of separation of variables we suppose that the solution is a product of eigenfunctions of the form R(r)Y(\theta,\phi). Hence, Laplace’s equation becomes

\displaystyle \frac{Y}{r^{2}}\frac{d}{dr}\bigg\{r^{2}\frac{d^{2}R}{dr^{2}}\bigg\}+\frac{R}{r^{2}\sin{\theta}}\frac{\partial}{\partial \theta}\bigg\{\sin{\theta}\frac{\partial Y}{\partial \theta}\bigg\}+\frac{R}{r^{2}\sin^{\theta}}\bigg\{\frac{\partial^{2}Y}{\partial \phi^{2}}\bigg\}=0. (2)

Furthermore, we can separate further the term Y(\theta, \phi) into \Theta(\theta)\Phi(\phi). Rewriting (2) and multiplying by r^{2}\sin^{2}{\theta}R^{-1}Y^{-1}, we get

\displaystyle \frac{1}{R}\frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}+\frac{\sin{\theta}}{\Theta}\frac{d}{d\theta}\bigg\{\sin{\theta}\frac{d\Theta}{d\theta}\bigg\}+\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=0.(3)

Bringing the radial and angular component to the other side of the equation and setting the azimuthal component equal to a separation constant -m^{2}, yielding

\displaystyle -\frac{1}{R}\frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}-\frac{\sin{\theta}}{\Theta}\frac{d}{d\theta}\bigg\{\sin{\theta}\frac{d\Theta}{d\theta}\bigg\}=\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=-m^{2}.

Solving the right-hand side of the equation we get

\displaystyle \Phi(\phi)=A\exp({+im\phi}). (4)

Now, we set the azimuthal component equal to m^{2} and carry out the derivative in the angular component to get the Associated Legendre equation:

\displaystyle \frac{d}{d\mu}\bigg\{(1-\mu^{2})\frac{d\Theta}{d\mu}\bigg\}+\bigg\{l(l+1)-\frac{m^{2}}{1-\mu^{2}}\bigg\}\Theta=0, (5)

where I have let \mu=\cos{\theta}. The solutions to this equation are known as the associated Legendre functions. However, instead of solving this difficult equation by brute force methods (i.e. power series method), we consider the case for which m^{2}=0. In this case, Eq.(5) simplifies to Legendre’s differential equation discussed previously. Instead of quoting the explicit form of the Legendre polynomials, we equivalently state the Rodrigues formula for these polynomials

\displaystyle P_{l}(\mu)=\frac{1}{2^{l}l!}\frac{d^{l}}{d\mu^{l}}(\mu^{2}-1)^{l}. (6)

However, the solutions to Eq.(5) are the associated Legendre functions, not polynomials. Therefore, we use the following to determine the associated Legendre functions from the Legendre polynomials:

\displaystyle {\Theta_{l}}^{m}(\mu) = (1-\mu^{2})^{\frac{|m|}{2}}\frac{d^{|m|}}{d\mu^{|m|}}P_{l}(\mu). (7)

For real solutions |m|\leq l and for complex solutions -l \leq m \leq l, thus we can write the solutions as series whose indices run from 0 to l for the real-values and from -l to l for the complex-valued solutions.

Now we turn to the radial part which upon substitution of -l(l+1) in place of the angular component, we get

\displaystyle \frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}+l(l+1)R=0,

and if we evaluate the derivatives in the first term we get an Euler equation of the form

\displaystyle r^{2}\frac{d^{2}R}{dr^{2}}+2r\frac{dR}{dr}+l(l+1)R=0. (8)

Let us assume that the solution can be represented as

\displaystyle R(r)=\sum_{j=0}^{\infty}a_{j}r^{j}.

Taking the necessary derivatives and substituting into Eq.(8) gives

\displaystyle r^{2}\sum_{j=0}^{\infty}j(j-1)a_{j}r^{j-2}+\sum_{j=0}^{\infty}2rja_{j}r^{j-1}+l(l+1)\sum_{j=0}^{\infty}a_{j}r^{j}=0. (9)

Simplifying the powers of r we find that

\displaystyle \sum_{j=0}^{\infty}[j(j-1)+2j+l(l+1)]a_{j}r^{j}=0.

