Principle of Least Action OR Hamilton's Principle

AI Thread Summary
The discussion centers on the relationship between the principle of least action and Hamilton's principle, with participants agreeing that Hamilton's principle generalizes the least action principle. It is noted that the least action principle can be viewed as a special case of Hamilton's principle when potential energy is constant. The conversation highlights that in celestial mechanics, potential energy is not constant, prompting questions about how orbits can be represented within this framework. There is also mention of the freedom to add a total time derivative of a function in the Lagrangian, which may affect the potential in orbital mechanics. Overall, the dialogue explores the nuances and applications of these principles in physics.
shehry1
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Are the principle of least action(http://astro.berkeley.edu/~converse/Lagrange/Kepler%27sFirstLaw.htm) and the hamilton principle 'exactly' the same? As far as I know, yes. How do I go from one to the other
 
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i believe 'Hamilton Principle' generalizes the 'Least Action principle'
 
It looks like the first equation that you refer to is a special case of Hamilton's principle, where the potential energy is taken to be constant. In the case of U = c, the Lagrangian is simply (1/2)mv^2 + c, and hamilton's principle becomes

\delta \int \frac{1}{2} mv^2 {\mathrm d}t = 0 \Longrightarrow \delta \int mv^2 {\mathrm d}t = 0.

Since dt = {dx}/{v}, this is equivalent to

\delta \int mv {\mathrm d}x = 0.

This is usually written in the form

\delta \int p {\mathrm d}q = 0

to emphasize that q is a generalized coordinate.
 
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dx said:
It looks like the first equation that you refer to is a special case of Hamilton's principle, where the potential energy is taken to be constant. In the case of U = c, the Lagrangian is simply (1/2)mv^2 + c, and hamilton's principle becomes

\delta \int \frac{1}{2} mv^2 {\mathrm d}t = 0 \Longrightarrow \delta \int mv^2 {\mathrm d}t = 0.

Since dt = {dx}/{v}, this is equivalent to

\delta \int mv {\mathrm d}x = 0.

This is usually written in the form

\delta \int p {\mathrm d}q = 0

to emphasize that q is a generalized coordinate.

I wondered whether that was the case. However, this being celestial mechanics, it is obvious that the potential is not constant. In fact, only one or two lines down, the energy is written as a sum of KE and Potential (as you would expect)
 
There's also a freedom of a total time derivative of a function of (q,t) in the Lagrangian, so maybe in the case orbits the potential can be written in this way? I'm not sure.
 
dx said:
There's also a freedom of a total time derivative of a function of (q,t) in the Lagrangian, so maybe in the case orbits the potential can be written in this way? I'm not sure.

http://en.wikipedia.org/wiki/Reduced_action
Check out the disambiguation
 
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