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7

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 ...

3

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

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 ...

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

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 ...

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 ...

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: $t(x) = \frac{x - x_0}{ v_0 ... 1 If the force is time dependent, you do not have energy conservation:$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$... 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 ... 1 There certainly are systems that are fully described by their energy-momentum conservation as a simple example let us take one-dimensional particle in a potential. The energy is conserved: $$\frac{m \dot{x}^2}{2} + V(x) = E$$ Differentiating this expression w.r.t. time one gets$$\frac{2 m \dot{x} \ddot{x}}{2} + \frac{\partial V(x)}{\partial x} \dot{x} = ... 1 The situation is similar to conservation of energy, momentum etc. in classical mechanics. Consider for instance motion of the mass$m$in the potential$V(r)$. In general case the energy is conserved, and the momentum is not. We know the expression for energy:$E=\frac{mv^2}{2}+V(r) = const$The equation of motion is$\frac{d\vec{v}}{dt}=-\nabla V(r)\$ ...

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