Math Stuff
Various engineering and physics type math problems solved.
Tuesday, November 7, 2017
Addendum to Symmetry of the Riemann-Christoffel Tensor
Symmetry of the Riemann-Christoffel Tensor
Property #1
The first property is the easiest to show and follows directly from the definition and that is that if we switch \(\alpha\) and \(\beta\) (the 3rd and 4th indices) you get the negative value. This is anti-symmetry and it looks like this $$ R^\gamma_{\cdot\delta\beta\alpha} = -R^\gamma_{\cdot\delta\alpha\beta} $$ If you look at the definition, it is evident. $$ \begin{equation} (\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma = R^\gamma_{\cdot\delta\alpha\beta}T^\delta \label{eq:commutator} \end{equation} $$ Switching the indices \(\alpha\) and \(\beta\) $$ \begin{equation} \eqalign { (\nabla_\beta\nabla_\alpha - \nabla_\alpha\nabla_\beta)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= -R^\gamma_{\cdot\delta\alpha\beta}T^\delta } \label{eq:commutator2} \end{equation} $$ Since this is true for an arbitrary tensor, \(T^\gamma\), we have $$ \begin{equation} R^\gamma_{\cdot\delta\beta\alpha} = -R^\gamma_{\cdot\delta\alpha\beta} \label{eq:sym1} \end{equation} $$Property #2
The next symmetric property will require a little more finesse. Now, I don't claim that this is the easiest or most intuitive way to prove this but it does work and it is the one that I came up with. Besides, if you have a better way then it will still be good to see an alternative as that seems to broaden one's understanding. And perhaps you could be so kind as to give me a few clues in the comments. Here we will show that the Riemann-Christoffel tensor with all indices lowered is symmetric if you swap the first two indices with the second two in order. In other words, \(\gamma\leftrightarrow\alpha\) and \(\delta\leftrightarrow\beta\). We start by going to the definition. $$ \begin{equation} \eqalign { R_{\gamma\delta\alpha\beta} &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:rlower} \end{equation} $$ It's clear that the last term \(\Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta}\) is symmetric with \(\{\alpha,\beta\}\leftrightarrow\{\gamma,\delta\}\) because the Christoffel symbols are symmetric in the last two indices. $$ \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} = \Gamma_{\epsilon,\alpha\gamma}\Gamma^\epsilon_{\delta\beta} $$ One down, three to go. The third term requires a few additional steps. $$ \eqalign{ \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} &= \left(S_{\omega\epsilon}\Gamma^\omega_{\gamma\beta}\right)\Gamma^\epsilon_{\alpha\delta} && \text{definition of lowered index} \\ &= \Gamma^\omega_{\gamma\beta} \left( S_{\omega\epsilon} \Gamma^\epsilon_{\alpha\delta} \right) && \text{commutativity/associativity} \\ &= \Gamma^\epsilon_{\gamma\beta} \Gamma_{\epsilon,\alpha\delta} && \text{lower the index of second term and rename the contracted index } \omega \rightarrow \epsilon \\ } $$ We see that the result is the same as the initial expression with the required indices swapped. Two down. The last part of showing this symmetry is a bit more harrowing as we will need to consider the remaining two terms in tandem. We start by rewriting the first two terms of the definition of \(R\) in terms of the definitions of the Christoffel symbols of the surface. The definitions are $$ \begin{equation} \eqalign{ \Gamma^\gamma_{\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S^\gamma} \cr \Gamma_{\gamma,\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S_\gamma} } \label{eq:christoffel} \end{equation} $$ [EDIT: I've amended this proof with a fully intrinsic version of this portion of the