As you can tell, there has been a bit of a hiatus. Haven't really been productive (beyond just working on my classes) in the meanwhile. But I'm back now! I suppose this will be a bit of a meta post, followed by some electromagnetic theory.
First of all, I'm working on a research project (along with principal investigator Dr. Thomas Siegert). The project pertains the impact of small solar system bodies (SSSBs) on gamma ray data from INTEGRAL. I'll probably talk a bit more about this once I move farther into the project and have a better grasp of things. I'm also giving a lecture on linear algebra at the San Diego Math Circle (SDMC). I plan on drawing inspiration from some of the stuff that I've exposited on here. In particular, I really want to show the students that linear algebra is a lot more than just Euclidean vectors and solving simultaneous linear equations. That lecture will be happening this Saturday (11/14). Some of the heuristic arguments in physics rub me in the wrong way (due to the lack of mathematical rigor), but I cannot deny the importance of being able to reason quickly with heuristics in physics (and in general, using ad hoc methods that aren't entirely rigorous is, in my opinion, both natural and fine in the path to a more rigorous solution). Here are some examples. The electric field and potential are related by \(\vec{E}=-\nabla\phi\). It follows that the divergence of the electric field is related to the Laplacian of the potential: \[\nabla\cdot\vec{E}=-\nabla^2\phi.\] Gauss' law tells us that \(\int_{S}{\vec{E}\cdot\mathrm{d}\vec{a}}=\frac{Q}{\epsilon_0}=\frac{1}{\epsilon_0}\int_{V}{\rho\ \mathrm{d}v}\), where \(\rho\) is the charge density. However, Gauss' theorem says that the integral of the flux over the surface is equal to the integral of the divergence in the volume. Equating the integrands gives us \(\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}\), hence \[\nabla^2\phi=-\frac{\rho}{\epsilon_0}.\] But when \(\rho=0\), as is the case in empty space where there are no charges, the potential function must satisfy Laplace's equation \[\nabla^2\phi=0.\] This equation is pretty important, and there is quite a bit of theory behind it, as it not only pops up here, but in other areas (such as heat transfer). I don't even remember much of the basic theory we learned about it in MATH 110, but there is one property of functions that satisfy Laplace's equation (called harmonic functions) that is relevant here. Theorem: Let \(f\colon U\to\mathbb{R}\) be a harmonic function over the set \(U\subset\mathbb{R}^3\). Then, the average value of \(f\) over any sphere contained in \(U\) is equal to the value of \(f\) at the center of that sphere. An analogous result holds in two dimensions as well, if \(U\subset\mathbb{R}^2\) and we replace spheres with circles. Proof (sketch): I know of two ways of thinking about this. Ironically, neither of them are entirely rigorous. One of them is something you'd expect to encounter from an applied math class (like MATH 110), and the other is a physical argument that applies just to electric potentials (provided in Purcell). The applied math-y way is to consider the Taylor expansions of \(f\) incremented and decremented by \(h\) in each variable, one at a time. Then, you add each pair (for instance, \(f(x+h,y,z)+f(x-h,y,z)\)). This eliminates mixed partials (at least the lower-order ones). Then, you can isolate the repeated second partial derivative terms (partial with respect to \(x\) and then \(x\), etc.) in each pair-sum. Plugging these expressions into Laplace's equation, one finds that neglecting the higher-order terms, the value of \(f\) is equal to the to the average of the values of function at the points that are incremented and decremented by \(h\) away from the center, in each direction. Heuristically, we can extend this to every direction in a full sphere