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If you have a mechanical system with $N$ particles, you'd technically need $n = 3N$ coordinates to describe it completely. But often it is possible to express one coordinate in terms of others: for example of two points are connected by a rigid rod, their relative distance does not vary. Such a condition of the system can be expressed as an equation that ...


24

Superficially, D'Alembert's Principle $$\tag{1} \sum_{j=1}^N ( {\bf F}_j^{(a)} - \dot{\bf p}_j ) \cdot \delta {\bf r}_j~=~0 $$ may look like a trivial consequence of Newton's 2nd law, but the devil is in the detail. Here the detail is the superscript $(a)$ on the force, which stands for applied forces. The term applied forces refers to that we have ...


17

Strength of materials is affected by defects. A perfect crystal of iron would be extremely strong. Once a crack starts, it is not so hard to make it advance one atom deeper. Think of tearing open a plastic bag. Much easier once the tear starts. Brittle materials can be easier to break because they stretch less. It is easier to tear a sheet of paper than a ...


15

Constraints are handled in Lagranian mechanics through either of two approaches: 1) The constraint equation is used to reduce the degrees of freedom of the system. For example, if a particle is constrained to the surface of a sphere, then the Lagrangian can be written entirely in terms of two generalized coordinates and their associated momenta (typically, ...


14

Let $Q$ denote the set of all possible configurations of the system (the configuration manifold). Consider a point $q_0\in Q$. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around $q_0$. Suppose that $q_0$ represents the position of the system under consideration at time $t_0$. ...


13

...what I would like to know is why we get a zero Hamiltonian. I suspect that this is due to the reparametrization invariance... Will this always happen? Why? Yes, it is due to reparameterization invariance. In other words, the zero-Hamiltonian result holds for any reparameterization-invariant action, not just for the relativistic particle. In this sense, ...


13

As a quick note, the equations of motion that come from that Lagrangian are $$\frac{d}{dt}\left(\dot q_1 + \dot q_2\right) = -2kq_1^3$$ $$\frac{d}{dt}\left(\dot q_2 + \dot q_1\right) = -2kq_2^3$$ The Dirac procedure for singular Lagrangians goes as follows: Step 1: Calculate the generalized momenta as usual $$p_1 \equiv \frac{\partial L}{\partial \dot q_1}...


12

Hints to the question (v1): We cannot resist the temptation to generalize the background spacetime metric from the Minkowski metric $\eta_{\mu\nu}$ to a general curved spacetime metric $g_{\mu\nu}(x)$. We use the sign convention $(-,+,+,+)$. Let us parametrize the point particle by an arbitrary world-line parameter $\tau$ (which does not have to be the ...


11

The question has been well-answered several times. I'll just add some geometrical context. In geometry, the holonomy group of a connection is the set of transformations an object can experience when it is parallel transported in a loop. Many constraints can be phrased in terms of forcing something to be parallel transported. If the associated holonomy ...


10

The fact that $p = \large \frac{\partial L}{\partial \dot{q}} = 0$ introduces a problem in the equivalence between Lagrangian and Hamiltonian representations. The idea is that the Hamiltonian representation plus the constraint $p = 0$ is equivalent to the Lagrangian representation The Lagrangian $L$ is a function of $q$ and $\dot q$, that is $L(q, \dot q)$...


10

Comments to the question (v2): To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint $$g ~:=~ \{f,H^{\prime}\}_{PB} +\frac{\...


10

I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space (TS) metric $G_{\mu\nu}(X)$ has sign convention $(-,+,\ldots,+)$, and $c=1=\hbar$. The NG Lagrangian density is $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}, $$ $$ {\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\...


10

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian is of the form "Lagrange multipliers times constraints"$^1$ $$ H~:=~\int_0^{\ell}\! d\sigma~{\cal H}, \qquad {\...


10

The correct argument for how reparametrization invariance implies vanishing Hamiltonian is as follows: Given a generic Hamiltonian action $$ S = \int \left(\dot{q}^i p_i - u^\alpha \chi_\alpha - H\right)\mathrm{d}t,$$ where $q,p$ are the canonical phase space variables, $\chi_\alpha$ are Hamiltonian constraints with Lagrange multipliers $u^\alpha$ and $H(q,p)...


10

It is natural to generalize to an Abelian $p$-form gauge field $$A~=~\frac{1}{p!} A_{\mu_1\mu_2\ldots\mu_p} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_p}\tag{1}$$ with $\begin{pmatrix} D \cr p \end{pmatrix}$ real component fields $A_{\mu_1\mu_2\ldots\mu_p}$ in a $D$-dimensional spacetime. I) Massless case: There is a gauge symmetry $$ \...