Now, since \displaystyle \sum_{j=0}^{\infty}a_{j}r^{j}\neq 0, this means that

\displaystyle j(j-1)+2j+l(l+1)=0. (10)

We can simplify this further to get

\displaystyle j^{2}+l^{2}+j+l=0.

Factoring out (j+l), we arrive at

\displaystyle (j+l)(j+l+1)=0 \implies j=-l ; j=-l-1.

Since this must be true for all values of l, the solution therefore becomes

\displaystyle R(r)= \sum_{l=0}^{\infty}Br^{-l}+Cr^{-l-1}. (11)

Thus, we can write the solution to Laplace’s equation as

\displaystyle \psi(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=0}^{\infty}[Br^{-l}+Cr^{-l-1}]\Theta_{l}^{m}(\mu)A\exp({im\phi}),  (12)

for real solutions, and

\displaystyle \psi(r,\theta,\phi)=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}[Br^{-l}+Cr^{-l-1}]\Theta_{l}^{m}(\mu)A\exp({im\phi}), (13)

for complex solutions.

Solution to Legendre’s Differential Equation

Typically covered in a first course on ordinary differential equations, this problem finds applications in the solution of the Schrödinger equation for a one-electron atom (i.e. Hydrogen). In fact, this equation is a smaller problem that results from using separation of variables to solve Laplace’s equation. One finds that the angular equation is satisfied by the Associated Legendre functions. However, if it is assumed that m=0 then the equation reduces to Legendre’s equation.


The equation can be stated as

\displaystyle (1-x^{2})\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+l(l+1)y(x)=0. (1)

The power series method starts with the assumption

\displaystyle y(x)=\sum_{j=0}^{\infty}a_{j}x^{j}. (2)

Next, we require the first and second order derivatives

\displaystyle \frac{dy}{dx}=\sum_{j=1}^{\infty}ja_{j}x^{j-1}, (3)


\displaystyle \frac{d^{2}y}{dx^{2}}=\sum_{j=2}^{\infty}j(j-1)a_{j}x^{j-2}. (4)

Substitution yields

\displaystyle (1-x^{2})\sum_{j=2}^{\infty}j(j-1)a_{j}x^{j-2}-2x\sum_{j=1}^{\infty}ja_{j}x^{j-1}+l(l+1)\sum_{j=0}^{\infty}a_{j}x^{j}=0, (5)

Distribution of the first terms gives

\displaystyle \sum_{j=2}^{\infty}j(j-1)a_{j}x^{j-2}-\sum_{j=0}^{\infty}j(j-1)a_{j}x^{j}-2ja_{j}x^{j}+l(l+1)a_{j}x^{j}=0, (6)

where in the second summation term we have rewritten the index to start at zero since the terms for which j=0,1 are zero, and hence have no effect on the overall sum. This allows us to write the sum this way. Next, we introduce a dummy variable. Therefore, let m=j-2\implies j=m+2\implies m+1=j-1. Thus, the equation becomes

\displaystyle \sum_{j=0}^{\infty}\bigg\{(j+2)(j+1)a_{j+2}-j(j-1)a_{j}-2ja_{j}+l(l+1)a_{j}\bigg\}x^{j}=0. (7)

In order for this to be true for all values of j, we require the coefficients of x^{j} equal zero. Solving for a_{j+2} we get

\displaystyle a_{j+2}=\frac{j(j+1)-l(l+1)}{(j+2)(j+1)}. (8)

It becomes evident that the terms a_{2},a_{3},a_{4},... are dependent on the terms a_{0} and a_{1}. The first term deals with the even solution and the second deals with the odd solution. If we let p=j+2 and solve for a_{p}, we arrive at the term a_{p-2} and we can obtain the next term a_{p-4}. (I am not going to go through the details. The derivation is far too tedious. If one cannot follow there is an excellent video on YouTube that goes through a complete solution of Legendre’s ODE where they discuss all finer details of the problem. I am solving this now so that I can solve more advanced problems later on.) A pattern begins to emerge which we may express generally as:

