Can you explain me what are the differences between the four following subjects?

  • analytical mechanics
  • rational mechanics
  • classical mechanics
  • theoretical mechanics
  • $\begingroup$ Rational mechanics is the development of mechanics according to a strict logical scheme, from certain hypotheses/axioms, from which general theorems (e.g. Liouville's theorem, existence of integrals of motion or Lyapunov exponents) or applications are deduced. In other words, it is classical mechanics as seen by mathematicians. en.wikipedia.org/wiki/Rational_mechanics $\endgroup$
    – Quillo
    Commented Jul 29, 2023 at 6:55

2 Answers 2


My understanding of the usage of these terms is obviously not perfect, but I can still try. I have to admit beforehand that I have never encountered the term ‘rational mechanics’, maybe someone more knowledgeable can expand on this.

Theoretical Mechanics

This is a term used to differentiate between experimental mechanics (bouncing little balls off each other) and theoretical mechanics (trying to derive equations how little balls bounce off each other). As such, it encompasses classical mechanics, analytical mechanics and rational mechanics.

Classical Mechanics

Classical Mechanics is used in two different contexts: First, it is the antonym to Quantum Mechanics. As such, we usually assume a strictly deterministic world ruled by certain differential equations (in one of the many formulations of classical mechanics, see below), as opposed to a quantum mechanical view where probability densities evolve according to the Schrödinger/Heisenberg equation ($i \hbar \partial_t \Psi = H \Psi$ or $\dot A = \frac{i}{\hbar}[ H , A ] + \partial_t A$).

On the other hand, the term ‘classical mechanics‘ is sometimes used to describe Newtonian Mechanics as opposed to the later developments by Euler, Lagrange, Hamilton and Jacobi. Newtonian mechanics rely on the equation $ F = \dot p = m a $ (assuming $\dot m = 0$) to describe the movement of a point particle of mass m. Since $a = \ddot x$, they usually require two integrations to solve for the trajectory of the particle.

Analytical Mechanics

Analytical Mechanics form the ‘other’ branch of classical mechanics and build upon Newtonian mechanics. It can be divided into three major steps: Lagrangian Mechanics, Hamiltonian Mechanics and mechanics based on the Hamilton-Jacobi equation.

Lagrangian Mechanics

Lagrangian Mechanics starts off with the definition of the Lagrangian $\mathcal{L}$ which acts as a measure for the difference between kinetic and potential energy. It is a function of the coordinates and velocities of all particles:

$$\mathcal{L}(q_1,q_2,\ldots,q_N,\dot q_1,\dot q_2,\ldots,\dot q_N,t) = T - U$$

Lagrange then postulated that the actual trajectories of the particles between time $t_1$ and time $t_2$ are these that minimise the action $S$ as defined by

$$ S = \int_{t_1}^{t_2} \mathcal{L} dt $$

which leads to a variation problem: Find $\{q_i,\dot q_i\}$ such that

$$ \delta S = \int_{t_1}^{t_2} \delta \mathcal{L} dt = 0$$

where $\delta X$ describes the variation of $X$ by its arguments (coordinates and velocities, in our case). As it happens, there is an equation that describes when a given quantity fulfills this requirement, namely the Euler-Lagrange equations. These are:

$$ \partial_{q_i} \mathcal{L} - \frac{\mathrm{d}}{\mathrm{d}t} \partial_{\dot q_i} \mathcal{L} = 0 $$

where $\partial_x = \frac{\partial}{\partial x}$ and I dropped the argumets of $\mathcal{L}$. Note that these equations are of first order in $\{q_i,\dot q_i\}$, as opposed to Newton’s $F = \ddot p$. Furthermore, note that I used $q_i$ to denote the coordinate(s) of the $i$-th particle rather than $x_i$: This is because Lagrangian mechanics makes it very easy to implement generalised coordinates. This is best shown by example:

Assume (in two dimensions) that you have a bolt of length $l$ fixed at the origin $(0,0)$ and a mass $m$ at the other end of the string. Furthermore assume that the bolt has always the same length. To then describe the movement of the mass using the standard coordinates, we need to introduce $y$ and $x$ and integrate each of them and do all sorts of ugly things and, most importantly, always have to take care that $y^2 + x^2 = l^2$. However, we notice that there is only one degree of freedom: the angle. By then introducing a generalised coordinate $q$, we can implement the requirement $y^2 + x^2 = l^2$ by simply not admitting any other coordinates. We set

$$ x = l \cos(q) \quad y = l \sin(q) $$

and can be sure that the bolt always has the same length. Assuming a constant gravitational potential (i.e. potential energy $m \cdot g \cdot x$), we can write

$$ \mathcal{L}(q,\dot q,t) = \frac{1}{2m} l \dot q^2 - m \cdot g \cdot l \cos(q) $$

and hence

$$ m \cdot g \cdot l \sin(q) - \frac{\textrm{d}}{\textrm{d}t} \frac{1}{m} l \dot q = 0 .$$

You might notice that there’s still a $\ddot q$ hidden there. That’s where Hamiltonian Mechanics comes in.

Hamiltonian Mechanics

Hamilton noticed that Lagrangian mechanics is still basically Newtonian mechanics with a nicer dress, but by applying a Legendre transformation to $\mathcal{L}$, we can actually get rid off $\dot q$ (and therefore $\ddot q$).

