1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Intuition for Hamilton's Principle

  1. Mar 17, 2016 #1
    Is there an intuitive way to understand why nature selects the path that minimizes the action? I've seen it proven that the Euler-Lagrange equations are equivalent to Newton's laws (at least in Cartesian coordinates). So I can understand it mathematically. But on a more common-sense level, what the heck is an action, and why is it always minimized or made stationary?

    Maybe another way is to pose the question historically. Why did they call it "action"?
  2. jcsd
  3. Mar 17, 2016 #2
    the action is extremum (not minimized) i.e. the value is such that it is stationary.
    how this action is related to Fermat's Principle? i do not know but pl. try to find out .
  4. Mar 17, 2016 #3
    Fermat's principle says that light traverses the path that takes the least time. The duration, like the action, is a functional of path, namely the product of the inverse velocity and length element integrated along the path. So Fermat's principle tells you to select the path that minimizes the duration functional. I don't know if there is any further connection.
  5. Mar 17, 2016 #4
    Somewhere i have read that the Hamilton's Principle or Principle of least action is something like a basic law of dynamics in classical world and a version of "quantum action" have been defined for the quantum dynamics.
    Well i do not know in detail but using calculus of variation we used to calculate the least path between two points on a sphereical surface.
    you have to dig deeper.
  6. Mar 19, 2016 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    That's a very deep question. The only resolution I know is to use the Feynman path-integral in quantum theory, which evaluates the socalled propagator, which is the probability amplitude for a particle starting at ##x'## at time ##t'## and ending at ##x## at time ##t##. Up to a normalization constant it reads
    $$U(t,x;t',x')=\int \mathrm{D}p \int_{(t',x')}^{(t,x)} \mathrm{D} x \exp \left (\frac{\mathrm{i}}{2 \pi} S[x,p] \right),$$
    where the action functional is given by
    $$S[x,p]=\int_{t'}^{t} \mathrm{d} t' [\dot{x} p-H(x,p)].$$
    The integral is over all trajectories in phase space where the momentum is totally unconstrained and in position space you always have ##x(t)=x## and ##x(t')=x'##.

    Now you can do a formal expansion in powers of ##\hbar##. If the action is rapidly changing the path integral will tend to vanish, because the integrand is a very rapidly oscillating sine/cosine like expression (the exponential with an imaginary argument). Thus the main contribution to the integral shoud be in the region, where the action is stationary under variations of the phase-space trajectory, and this is precisely the trajectory of the classical particle.

    For a macroscopic object the approximation to take the leading order of the path integral is very good, because ##\hbar## is very small compared to the typical values of the action of the macroscopic object. That explains why we observe the particle as moving along the classical trajectory, and this explains why it is described as the stationary point of the classical action functional.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted