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I've tried all sorts of variations on the Euler-Lagrange equations (after changing to cylindrical coordinates), but am getting nowhere. Part of the problem is no doubt that it's been a while since I did this in college - but more problematic is the fact that this is a problem I just want to solve, rather than being a homework problem - so I don't KNOW if the answer is easily solvable.

It seems like it should be a very straightforward problem - one that's both interesting and easy to define. Yet I can find no examples of it being worked out anywhere (my old textbook comes closest by finding the geodesic on a cylinder is a helix), either in books or on-line. Maybe that's because it's not all that trivial?

So far what I believe to be my best approach was the following:

-Define the system in cylindrical coordinates.

-A line element ds = sqrt( dr^2 + r^2 dphi^2 + dz^2 ).

-Take out dphi from the square root, so that ds = sqrt( r'^2 + r^2 + z'^2) * dphi, such that r' = dr/dphi and z' = dz/dphi

-Attempt to minimize s = Integral(ds), so that in Euler-Lagrange, f = sqrt( r'^2 + r^2 + z'^2).

-Apply the surface constraint z = A / r^2. So z - A / r^2 = 0, which means that z' = -2Ar' / (r^3).

-Then f = sqrt( r'^2 + r^2 + 4A^2r'^2/(r^6) )

-The "second form" of the Euler equation (as defined in Thornton and Marion, chapter 6.4) is:

df/dphi - d/dphi (f - y' * df/dy') = 0

This is desirable of course because f does not depend explicitly on phi, so that df/dphi = 0, which means

f - y' * df/dy' = constant

-Plug in my value of f and differentiate, then cancel terms, to get:

r^2 = constant * sqrt( r^2 + r'^2 + 4a^2*r'^2 / r^6 )

Mathematica doesn't seem to be able to solve this - so my guess is that either I did it wrong or it's just not something that is easily solved.

My questions:

-Is this easily solved?

-If so, how do I do it? Or what did I do wrong?

-Even better, can you point me to an existing solution? (I'd like to know HOW it was solved, though I'd settle for simply the correct answer).

If it's not easily solved, but there IS a solution (I'm sure SOME mathematician's done it), the solution is plenty enough for me. Ultimately I would like to apply the solution to another project I'm working on, so knowing the answer is what I really need (it just would be personally satisfying to know HOW to solve it, but not strictly necessary)