proof. If this extrinsic argument gives you the willies, you can go here Where \(\vec{S_\gamma}\) is the covariant basis of the surface. Rewriting the first two terms with these definitions and expanding via the product rule $$ \begin{equation} \eqalign { \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} &= \pard{}{S^\alpha} \left( \pard{\vec{S_\beta}}{S^\delta} \cdot \vec{S_\gamma} \right) - \pard{}{S^\beta} \left( \pard{\vec{S_\alpha}}{S^\delta} \cdot \vec{S_\gamma} \right) \cr &= \frac{\partial^2 \vec{S_\beta}}{\partial S^\alpha \partial S^\delta} \cdot \vec{S_\gamma} + \bbox[5px,border:2px solid red]{\pard{\vec{S_\beta}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\alpha}} - \frac{\partial^2 \vec{S_\alpha}}{\partial S^\beta \partial S^\delta} \cdot \vec{S_\gamma} - \bbox[5px,border:2px solid blue]{\pard{\vec{S_\alpha}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\beta}} } \label{eq:first2terms} \end{equation} $$ Exchanging indices \(\gamma\leftrightarrow\alpha\) and \(\delta\leftrightarrow\beta\) in the term boxed in blue yields the same two terms in reverse order. So, enough of that one. The term boxed in red requires the realization that because of their definition the partial derivatives of the basis vectors have the property that \( \pard{\vec{S_\omega}}{S^\epsilon} = \pard{\vec{S_\epsilon}}{S^\omega} \). So, $$ \eqalign { \pard{\vec{S_\beta}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\alpha} &= \pard{\vec{S_\delta}}{S^\beta} \cdot \pard{\vec{S_\alpha}}{S^\gamma} } $$ Which is the same as swapping the required indices. Lastly, we examine the remaining two terms. "Factoring" out the derivative with respect to \( S^\delta \) gives us $$ \frac{\partial^2 \vec{S_\beta}}{\partial S^\alpha \partial S^\delta} \cdot \vec{S_\gamma} - \frac{\partial^2 \vec{S_\alpha}}{\partial S^\beta \partial S^\delta} \cdot \vec{S_\gamma} = \pard{}{S^\delta} \left( \pard{\vec{S^\beta}}{S^\alpha} - \pard{\vec{S^\alpha}}{S^\beta} \right) $$ By the same argument as the previous step, the difference inside the parentheses is identically zero. Now that all the terms are accounted for, we can state that the Riemann-Christoffel tensor is symmetric in \(\{\alpha,\beta\}\leftrightarrow\{\gamma,\delta\}\).Property #3
Now that we have the above symmetries, showing the anti-symmetry of \( \gamma \leftrightarrow \delta \) is just 3 swaps away. $$ \eqalign { R_{\delta\gamma\beta\alpha} &= R_{\beta\alpha\delta\gamma} && \text{Property #2} \cr &= -R_{\beta\alpha\gamma\delta} && \text{Property #1} \cr &= -R_{\gamma\delta\alpha\beta} && \text{Property #2} \cr } $$Values in the 2 Dimensional Riemann-Christoffel Tensor
The symmetries greatly restrict the degrees of freedom of the values in the tensor. Wikepedia tells me that the degrees of freedom from a "simple calculation" can be found to be $$ N = \frac{n^2(n^2 - 1)}{12} $$ In our case, \( n = 2 \) so we would expect one independent component. Here are all 16 values where \( x \) is the only value we can choose. $$ \eqalign { R_{1212} &= x \cr R_{1221} &= -x \cr R_{2112} &= -x \cr R_{2121} &= x \cr &\text{All other elements are 0} } $$Sunday, November 5, 2017
Lowering the Index on the Riemann-Christoffel Tensor
Sunday, March 19, 2017
Portland to New York Using only Gravity (Part 3)
Friday, March 10, 2017
Underground from Portland to New York (Part 2)
Saturday, March 4, 2017
Underground from Portland to New York (Part 1)
The Brachistochrone