9

Legendre transformation. OP's question contains some subtle aspects that we would like to illuminate. Note that a Legendre transformation (singular or not) should be an involutive operation. (It would therefore be inconsistent to propose, say, that a Lagrangian, that leaves everything undetermined, should correspond to a Hamiltonian, that fixes everything. ...


8

Let us consider a non-relativistic Newtonian problem of $N$ point particles with positions $$ {\bf r}_i(q,t), \qquad i\in\{1, \ldots, N\},\tag{1}$$ with generalized coordinates $q^1, \ldots, q^n$, and $m=3N-n$ holonomic constraints. Let us for simplicity assume that the applied force of the system has generalized (possibly velocity-dependent) potential $...


8

I) For a general Lagrangian $L(q,v,t)$, the Legendre transformation may be singular, i.e. the velocities $v^i$ in the momentum relations $$\tag{1} p_i~:=~\frac{\partial L(q,v,t)}{\partial v^i}$$ cannot be isolated. How to perform a singular Legendre transformation to achieve the corresponding Hamiltonian formulation goes under the name Dirac-Bergmann ...


8

The problem is that the Legendre transformation from 4-velocity to 4-momentum is singular: The 4 components of the 4-momentum $p_{\mu}$ are constrained to live on the mass-shell $$p_{\mu}g^{\mu\nu}p_{\nu}~=~\pm m^2. \tag{A}$$ Here the $\pm$ refers to the choice of Minkowski sign convention $(\pm,\mp,\mp,\mp)$. Therefore it is a constrained system. The 4-...


8

This is a good but quite broad question. Let us suppress position dependence $q^i$, $i\in\{1, \ldots, n\}$, and explicit time dependence $t$ in the following to keep the notation simple. Given a Lagrangian $L(v)$, several things could go wrong if we try to perform a Legendre transformation in order to construct a Hamiltonian $H(p)$. Define for later ...


8

Here's another way: Suppose that your lagrangian has the following property, for any $\theta$ (it could be a function of time $t$): \begin{equation}\tag{1} L(q, \, \theta \, \dot{q}) = \theta \, L(q, \, \dot{q}). \end{equation} This implies that the action \begin{equation}\tag{2} S = \int_{t_1}^{t_2}L(q, \, \dot{q}) \, dt \end{equation} is invariant under a ...


8

Here on SE, you may already find many answers to your question. Even if most of them are correct, I feel that a plain and correct answer is still missing. Where plain does not mean non-rigorous. But mathematical rigor is not the same as introducing differential manifolds. One could keep making confusing arguments even after introducing fiber bundles. If ...


7

There is a simple reason why we can consider variations on the whole $A$ rather than the quotient $C=A/G$ and the reason is following: all configurations that are $G$-equivalent have the same value of the action $S$. That's what we mean by the statement that the theory has the symmetry $G$. So the variation of the action $S$ in the directions that are ...


7

I) Let us suppress position dependence $q^i$ and explicit time dependence $t$ in the following, and also assume that the Lagrangian $L=L(v)$ is a smooth function of the velocities $v^i$, where $i=1, \ldots, n$. The Hessian matrix is defined as $$ H_{ij}~:=~\frac{\partial^2 L}{\partial v^i \partial v^j}.\tag{1}$$ Let us consider an open neighborhood$^1$ $V$...


7

Off-shell, meaning without assuming the Lagrange equations and the constraints, the Lagrange multipliers $\lambda^a(t)$ does by definition not depend$^1$ on the dynamical variables $q^j(t)$. (Hence, in the context of point mechanics, which we assume here, the Lagrange multipliers $\lambda^a(t)$ depend on time $t$. Correspondingly, in the context of field ...


7

Basically, the multiplier method is a way to encode the constraint information of the system directly into the Lagrangian so that you don't have to worry about screwing up the physical requirements of the problem when you solve the equations of motion. In other words, instead of solving the equations of motion and constraining the results, you're ...


7

Yes. There is a standard way to generalize to field theory. Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$. To obtain the ...


7

The principle of Least (Stationary) Action (aka Hamilton's Principle) is derived from Newton's axioms plus D'Alembert's principle of virtual displacements. Because D'Alembert's principle allows to account for the (reactions of the) bonds between the components of a system in a transparent way, the Lagrangian and Hamiltonian formulations are possible. Note1:...


7

The Hamiltonian is undefined. Converting a Lagrangian to a Hamiltonian requires: Finding $p$ Writing $H=p\dot q-L$ Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $\dot q$. For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,\dot q)$ coordinate system and the $(q,p)$ coordinate ...


7

The problem here is that, because there exist constraints of the form $f(q,\,p)=0$, the phase space coordinates of the usual Hamiltonian formulation aren't independent. I'm not sure how you encountered this Lagrangian, but this issue is a common hiccup in electromagnetism and (if you'll pardon a more obscure example) BRST quantisation. The good news is you ...


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