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Let's do what Heidar says and write it with indices, and identify the Lagrangian. $$L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m$$ where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index. Then since there are no bare $A_i$ sitting by ...

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There are two differnt levels to see this connection. Formally, you can derive a Fokker-Planck equation from the Boltzmann equation and do a Wick rotation on the time variable. This can be seen as a mathematical curiosity presently. But there is a more relevant way to recover this and is given by a formulation of the quantum Boltzmann equation. There is a ...

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1) Let us assume that the system is classical and has a Lagrangian formulation $$S[\phi]~=~ \int \mathbb{L}, \qquad\qquad \mathbb{L} ~=~{\cal L} ~dx^0 \wedge \ldots\wedge dx^{d-1},$$ in terms of a Lagrangian density ${\cal L}={\cal L}(\phi,\partial\phi)$ that doesn't depend explicitly on the $d$-dimensional spacetime coordinates $x^{\mu}$. 2) Noether's ...

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I'll try to rephrase the question. If a planet only experiences one external force directed toward the Sun, why does it orbit instead of crashing into the Sun? This question falls under the category of "uniform circular motion," so if you want more detail, you can look up that chapter in a physics text. The main idea is the following: The force of gravity ...

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(1) Wikipedia (2) The equation $$\frac{d^2 x^\alpha}{d\tau^2}=0$$ is the equation of motion of a free particle in special relativity. It is in a sense a generalization of Newton's first law. It says that free particles move in straight lines (recall that second derivatives annihilate lines). Since in special relativity the Christoffel symbols are trivially ...

