I'm reading Arnold's "Mathematical Methods of Classical Mechanics" but I failed to find rigorous development for the allowed forms of Hamiltonian.

Space-time structure dictates the form of Hamiltonian. Indeed, we know how the free particle should move in inertial frame of references (straight line) so Hamiltonian should respect this.

I know how the form of the free particle Lagrangian can be derived from Galileo transform (see Landau's mechanics).

I'm looking for a text that presents a rigorous incorporation of space-time structure into Hamiltonian mechanics. I'm not interested in Lagrangian or Newtonian approach, only Hamiltonian. The level of the abstraction should correspond to the one in Arnolds' book (symplectic manifolds, etc).

Basically, I want to be able to answer the following question: "Given certain metrics, find the form of kinetic energy".

  • $\begingroup$ @Yrogirg: Why do you talk about spacetime when you are talking about Hamiltonian mechanics? Hamiltonian mechanics is purely classical (i.e. non-relativistic) as far as I know. $\endgroup$
    – Zhen Lin
    Jul 19, 2012 at 7:39
  • $\begingroup$ @Zhen Lin Classical physics has it's own space time, though quite special. And it defines the form of a Hamiltonian of a free particle (kinetic energy). As for the relativity theory, I just hoped that in such a manner I'll learn how Hamiltonian mechanics is generalized for that case. I'm curious about it either. $\endgroup$
    – Yrogirg
    Jul 19, 2012 at 8:18
  • $\begingroup$ @ZhenLin Really the relativistic motion of a charged particle in an electromagnetic field is Hamiltonian. $\endgroup$
    – Giuseppe Tortorella
    Jul 19, 2012 at 14:02
  • $\begingroup$ @GiuseppeTortorella OK, but I was thinking about the motion of a massive particle under gravity. $\endgroup$
    – Zhen Lin
    Jul 19, 2012 at 14:34
  • $\begingroup$ @GiuseppeTortorella If you write the hamiltonian for the full system: EM field+particle, then the motion of the particle is hamiltonian. $\endgroup$
    – Shaktyai
    Jul 22, 2012 at 15:58

4 Answers 4


From reading the comments to the question, I think that a partial answer could be to show the Hamiltonian character of the relativistic dynamics of a material particle in an electromagnetic field. If this interpretation of the question isn't correct then at least I hope to help in finding out its true interpretation.

Let $(M,g)$ be a Lorentzian $4$-manifold and $F$ the closed $2$-form on $M$ describing the electromagnetic field. Let $H\in C^{\infty}(T^\ast M)$ be the kinetic energy defined by $H(\nu)=\frac{1}{2}g(g^\sharp\nu,g^\sharp\nu).$ It easily proved that if $\omega_0$ denotes the canonical symplectic form then $\omega_0+(\tau_M^\ast)^\ast F$ is also a symplectic form on $T^\ast M.$ (Here $\tau_M^\ast:T^\ast M\to M$ is the projection on the base.)

Then the motions are the projection on the base of the integral curves for the Hamiltonian vector field $X_H=\omega_F^\sharp(dH).$


I tried the following: assuming that the metric only depends on the coordinates, $g(q)$, and the energy only depends on the the coordinates and the spatial momenta, $p_0 (q,p_i)$. I then considered variations like $$ \delta S=\int \delta \theta=\int \delta(g_{ab}p^a dq^b)=\int \delta(g_{00}p^0dq^0+g_{0i}(p^idq^0+p^0dq^i)+g_{ij}p^idq^j) $$ Then just remember for variations of metric components $$ \delta g_{ab}=\frac{\partial g_{ab}}{\partial q^c}\delta q^c=\frac{\partial g_{ab}}{\partial q^0}\delta q^0 + \frac{\partial g_{ab}}{\partial q^i}\delta q^i $$ and $$ \delta p^0=\frac{\partial p^0}{\partial q^c}\delta q^c+\frac{\partial p^0}{\partial p^k}\delta p^k=\frac{\partial p^0}{\partial q^0}\delta q^0 + \frac{\partial p^0}{\partial q^i}\delta q^i +\frac{\partial p^0}{\partial p^k}\delta p^k $$ You will want to isolate the coefficients of $\delta q^0$ and $\delta q^i$ and $\delta p^i$. These are Hamilton's Equations. You will need to integrate by parts when you have derivative's of variations remembering the the variations are zero at the end points. For example: $$ g_{00}p^0 d(\delta q^0)=d(g_{00}p^0q^0)-d(g_{00}p^0)\delta q^0 $$ I think this will work.

