Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The action is defined as $S = \int_{t_1}^{t_2}L \, dt$ where $L$ is Lagrangian.

I know that using Euler-Lagrange equation, all sorts of formula can be derived, but I remain unsure of the physical meaning of action.

share|improve this question
1  
You see! they are also unsure :D Only user2781942 knows. I suggest you dig very rusty books, the rustier the better - there you will find some wisdom on the subject in question. What stops you to open Lagrange and Euler and READ? DO IT. Are you already doing it? –  Asphir Dom Apr 25 at 0:01
    

6 Answers 6

up vote 2 down vote accepted

The action has no immediate physical interpretation, but may be understood as the generating function for a canonical transformation; see e.g., http://en.wikipedia.org/wiki/Hamilton-Jacobi_equation

share|improve this answer

The Hamiltonian H and Lagrangian L which are rather abstract constructions in classical mechanics get a very simple interpretation in relativistic quantum mechanics. Both are proportional to the number of phase changes per unit of time. The Hamiltonian runs over the time axis (the vertical axis in the drawing) while the Lagrangian runs over the trajectory of the moving particle, the t’-axis.

enter image description here

The Illustration shows the relativistic de Broglie wave in a Minkowski diagram. The triangle represents the relation between the Lagrangian an the Hamiltonian, which holds in both relativistic and non-relativistic physics.

$$L ~=~pv-H$$

The Hamiltonian counts the phase-changes per unit of time on the vertical axis while the term pv counts the phase-changes per unit on the horizontal axis representing distance: v is the distance traveled per unit of time while p is proportional with the phase-changes per unit of distance, hence the term pv.

The Action can now be seen as being proportional to the total number of phase changes over the trajectory of the particle. The principle of least action is thus equivalent to the principle of least phase change. In the theory of special relativity the latter is equivalent to the principle of least proper time since the 'proper time' as experienced by the particle is proportional to the number of phase changes over the trajectory.

Hans

share|improve this answer
    
+1 This is why I love this place, Perhaps you could point me to some elaboration on this? –  Prathyush Oct 18 '12 at 18:27
    
Do you have a Natural extension of this way of thought for fields? –  Prathyush Oct 18 '12 at 18:38
    
This interpretation of the Lagrangian is closely related to the path integral formulation, isn't it? –  Brian Bi Apr 28 at 6:46
    
This is fascinating - I've never seen this explanation before. Does it still hold for non-inertial particles (i.e. when there's a force involved), and for entangled particles, fields etc.? –  Nathaniel Aug 5 at 13:21
1  
@Nathaniel: It holds for non-inertial particles, yes, since we can derive the equation-of-motion from the Lagrangian. You can find images similar to the one above for particles subject to the Lorentz force here: physics-quest.org/Book_Lorentz_force_from_Klein_Gordon.pdf (A chapter of my book) –  Hans de Vries Aug 5 at 21:21

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$:

  1. The off-shell action $S[q;t_i,t_f]$,

  2. The (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$, and

  3. Hamilton's principal function $S(q,\alpha, t).$

For their definitions and how they are interrelated, see e.g. this Phys.SE answer. (Here the words on-shell and off-shell refer to whether the equations of motion (eom) are satisfied or not, see the general definition)

II) OP is apparently thinking of the first option: The off-shell action

$$S[q;t_i,t_f]~=~\int_{t_i}^{t_f}\! dt ~L,$$

which may be evaluated along (possibly virtual) paths $q:[t_i,t_f]\to\mathbb{R}$, which do not necessarily satisfy Euler-Lagrange equations (=eom). The Lagrangian $L$ is typically the difference between the kinetic and potential energy, but we warn that this need not be the case, cf. e.g. this Phys.SE post and links therein.

III) One may ask: Why do we consider virtual/unphysical paths that do not necessarily satisfy eom?

Answer: For at least two reasons:

  1. One cannot derive Euler-Lagrange equations without allowing virtual paths, cf. the principle of stationary action.

  2. In quantum mechanics, the virtual paths contribute to the path integral as quantum fluctuations, and have physical consequences. (They are e.g. responsible for the Van Vleck determinant in the semiclassical approximation via Gaussian integration.)

share|improve this answer

I agree with Arnold, more or less, confining our attention to classical dynamics. In quantum mechanics (QM) and field theory (QFT), however, the action is the natural logarithm of the probability amplitude to propagate a system from an initial configuration of particles in QM or fields in QFT. Feynman exploited a comment by Dirac in his QM book that, paraphrasing, the exponential of $-i \hbar S$ is related to the propagation probability amplitude.

This is somewhat less satisfying than, say, the interpretation of potential or kinetic energy. But at least it's something.

share|improve this answer

Some time after Newton described the laws of nature in terms of an instantaneous relationship, others noticed that the history, rather than the instantaneous state, of a system could, in at least one case, be described by saying that it obeyed a certain relationship: a particular function describing the history must always be the one that (a) starts and ends with the observed values and (b) has the lowest value of that function.

This is the exact opposite perspective from viewing nature at an instant.

The particular case was the path (history) of a light beam through two different media. The function that was minimized was T: the time for the beam to get from A to B. They said, "There are an infinite number of possible paths; the law of nature that applies in this case is that the function T is the lowest of all possible paths."

This led instinctively to the question of whether this mightn't be a specific case of a more general formulation of the law of nature, equivalent to Newton's: In any system, not just light beams, there is some function (like time of travel) which can be discovered which is minimized. Nature will always choose the trajectory where this function is minimized.

By definition, this function, if it exists, is the action. (There can be more than one). For classical mechanics, the function is the integral of the difference between the potential and kinetic energies, but it is perverse and obfuscating to to take the latter as the definition of the action. The principle of least action is much more general than that. For example, it applies to quantum physics.

Action is intuitively the thing that is minimized, like the propagation time of a light beam, or average potential energy minus kinetic energy of a body skating across a hilly surface from A to B, in every history, every trajectory.

If you know relational databases, Newton framed the law of nature as a SELECT of attributes describing the state of the system (where was the particle at time t and what was its momentum), and the later scientists chose instead a GROUP function of all the intermediate states between the beginning and end of the thought experiment.

share|improve this answer
    
The only person who understood question and as a result could give correct answer. +1. –  Asphir Dom Apr 24 at 23:57

The action is a functional of not-yet-defined functions $q(t)$ and $\dot q(t)$ such that its minimum (or a stationary condition $\delta S =0$) determines a family of possible real motions of a physical system as differential equation general solutions. The final choice of one real motion out of this family is determined with giving some endpoints (or more often - initial conditions). It fixes the arbitrary constants and yields a unique curve.

share|improve this answer

protected by Qmechanic Aug 6 at 6:09

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.