\displaystyle a_{p-2n}=\frac{(-1)^{n}(2p-2n)!}{(n!)(2^{p})(p-n)!(p-2n)!}. (9)

Now, for even terms 0\leq n \leq \frac{p}{2} and for odd terms 0 \leq n \leq \frac{p-1}{2}. Thus for the even solution we have

\displaystyle P_{p}(x)=\sum_{n=0}^{\frac{p}{2}}\frac{(-1)^{n}(2p-2n)!}{(n!)(2^{p})(p-n)!(p-2n)!}, (10.1)

and for the odd solution we have

\displaystyle P_{p}(x)=\sum_{n=0}^{\frac{p-1}{2}}\frac{(-1)^{n}(2p-2n)!}{(n!)(2^{p})(p-n)!(p-2n)!}. (10.2)

These two equations make up the even and odd solution to Legendre’s equation. They are an explicit general formula for the Legendre polynomials.  Additionally we see that they can readily be used to derive Rodrigues’ formula

\displaystyle \frac{1}{2^{p}p!}\frac{d^{p}}{dx^{p}}\bigg\{(x^{2}-1)^{p}\bigg\}, (11)

and that we can relate Legendre polynomials to the Associated Legendre function via the equation

\displaystyle P_{l}^{m}(x)= (1-x^{2})^{\frac{|m|}{2}}\frac{d^{|m|}P_{l}(x)}{dx^{|m|}}, (12)

where I have let p=l so as to preserve a more standard notation.  This is the general rule that we will use to solve the associated Legendre differential equation when solving the Schrödinger equation for a one-electron atom.



Solution to the three-dimensional Heat Equation

After taking a topics course in applied mathematics (partial differential equations), I found that there were equations that I should solve since I would later see those equations embedded into other larger-scale equations. This equation was Laplace’s equation (future post). Once I solved this equation, I realized that it becomes a differential operator when acted upon a function of at least two variables. Thus, I could solve equations such as the Schrödinger equation using a three-dimensional laplacian in spherical-polar coordinates (another future post) and the three-dimensional heat equation. I will be solving the latter.


The heat equation initial-boundary-value-problem is therefore

\frac{\partial u}{\partial t}=\frac{1}{c\rho}\nabla^{2}u+Q(x,y,z), (1.1)

subjected to the boundary conditions

u(0,0,0,t)=0, (1.2)

and the initial condition

u(x,y,z,0)= \xi(x,y,z). (1.8)

Now, this solution is not specific to a single thermodynamic system, but rather it is a more general solution in a mathematical context. However, I will be appropriating certain concepts from physics for reasons that are well understood (i.e. that time exists on the interval I=0\leq t < +\infty ). It is nonphysical or nonsensical to speak of negative time.

To start, consider any rectangular prism in which heat flows through the volume, from the origin to the point (X,Y,Z).At a time t=0, the overall heat of the volume can be regarded to be a function \xi(x,y,z). After a time period \delta t=t has passed, the heat will have traversed to the point (X,Y,Z) from the origin (think of the heat traveling along the diagonal of a cube). The goal is to find the heat as a function of the three spatial components and a single time component.  Furthermore, the boundary conditions maintain that at the origin and the final point,  the heat vanishes. In other words, these points act as heat sinks. (A sink is a point where energy can leave the system.)

To simplify the notation, we define \textbf{r}=x\hat{i}+y\hat{j}+z\hat{k}. Thus the initial-boundary-value-problem becomes

\frac{\partial u}{\partial t}=\frac{1}{c\rho}\frac{\partial^{2}u}{\partial \textbf{r}^{2}}, (2.1)

where u\rightarrow u(\textbf{r},t). Also, this definition also reduces the three-dimensional laplacian to a second-order partial derivative of u. The boundary and initial conditions are then


u(R,t)=0, (2.3)

u(\textbf{r},0)=\xi(\textbf{r}). (2.2)

Now, we assume that the solution is a product of eigenfunctions of the form

u(\textbf{r},t)=\alpha(\textbf{r})\Gamma(t). (3)