  • $\begingroup$ To make it clearer, you assume that the system moves along geodesics and derive Hamiltonian equations? $\endgroup$
    – Yrogirg
    Jul 22, 2012 at 19:51
  • $\begingroup$ The variations $\delta q$ and $\delta p$ can be arbitrarily small, so their coefficients must vanish if the variation of the action is to vanish. Minimizing the action yields geodesics. $\endgroup$
    – kηives
    Jul 22, 2012 at 20:03
  • $\begingroup$ What you've done you showed how to derive equations of motion of a free particle given the metrics. Am I right? It is very close to the original question, though the question was asking to find the kinetic energy, not the equations of motion. And by the way, what's your view on how to find $g$ for a system of particles? $\endgroup$
    – Yrogirg
    Jul 24, 2012 at 19:03
  • $\begingroup$ What do you mean by $g$? Well, what I wrote has the equations of motion in it because it gives Hamilton's equations, so you will get equations for $\dot{p}$ and $\dot{q}$ among the equations. $\endgroup$
    – kηives
    Jul 29, 2012 at 15:55

UPD There was an error concerning the place of $\frac{1}{2}$. fixed it.

First of all, now I see that this question fits better to physics.SE. I thought the answer will be well known, standard and taken for granted, but it seems it's not the case.

For a classical system I believe I have found how does classical 3D space metrics defines the form of kinetic energy. So what I have understood by now.

The configuration space $Q$ of a single particle permits a metrics on it as

$$g = m g_{E}$$

where $m$ is particle's mass and $g_{E}$ is the Euclidean metric tensor. Then the kinetic part of Hamiltonian is given by

$$T = \frac{1}{2} g^{-1}$$

whereas the kinetic part of Lagrangian is

$$T = \frac{1}{2} g$$

See my question at MSE on how the inverse metrics is connected to the Legendre transform of quadratic forms.

The configuration space of N-particle system is given by

$$Q = Q_1 \times Q_2 \times \ldots \times Q_N$$

where $Q_i$ is the configuration space of $i$-th particle. Since $Q_i$ are all Riemannien manifolds, then $Q$ itself has a natural structure of a Riemannien manifold, see my corresponding question at Math.SE. This new metrics agrees with the view of kinetic energy of the system as a sum of kinetic energies of it's subsystems. The metric tensor on $Q$ defines the kinetic energy

As for the relativistic case things are harder.

To see how the formalism works I studied double pendulum as a practical example.

Let $Q_1$ and $Q_2$ be first defined in Cartesian system $(x_1, y_1)$, $(x_2, y_2)$. Euclidean metric in both of them $g_E$ is given as

$$g_E = dx \otimes dx + dy \otimes dy $$

Let's do the variable substitution:

$$ \begin{cases} x_1 = l_1 \sin \theta_1, \\ y_1 = - l_1 \cos \theta_1, \\ \end{cases} $$

$$ \begin{cases} x_2 = l_1 \sin \theta_1 + l_2 \sin \theta_2, \\ y_2 = - l_1 \cos \theta_1 - l_2 \cos \theta_2, \\ \end{cases} $$

In new coordinates $g_1$ for $Q_1$ and $g_2$ for $Q_2$ are:

$$g_1 = m_1 l_1^2 \, d \theta_1 \otimes d \theta_1$$

$$ \begin{multline} g_2 = m_2 \left[ l_1^2 \, d \theta_1 \otimes d \theta_1 + l_2^2 \, d \theta_2 \otimes d \theta_2 + \right .\\ \left. l_1 l_2 \left(\cos(\theta_1 - \theta_2)\right)(d \theta_1 \otimes d \theta_2 + d \theta_2 \otimes d \theta_1) \right] \end{multline} $$

