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How and why would a particle take the shortest path?

$L=KE-PE$? What's the $KE-PE$ mean in English?

I understand the 'mechanics' but not the idea itself.

Please explain simply, I do know Calculus but the simpler the better.

Please don't say because that's how nature is, because I get that a lot.

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Possible duplicates: physics.stackexchange.com/q/9/2451 and links therein. –  Qmechanic Jun 30 '13 at 7:00

2 Answers 2

up vote 2 down vote accepted

Borrowing from the new GR book by Anthony Zee, imagine that a boy on the beach, $x_1$ meters from water, wants to get as quickly as possible to a drowning girl in the water, $x_2$ meters from the beach, while the separation of the two people in the other direction parallel to the water-sand boundary is $y$. He won't pick the straight path because he's slower in the water, so the path through the water will be closer to orthogonal (to the water-sand boundary) than the path through the sand because it's "primarily" important to minimize the distance in the water.

You want to minimize the total time, which is $$t=t_1+t_2 = s_1/v_1+s_2/v_2, \quad v_1\gt v_2$$ By demanding that this is minimal – i.e. an extremum, and you said that you know how to derive the mathematical things – you may calculate the optimum angles of the two paths out of $\delta t = 0$. This was how the minimization principle first appeared – for the closely analogous propagation of light and Snell's law or refraction.

It turns out that not only the trajectories of light and point masses but (almost) all possible laws of classical physics may be derived from such a minimization procedure, from $\delta S = 0$, where $S$ is the action. Almost always, $S$ is the time integral $\int dt \,L$ where $L$ is the Lagrangian.

In your example of simple mechanics, $L=KE-PE$ where $KE,PE$ are kinetic and potential energy, respectively. So $L=mv^2-V(x)$ is similar to the total energy $H$ (also known as the Hamiltonian) except that there is an extra minus sign in front of the potential (but not kinetic) term.

This works because $\delta S = 0$, after integrating by parts, gives you $m\ddot x = F = -V'(x)$, the spatial derivative of the potential energy. When moved to the left hand side, both terms have a plus sign, which is OK even though the relative sign of $PE,KE$ was minus because one of the minus signs disappears after we integrate over time by parts.

See also

The meaning of action

and

Why the Principle of Least Action?

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The best answer so far, and I didn't think that I could really top it either. I have an understanding, but this guy really nailed it.

http://answers.yahoo.com/question/index?qid=20080620100531AAEgOhQ

Good Luck!

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It would be good if you could quote a relevant snippet from that answer and if possible also summarize it in your own words. –  Eugene Seidel Jun 30 '13 at 8:13

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