To this end, we introduce the ‘canonically conjugated momentum‘ $p_j = \partial_{\dot q_j} \mathcal{L}$ and the Hamiltonian $\mathcal{H}$ which is a function of $q$, $p$ (and, rarely, $t$), defined by:

$$\mathcal{H}(q_1,q_2,\ldots,q_N,p_1,p_2,\ldots,p_N,t) = \sum_i \dot q_i p_i - \mathcal{L} \quad = T + U.$$

You might want to verify that $\mathcal{H}$ does not depend on $\dot q_i$, but only on the canonically conjugated coordinates and their momentums $\{ q_i , p_i \}$. We can then rewrite the Euler-Lagrange equations as follows:

$$ \dot q_i = \partial_{p_i} \mathcal{H} \quad ; \quad \dot p_i = - \partial_{q_i} \mathcal{H} \quad .$$

You can memorise these by setting $H(q,p,t) = T(p) + U(q)$, the second term then becomes ‘something like’ $\nabla U = -F = - \dot p_i$.

These equations are still asymmetric (hence the rule above), but by introducing Poisson brackets:

$$ \{ A , B \} = \sum_i [ \partial_{q_i} A \partial_{p_i} B - \partial_{p_i} A \partial_{q_i} B ] $$

we can actually fix that. Observe:

$$ \dot q_i = \{ q_i , H \} \quad ; \quad \dot p_i = \{ p_i , H \} $$

since $\partial_{q_i} p_j = 0 \quad \forall i,j$.

Furthermore, we can now leap to quantum mechanics with relative ease: Simply add a ‘hat’ to $\mathcal{H}$ and replace $\{ \cdot , \cdot \}$ by $-\frac{i}{\hbar} [ \cdot , \cdot ]$, where $[ A , B ] = AB - BA$ denotes the commutator :)

If you feel particularly nasty, look up the Hamilton-Jacobi equation, it is really not nice (or the most beautiful thing ever seen, depending on your POV).

  • 1
    $\begingroup$ Great answer! I would like to add that analytical mechanics, as defined here, is still an evolving subject. The current frontiers of the work revolves around understanding the "geometry" hidden behind the equations (which usually means trying to do things without explicit coordinates, or with all coordinate choices being equal; this has practical bearing on using such methods in general relativistically relevant problems). As a side branch, there is also a lot of interest in finding numerically tractable forms, which are intrinsically stable and accurate. $\endgroup$
    – genneth
    Commented Oct 12, 2012 at 20:33
  • $\begingroup$ Cool, so far I regarded it as rather complete following the leap to quantum mechanics. But then I didn’t do GR until about two weeks ago…. Feel free to edit in your additions in the text above :) $\endgroup$
    – Claudius
    Commented Oct 12, 2012 at 20:36
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    $\begingroup$ That's okay -- I'm happy for my thoughts to remain as a comment; perfectly appropriate. By the way, part of the great tensions in QM regarding the nature of time, measurement vs unitary evolution, etc. can be traced to a Lorentz non-covariant formulation of Hamiltonian mechanics; the same inelegance is found there, but because the meaning is so obvious, it's easy to live with. In my opinion, one of the more interesting recent developments has been the (more widespread) recognition that there is a covariant version, whose extension into quantum or discrete/numerical setting is far more natural. $\endgroup$
    – genneth
    Commented Oct 12, 2012 at 20:41
  • $\begingroup$ Huh, I'm used to hearing "classical mechanics" being used to describe the developments by Euler, Lagrange, Hamilton and Jacobi, and to distinguish them from Newtonian mechanics. I've never heard the usage the other way around. $\endgroup$
    – David Z
    Commented Oct 13, 2012 at 0:10
  • $\begingroup$ @DavidZaslavsky I would assign ‘canonical mechanics’ to these developments (and usually call everything deterministic classical, as opposed to quantum mechanics). But usage surely varies both from department to department and from language to language. $\endgroup$
    – Claudius
    Commented Oct 13, 2012 at 9:06

Analytical mechanics is a branch of classical mechanics that is not vectorial mechanics (original Newton's work). Analytical mechanics uses two scalar properties of motion, the kinetic and potential energies, instead of vector forces, to analyse the motion. Analytical mechanics includes Lagrangian mechanics, Hamiltonian mechanics, Routhian mechanics...

Theoretical mechanics is a branch of mechanics which employs mathematical models and abstractions of physics to rationalize, explain and predict mechanical phenomena. This is in contrast to experimental mechanics, which uses experimental tools to probe these phenomena.

Rational mechanics is a branch of theoretical mechanics characterized by a purely axiomatic approach, where some few axioms are selected and then the rest of the theory logically derived as theorems and corollaries. This branch is usually more mathematically oriented than others.

Classical mechanics is that branch of mechanics that ignores quantum effects. Classical mechanics can be either relativistic or non-relativistic, although in older literature classical mechanics often means pre-relativistic classical mechanics.

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    $\begingroup$ Thanks for your clear and concise answer :) I'm sorry I can't upvote you (too low rep). $\endgroup$ Commented Oct 15, 2012 at 20:14
  • $\begingroup$ Thank you for the comment. If the response was useful that is much better than any vote :-). Don't worry about votes $\endgroup$
    – juanrga
    Commented Oct 16, 2012 at 10:53
  • 1
    $\begingroup$ Very to the power infinity nice explanation. $\endgroup$ Commented Feb 14, 2021 at 11:39

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