Taking the respective derivatives and dividing by the assumed form of the solution, we get

\frac{\Gamma^{\prime}(t)}{\Gamma(t)}=\frac{1}{c\rho}\frac{\alpha^{\prime\prime}(\textbf{r})}{\alpha(\textbf{r})}+Q(\textbf{r}). (4)

Now, Eq.(4) can be equal to three different values, the first of which is zero, but this solution does not help in any way,  nor is it physically significant since it produces a trivial solution. The second is \lambda^{2}>0, \lambda \neq 0, so we can apply to the time dependence equation which then becomes

\frac{\Gamma^{\prime}(t)}{\Gamma(t)}=\lambda^{2}, (5.1)

whose solution is

\Gamma(t)=\Gamma_{0}\exp({\lambda^{2}t}). (5.2)

The third case where \lambda^{2}<0, \lambda \neq 0 allows us to write the spatial equation upon rearrangement as

\alpha^{\prime\prime}(\textbf{r})+c\rho(\lambda^{2}+Q)\alpha(\textbf{r})=0, (6.1)

whose solution is

\alpha(\textbf{r})=c_{1}\cos({\sqrt[]{c\rho(\lambda^{2}+Q)}\textbf{r}})+c_{3}\sin({\sqrt[]{c\rho(\lambda^{2}+Q)}\textbf{r}}), (6.2)

where c_{3}=-c_{2}. Next we apply the boundary conditions. Let \textbf{r}=0 in \alpha(\textbf{r}):

\alpha(0)=c_{1}+0=0 \implies c_{1}=0 \implies \alpha(\textbf{r})=C\sin(\sqrt[]{c\rho(\lambda^{2}+Q)} \textbf{r}). (7.1)

and let \textbf{r}=R in (7.1) to get

\alpha(R)\implies \sin({\sqrt[]{c\rho(\lambda^{2}+Q)}R})=0\implies \sqrt[]{c\rho(\lambda^{2}+Q)}R^{2}=(n\pi)^{2}.

Solving for \lambda^{2} above gives

\lambda^{2}=\frac{1}{c\rho}\bigg\{\frac{n\pi}{R}\bigg\}^{2}-Q. (8)

Substituting into  Eq.(7.1) and simplifying gives

\alpha(\textbf{r})=C\sin\bigg\{\frac{n\pi}{c\rho}\bigg\}^{2}\bigg\{\frac{\textbf{r}}{R}\bigg\}. (9)

To find as many solutions as possible we construct a superposition of solutions of the form

u(\textbf{r},t)=u_{l}(\textbf{r},t)=\sum_{l=0}^{\infty}\alpha_{l}(\textbf{r})\Gamma(t). (10)

Rewriting \alpha(\textbf{r}) as \alpha_{l}(\textbf{r}) and using the solution for the time dependence, and also let the coefficients form a product equivalent to the indexed coefficient A_{l}, we arrive at the solution for the heat equation:

u(\textbf{r},t)=\sum_{l=0}^{\infty}A_{l}\sin\bigg\{\frac{l\pi}{c\rho}\bigg\}^{2}\bigg\{\frac{\textbf{r}}{R}\bigg\}\exp(\lambda^{2}t). (11)

Now suppose that t=0 in Eq.(11). In this case we get the initial heat distribution given by \xi(\textbf{r}):

\xi(\textbf{r})=\sum_{l=0}^{\infty} A_{l}\sin\bigg\{\frac{l\pi}{c\rho}\bigg\}^{2}\bigg\{\frac{\textbf{r}}{R}\bigg\}. (12)


Deriving the speed of light from Maxwell’s equations

We are all familiar with the concept of the speed of light. It is the speed beyond which no object may travel. Many seem to associate Einstein for the necessity of this universal constant, and while it is inherent to his theory of special and general theories of relativity, it was not necessarily something he discovered. It is, in fact, a consequence of the Maxwell equations from my first post. I will be deriving the speed of light quantity using the four field equations of electrodynamics, and I will explain how Einstein used this fact to challenge Newtonian relativity in his theory of special relativity (I am not as familiar with general relativity).  The reason for this post is just to demonstrate the origin of a well-known concept; the speed of light.