around the center point. Physically, we consider a point charge \(P\) with charge \(Q\) and a sphere \(\Omega\) that has a charge \(q\) distributed uniformly over it, such that the center of \(\Omega\) is a distance \(R\) from \(P\). We can compute how much work it takes to construct this configuration, in two different ways. The first way is to note that due to the shell theorems, outside of \(\Omega\), it is electromagnetically indistinguishable from a point charge with charge \(q\) at its center. Hence, the work required to assemble the configuration is the same as the work required to bring two point charges together (or more precisely, bring \(P\) in from infinity). This is simply \(\frac{Qq}{4\pi\epsilon_0R}\). On the other hand, instead of bringing \(P\) in from infinity, we can bring \(\Omega\) in from infinity. Then, the work done must be equal to the average potential over \(\Omega\) times the total charge of \(\Omega\). But clearly, the work done doesn't change from us thinking about it this way! So we still have a work of \(\frac{Qq}{4\pi\epsilon_0R}\). This means that the average potential over \(\Omega\) is simply \(\frac{q}{4\pi\epsilon_0R}\), which is precisely the potential at the center of \(\Omega\) due to \(P\). So the electric potential function satisfies the theorem in empty space that does not contain charges, where there is only one point charge in the universe. Superposition gives the general result for arbitrary charge distributions in the universe. A simple corollary of this result is that no harmonic function can have local extrema (or at least, extrema will be on the boundary of the domain over which the function is harmonic). This leads itself naturally into Earnshaw's theorem, which states that there exists no configuration that provides a stable equilibrium for a point charge. To see this, observe that if there was a point in space where there was a stable equilibrium, the potential there must be a local minimum, a contradiction. Alternatively, the electric field at such a point would have to have a negative divergence. Gauss' law then says a negative must exist at that point, contradicting our assumption that we were considering empty space. There's quite a bit more to talk about, but I'll end it here for now. I've procrastinated quite a bit and I need to catch up on other stuff!
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The AM-GM inequality is a well-known result, and often a pretty nifty tool. Less well-known is the QM-AM inequality. These two inequalities make up the more general mean-inequality chain. The quadratic mean (QM), sometimes called root-mean-square (RMS), is defined by
\[\sqrt{\frac{x_1^2+...+x_n^2}{n}}.\] The QM-AM inequality then states \[\sqrt{\frac{x_1^2+...+x_n^2}{n}}\geq\frac{x_1+...+x_n}{n}.\] There are two nice ways to prove QM-AM. The first, which I will get out of the way, is by Cauchy-Schwarz. That inequality implies that \[n(x_1^2+...+x_n^2)=(1^2+...+1^2)(x_1^2+...+x_n^2)\geq(x_1+...+x_n)^2.\] Rearranging this yields \[\frac{x_1^2+...+x_n^2}{n}\geq\left(\frac{x_1+...+x_n}{n}\right)^2.\] Both sides are clearly nonnegative, so we can take the square root to obtain the result: \[\sqrt{\frac{x_1^2+...+x_n^2}{n}}\geq\frac{x_1+...+x_n}{n}.\] Cauchy-Schwarz also implies that the equality condition is precisely when the \(x_i\) are all equal. This is clean, but it is pretty opaque. Mostly because Cauchy-Schwarz is not very intuitive (beyond the vector dot product geometric interpretation, do you really have a feel for why the numbers themselves must satisfy Cauchy-Schwarz?). Lucky for us, there is another, arguably more natural, way to derive QM-AM. Suppose I gave you a list of four numbers at McDonald's. For each number, you are allowed to select one of the four types of chicken nuggets box that number of times (and you can only select each box type twice). For