Principle of Least Action OR Hamilton's Principle

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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 [tex]U = c[/tex], the Lagrangian is simply [tex](1/2)mv^2 + c[/tex], and hamilton's principle becomes

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

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

[tex]\delta \int mv {\mathrm d}x = 0[/tex].

This is usually written in the form

[tex]\delta \int p {\mathrm d}q = 0[/tex]

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 [tex]U = c[/tex], the Lagrangian is simply [tex](1/2)mv^2 + c[/tex], and hamilton's principle becomes

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

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

[tex]\delta \int mv {\mathrm d}x = 0[/tex].

This is usually written in the form

[tex]\delta \int p {\mathrm d}q = 0[/tex]

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.