We start with Maxwell’s equations of electrodynamics

\nabla \cdot \textbf{E}=\frac{\rho}{\epsilon_{0}}, (1)

\nabla \cdot \textbf{B}=0, (2)

\nabla \times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}, (3)

\nabla \times \textbf{B}=\mu_{0}\textbf{j}+\mu_{0}\epsilon_{0}\frac{\partial \textbf{E}}{\partial t}. (4)

Now, we let \rho =0, which means that the charge density must be zero, and we also let the current density \textbf{j}=0. Moreover, note that the form of the wave equation as

\frac{\partial^{2} u}{\partial t^{2}}=\frac{1}{v^{2}}\nabla^{2}u. (5)

This equation describes the change in position of material in three dimensions (choose whichever coordinate system you like) propagating through some amount of time, with some velocity v.

After making these assumptions, we arrive at

\nabla \cdot \textbf{E}=0, (6)

\nabla \cdot \textbf{B}=0, (7)

\nabla \times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}, (8)

\nabla \times \textbf{B}=\mu_{0}\epsilon_{0}\frac{\partial \textbf{E}}{\partial t}. (9)

Also note the vector identity\nabla \times (\nabla \times \textbf{A})=\nabla(\nabla\cdot\textbf{A})-\nabla^{2}\textbf{A}. Now, take the curl of Eqs.(8) and (9), and we get

\frac{1}{\mu_{0}\epsilon_{0}}\nabla^{2}\textbf{E}=\frac{\partial^{2}\textbf{E}}{\partial t^{2}}, (10)


\frac{1}{\mu_{0}\epsilon_{0}}\nabla^{2}\textbf{B}=\frac{\partial^{2}\textbf{B}}{\partial t^{2}}, (11)

where we have used Eqs. (6), (7), (8), and (9) to simplify the expressions. Eqs. (10) and (11) are the electromagnetic wave equations. Note the form of these equations and how they compare to Eq. (5). They are identical, and upon inspection one can see that the velocity with which light travels is

\frac{1}{c^{2}}=\frac{1}{\mu_{0}\epsilon_{0}} \implies c=\sqrt[]{\mu_{0}\epsilon_{0}}, (12)

where \mu_{0} is the permeability of free space and \epsilon_{0} is the permittivity of free space.

Most waves on Earth require a medium to travel. Sound waves, for example are actually pressure waves that move by collisions of the individual molecules in the air.  For some time, light was thought to require a medium to travel. So it was proposed that since light can travel through the vacuum of space, there must exist a universal medium dubbed “the ether”. This “ether” was sought after most notably in the famous Michelson-Morley experiment, in which an interferometer was constructed to measure the Earth’s velocity through this medium. However, when they failed to find any evidence that the “ether” existed, the new way of thinking was that it didn’t exist. It turned out that light doesn’t need a medium to travel through space. Technically-speaking, space itself acts as the medium through which light travels.

 In Newtonian relativity, it was assumed that time and space were separate constructs and were regarded as absolute. In other words, it was the speed that changed. What this meant is that even as speeds became very large, space and time remained the same. What Einstein did was that he saw the consequence of Maxwell’s equations and regarded this speed as absolute, and allowed space and time (really spacetime) to vary. In Einstein’s theory of special relativity, as one approaches the speed of light, time slows down, and objects become contracted. These phenomena are known as time dilation and length contraction:

\delta t = \frac{\delta t_{0}}{\sqrt[]{1-v^{2}/c^{2}}}, (13)

\delta l = l_{0}\sqrt[]{1-v^{2}/c^{2}}. (14)

These phenomena will be discussed in more detail in a future post. Thus, Maxwell’s formulation of the electrodynamic field equations led Einstein to change the way we perceive the fundamental concepts of space and time.