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Since $\mathbf{h} + \mathbf{s} = \mathbf{r}$, there's no need to use $\mathbf{h}$ or any of its time derivatives. Physics doesn't care about something like $\dot{m}$. It cares about things like, "What's the energy?" and "What's the force?". But the kinetic energy of the bob is $\frac{1}{2}mv^2$ regardless of $\dot{m}$. The energy in the stretched spring is $... 3 Orders for differential equations refer to the largest number of derivatives that are taken with respect to the independent variable. In partial differential equations, there can be multiple independent variables, so you have to specify which one you're talking about -- hence the "in time" part. Thus, first-order in time means just one derivative with ... 3 Can someone explain why the time-independent Schrödinger equation isn't an eom? The TISE is an eigenvalue equation due to applying separation of variables to the TDSE; it is an equation for the spatial function alone. Can someone explain in what sense exactly is the time-dependent Schrödinger equation an equation of motion? A Lagrangian (density) ... 3 That is not the hamilton-jacobi equation. What you have above is an equality between the time-ind. Hamiltonian and the energy of a particle with generalized postion q and momentum p. Energy is conserved in some systems, but it depends on the exact mathematical form of the hamiltonian and has nothing to do with this equation as far as I know. 2 Suppose you are holding an apple. You obviously know where the apple is, and as long as you don't drop the apple you can predict it's future position. But suppose you now drop the apple and you want to predict where it will be in one second, two seconds or even ten seconds if you're standing on a tall building. As soon as you drop the apple it starts ... 2 As Timtam pointed out, this is not the Hamilton-Jacobi Equation. If you say, quantum mechanically, this does not even make any sense because, there the Hamiltonian is an operator which has to act on some eigenstate to give the energy of that state. Effectively, energy of the different states becomes the eigenvalue of the Hamiltonian. As far as conservation ... 2 At the moment the force F' starts to act, you will still have gravity. There is no net force, thus no deceleration. If you this opposing force is indeed the only force, it will indeed take 10 seconds to reach zero velocity, since now you get$v=u_0 -a t$And$u_0=a T$2 For example, the Dirac-Lorentz equation. 2 Simple answer, your equation has only four components, but there is one equation of motion for every field variable in the system which is more than four, so the answer is no. 2 This is actually a question for mathematical forum - solving linear non-homogenous differential equations. You have to find solutions for homogenous ($C=0$) and non-homogenous ($C\ne0$) equation. Non-homogenous solution is obviously $$x_\text{N} = \frac{C}{A},$$ while homogenous solution in $$x_\text{H} = x_0 \exp[-\frac{A}{B}t].$$ The whole solution ... 2 If the force is time dependent, you do not have energy conservation:$m \frac{dV}{dt}= F(t)d(\tfrac{1}{2}mV^2)=F V dt=F dx$For a time dependant potential:$dU=\mathbb{grad}(U(t))dx + \frac{∂U}{∂t}dt $But$F=-\mathbb{grad}(U(t)) \Rightarrow Fdx=-\mathbb{grad}(U(t))dx =-dU+\frac{∂U}{∂t}dt $So you get:$d(\tfrac{1}{2}mV^2+U(t))=\frac{∂U}{∂t}dt$... 2 It's not really a circular reference, it's an ordinary differential equation: $$\frac{d^2y}{dx^2}=f(y)$$ In particular, it is a general 2nd order ODE, which, for some functions$f(y)$, has known analytic solutions. One way to think of this is a mass on a spring: (source: Julius O Smith, Standorf MUS420/EE367A Supplementary: The Laplace Transform) If ... 1 Assuming that the pressure change across the cone is small (e.g., no significant density changes for the flowing gas), use the continuity equation. With constant density, this simplifies to$A_1 \cdot v_1 = A_2 \cdot v_2$, where$A$is the cross sectional area of the flow stream and$v$is the velocity of the flow stream. If you additionally need the ... 1 Intuitively, I can offer two hopefully helpful ways to think about a purely time-dependent term. First, since such a term depends on neither positions nor velocities, it has no spatial gradient, so there is no corresponding force. Neither the potential nor the kinetic energy are changed if we displace the system or change its velocities. So if I may be ... 1 Planets do not move in fixed orbits, they move in very finely balanced stable orbits. What you see in the sky today is the result of 4 billion years of thumping, bumping, gravitational games and a lot of stuff falling into the Sun. It's all a very delicate balance of velocity, vector and gravity and it's not all that difficult to upset. While Jupiter won't ... 1 The proper constructions resembling quantum mechanic's formalism does exist in classical mechanics, but it goes a bit beyond lagrangian formalism. In classical mechanics, you can represent a system by a phase space with points corresponding to states of the system. Now, functions over that phase space form a symplectic Lie algebra together with the Poisson ... 1 Potential energy is always measured relative to something, just as kinetic energy is measured in a particular frame of reference. That being the case, I could always make$T=V$by choosing the reference in a certain way - but it will not generally be true. For example, I can set$T=0$at the top of a tower, or at ground level. Both are valid - it's really ... 1 Let me begin with an analogy to high school calculus. We know that a function$f(x)$has a stationary point$x_0$if at that point $$\left.\frac{df}{dx}\right|_{x_0}=0$$ There are three possibilities: minimum, maximum and saddle. This is called the first derivative test. To test for these conditions, we use the second derivative test. We check $$\left.\frac{... 1 I think the author refers to the order of the differential equations describing the dynamical system that means the order of time derivative in the equation. How is it related to dissipation? To get answer and understand it one has to be able to solve following differential equations:$$\frac{dx}{dt}=-\alpha x\frac{d^2x}{dt^2}=-\alpha x$$The answer ... 1 Starting as Pieter Geerkens, the equations of motion in 2-D a parabolic system: x(t) = x_0 + v_x t = x_0 + v_0 \cos(\alpha) t and int the y axis: y(t) = y_0 + v_0\sin(\alpha)t + \frac{1}{2}g t^2 where g \sim -9.8 is the gravitational acceleration. Solvinf for t in the first equation as Pieter did: t(x) = \frac{x - x_0}{ v_0 \cos(\alpha)}... 1 First consider parametric equations for x and y as a function of t. Now, for each x-value solve for t = (x-x_0) / (v_0 \cdot \cos(\theta\,)). Then solve for the corresponding y = y_0 + v_0 \cdot \sin(\theta\,) \cdot t - 4.9 \cdot t ^ 2. 1 Note that the equation you give for position only applies under conditions of constant acceleration. That is you already have a particle under acceleration. Now, by definition$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$\$ In other words you care about acceleration and velocity because the position of the particle depends ...

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