To obtain the form of kinetic energy in the Lagrangian setting:

$$ \begin{multline} T(\dot \theta_1, \dot \theta_2) = \frac{1}{2} g\left( \dot \theta_1 \frac{\partial}{\partial \theta_1} + \dot \theta_2 \frac{\partial}{\partial \theta_2}, \dot \theta_1 \frac{\partial}{\partial \theta_1} + \dot \theta_2 \frac{\partial}{\partial \theta_2} \right) = \frac{1}{2} (g_1 +g_2) = \\ \frac{1}{2} m_1 l_1^2 \dot \theta_1 \dot \theta_1 + \frac{1}{2} m_2 \left[ l_1^2 \dot \theta_1^2 + l_2^2 \dot \theta_1^2 + 2 l_1 l_2 \left(\cos(\theta_1 - \theta_2)\right) \dot \theta_1 \dot \theta_2 \right] \end{multline} $$

Up to this point all the calculations were conducted by pure algebraical tossing of symbols that I did by hand. No drawings or whatsoever, actually that was the goal --- given the metrics do everything further as a machine, blindly. Unfortunately I don't know yet the easy way to cast kinetic energy in Hamiltonian form. One can do a Legendre transform or, as I did resort to CAS to find the inverse matrix:

$$ \begin{multline} T(p_{\theta_1}, p_{\theta_2}) = \\ \frac{1}{2} \begin{bmatrix} p_{\theta_1} & p_{\theta_2} \end{bmatrix} \begin{bmatrix} (m_1 + m_2) l_1^2 && m_2 l_1 l_2 \left(\cos(\theta_1 - \theta_2)\right) \\ m_2 l_1 l_2 \left(\cos(\theta_1 - \theta_2)\right) && m_2 l_2^2 \end{bmatrix}^{-1} \begin{bmatrix} p_{\theta_1} \\ p_{\theta_2} \end{bmatrix} = \\ = \frac{ \left(m_2 + m_1 \right) l_1^2 p_{\theta_2}^2 + l_2 ^ 2 m_2 p_{\theta_1}^2 - 2 m_2 l_1 l_2 p_{\theta_1} p_{\theta_2} \cos \left(\theta_2 - \theta_1\right) }{ 2\left( l_1^2 l_2^2 m_2^2 + l_1^2 l_2^2 m_1 m_2 - l_1^2 l_2^2 m_2^2 \cos \left(\theta_2 - \theta_1 \right)^2 \right) } \end{multline} $$


If you are interesting in the Hamiltonian for a free particle in a static (or stationary) metric, then it is easy to find it. Although, it is worth to remember that the choice of world (evolution) time is not completely unique. Below is the very formal solution.

Lagrangian mechanics

Firstly, we consider the action of a single particles in an external static (or stationary) metric. For the sake of simplicity, let's put the mass of the particle $m=1$, then $$ S=-\int\mathrm{d}s=-\int\sqrt{g_{\mu\nu}\,\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}} =-\int\sqrt{g_{00}+2g_{i0}v^{i}+g_{ij}v^{i}v^{j}}\mathrm{d} t,\qquad(1) $$ where $v^{i}=\dot{x}^{i}$. The action (1) implies that the Lagrangian has the form: $$ L\left( x,v\right) =-\sqrt{g_{00}+2g_{i0}v^{i}+g_{ij}v^{i}v^{j}} $$ or $L=-\mathrm{d}s/\mathrm{d}t$ for short. The canonic momentum takes the form: $$ p_{i}=\frac{\partial L}{\partial v^{i}}=-\frac{g_{i0}+g_{ij}v^{j}}{\sqrt{g_{00}+2g_{k0} v^{k}+g_{mn}v^{m}v^{n}}}.\qquad(2) $$ If one can find the solution $v^{i}\left( p,x\right) $ of the equation (2) then the Hamiltonian can be obtained in the standard way: $$ H\left( x,p\right) =p_{i}v^{i}\left( p,x\right) -L\left( x,v\left( p,x\right) \right) .\qquad(3) $$ However, it is worth to show here another way to find the Hamiltonian.