instance, if the set I gave you was \(\{2,3,5,8\}\). You select a number, for instance 5, and then select a box type (out of 4-piece, 6-piece, 10-piece, and 20-piece) for instance 10-piece, and buy five 10-pieces to get 50 chicken nuggets. How would you maximize the number of chicken nuggets you get? This is pretty obvious. The greedy algorithm is optimal. We pair the highest box numbers with the highest purchase numbers. This gives rise to what is called the rearrangement inequality. If \(a_i\leq a_j\) and \(b_i\leq b_j\) iff \(i\leq j\), then we have that \(\sum_{i=1}^{n}{a_ib_i}\) is the maximal dot product of vectors with components \(a_i\) with vectors with components \(b_i\). In other words, \[\sum_{i=1}^{n}{a_ib_i}\geq\sum_{k=1}^{n}{a_kb_{\sigma(k)}},\] where \(\sigma\) is a permutation on \(\{1,...,n\}\). There is nothing complicated going on here. Just a super intuitive application of the greedy algorithm. Any third-grader can follow that. Now, it follows that \[a_1b_1+...+a_nb_n\geq a_1b_1+...+a_nb_n\] \[a_1b_1+...+a_nb_n\geq a_1b_2+...+a_nb_1\] \[\vdots\] \[a_1b_1+...+a_nb_n\geq a_1b_{n-1}+...+a_nb_{n-2},\] where in each line we cycle through the order of \(b_i\). Now, observe what happens when we add all \(n\) of the inequalities together. We obtain: \[n\sum_{i=1}^{n}{a_ib_i}\geq\left(\sum_{k=1}^{n}{a_k}\right)\left(\sum_{j=1}^{n}{b_j}\right).\] Perhaps you see where this is going now. Suppose that \(a_i=b_i=x_i\). Then \[n(x_1^2+...+x_n^2)\geq(x_1+...+x_n)^2.\] Now, the rest of the proof follows as before with Cauchy-Schwarz. A practical application of QM-AM is that it immediately establishes that the RMS speed is greater than the average speed of molecules in a gas. This avoids the use of the Maxwell-Boltzmann distribution. Deriving RMS speed can be (and is traditionally) done without Maxwell-Boltzmann, but doing it with Maxwell-Boltzmann is fun because it presents an interesting application of Feynman's trick. Perhaps I'll make a main page post about that. Here is an interesting problem that was walked through in my homework for physics class (PHYS 4B at UCSD).
Problem: Consider water (or some inviscid fluid) rotating in a large bucket with constant angular velocity, \(\omega\) (about the axis through the center of the bucket). What shape does the surface of the water take? The solution was unfortunately given away by the structure of the problem in the homework (it was split up into parts). Nonetheless, I think that it is a very cool problem and worth expositing. The key idea is Bernoulli's principle. Solution: We leave the lab frame and enter a frame, which we will call \(S\) that rotates along with the water in the bucket. Observe that this frame is non-inertial, since it accelerates with respect to the lab frame. So we perceive fictitious forces in \(S\). Importantly, a packet of water at a constant distance from the axis will stay at that constant distance and the centrifugal force it feels is actually the centripetal force that keeps it in uniform circular motion. This means that the force vector in our frame has a clean form in cylindrical coordinates, \((r,\theta,z)\). Let the axis of rotation hit the floor of the bucket at the origin, \((0,0,0)\). The force on a packet of water is thus given by \[\vec{F}=mr\omega^2\hat{e}_r-mgz\hat{e}_z.