Electron Scattering of a Step Potential

The following post was initially one of my assignments for an independent study in modern physics in my penultimate year as an undergraduate. While studying this problem the text that I used to verify my answer was:

R. Eisberg and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. John Wiley & Sons. 1985. 6. 

One of the hallmarks of quantum theory is the Schrödinger equation. There are two forms: the time-dependent and the time-independent equation. The former can be turned into the latter by way of assuming stationary states in which case there is no time evolution (i.e. the time derivative \partial_{t} \Psi(x,t)=0. \partial_{t}\Psi(x,t) denotes the derivative of the wavefunction \Psi(x,t) with respect to time.)

The wavefunction describes the state of a system, and it is found by solving the Schrödinger equation. In this post, I’ll be considering a step potential in which

V(x) = V_{0},  x < 0,  (1.1)

V(x) = 0, x > 0.  (1.2)

After solving for the wavefunction, I will calculate the reflection and transmission coefficients for the case where the energy of the electron is less than that of the step potential E < V_{0}.

First, we assume that we are dealing with stationary states, by doing so we assume that there is no time-dependence. The wavefunction \Psi(x,t) becomes an eigenfunction (eigen– is German for “characteristic” e.g. characteristic function, characteristic value (eigenvalue), and so on), \psi(x). There are requirements for this eigenfunction in the context of quantum mechanics: eigenfunction \psi(x) and its first order spatial derivative \frac{d\psi(x)}{dx} must be finite, single-valued, and continuous. Using the wavefunction, we write the time-independent Schrödinger equation as

-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x) = E\psi(x), (2)

where \hbar is the reduced Planck’s constant \hbar \equiv h(2\pi)^{-1}, m is the mass of the particle, V(x) represents the potential, and E is the energy.

Now, in electron scattering there are two cases regarding a step potential: the case for which E < V_{0} and the case for which E>V_{0} The focus of this post is the former case. In such a case, the potential is given mathematically by Eqs. (1.1) and (1.2) and can be depicted by the image below:


Image Credit:

The first part of this problem is to solve for \psi(x) when x<0. Then Eq.(2) becomes

\frac{d^{2}\psi_{I}(x)}{dx^{2}}=-\kappa_{I}^{2}\psi_{I}(x), (3)

where \kappa_{I}^{2}\equiv \frac{-2mE}{\hbar^{2}}. Eq.(3) is a second order linear homogeneous ordinary differential equation with constant coefficients and can be solved using a characteristic equation. We assume that the solution is of the general form


which upon taking the first and second order spatial derivatives and substituting into Eq.(3) yields


We can factor out the exponential and recalling that the graph of e^{x} \neq 0 (except in the limit x\rightarrow -\infty)we can then conclude that


Hence, r = \pm i\kappa_{I}. Therefore, we can write the solution Schrödinger equation in the region x < 0 as

\psi_{I}(x) = A\exp{+i\kappa_{I}x}+B\exp{-i\kappa_{I}x}. (4)

This is the eigenfunction for the first region. Coefficients A and B will be determined later.

Now we can use the same logic for the Schrödinger equation in the region x > 0 where V(x)=V_{0}:

\frac{d^{2}\psi_{II}(x)}{dx^{2}}= -\kappa_{II}^{2}\psi_{II}(x), (5)

where \kappa_{II}\equiv \frac{\sqrt[]{2m(E-V_{0})}}{\hbar}. The general solution for this region is

\psi_{II}(x) = C\exp{+i\kappa_{II}x}+D\exp{-i\kappa_{II}x}. (6)

Now the next step is taken using two different approaches: the first using a mathematical argument and the other from a conceptual interpretation of the problem at hand.  The former is this: Suppose we let x\rightarrow \infty. What results is that the first term on the right hand side of Eq.(6) diverges (i.e. becomes arbitrarily large). The second term on the right hand side ends up going to zero (it converges). Therefore, D remains finite, hence D\neq 0. However, in order to suppress the divergence of the first term, we let C=0. Thus we arrive at the eigenfunction for x>0

\psi_{II}(x) = D\exp{-i\kappa_{II}x}. (7)

The latter argument is this: In the region x < 0, the first term of the solution represents a wave propagating in the positive x-direction, while the second denotes a wave traveling in the negative x-direction. Similarly, in the region where x > 0, the first term corresponds to a wave traveling in the positive x-direction. However, this cannot be because the energy of the wave is not sufficient enough to overcome the potential. Therefore, the only term that is relevant here for this region is the second term, for it is a wave propagating in the negative x-direction.

