36
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 ...
29
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 ...
18
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 ...
18
...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, ...
17
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
$$p_1 \equiv \frac{\partial L}{\partial \dot q_1}...
15
Let $Q$ denote the set of all possible configurations of the system (the configuration manifold). Consider a point $q_0\in Q$. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around $q_0$.
Suppose that $q_0$ represents the position of the system under consideration at time $t_0$. ...
13
The question has been well-answered several times. I'll just add some geometrical context.
In geometry, the holonomy group of a connection is the set of transformations an object can experience when it is parallel transported in a loop. Many constraints can be phrased in terms of forcing something to be parallel transported. If the associated holonomy ...
12
Hints to the question (v1):
We cannot resist the temptation to generalize the background spacetime metric from the Minkowski metric $\eta_{\mu\nu}$ to a general curved spacetime metric $g_{\mu\nu}(x)$. We use the sign convention $(-,+,+,+)$.
Let us parametrize the point particle by an arbitrary world-line parameter $\tau$ (which does not have to be the ...
12
This is a good but quite broad question. Let us suppress position dependence $q^i$, $i\in\{1, \ldots, n\}$, and explicit time dependence $t$ in the following to keep the notation simple.
Given a Lagrangian $L(v)$, several things could go wrong if we try to perform a Legendre transformation in order to construct a Hamiltonian $H(p)$. Define for later ...
11
Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective:
The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian is of the form "Lagrange multipliers times constraints"$^1$
$$ H~:=~\int_0^{\ell}\! d\sigma~{\cal H}, \qquad {\...
11
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
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
$$g ~:=~ \{f,H^{\prime}\}_{PB} +\frac{\...
10
I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space (TS) metric $G_{\mu\nu}(X)$ has sign convention $(-,+,\ldots,+)$, and $c=1=\hbar$. The NG Lagrangian density is
$$\begin{align}{\cal L}_{NG}
~:=~&-T_0\sqrt{{\cal L}_{(1)}}, \cr
{\cal L}_{(1)}~:=~&-\det\left(\partial_{\...
10
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 ...
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
$$ \...
9
Actually, at least for a single point subjected to a friction force $F= -\gamma v$ and other forces associated with a potential $U(t,x)$ there exists a Lagrangian: $${\cal L}=e^{t\gamma/m}\left(\frac{m}{2}\dot{x}^2 -U(t,x)\right)\:.$$
The point is that this Lagrangian is not of the form $T-U$, nevertheless it gives rise to the correct equation of motion, the ...
9
Legendre transformation. OP's question contains some subtle aspects that we would like to illuminate. Note that a Legendre transformation (singular or not) should be an involutive operation. (It would therefore be inconsistent to propose, say, that a Lagrangian, that leaves everything undetermined, should correspond to a Hamiltonian, that fixes everything. ...
9
Here's another way:
Suppose that your lagrangian has the following property, for any $\theta$ (it could be a function of time $t$):
\begin{equation}\tag{1}
L(q, \, \theta \, \dot{q}) = \theta \, L(q, \, \dot{q}).
\end{equation}
This implies that the action
\begin{equation}\tag{2}
S = \int_{t_1}^{t_2}L(q, \, \dot{q}) \, dt
\end{equation}
is invariant under a ...
8
Let us consider a non-relativistic Newtonian problem of $N$ point particles with positions
$$ {\bf r}_i(q,t), \qquad i\in\{1, \ldots, N\},\tag{1}$$
with generalized coordinates $q^1, \ldots, q^n$, and $m=3N-n$ holonomic constraints.
Let us for simplicity assume that the applied force of the system has generalized (possibly velocity-dependent) potential $...
8
Basically, the multiplier method is a way to encode the constraint information of the system directly into the Lagrangian so that you don't have to worry about screwing up the physical requirements of the problem when you solve the equations of motion. In other words, instead of solving the equations of motion and constraining the results, you're ...
8
I) For a general Lagrangian $L(q,v,t)$, the Legendre transformation may be singular, i.e. the velocities $v^i$ in the momentum relations
$$\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}$$
Here the $\pm$ refers to the choice of Minkowski sign convention $(\pm,\mp,\mp,\mp)$. Therefore it is a constrained system. The 4-...
8
These conditions are not equivalent, only under several assumptions. A good reference are chapters 1 and 3 of Henneaux' and Teitelboim's "Quantization of gauge systems".
The "proper" definition of a gauge theory that relies neither on a Hamiltonian nor on a Lagrangian formalism explicitly is that the solutions $q(t)$ to the equations of motion contain some ...
8
Setup. Let there be given an $m$-dimensional manifold $M$ with coordinates $(x^1, \ldots, x^m)$. Let there be given an $n$-dimensional physical submanifold $N$ with physical coordinates $(y^1, \ldots, y^n)$. Let there be given $m-n$ independent constraints
$$ \chi^1(x)~\approx~ 0,\quad \ldots,\quad \chi^{m-n}(x)~\approx~ 0, \tag{1}$$
which defines the ...
8
For completeness: There is also a notion of semi-holonomic constraints.
Recall that a holonomic constraint$^1$
$$f(q,t)~=~0\tag{H}$$
only depends on the generalized coordinates$^2$ $q^j$ and time $t$, but not the generalized velocities $\dot{q}^j$.
A non-holonomic constraint is unsurprisingly a constraint that is not holonomic.
A semi-holonomic/Pfaffian ...
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
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 problem lies in what we learn about good old constrained dynamics from traditional Dirac approach is not complete and is somehow inconsistent, and the above is one example of this. This was the message of Pitts' paper mentioned in the question above, who reviewed a bunch of previous work on this very matter. I will mention couple of references from that ...
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
1. Brachistochrone
\begin{equation}
\boxed{\:
\begin{matrix}
x\left(\theta\right) = R\left(\theta-\sin \theta\right)\\
y\left(\theta\right) = R\left( 1-\cos \theta\right)
\end{matrix}\:}
\tag{b-01}
\end{equation}
\begin{equation}
\omega= \dfrac{\,\theta \,}{t}=\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\sqrt{\dfrac{\,g\,}{R}} =\text{constant}
\tag{b-02}
\end{...
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