\] That is, the net force felt in this frame is a combination of the centrifugal force and weight. Since the centrifugal force is the centripetal force, which proportional to the mass, we can apply the equivalence principle. That is, the laws of physics do not change in our frame, as long as we account for an extra fictitious force that is proportional to mass. One may question why we take the radial component of our force (which is the centrifugal force here) to be positive. The reason is because in \(S\), each packet of water is stationary! So it does not have an inward radially-directed centripetal force. The centrifugal force happens to actually be directed outward (but still radially). This occurs to yield equilibrium in \(S\). A free body diagram will be balanced with the outward centrifugal force, weight, and contact force normal to the surface. I won't say more here, because seeing things this way actually leads to a simpler solution of the problem which is not what I am trying to demonstrate at the moment! Going back to our problem, now that we have a force function, we can come up with a potential \(\varphi\) such that \(\vec{F}=-m\vec{\nabla}\varphi\) (observe the similarities with other forms of potential, such as electric potential, namely that we keep mass out of \(\varphi\) as we would keep charge out of the electric field). This is just a standard exercise in basic calculus. We have that \[\frac{\partial\varphi}{\partial r}=-r\omega^2\Rightarrow\varphi=\int{-r\omega^2\textrm{ d}r}=-\frac{1}{2}\omega^2r^2+A(\theta,z),\] \[\frac{\partial\varphi}{\partial z}=g\Rightarrow\varphi=\int{g\textrm{ d}z}=gz+B(r,\theta).\] From equating these two expressions for \(\varphi\), we have that \(A(\theta,z)=gz+C\) and \(B(r,\theta)=-\frac{1}{2}\omega^2r^2+C\). Putting everything together, we have \[\varphi(r,z)=gz-\frac{1}{2}\omega^2r^2+C,\] for some constant \(C\). The choice of \(C\) is irrelevant, since a potential by itself is meaningless – only a potential difference has an interpretation. WLOG, take the potential at the origin to be zero, so that \(C=0\). Next, consider a tube along a radial direction on the surface of the water. By Bernoulli's principle in the lab frame, we have that \[p+\frac{1}{2}\rho v_{\textrm{tube}}^2+\rho gz=C,\] where \(p\) is the pressure, \(\rho\) is the density of water, \(v_{\textrm{tube}}\) is the speed of flow in the tube, and \(C\) is some constant. But for \(v_{\textrm{tube}}\) to be nonzero, a fluid packet must have some radial component of velocity (again, remember that we are now in the lab frame). But this is not true since the water is just rotating in the bucket! So Bernoulli's principle reduces to the statement \[p+\rho gz=C.\] Now, let us add and subtract \(\frac{1}{2}\rho\omega^2r^2\) to the LHS of the above equation. We obtain \[p+\frac{1}{2}\rho\omega^2r^2+\rho gz-\frac{1}{2}\rho\omega^2 r^2=C.\] Now \(\omega r=v\), the true speed of any packet of water rotating in the bucket. And the last to terms are just \(\rho\varphi\)! So we obtain a modified form of Bernoulli's principle: \[p+\frac{1}{2}\rho v^2+\rho\varphi=C.\] This statement could also have been derived by understanding what exactly Bernoulli's principle asserts. In particular, it states that the sum of pressure, kinetic energy density, and potential energy density is an invariant. We could have just modified the potential energy density term by introducing \(\varphi\). More implicitly, what's going on here is that we are using the equivalence principle. That is, we are saying that the combination of the true gravitational force and the centrifugal force is indistinguishable from a very strange modified gravitational force in \(S\). Furthermore, since in \(S\) we have \(v=0\), our modified Bernoulli's principle now reads \[p+\rho\varphi=C.\] Now we take a point on the surface, say \((0,\theta_0,z_0)\). For elegance, let this be the point on the surface that is also on the axis of rotation. For any point on the surface, we must have that \(p=p_0\), the atmospheric pressure! In particular, we have \[p_0+\rho\varphi(0,z_0)=C.\] And now we are done. Because the surface of the water is just the locus of points where the pressure is equal to atmospheric pressure. In particular, we will have \[p_0+\rho\varphi(r,z)=p_0+\rho\varphi(0,z_0).\] Rearranging this gives \[\varphi(r,z)=\varphi(0,z_0)\Rightarrow z=\frac{1}{2g}\omega^2r^2+\frac{1}{g}\varphi(0,z_0).