We now determine the coefficients A and B. Recall that the eigenfunction must satisfy the following continuity requirements

\psi_{I}(x)=\psi_{II}(x), (8.1)

\psi^{\prime}_{I}(x)=\psi^{\prime}_{II}(x), (8.2)

evaluated when x=0. Doing so in Eqs. (4) and (7), and equating them we arrive at the continuity condition for \psi(x)

A+B = D. (9)

Taking the derivative of \psi_{I}(x) and \psi_{II}(x) and evaluating them for when x=0 we arrive at

A-B = \frac{i\kappa_{II}}{i\kappa_{I}}D. (10)

If we add Eqs. (9) and (10) we get the value for the coefficient A in terms of the arbitrary constant D

A =\frac{D}{2}(1+\frac{i\kappa_{II}}{i\kappa_{I}}. (11)

Conversely, if we subtract (9) and (10) we get the value for B in terms of D

B = \frac{D}{2}(1-\frac{i\kappa_{II}}{i\kappa_{I}}. (12)


Now the reflection coefficient defined as

R = \frac{B^{*}B}{A^{*}A}, (13)

which is the probability that an incident electron (wave) will be reflected. On the other hand, the transmission coefficient is the probability that an electron will be transmitted through the potential (e.g. barrier potential).  These two also must satisfy the relation

R+T =1. (14)

This means that the probability that the electron will be reflected or transmitted is 100%. Therefore, to evaluate R (the reason why I don’t calculate T will become apparent shortly), we take the complex conjugate of A and B and using them in Eq.(13) we get

R = \frac{\frac{D}{2}(1+\frac{\kappa_{II}}{\kappa_{I}})\frac{D}{2}(1-\frac{\kappa_{II}}{\kappa_{I}})} {\frac{D}{2}(1-\frac{\kappa_{II}}{\kappa_{I}}) \frac{D}{2}(1+\frac{\kappa_{II}}{\kappa_{I}})} = 1. (15)

What this conceptually means is that the probability that the electron is reflected is 100%. This implies that it is impossible for an electron to be transmitted through the potential for this system.


Basic Equations of Ideal One-Fluid Magnetohydrodynamics (Part II)

Continuing with the derivation of the ideal one-fluid MHD equations, the next equation governs the motion of a parcel of fluid (in this case plasma). This momentum equation stems from the Navier-Stokes’ equation. The derivation of this equation will be reserved for a future post. However, the solution of this equation will not be attempted. (Incidentally, the proof of the existence and uniqueness of solutions to the Navier-Stokes’ equation is one of the Millennium problems described by the Clay Institute of Mathematics.)

The purpose of this post is to derive the momentum equation from the Navier-Stokes’ equation.


The Navier-Stokes’ equation has the form

\frac{\partial \textbf{v}}{\partial t} + (\nabla \cdot \textbf{v})\textbf{v} = \textbf{F} - \frac{1}{\rho}\nabla P+\nu \nabla^{2}\textbf{v}, (1)

where \textbf{F} represents a source of external forces, \textbf{v} is the velocity field, \nabla P is the pressure gradient, \rho is the material density, \nu is kinematic viscosity, and \nabla^{2}\textbf{v} is the laplacian of the velocity field. More specifically, it is a consequence of the viscous stress tensor whose components can cause the parcel of fluid to experience stresses and strains.