The brachistochrone is a classic problem put forth by Johann Bernoulli and solved by some of the mathematical heavyweights of the day. The basic premise of the problem is that given two points connected by a curve, find the shape of the curve such that a particle moving only under the influence of gravity would travel along the curve in the minimum time. Somewhat surprisingly, this curve is not a straight line. Even Galileo figured this out in 1638 though mistakenly thought the curve was a circular arc. It turns out that the curve is a cycloid but that's not what we're going to prove here.Going Underground
Lets say we want to build a transportation tunnel from Portland, Oregon to New York, New York. We've invented a frictionless surface and we can evacuate the tunnel so there won't be any air resistance. We want to go completely green and only use gravity for travel. Oh, and to add to our perfectly realistic assumptions, we live on an earth of uniform density and of perfectly spherical shape. What is the shape of the tunnel that will get us across in the shortest amount of time? In similar fashion to solving the Brachistochrone problem, we must minimize the functional $$ I=\int_{t0}^{t1}dt = \int_a^b \frac{ds}{v} $$ To get started we'll employ some basic physics. Because there is no energy added to the system, the sum of kinetic and potential energies will be a constant throughout the trip. The kinetic energy, \(T\), is familiar and we can write it directly as $$ T = \frac{1}{2}m_Tv^2 $$ where \(m_T\) is the mass of the train. The potential energy, \(V\), is not quite as straightforward as it is not the familiar \(V=mgh\) we've seen in our entry level physics class. To derive it we first need to find the forces involved. The force of gravity is $$ \vec{F} = -G\frac{m_Em_T}{r^2} \uvec{r} $$ where \(m_E\) is the effective mass of the earth when we are somewhere inside of it at a distance \(r\) from the center. This mass is all the mass that is located less than or equal to a distance \(r\) from the center. So $$ m_E = \frac{4}{3}\rho\pi r^3 $$ where \(\rho\) is the (uniform!) density of the earth. Plugging that back into our force equation gives $$ \vec{F} = -G\frac{\left(\frac{4}{3}\rho\pi r^3\right)m_T}{r^2} \uvec{r} = -\frac{4}{3}G\rho\pi m_T r \uvec{r} $$ (Note the negative sign since the force is directed opposite the radial direction.) Since we know the acceleration due to gravity is \(g = 9.8 \frac{m}{s^2} \) at the surface, \(r = R\) we make the substutition of $$ \eqalign{ g &= \frac{4}{3}G\pi\rho R \cr \vec{F} &= -\frac{m_T g}{R}r\uvec{r} } $$ As an aside, here it is interesting to note that inside the earth, the force actually increases linearly with the distance \(r\). This is because the mass increases as \(r^3\) even though the gravity is an inverse-square law.Now to compute the integral for potential energy, we will choose our reference (\(V=0\)) point to be \(r = R\). The line integral is along the radius to a distance \(r\) from the center. $$ V = -\int_R^r -\frac{m_T g}{R} x dx = \frac{m_Tg}{2R}(r^2 - R^2) $$ Since the total energy \(T+V=0\) at the start and no energy is added to the system we can set the total energy to 0 and solve for the speed as a function of \(r\). $$ \eqalign{ T+V &= \frac{1}{2}m_Tv^2 + \frac{m_Tg}{2R}(r^2 - R^2) = 0 \cr v &= \sqrt{\frac{g}{R}(R^2-r^2)} } $$ We are now at a point where we can write the functional we wish to minimize as $$ \eqalign{ I &= \int_{t0}^{t1}dt = \int_a^b \frac{ds}{v} \cr &= \int_a^b\frac{\sqrt{(dr)^2+(rd\theta)^2}}{\sqrt{\frac{g}{R}(R^2-r^2)}} } $$ To be continued in part deux.
Monday, February 27, 2017
Derivation of Divergence in Cylindrical Coordinates
The vector field in cylindrical coordinates is given by $$ \vec{F} = F_{\rho} \unitv{\rho} + F_{\phi} \unitv{\phi} + F_{z} \unitv{z} $$ We place our tiny cube-like volume in the field \(\vec{F}\) at some point \(\vec{P}=(\rho,\phi,z)\).
