# Tag Info

21

There are several reasons for using the Hamiltonian formalism: 1) Statistical physics. The standard thermal states weight pure states according to $Prob(state) \propto e^{-H(state)/k_BT}$. So you need to understand Hamiltonians to do stat mech in real generality. 2) Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to ...

14

Some more comments to add to user1504's response: For a system with configuration space of dimension $n$, Hamilton's equations are a set of $2n$, coupled, first-order ODEs while the Euler-Lagrange equations are a set of $n$ uncoupled, second-order ODEs. In a given problem it might be easier to solve the first order Hamilton's equations (although sadly, I ...

6

First of all, Lagrangian is a mathematical quantity which has no physical meaning but Hamiltonian is physical (for example, it is total energy of the system, in some case) and all quantities in Hamiltonian mechanics has physical meanings which makes easier to have physical intuition. In Hamiltonian mechanics you have canonical transformations which allows ...

4

Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ...

3

Kepler's 3rd Law: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. If we take Jupiter's semimajor axis of 5.2 AU (avg value), then objects at 2.5 AU, 2.82 AU, 2.95 AU, 3.25 AU have orbital periods shorter than Jupiter by a factor of 3:1, 5:2, 7/3 etc. The fact that the ratios for the deepest ...

3

Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where $T$ is quadratic in ...

2

If we have some coordinates $q_i$ and some momenta $p_i$, then a generator of a transformation is defined as a function $g(q_i, p_i)$. By definition, this generates the transformation $$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}$$ $$p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$ So if we want the generator of translations, we want ...

2

This is similar, but not quite identical to Newton's cradle, with the difference being the heavy objects placed on the middle coins. To explain things, first consider the simpler case where there is no heavy object on top of the coins, and suppose the 5 nonmoving coins in "frame 1" are separated by a distance $L$. When two objects of mass $m$ and ...

2

I) In case of a point particle with mass $m$ (and no moment of inertia), the best one can do seems to be to model the friction/drag via a Rayleigh dissipation function ${\cal F}(v^2)$ with a friction/drag force $$\tag{1} {\bf F}_f~:=~-\frac{\partial {\cal F}(v^2)}{\partial {\bf v}} ~=~-2{\cal F}^{\prime}(v^2){\bf v},$$ i.e. the Lagrange equations read ...

2

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$: The off-shell action $S[q;t_i,t_f]$, The (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$, and Hamilton's principal function $S(q,\alpha, t).$ For their definitions and how they are related, see e.g. this Phys.SE answer. II) OP's ...

2

The problem with the phase space flow in Hamiltonian mechanics is that the flow itself is non-dynamical, that is, the flow is immediately defined for a given Hamiltonian, so there is no independent equation governing its evolution. Thus, Liouville equation is simply a transport of a scalar variable in a given flow. So, dimensional analysis of the flow ...

2

We start from the Lagrangian in spherical coordinates $$L=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+r^{2}\sin^{2}\theta\dot{\phi}^{2})+mgr\cos\theta$$ The length of the string is $d$ and the system is a constrained one with $|\vec{r}|=d$. Now, the constraint that is associated with a multiplier $\lambda$ is given by $c(r)=r-d=0$. At this point we have ...

2

Hints to the question (v1): Let us parametrize the problem wrt. an arbitrary world-line parameter $\tau$ (which does not have to be the proper time). The Lagrange multiplier $\lambda=\lambda(\tau)$ depends on $\tau$, but it does not depend on the canonical variables $x^{\mu}$ and $p_{\mu}$. Similarly, $x^{\mu}$ and $p_{\mu}$ depend only on $\tau$. The ...

2

One way to see the relationship of Hamilotian classical mechanics and Quantum mechanics is not to look for a direct translation of Hamiltionian -> Quantum Hamiltionian (which exists: Geometric Quantization), but consider the reverse relationship. Given a Hamiltion operator and evaluating it on wave functions of the form $e^{\frac{i}{\hbar} \phi}$ (which can ...

2

The canonical (Hamiltonian) formalism offers one of the main paths for quantizing gravity. General Relativity can be expressed in terms of the ADM 3+1 decomposition of spacetime: http://en.wikipedia.org/wiki/ADM_formalism And Hamiltonian's underlie quantum mechanics: http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics) Not only does this provide ...