Defining the magnetic force per unit volume as

\textbf{F}=\frac{\textbf{J}\times \textbf{B}}{\rho}, (2)

where we recall that \textbf{J} is the current density defined by Ohm’s law in a previous post, and also recall from Basic Equations of Ideal One-Fluid Magnetohydrodynamics (Part I) the equation

\nabla \times \textbf{B}=\mu_{0}\textbf{J}, (3)

if we solve for the current density \textbf{J}, we get

\textbf{J}=\frac{1}{\mu_{0}\rho}[(\nabla \times \textbf{B})\times \textbf{B}], (4)

where \mu_{0} is the permeability of free space. Now we invoke the vector identity

[(\nabla \times \textbf{B})\times \textbf{B}]=(\nabla \cdot \textbf{B})\textbf{B}-\nabla \bigg\{\frac{B^{2}}{2}\bigg\}. (5)

At this point, we assume that we are dealing with laminar flows in which case the term \nu=0. The Navier-Stokes’ equation becomes the Euler equation at this point. (Despite great understanding of classical mechanics, one phenomena for which we cannot account is turbulence and sources of friction, so this assumption is made out of necessity as well as simplicity. For processes in which turbulence cannot be neglected the best we can do in this regard is to parameterize turbulence in numerical models.) Using the vector identity as well as making use of our assumption of laminar flows, ideal one-fluid momentum equation is

\frac{\partial \textbf{v}}{\partial t}+ (\nabla \cdot \textbf{v})\textbf{v}=-\frac{1}{\rho}\nabla \bigg\{P + \frac{B^{2}}{2\mu_{0}}\bigg\}+\frac{(\nabla \cdot \textbf{B})\textbf{B}}{\mu_{0}\rho}, (6)

where the additive term to the pressure \frac{B^{2}}{2\mu_{0}} is the magnetic pressure exerted on magnetic field lines and the additional term \frac{(\nabla \cdot \textbf{B})\textbf{B}}{\mu_{0}\rho} is the magnetic tension acting along the magnetic field lines.



Basic Equations of Ideal One-Fluid Magnetohydrodynamics (Part I)

The field in which the interaction of electrically conducting fluids (i.e. plasmas) and magnetic fields are studied is called magnetohydrodynamics (MHD). As an undergraduate, I investigated the MHD processes occurring in the core of Jupiter. The project had two components: calculation of the magnetic field by coding via MATLAB and investigating the magnetohydrodynamics of the metallic hydrogen in the jovian interior. 

The purpose of this post is to derive the the induction equation of MHD from first principles as well as to describe one of my undergraduate research topics.


We start with Maxwell’s equations of electrodynamics:

\nabla \cdot \textbf{E}=\frac{\rho}{\epsilon_{0}}, (1)

\nabla \cdot \textbf{B}=0, (2)

\nabla \times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}, (3)

\nabla \times \textbf{B}=\mu_{0}\textbf{J}+\mu_{0}\epsilon_{0}\frac{\partial \textbf{E}}{\partial t}. (4)

In the above equations, \textbf{E} is the electric field, \textbf{B} is the magnetic field, \rho is the charge density, \epsilon_{0} and \mu_{0} are the permittivity and permeability of free space, respectively, and \nabla is the del differential operator.  In the context of ideal MHD, we assume that the displacement current (the second term in Eq.(4)) becomes negligible since the velocities considered are believed to be non-relativistic. Therefore, Eq.(4) becomes

\nabla \times \textbf{B}=\mu_{0}\textbf{J}. (5)

A few ancillary expressions are required at this point, namely: Ohm’s law \textbf{J}=\sigma \textbf{E} and the Lorentz force \textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B}). Combining these two gives

\textbf{J}=\sigma(\textbf{E}+\textbf{v}\times\textbf{B}). (6)

Solving for \textbf{J} in Eq.(5) and substitution into Eq.(6) and then solving for the electric field \textbf{E} yields

\textbf{E}=\frac{1}{\mu_{0}\sigma}(\nabla \times \textbf{B})-(\textbf{v}\times\textbf{B}), (7)

Taking the curl of the electric field above, while noting the identity \nabla \times (\nabla \times \textbf{A})=\nabla(\nabla \cdot \textbf{A})-\nabla^{2}\textbf{A} we get the induction equation

\frac{\partial \textbf{B}}{\partial t}= (\textbf{v}\times \textbf{B})+\lambda \nabla^{2}\textbf{B}, (8)

where \lambda is the magnetic diffusivity.

The induction equation is the first of the basic equations of ideal one-fluid MHD.





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