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Sameer
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Particle Path
« on: Dec 9th, 2006, 2:04pm » |
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Find the path on which a particle in the absence of friction, will slide from one point to another in the shortest time under the action of gravity.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
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Whiskey Tango Foxtrot
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Re: Particle Path
« Reply #1 on: Dec 10th, 2006, 11:44am » |
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This should be path independent and therefore all we need is a minimal value of the function. First derivative should do the trick. Any information about what kind of surface we are dealing with? Or are we supposed to generalize?
The answer will always be a straight line if we are allowed to constrain the surface to be flat. I guess this isn't what you want then. If the particle must be in contact with the curved surface at all times, the answer is just the line integral along the surface from one point to the other.
All equations and processes for finding geodesics should also work fine.
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| « Last Edit: Dec 10th, 2006, 11:53am by Whiskey Tango Foxtrot » |
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towr
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Re: Particle Path
« Reply #2 on: Dec 10th, 2006, 12:10pm » |
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There is one speediest path between two points. You want to gain speed quickly at the start, such that you can travel the remainder of the path quickest.
A straight line is not the fastest, unless you only need to go straigth down.
I forgot how to do the actual optimizing, but I know the curve you're supposed to get. It can be fun finding approximations via genetic algorithms (as test case).
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ThudnBlunder
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Re: Particle Path
« Reply #3 on: Dec 10th, 2006, 12:25pm » |
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It is also the curve down which a particle placed anywhere on it will slide (without friction) to the bottom in the same amount of time.
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Whiskey Tango Foxtrot
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Re: Particle Path
« Reply #4 on: Dec 10th, 2006, 12:39pm » |
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The way the question was phrased, it must be a straight line. If the particle is on a surface, it has a defined plane of existence. For a three dimensional space, it must have one and only one free variable. If we could change the surface to redefine the path in the other two dimensions, we could pick the path to be a cycloid, which is always the fastest, and I think is what you were referring to.
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Grimbal
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Re: Particle Path
« Reply #5 on: Dec 10th, 2006, 12:50pm » |
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I know the answer but I won't tell  .
A teacher of mine expressed the problem so that you want to build a zero-energy tram line. To start the tram you just push it a bit so it accelerates down a slope. Later it goes up another slope and reaches zero speed just in front of the next station. All stations are at the same altitude. So the question was: what is the best curve to minimize the travel time? The more you go down the faster you go, but also, the longer you have to travel.
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Sameer
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Re: Particle Path
« Reply #6 on: Dec 10th, 2006, 12:57pm » |
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Yes, you guys are on the right track. And yes it is not a straight line. And yes it is sliding not falling. Since you all know what I am talking about, how about some math to prove it  And of course now you all know what it is, it is popularly known as "Brachistochrone" problem proposed by John Bernoulli in 1696. The word comes from Brachstos meaning shortest and chronos meaning time.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Icarus
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Re: Particle Path
« Reply #7 on: Dec 10th, 2006, 1:00pm » |
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You could look it up in any book on the calculus of variations. I too didn't answer because I am familiar with the problem, as it is one of the great classics.
Grimbal - Your tram question is actually a different one than this. The Brachistochrone problem does not specify that the end points be at the same level, nor that the "tram" must come to rest at the other end. It is often expressed by stating the particle is a bead sliding along a well-oiled wire (so friction is negligible) from one point to another. The question is: what shape of wire will result in the shortest transit time?
Since this problem is local, it is assumed that gravity is a constant parallel force field. The tram question is usually done on global distances, so in its case, gravity is radial from a point and obeys the inverse square law.
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| « Last Edit: Dec 10th, 2006, 1:09pm by Icarus » |
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Whiskey Tango Foxtrot
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Re: Particle Path
« Reply #8 on: Dec 10th, 2006, 1:09pm » |
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So it is the Brachistochrone problem, then. I've done this one before, too. Disregard my posts. 
And Icarus is about to hit the 4000 post mark. Time for an oil change.
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| « Last Edit: Dec 10th, 2006, 1:10pm by Whiskey Tango Foxtrot » |
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"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forgo their use." - Galileo Galilei
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Sameer
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Re: Particle Path
« Reply #9 on: Dec 10th, 2006, 2:03pm » |
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on Dec 10th, 2006, 1:00pm, Icarus wrote:You could look it up in any book on the calculus of variations. I too didn't answer because I am familiar with the problem, as it is one of the great classics.
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Yea I know the solution, just want to see if we have different approaches to get the solution!!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Grimbal
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Re: Particle Path
« Reply #10 on: Dec 11th, 2006, 5:42am » |
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Actually, I don't remember how it was solved, even if the point of the whole lesson was to teach us how to do it.
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| « Last Edit: Dec 11th, 2006, 8:13am by Grimbal » |
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ThudnBlunder
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Re: Particle Path
« Reply #11 on: Dec 11th, 2006, 6:02am » |
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Brachistochrone = Tautochrone = Cycloid, loosely speaking.
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towr
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Re: Particle Path
« Reply #12 on: Dec 11th, 2006, 6:04am » |
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on Dec 11th, 2006, 5:42am, Grimbal wrote:| Actually, I don't remember how it was solved, even if the point of the whole thing was to tech us how to do it. |
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Yeah, same here. And it was only two years ago for me.
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Sameer
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Re: Particle Path
« Reply #13 on: Dec 11th, 2006, 10:08am » |
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on Dec 11th, 2006, 6:04am, towr wrote:
Yeah, same here. And it was only two years ago for me.
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Well then it is good I posted it, so you can now exercise your gray cells and get the intended cycloid solution!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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towr
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Uberpuzzler
Some people are average, some are just mean.
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Re: Particle Path
« Reply #14 on: Dec 11th, 2006, 11:19am » |
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on Dec 11th, 2006, 10:08am, Sameer wrote:| Well then it is good I posted it, so you can now exercise your gray cells and get the intended cycloid solution!! |
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Hmm, yes, but it would just consist of digging out my textbook and notes, and finding how to use the calculus of variation.
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Icarus
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Re: Particle Path
« Reply #15 on: Dec 11th, 2006, 7:34pm » |
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Basic tool of the calculus of variations: You want to extremize F: S --> R, where S is some vector space of functions into Rn. If g is an extremizing member of S and v is any other member, then define f: R --> R by f(t) = F(g+tv). if F is well-behaved, the f is differentible and f'(0) = 0. The latter condition translates into a differential equation in g.
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I wonder: Which is larger
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