First we consider the flux passing through the surfaces normal to the \(z\) direction (the top and bottom faces of the cube). Since the distances involved are small, we can approximate the field passing through these as $$ F_{top} = F_z(\rho,\phi,z + \Delta z) \approx F_z(\rho,\phi,z) + \pard{F_z}{z}(\rho,\phi,z)\frac{\Delta z}{2} $$ It should be noted that the other components of the field don't play a part here because they are parallel to these surfaces so nothing going in the, say, \(\rho\) direction leaves via the \(z\) direction. Similarly for the bottom surface, $$ F_{bottom} = F_z(\rho,\phi,z - \Delta z) \approx F_z(\rho,\phi,z) - \pard{F_z}{z}(\rho,\phi,z)\frac{\Delta z}{2} $$ The areas of these surfaces are equal and independent of \(z\). $$ A_z = \rho\Delta\phi\Delta\rho $$ Therefore, the flux through these z-surfaces is (remembering that the flux through the bottom surface is into the volume and so is negative) $$ \Psi_z = F_{top}A_z - F_{bottom}A_z = (F_{top} - F_{bottom})A_z $$ $$ \Psi_z = (F_z + \pard{z}{F_z}\frac{\Delta z}{2} - (F_z - \pard{F_z}{z}\frac{\Delta z}{2}))\rho\Delta\phi\Delta\rho $$ $$ \Psi_z = \pard{F_z}{z}\rho\Delta \phi \Delta\rho \Delta z $$ Recognize that the term \( \rho\Delta \phi \Delta\rho \Delta z \) is the differential volume in cylindrical coordinates, so we can write the previous equation as $$ \begin{equation} \label{eq:fluxz} \Psi_z = \pard{F_z}{z}\Delta V \end{equation} $$ Where \(\Delta V\) is the differential volume element. I should note that if you were to do this derivation in Cartesian coordinates, the x and y component derivations would be identical (except for the subscripts). But in cylindrical coordinates, things get a little more interesting from here on out. Looking now at the \(\phi\) direction we consider the front and back faces of the volume. (I will omit the function arguments to the partial derivatives in the service of brevity with the understanding that we are talking about the point \(P=(\rho,\phi,z)\)). $$ F_{back} = F_\phi(\phi + \Delta \phi) \approx F_\phi + \pard{F_\phi}{\phi}\frac{\Delta \phi}{2} $$ $$ F_{front} = F_\phi(\phi - \Delta \phi) \approx F_\phi - \pard{F_\phi}{\phi}\frac{\Delta \phi}{2} $$ $$ A_\phi = \Delta\rho\Delta z $$ Because the positive \(\phi\) direction is toward the back of the volume, $$ \Psi_{\phi} = (F_{back} - F_{front})A_{\phi} $$ $$ \Psi_{\phi} = (F_\phi + \pard{F_\phi}{\phi}\frac{\Delta \phi}{2} - (F_\phi - \pard{F_\phi}{\phi}\frac{\Delta \phi}{2}))\Delta\rho\Delta z $$ $$ \Psi_{\phi} = \pard{F_\phi}{\phi}\Delta\phi\Delta\rho\Delta z $$ The term \(\Delta\phi\Delta\rho\Delta z\) is almost the differential volume. And maybe because we know where we are headed, we bend the equation to our will $$ \Psi_{\phi} = \pard{F_\phi}{\phi}\Delta\phi\Delta\rho\Delta z = \pard{F_\phi}{\phi}\frac{\rho}{\rho}\Delta\phi\Delta\rho\Delta z $$ $$ \begin{equation} \label{eq:fluxphi} \Psi_\phi = \frac{1}{\rho}\pard{F_\phi}{\phi}\Delta V \end{equation} $$ Now we try to handle the \(\rho\) component. We need to be careful here. But first things first. The