\] This, in fact, fits the cylindrical form of a paraboloid. \(\square\) It is interesting to note that the solution is entirely determined by energy conservation and the equivalence principle. But any problem that can be solved with energy conservation can also be solved directly by Newton's laws. So it is in fact possible to find a simpler solution, which I mentioned before. The idea there is to simply note that an observer in \(S\) would see that the system is in static equilibrium. I'll leave that solution to the reader. Continuing on an MIT pset, we have the following equation for fluid flow. Let \(\mathbf{F}(x,y,t)=\rho (x,y,t)\mathbf{v}(x,y,t)\), where \(\rho\) is density and \(\mathbf{v}\) is velocity. Then,
\[\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{F}=0.\] This is the so-called equation of continuity. As it turns out, it is simply equivalent to the conservation of mass. We can see this by integrating the equation over some region \(R\). We obtain \[\iint_{R}{\frac{\partial\rho}{\partial t}\textrm{ d}A}=-\iint_{R}{\nabla\cdot\mathbf{F}\textrm{ d}A}.\] But now, using Green's theorem in normal form, the RHS becomes a line integral over \(C\), which is the positively-oriented boundary of \(R\): \[\iint_{R}{\frac{\partial\rho}{\partial t}\textrm{ d}A}=-\oint_{C}{M\textrm{ d}y-N\textrm{ d}x},\] where \(M\) and \(N\) are the \(x\) and \(y\) components of \(\mathbf{F}\), respectively. Observe that the RHS is now the negative of the fluid flux through \(C\). That is, the change in pressure over a region is equal to the negative of the flux through the boundary of that region. Obviously. If mass wants to leave a region, it must pass through the boundary of that region. It cannot simply "vanish". Mass is conserved. In a previous blog post, we talked about the convective derivative, and how it is used in defining incompressible flow. In particular, a flow is incompressible when the convective derivative of the pressure is zero. We can combine this with the equation of continuity to find another necessary and sufficient condition for flow incompressibility. We have \[\begin{split} \frac{D\rho}{Dt}&=\frac{\partial\rho}{\partial t}+\mathbf{v}\cdot\nabla\rho\\ &=0. \end{split}\] But by rearranging the product rule, we have \(\mathbf{v}\cdot\nabla\rho=\nabla\cdot(\rho\mathbf{v})-\rho\nabla\cdot\mathbf{v}\). So our convective derivative becomes \[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})-\rho(\nabla\cdot\mathbf{v})=0.\] The first two terms sum to zero by the equation of continuity, which leaves us with \(\boxed{\nabla\cdot\mathbf{v}=0}\). Flow is incompressible only when the divergence of the velocity is zero. Sorry for the inactivity. I have just started my studies at the University of California, San Diego.
Here is a method I've learned from Introductory Classical Mechanics to solve a certain system of ODEs. Consider a system of ODEs of the form: \[\left\{\begin{array}{ll}\ddot{x}+\omega^2(ax+by)=0\\\ddot{y}+\omega^2(cx+dy)=0\\\end{array}\right.\] Such a system often arises from coupled oscillations. To solve such equations, we suppose there exists solutions of the form \(x=Ae^{i\alpha t}\) and \(y=Be^{i\alpha t}\). Substituting these solutions into the the differential equations, we obtain the system of equations \[\left\{\begin{array}{ll}(a\omega^2-\alpha^2)A+b\omega^2B\\c\omega^2A+(d\omega^2-\alpha^2)B\\\end{array}\right.