Hamilton-Jacobi equation

Let's consider the Hamilton-Jacobi equation: $$ g^{ik}\frac{\partial S}{\partial x^{i}}\frac{\partial S}{\partial x^{k}% }=1,\quad\Rightarrow\quad g^{00}\frac{\partial S}{\partial x^{0}}% \frac{\partial S}{\partial x^{0}}+2g^{0i}\frac{\partial S}{\partial x^{0}% }\frac{\partial S}{\partial x^{i}}+g^{ik}\frac{\partial S}{\partial x^{i}% }\frac{\partial S}{\partial x^{k}}=1.\qquad(4) $$ For a static (or stationary) metric, this equation has an integral $\partial S/\partial x^{0}$ which formally coincides the covariant $0$-component of the 4-momentum, so that the canonic momentum is $\partial S/\partial x^{i}$: $$ p_{0}=-\frac{\partial S}{\partial x^{0}},\quad p_{i}=\frac{\partial S}{\partial x^{i}}. $$ Thus the Hamilton-Jacobi equation reads as follows: $$ g^{00}p_{0}^{2}-2p_{0}h+T-1=0, $$ where $$ h=g^{0i}p_{i},\quad T=g^{ik}p_{i}p_{k}. $$ The solution of the Hamilton-Jacobi with respect to $p_{0}$ yields the desired Hamiltonian: $$ H\left( p,x\right) =\frac{1}{g^{00}}\left[ h+\sqrt{h^{2}+g^{00}\left( 1-T\right) }\right] =\frac{g^{0i}}{g^{00}}p_{i}+\frac{\sqrt{1-\gamma ^{ik}p_{i}p_{k}}}{\sqrt{g^{00}}},\qquad(5) $$ where I introduced the 3-dimensional contravariant metric tensor: $$ \gamma^{ij}=g^{ij}-\frac{g^{i0}g^{j0}}{g^{00}}, $$ which has the following property: $$ \gamma^{ni}g_{nj}=\delta_{j}^{i},\quad\gamma^{ij}g_{i0}=-\frac{g^{j0}}{g^{00}% },\quad\gamma^{ij}g_{i0}g_{j0}=g_{00}-\frac{1}{g^{00}}.\qquad(6) $$ It is easy to show that the Hamiltonian (5) coincides with that of (3). Let's consider the following Hamilton's equation: $$ \dot{q}^{k}=\frac{\partial H}{\partial p_{k}}=\frac{g^{0k}}{g^{00}}% -\frac{\gamma^{ik}p_{i}}{\sqrt{g^{00}}\sqrt{1-\gamma^{ik}p_{i}p_{k}}}% .\qquad(7) $$ One can see that expression (7) is the desired solution of the equation (2). In fact, the contraction $\gamma^{ij}p_{i}p_{j}$ with canonical momentum (2) and utilizing the properties (6) result in: $$ \frac{1}{g^{00}\left( 1-p_{i}\gamma^{ij}p_{j}\right) }=g_{00}+2g_{k0}% v^{k}+g_{mn}v^{m}v^{n}. $$ Using the equation (2) we find: $$ g_{i0}+g_{ij}v^{j}=-p_{i}\sqrt{g_{00}+2g_{k0}v^{k}+g_{mn}v^{m}v^{n}}% =-\frac{p_{i}}{\sqrt{g^{00}\left( 1-p_{i}\gamma^{ij}p_{j}\right) }}, $$ hence: $$ v^{j}=-\gamma^{ij}g_{i0}-\frac{\gamma^{ij}p_{i}}{\sqrt{g^{00}\left( 1-p_{i}\gamma^{ij}p_{j}\right) }}=\frac{g^{j0}}{g^{00}}-\frac{\gamma ^{ij}p_{i}}{\sqrt{g^{00}\left( 1-p_{i}\gamma^{ij}p_{j}\right) }}. $$ The last step is to calculate the Lagrangian: $$ p_{k}\dot{q}^{k}=\frac{g^{0k}p_{k}}{g^{00}}+\frac{\sqrt{1-\gamma^{ik}% p_{i}p_{k}}}{\sqrt{g^{00}}}-\frac{1}{\sqrt{g^{00}}\sqrt{1-\gamma^{ik}% p_{i}p_{k}}}=H-\frac{1}{\sqrt{g^{00}}\sqrt{1-\gamma^{ik}p_{i}p_{k}}}, $$ hence: $$ L=p_{k}\dot{q}^{k}-H=-\frac{1}{\sqrt{g^{00}}\sqrt{1-\gamma^{ik}p_{i}p_{k}}% }=-\sqrt{g_{00}+2g_{k0}v^{k}+g_{mn}v^{m}v^{n}}. $$

  • $\begingroup$ Sorry, I need time, I hope to give my response in several days, before the weekend passes. $\endgroup$
    – Yrogirg
    Aug 16, 2012 at 8:14

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