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Lagrangian mechanics - Euler Lagrange Equation

by brisingr7
Tags: functional, lagrange equation
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brisingr7
#1
Jan27-11, 12:11 PM
P: 3
Euler Lagrange Equation : if y(x) is a curve which minimizes/maximizes the functional :

F[tex]\left[y(x)\right][/tex] = [tex]\int^{a}_{b} [/tex]f(x,y(x),y'(x))dx
then, the following Euler Lagrange Differential Equation is true.

[tex]\frac{\partial}{\partial x}[/tex] - [tex]\frac{d}{dx}(\frac{\partial f}{\partial y'})[/tex]=0

Well....
I don't understand why the function f has only three variables x, y(x) and the derivative of that.
what about y'' or y[tex]^{(3)}[/tex]? I think it could be possible.(physically) All files related to this topic states that the function f as a function of variables x, f(x), and f'(x).
i.e. can function f be like : f(x,y(x),y'(x),y''(x),.....) ??? or.... is it unnecessary to think about the second derivative and furthermore?
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brisingr7
#2
Jan27-11, 12:22 PM
P: 3
oops!
the partial symbol should be on the left!
also, the lower/upper limits should change seats with the integral....
arildno
#3
Jan27-11, 12:35 PM
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For variational problems in general maths, you are perfectly correct. You can perfectly well create higher-order "Euler-Lagrange"-like equations!
What they mean, however, is quite another thing!

For mechanics problems though, it is sort of axiomatic that forces are, at most, only dependent on an object's position&velocity (0th and 1st derivatives), and not on higher-order derivatives.

Thus, it is sufficient (given this "axiom") for the variational formulation to use an integrand merely dependent on x, y and y'.


Exceptions probably exist, though, but I'm not familiar with them.

AlephZero
#4
Jan27-11, 04:50 PM
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Lagrangian mechanics - Euler Lagrange Equation

Translate this back into "pre-Euler-Lagrange mechanics", as a statement about the work done on a system and its potential and kinetic energy.

Those three quantities are (almost always!) functions of position and velocity, and nothing else.
randys006
#5
Jan27-11, 06:23 PM
P: 7
The normal "baby-steps" into this are by defining a conservative system as one that you can identify a potential for, and the conditions for defining a potential is that your forces only depend on x, y, and y'. However, the formal definition seems to involve circular reasoning: the Lagrangian formalism only applies to conservative systems, and a conservative system is (formally) defined as one that follows Lagrangian mechanics. It isn't actually circular reasoning, it is just that the two are defined simultaneously, and the mathematical relationship between force and potential naturally attaches the f(x, y, y') limitation.

The Work-Energy theorem from baby physics is nothing more than the integral of the Lagrangian.

Hope that helps!
Randy
lalbatros
#6
Jan28-11, 12:01 AM
P: 1,235
brisingr7,

There is no real need to introduce a y" in the Lagrangian.
If you needed that, you could just introduce an additional variable:

z = y'

and write a new Lagrangian including this additional variable:

L(x,y,y',z,z')

Additional variables keep the formalism the same and handles the extra y" you are thinking of.
brisingr7
#7
Jan28-11, 10:39 AM
P: 3
Wow! thanks!
I think I get it!


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