\] We wish to find nontrivial solutions \(A,B\) (which in this case means nonzero solutions). It is easier to see the condition for the independent nontrivial solution if we write the system in matrix form: \[\begin{bmatrix} a\omega^2-\alpha^2 & b\omega^2 \\ c\omega^2 & d\omega^2-\alpha^2 \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] Let \(\mathbf{M}\) be the first matrix and \(\mathbf{N}\) be the second matrix in the above equation. Then, if \(\mathbf{M}\) is invertible, then multiplying through with its inverse would yield \(\mathbf{N}=0\), the solutions that we wish to avoid. Hence, if nontrivial solutions are to exist, \(\mathbf{M}\) cannot be invertible. The inverse of \(\mathbf{M}\) is the adjugate matrix of \(\mathbf{M}\) divided by the determinant of \(\mathbf{M}\). So if we wish for this to not exist, the determinant of \(\mathbf{M}\) must be zero. That is, \[\begin{vmatrix} a\omega^2-\alpha^2 & b\omega^2 \\ c\omega^2 & d\omega^2-\alpha^2 \end{vmatrix}=\alpha^4-(a+d)\omega^2\alpha^2+(ad-bc)\omega^4=0.\] We can view this multivariate polynomial as biquadratic in \(\alpha\). We solve the biquadratic by letting \(\beta=\alpha^2\) and then using the quadratic formula. \[\beta=\frac{1}{2}\left[a+d\pm\sqrt{(a-d)^2+4bc}\right]\omega^2.\] Hence, we have four solutions for \(\alpha\): \[\alpha=\pm\frac{\omega}{\sqrt{2}}\sqrt{a+d\pm\sqrt{(a-d)^2+4bc}}.\] Suppose that these solutions are \(\pm R\omega\) and \(\pm S\omega\). We can then obtain \[A=-\frac{b}{a-R^2}B\] \[A=-\frac{b}{a-S^2}B,\] We can now use the principle of superposition to write the general solution as \[\begin{bmatrix} x \\ y \end{bmatrix}=A_1\begin{bmatrix} 1 \\ -\frac{b}{a-R^2} \end{bmatrix}e^{iR\omega t}+A_2\begin{bmatrix} 1 \\ -\frac{b}{a-R^2} \end{bmatrix}e^{-iR\omega t}+A_3\begin{bmatrix} 1 \\ -\frac{b}{a-S^2} \end{bmatrix}e^{iS\omega t}+A_4\begin{bmatrix} 1 \\ -\frac{b}{a-S^2} \end{bmatrix}e^{-iS\omega t}\] Where the coefficients \(A_i\) are determined by initial conditions. Depending on the \(A_i\), we can condense the exponentials into, for instance, sinusoidal functions. This sort of system of ODEs appears most often with coupled oscillation. This post was way overdue. Sorry. UC San Diego is keeping me busy. I'll grind this weekend to catch up on stuff on this website. It's a shame that quantum and relativistic corrections are necessary. Though perhaps it's self-entitled of me to wish that the universe would be simple enough for me to understand it.
It turns out that current flows in such a way that power loss in resistors is minimized in parallel. This claim comes from problem 2 from a pset here. Consider two resistors in parallel. The current into a terminal is \(I\). Let the current the resistors be \(I_1\) and \(I_2\), and let the resistances be \(R_1\) and \(R_2\), respectively. Then by conservation of charge: $$I=I_1+I_2,$$ and since power dissipated through a resistor is \(I^2R\), we have a total power loss of: $$P(I_1,I_2)=I_1^2R_1+I_2^2R_2.$$ Now we have a simple optimization problem. We must minimize the function above subject to the constraint \(S(I_1,I_2)=I_1+I_2=I\). We proceed with Lagrange multipliers: $$\nabla P=\lambda\nabla S\Rightarrow\left<2I_1R_1,2I_2R_2\right>=\left<\lambda,\lambda\right>.$$ From this, we immediately obtain: $$I_1R_1=I_2R_2,$$ which if Ohm's law holds, is simply \(V\), the potential difference across the resistors! This physically makes sense. Potential difference is constant over resistors in parallel since all of the resistors of coinciding terminals and potential difference is path-independent. The calculation above easily generalizes to an arbitrary number of resistors. I am awaiting a response on StackExchange for a greater physical insight into why currents arrange themselves such that power dissipation is minimized. This seems to be a manifestation of a common theme throughout physics. For instance, Fermat's principle from optics states that light will always choose a path that minimizes travel time. This is an interesting principle, but it is actually due to the Huygens principle, in which waves are framed as propagating by new wavelets emanating from each wavefront. I am simply wondering if there is a similar underlying principle behind why currents travel so that power dissipation is minimized in parallel resistors. Here are two problems in a pset from MIT 18.02 (multivariable calculus).
