# Tag Info

70

What you're seeing is a moiré pattern caused by beating between the regular grid of the screen's pixels and the regular grid of your camera sensor's pixels. The departure of the pattern from straight lines is likely due to a slight deformation away from planarity in the screen. You can test for this by keeping the camera fixed on a tripod and mildly ...

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It isn't a problem because two of the eight equations are constraints and they're not quite independent from the remaining six. The constraint equations are the scalar ones, $${\rm div}\,\,\vec D = \rho, \qquad {\rm div}\,\,\vec B = 0$$ Imagine $\vec D=\epsilon_0\vec E$ and $\vec B=\mu_0\vec H$ everywhere for the sake of simplicity. If these equations are ...

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On the other hand, it requires only two initial conditions x(0) and x˙(0), to obtain the function x(t) by solving Newton's equation For notational simplicity, let $$x_0 = x(0)$$ $$v_0 = \dot x(0)$$ and then write your equations as $$x(t) = x_0 + v_0t + \ddot x(0)\frac{t^2}{2!} + \dddot x(0)\frac{t^3}{3!} + \cdots$$ $$m\ddot x(t) = F(x,\dot x,t)$$ Now,...

29

If $D^n$ denotes the $n$th derivative and $D^{-n}$ the $n$th integral, then we have that, $$D^n f(t) = D^m[D^{-(m-n)}f(t)]$$ providing $m \geq \lceil{n}\rceil$. For our half derivative, we choose $n=1/2$, and $m=2$, in which case we have, $$D^{1/2}f(t) = D^2[D^{-(3/2)}f(t)]$$ There is a general formula for the $n$th integral of a function, one of my ...

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This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions. The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can ...

28

First of all, it's not true that all important differential equations in physics are second-order. The Dirac equation is first-order. The number of derivatives in the equations is equal to the number of derivatives in the corresponding relevant term of the Lagrangian. These kinetic terms have the form $${\mathcal L}_{\rm Dirac} = \bar \Psi \gamma^\mu \... 27 You lack complete knowledge of the system you're asking about. You only know about the jar. And if the complete system is the jar then yes, it has always been empty because there's nothing to interact with it. But if you suppose there are things which may have been in the jar, now those are added to the system. That puddle of water on the floor, was it ... 26 There are two famous cases in classical mechanics that fail to be deterministic. The first, and most famous, is Norton's Dome, which corresponds to a system with a force of the form$$F = \sqrt{r} $$There are more details on the Wikipedia article (it's usually described as the result of a reaction force from a surface with a certain shape), but the basic ... 24 I) Let us just for fun generalize OP's question to n spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic (-,+,\ldots,+) notation with c=1, where \mu,\nu\in\{0,\ldots,n-1\} denote ... 24 One can rewrite any pde of any order as a system of first order pde's, hence the assumption behind question is somewhat questionable. Also there exist first order PDE's of relevance to physics (Dirac equation, Burgers equation, to name just two). However, it is common that quantities in physics appear in conjugate pairs of potential fields and their ... 24 You need to be more precise about exactly what problem you're solving and what the inputs are. But if you're considering the general problem of what electromagnetic fields are produced by a given configuration of electric charge and current over spacetime, then the general solution is given by Jefimenko's equations. 19 What follows is certainly not a comprehensive answer addressing all of your concerns. It is an answer to the question is there a way to see something clearly pathological like superluminal signals in the heat equation? I would argue that yes, there is. The general solution to the initial value problem T(x,0) = T_0(x) for the heat equation on the real ... 18 The issue is that both the screen you're taking a picture of and your camera's sensors are discrete: they are made up of pixels and those pixels don't always line up. For example, if the pixels on the screen are slightly smaller than the camera's pixels, then each camera pixel will pick up a screen pixel plus a little bit of the next pixel. As you go from ... 16 Here we will for simplicity limit ourselves to systems that have an action principle. (For fundamental and quantum mechanical systems, this are often the case.) Let us reformulate OP's question as follows: Why does the Euler-Lagrange equations of motion for a relativistic (non-relativistic) system have at most two spacetime-derivatives (time-derivatives),... 16 There's nothing wrong with the first order wave equation mathematically, but it's just a little boring. If you want to use this equation to describe waves, it basically amounts to having a 1d solid with speed of sound v for left moving waves (say) and speed of sound 0 for right moving waves. It wouldn't surprise me if such a thing could be constructed (... 16 I am no mathematician and am a little afraid that my answer is too simple to be true, but here goes: I use Fourier transforms to define the fractional derivative. x(\omega) is defined such that$$ x(t) = \int_{-\infty}^\infty \, \frac{\text{d}\omega}{2\pi} \text{e}^{i \omega t} \, x(\omega) \, .$$Then any integer derivatives is$$ \frac{\text{d}^n}{\...