vector field is defined similarly to previous components. $$ F_{right} = F_\rho(\rho + \Delta \rho) \approx F_\rho + \pard{F_\rho}{\rho}\frac{\Delta \rho}{2} $$ $$ F_{left} = F_\phi(\rho - \Delta \rho) \approx F_\rho - \pard{F_\rho}{\rho}\frac{\Delta \rho}{2} $$ Here is where it gets tricky. The area definitions are different because the two surfaces are not equal in size and so must be treated differently. $$ A_{right} = (\rho+\frac{\Delta\rho}{2})\Delta\phi\Delta z $$ $$ A_{left} = (\rho-\frac{\Delta\rho}{2})\Delta\phi\Delta z $$ First computing the flux on the right side. $$ \Psi_{right} = F_{right}A_{right}=(F_\rho + \pard{F_\rho}{\rho}\frac{\Delta \rho}{2})((\rho+\frac{\Delta\rho}{2})\Delta\phi\Delta z) $$ $$ \Psi_{right} = (F_\rho + \pard{F_\rho}{\rho}\frac{\Delta \rho}{2})(\rho\Delta\phi\Delta z + \frac{1}{2}\Delta\rho\Delta\phi\Delta z) $$ $$ \Psi_{right} = F_\rho\rho\Delta\phi\Delta z+\frac{1}{2}F_\rho\Delta\rho\Delta\phi\Delta z + \frac{1}{2}\pard{F_\rho}{\rho}\rho\Delta\rho\Delta\phi\Delta z + \frac{1}{4}\pard{F_\rho}{\rho}\Delta\rho^2\Delta\phi\Delta z $$ Similarly on the left hand side $$ \Psi_{left} = F_{left}A_{left}=(F_\rho - \pard{F_\rho}{\rho}\frac{\Delta \rho}{2})((\rho-\frac{\Delta\rho}{2})\Delta\phi\Delta z) $$ $$ \Psi_{left} = (F_\rho - \pard{F_\rho}{\rho}\frac{\Delta \rho}{2})(\rho\Delta\phi\Delta z - \frac{1}{2}\Delta\rho\Delta\phi\Delta z) $$ $$ \Psi_{left} = F_\rho\rho\Delta\phi\Delta z-\frac{1}{2}F_\rho\Delta\rho\Delta\phi\Delta z - \frac{1}{2}\pard{F_\rho}{\rho}\rho\Delta\rho\Delta\phi\Delta z + \frac{1}{4}\pard{F_\rho}{\rho}\Delta\rho^2\Delta\phi\Delta z $$ Computing \(\Psi_\rho\) (finally, after all that algebra) by subtracting and cancelling terms $$ \Psi_\rho = \Psi_{right} - \Psi_{left} $$ $$ \Psi_\rho = F_\rho\Delta\rho\Delta\phi\Delta z + \pard{F_\rho}{\rho}\rho\Delta\rho\Delta\phi\Delta z $$ We make the same modification to the first term that we did with \(\Psi_\phi\) and substituting \(\Delta V\) $$ \begin{equation} \label{eq:fluxrho} \Psi_\rho = \frac{1}{\rho}F_\rho\Delta V + \pard{F_\rho}{\rho}\Delta V \end{equation} $$ This last equation is correct but we normally don't see it written this way, so we factor out a \(\frac{1}{\rho}\) and invoke the reverse chain rule. $$ \Psi_\rho = \frac{1}{\rho}(F_\rho + \rho\pard{F_\rho}{\rho})\Delta V = \frac{1}{\rho}\pard{}{\rho}(\rho F_\rho)\Delta V $$ And so the total flux through our tiny volume is $$ \Psi_{total} = \Psi_\rho + \Psi_\phi + \Psi_z = \left(\frac{1}{\rho}\pard{}{\rho}(\rho F_\rho) + \frac{1}{\rho}\pard{F_\phi}{\phi} + \pard{F_z}{z}\right)\Delta V $$ The divergence is the ratio of the flux through a closed surface as the volume of that closed surface approaces 0. $$ \nabla\cdot\vec{F} = \lim\limits_{\Delta V \to 0} \frac{\Psi_{total}}{\Delta V} = \frac{1}{\rho}\pard{}{\rho}(\rho F_\rho) + \frac{1}{\rho}\pard{F_\phi}{\phi} + \pard{F_z}{z} $$
The derivation of divergence in spherical coordinates procedes along similar lines as the \(\phi\) and \(\rho\) terms. It's a little more complicated because of the added \(\sin\theta\) terms.