2

I) The restricted$^1$ transformation (RT) $$\tag{1} (q,p)~\longrightarrow~ (Q,P) ~:=~(q, \sqrt{p} - \sqrt{q})$$ of OP's professor with inverse RT $$\tag{2} (Q,P)~\longrightarrow~ (q,p) ~:=~(Q, (P+ \sqrt{Q})^2) ,$$ and with Hamiltonian $H=\frac{p^2}{2}$ and Kamiltonian $K=\frac{p^3}{3}$ is indeed interesting. Apparently we should assume that $p,q,Q\geq ... 2 The definition of temperature in the kinetic theory of gases emerges from the notion of pressure. Fundamentally, the temperature of a gas comes from the amount, and the strength of the collisions between molecules or atoms of a gas. The first step considers an (elastic) impact between two particles, and writes$\Delta p = p_{i,x} - p_{f,x} = p_{i,x} - ( - ...

2

Let there be $n$ coordinates $q^j$. Ref. 1 is discussing in Section 2.4 a type of non-holonomic constraints that is known as semi-holonomic constraints. However we interpret OP's question (v2) as mostly being about counting independent degrees of freedom in constrained systems, and not so much about semi-holonomic constraints per se. Therefore, to gain ...

1

Your solution looks fine to me. Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that $\theta$ is not the vertical angle, but lies on the plane tangent to the sphere at point B. There are two kinds of rotational ...

1

The Hamiltonian can be used to describe an evolution of the "density in phase" of a system of N bodies. The density in phase is a conserved quantity for a system in equilibrium by Liouville's Theorem. The position and momenta can describe any general intensive parameter. Gibbs used this approach to derive statistical mechanics. This approach of the concept ...

1

This is a fact about the hamiltonian compared to the lagrangian which I find not trivial (and worth to keep in mind). Suppose that the lagrangian $L$ and hamiltonian $H$ are cyclic with respect to some coordinate $q_1$. Then we have a theorem (cfr. [1]): The evolution of the other coordinates $q_2,...,q_n$ is the one of a system with $n-2$ independent ...

1

The original coordinates satisfy the equations of motion when the integral of $p\, \dot{q} - H(p,q)$ is minimized, and the new coordinates satisfy the equations of motion when the integral of $P\, \dot{Q} - K(P,Q)$ is minimized. There is no requirement that $H$ and $K$ be numerically equal. The transformation is canonical if the Poisson bracket remains ...

1

The problem is that you assume the system is in equilibrium in your first line. Apparently the pendulum is not in equilibrium if the dot product of gravity and motion is not zero. But actually d'alembert principle states the following for general cases, $$\sideset{}{}\sum_{i}(F_i-\dot{p})\cdot \delta x_i=0$$ So we have ...

1

It took me quite some time to clearly understand the experiment you're describing. Actually, pouring a full bottle in a container is a quite intriguing thing. Consider the following starting configuration : This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric ...

1

Both approaches are equally correct in this case. $F = mv^2/R$ is just a consequence of the law for rotational motion, which says $\tau = I\alpha$ (Torque = Moment of Inertia * Angular acceleration). The former formula may be used in case the objects in consideration are point masses. But the latter, more general version of the formula is applicable for ...

1

The Euler Lagrange equations just give the differential equations that determine the motion of the object. The end points are boundary conditions for the differential equation. The differential equation which the brachistochrone curve satisfies will have its constants fixed so that it reduces to a line when you give it the boundary conditions of the object ...

1

It's not missing, it's in the $\ddot{R}(t)$ matrix. It doesn't show up on its own when you do the calculation with matrices instead just vectors however. Vector equations The first equation should be written $$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$ (just reverse where you have $A$ and $B$) since given the position in frame $B$ you ...

1

When the (inertial) mass is zero, then the acceleration can be non-zero for zero force. This is similar, conceptually, to what has been discussed recently regarding an ideal conductor. Consider Ohm's Law: $$V = IR$$ Now, what if $R = 0$ as is the case with an ideal conductor? Clearly, the voltage must be zero for any current. The current through the ...

1

There is a thresh-hold below which the ball will not bounce determined by the mass of the ball and gravitational acceleration (force of gravity on the ball in other words) In order for the ball to bounce (meaning physically lose contact with the surface of the ground) the compressive force created by the impulse of impact (the point at which velocity is ...

1

Consider the positive quantity $X = (\omega - \omega_0)^2 (\omega + \omega_0)^2$. Let us make the approximation $\omega \approx \omega_0$: In the first factor: we get $X_1 = 0$ and the relative error $\epsilon _X = |\frac{X_1-X}{X}| = 100 \%$. In the second factor: we get $X_2 = 4 \omega _0^2 (\omega - \omega_0)^2$ and \$\epsilon _X = | \frac{X_2-X}{X}| = ...

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