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Your equation is a special case of Riccati equation: $$y'=q_0(x) + q_1(x)y + q_2(x)y^2\!$$ with $q_0(x)=q_1(x)=0$ and a constant $q_2(x)$: $$y'= ky^2\!$$ There are lots of applications for the main Riccati equation in physics, and some of them can be reduced to the special case of $y'= ky^2\!$. (Although, explosive behavior is usually avoided and means ...

13

If I have $n$ objects (say, reactants) colliding, the rate of collisions will be roughly proportional to $n^2$. If the population grows by a fixed amount with each collision, we would find this law. See the rate equation. I think that growth due to sexual reproduction might fit here as well, but I'm not familiar enough with population biology to say.

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Just looking at physics, if the initial problem has a solution and you rotate the plate of $\pi/2$ you obtain another solution with boundary conditions rotated of $\pi/2$. Perform the same procedure two other times and you end up with four solutions with corresponding four different boundary conditions rotated of $0$, $\pi/2$, $\pi$, $3/2 \:\pi$ ...

12

As you have noted, to solve the differential equation describing the system you need initial conditions. The condition "jar is empty" does not have a unique solution without specifying the conditions.

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The reason for equations of physics, being of at most second order, is due to the so-called Ostrogradskian instability. (see paper by Woodard). This is a theorem, which states that equations of motion with higher-order derivatives are in principle unstable or non-local. This is easily shown using the Lagrangian and Hamiltonian formalism. The key point is ...

11

Well, simpler than recognising a differential equation will be to recognise the quantity it describes. Quickly plugging in Solve[y'[t]==k y[t]^2,y[t],t] into wolfram alpha gives $$y(t)=\frac{1}{c-k\ t},$$ which explodes at $t=c/k$. Now let's think of physical quantities which diverge as $\frac{1}{t}$. The Coulomb potential $$\phi(r) = \frac{Q}{4 \pi \... 11 There are a few reasons I can think of: (1) The second order system is that it is time-reversible. If you let t\to-t, you get$$ \frac{\partial^2f}{\partial(-t)^2}=\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2} $$whereas the first order system has$$ \frac{\partial f}{\partial(-t)}=-\frac{\partial f}{\partial t}=\pm v\frac{\partial ...

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I'm not a 100% this will address the question, so this might be more of a long comment than an answer. My main point is, perhaps, that given a Hamiltonian, you'll still want to specify the dynamics to do simulations. Even the Ising model can be simulated in several different ways, all satisfying detailed balance, where the relaxation towards equilibrium is ...

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The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ...

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Firstly, there are a few issues with a time-dependent potential, $V(x,t)$. Namely, if we apply Noether's theorem, the conservation of energy may not apply. Specifically, if under a translation, $$t\to t +t'$$ the Lagrangian $\mathcal{L}=T-V(x,t)$ changes by no more than a total derivative, then conservation of energy will apply, but this resricts the ...

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Dimensionless equations have the advantage that they work for any value of the parameters. They are scale invariant. So the solution in terms of a single dimensionless variable applies to all values of $D$ and $t$. It also allows the definition of characteristic values for the dynamic variables. In your example, one could say $u_0$ = \$\frac{q}{\sqrt{...

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Let's say your goal is to describe the shape of some object, such as a box. You could create a completely arbitrary ruler and measure the three axes of the box, coming out for example with lengths of 11.72, 23.44, and 35.16 of your arbitrary ruler units. Or you might look at your results more closely and think hmm, something is going on here, since the ...

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This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...

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