Problem 2: Let \(f(x,y,z,t)\) be a smooth function, and let \(\nabla f=\left<f_x,f_y,f_z\right>\) be the gradient in space variables only. Let \(\mathbf{r}=\mathbf{r}(t)=\left<x(t),y(t),z(t)\right>\) be a smooth curve, and \(\mathbf{v}=\mathbf{r}'(t)\); and suppose we use the notation \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\textrm{d}}{\textrm{d}t}f(\mathbf{r}(t),t)\). Use the Chain Rule to show that \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\). Solution: We have: \[f(\mathbf{r}(t),t)=f(x(t),y(t),z(t),t)\] By the Chain Rule: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial x}\frac{\textrm{d}x}{\textrm{d}t}+\frac{\partial f}{\partial y}\frac{\textrm{d}y}{\textrm{d}t}+\frac{\partial f}{\partial z}\frac{\textrm{d}z}{\textrm{d}t}+\frac{\partial f}{\partial t}\] Which becomes: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial t}+\left<\frac{\textrm{d}x}{\textrm{d}t},\frac{\textrm{d}y}{\textrm{d}t},\frac{\textrm{d}z}{\textrm{d}t}\right>\cdot\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>\] Which is: \[\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\] As desired. \(\square\) This function, \(\frac{\textrm{D}f}{\textrm{D}t}\), is called the convective derivative or the material derivative. In fact, there are quite a few names for this. It is important to realize that \(\mathbf{r}\) defines a path or trajectory through space. The function \(f\) then describes something that is changing along a trajectory with time. The next problem makes this clear, letting \(f=\rho\), the density of a fluid. When \(\rho\) is constant in \(t\), the flow is termed steady. Unsteady flow, as one can imagine by this definition, must be enormously complicated, and it includes phenomena such as turbulence. In the case of steady flow, each trajectory is called a streamline. A fluid flow is called incompressible if the convective derivative of \(\rho\) is zero. In steady flow, this means that there is no pressure change along a streamline (which makes sense!). Problem 3a: Suppose that the density function depends only on time \(t\) but is constant in the space variables \((x,y,z)\), that is, \(\rho=\rho(t)\). Then show that the flow is incompressible if and only if the density \(\rho(t)\) is constant in all the variables \((x,y,z,t)\) (in other words, the flow must be steady). Solution: We want: \[\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho=0\] But since \(\rho\) does not depend on spatial variables, \(\nabla\rho=0\) and \(\frac{\partial \rho}{\partial t}=\frac{\textrm{d}\rho}{\textrm{d}t}\). Hence: \[\frac{\textrm{d}\rho}{\textrm{d}t}=0\] Integrating both sides WRT \(t\): \[\int{\frac{\textrm{d}\rho}{\textrm{d}t}\textrm{ d}t}=\rho(t)=C\] Hence, the flow is steady. \(\square\) Problem 3b: Next suppose instead that the density depends only on the space variables \((x,y,z)\) but not (explicitly) on \(t\), so that \(\rho=\rho(x,y,z)\). An incompressible flow in this case is called stratified. Use the result of problem 2 to give the condition on \(\rho\) and \(\mathbf{v}\) for stratified flow. Solution: In this case: \[\mathbf{v}\cdot\nabla\rho=0\] So the velocity is always orthogonal to direction in which \(\rho\) changes the most. But recall that \(\nabla\rho\) is always orthogonal to the contour surfaces of \(\rho\). It follows that the velocity must always be parallel, and hence tangential to, surfaces of equal density. \(\square\) Both gravity and electric force satisfy an inverse square law, and thus, both forces satisfy the shell theorem. That is, in a uniformly massive (or uniformly charged) spherical shell, there is no net force on any massive (or charged) object at any location within the shell.
Is this a unique property of inverse square functions? Or are there other functions that obey this? I reckon that I'll have to solve a differential equation of some sort (or perhaps an integral equation). My gut tells me that this is a unique property of inverse square functions (what sort of differential equation would be satisfied by inverse square functions and another class of functions?). I'll be investigating this further soon. School's out; the fun begins. I was up till 3:30. There was no way that I would sleep before the problem would.
