# Tag Info

0

Building on the responses from ACuriousMind and Gennaro Tedesco, I will make an attempt to provide a satisfactory, though not mathematically rigorous, answer. Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Yes, any invertible ...

2

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Answer: any function that produces the equations of motion under some sort of rules that you state is an allowed function to describe the dynamics. In particular any function that you can ...

1

In principle one can always write down the differential equations for a system of $N$ particles and attempt to solve them: as you pointed out, there is no general solution if usually $N>2$. As such, nevertheless the need to describe features of general systems remains. The key point here is understanding that, as a matter of fact, whenever dealing with ...

2

A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events: $$S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu ... 1 The super-Poisson bracket follows from a super-version of the Dirac-Bergmann or the Faddeev-Jackiw procedure. Diligent care must be taken to achieve consistent sign conventions when dealing with Grassmann-odd variables, see e.g. my Phys.SE answer here. The singular Legendre transformation for fermions is also discussed in my Phys.SE answer here. In OP's ... 0 Let us suppress explicit time dependence t from the notation in the following. Hamilton's eqs. are the Euler-Lagrange (EL) eqs. for the so-called Hamiltonian Lagrangian$$\tag{1} L_H(q,\dot{q},p)~:=~ p_i\dot{q}^i-H(q,p).$$In other words, the solutions to Hamilton's eqs. are stationary points for the Hamiltonian action$$\tag{2} S_H[q,p]~:=~\int \! ...

4

Yes, there exists a Legendre transformation from $g(p)$ to $f(x)$: $$f(x)=p(x)x-g(p(x))$$ with $x=dg/dp$. Here the notation $p(x)$ means $p$ written in terms of $x$. In your case, the Hamiltonian is a function of $p$ and you are transforming it to a function of $\dot{q}$, so you must use Hamilton's equation to get the velocity: $$\dot{q}_i=\frac{\partial ... 2 First of all, the hamiltonian contains the coordinates q_i and their momenta p_i. You have to calculate the velocities \dot{q}_i. For that, you'll need the Hamilton-Jacobi equations$$\dot{q}_i = \frac{\partial H}{\partial p_i}$$The Legendre transform, as noted in the comments, is involutive, so the lagrangian is just the Legendre transform of the ... 0 It's been some time since you asked this, but I'll give it a shot. The biggest problems in curved spacetime are defining exactly what you mean by a Hamiltonian, and what you mean by a Poisson bracket. Let's say you're dealing with a real scalar field \phi which minimizes some action$$ W = \int_{\mathcal{M}}\mathcal{L}(\phi,\partial_\mu \phi,x) d^4 x ...

2

Imagine this situation: at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity. at time t=T, we make the current to be I, thus we have a $\mathbf{B}$ field, and $\mathbf{A}$ field. during this process, $\mathbf{A}$ is build up from zero to some value, therefore we have induced electric field ...

1

Comments to the question (v3): The $X^-$ coordinate has (a part from a zero mode) been integrated out in the light-cone (LC) formalism. The above mentioned LC Hamiltonian cannot fully address questions about the $X^-$ coordinate. To get the well-known expansion of $X^-$ as a sum of zero and oscillator modes including the sought-for $\alpha^-_0$ mode term, ...

Top 50 recent answers are included