As it turns out, the integral equation that I discussed earlier is well-known. It is an example of a linear Volterra equation of the first kind. \[f(c)=\int_{a}^{c}{K(c,x)\rho(x)\textrm{ d}x}\] The function \(K\) is known as the kernel. In our case, the kernel is: \[K(c,x)=c+b-x\] The idea here is to use the Leibniz integral rule (which follows from Fundamental Theorem). By differentiating both sides with respect to \(c\), we obtain: \[\frac{\textrm{d}}{\textrm{d}c}\left(\int_{c}^{\infty}{(c+b-x)\rho(x)\textrm{ d}x}\right)=\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}-b\rho(x)+\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Observe that if \(G(c, R)=\int_{c}^{R}{(c+b-x)\rho(x)\textrm{ d}x}\) is well-behaved, then we may interchange the limit and derivative above to obtain: \[\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}=\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\frac{\textrm{d}R}{\textrm{d}t}\right]}=0\] As cited, the conditions we want for \(G\) for this manipulation to be valid are:
Anyway, we have: \[b\rho(c)=\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Let \(P'=\rho\). Then, by the Fundamental Theorem: \[bP'(c)=\lim_{R\rightarrow\infty}{P(R)}-P(c)\] Let \(L=\lim_{R\rightarrow\infty}{P(R)}\). Then, the equation above rearranges to: \[P'+\frac{1}{b}P=\frac{L}{b}\] This is simply a first-order linear ODE. We solve this by using an integrating factor of \(\exp{\int{\frac{1}{b}\textrm{ d}{x}}}\). This yields: \[P(x)=L+Ce^{-x/b}\] Differentiating this, we obtain: \[\boxed{\rho(x)=Ce^{-x/b}}\] Obviously with \(C>0\). Uniform convergence is forced by the fact that this solution satisfies our Volterra equation. It can probably also be proven with \(\epsilon\)-\(\delta\) calculations. That's for another time. And I'm done with mechanics for this weekend I think. Here's what I do know.
For brevity, let \(K=c+b\). Let \(u=\rho(x)\) and \(\textrm{d}v=K-x\textrm{ d}x\). Then, integrating by parts: \[\int{(K-x)\rho(x)\textrm{ d}x}=\rho(x)\left(Kx-\frac{1}{2}x^2\right)-\int{\left(Kx-\frac{1}{2}x^2\right)\rho'(x)\textrm{ d}x}\] We continue integrating by parts, letting the polynomial in each new integrand be \(\frac{\textrm{d}v}{\textrm{d}x}\), and the derivatives of \(\rho(x)\) be \(u\). This then yields: \[\int{(K-x)\rho(x)\textrm{ d}x}=\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}\] Ok Andrew, cool. Now what? Ah, simply observe. Our desired integral is the improper integral evaluated from \(c\) to \(\infty\). This means that the limit of the antiderivative (above) as \(x\) approaches \(\infty\) must converge! In other words: \[\lim_{x\rightarrow\infty}{\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}}\] Must converge! For this to occur, every single term must converge! But observe that the polynomials in each term of the summation tend to \(\infty\), hence for convergence to occur, every derivative of \(\rho(x)\), along with \(\rho(x)\) itself, must tend to \(0\) (this follows plainly from L'Hôpital's rule). In other words, we have derived: \[\lim_{x\rightarrow\infty}{\rho^{(n)}(x)}=0\textrm{ }\forall n\in\mathbb{N}_0\] This in and of itself hints at possible functions. For instance, an exponential decay function, or something of the form \(Ax^n\) for some \(n<0\). Ok sure, we can substitute these functions into our original integral equation and find that one of them actually works, but I want a little more insight. And rigor. And what sort of lame solution is that? Deriving one property and guessing and checking? Ew. The next thing I will use is the Leibniz integral rule. That is: \[\frac{\textrm{d}}{\textrm{d}x}\left(\int_{a(x)}^{b(x)}{f(x,t)\textrm{ d}t}\right)=f(x,b(x))\cdot\frac{\textrm{d}}{\textrm{d}x}b(x)-f(x,a(x))\cdot\frac{\textrm{d}}{\textrm{d}x}a(x)+\int_{a(x)}^{b(x)}{\frac{\partial}{\partial x}f(x,t)\textrm{ d}t}\] While this looks complicated, it just follows from the Fundamental Theorem of Calculus. And while I'd love to continue this post with the application of this rule, I should probably do my history for once. The next post will include this, and discussion of uniform continuity, and uniform and pointwise convergence. Stay tuned. |
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