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Calculus of Variation - Classical Mechanics

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Astrum
#1
May19-13, 12:13 AM
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I'm reading Classical Mechanics (Taylor), and the 6th chapter is a basic introduction to calculus of variations. I'm super confused

I've tried to go to other sources for an explanation, but they just make it even worse!

So, let me see if I can get some help here.

[tex]\int^{x_{2}}_{x_{1}} f(y(x), y'(x), x)dx[/tex] - the integration of a function of three variables. y(x) is an as yet unknown curve. I understand that although f(y, y', x) is a function of three variables, it is only dependent on one variable, x. (where do these come from, exactly?)

Taylor then defines Y(x) = y(x) + η(x) is the WRONG path, where y(x) is the correct one. η is the variation of Y(x) from y(x). - why do we need to introduce the INCORRECT path?

Next, he introduces α into Y(x) = y(x) + αη(x). If we set α = 0, we will have Y(x) = y(x) - why do we need α?

Our integral now becomes: [tex]\int^{x_{2}}_{x_{1}} f(y(x) + αη(x), y'(x) + αη'(x), x)dx[/tex] - we're assuming that α is equal to 0? I'm not sure I 100% understand this step.

We need to check that [itex]\frac{dS}{d\alpha} = 0[/itex]- is this to check that α is a constant? or used as a way of making sure α = 0?

Take partial derivative: [tex]\frac{\partial f ((y(x) + αη(x), y'(x) + αη '(x), x)}{\partial \alpha}= \eta \frac{\partial f}{\partial \alpha}+ \eta ' \frac{\partial f}{\partial y'} [/tex] - because of the chain rule


[tex]\frac{dS}{dα}=\int^{x_{2}}_{x_{1}}\frac{\partial f}{\partial α}dx = 0[/tex] -

Next he works some voodoo magic by using integration by parts on the integral. I haven't worked this step out myself, but I assume it's straight forward.

So, in the end, we get: [tex]\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}= 0[/tex]

So, I'm pretty lost. I think it would help if I understood the idea of what we're really doing here. This is essential arc length along the shortest curve, but all the additional variables and what not are confusing the hell out of me.

Sorry for the long post, but any help is much appreciated. I'm so desperate, I'm offering a reward of one (1) virtual cookie to the first helpful post.

NB - I put this in the Classical Physics section, because I'm more concerned with how this is used in mechanics right now. Although I'm interested, in what course is Calculus of Variation taught in at a rigorous level?
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jedishrfu
#2
May19-13, 12:24 AM
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This writeup on the Brachistochrone problem may help:

http://www.hep.caltech.edu/~fcp/math...alCalculus.pdf
WannabeNewton
#3
May19-13, 12:30 AM
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What part of Taylor is this exactly? Is it related to a problem? Or is it the derivation of the Euler-Lagrange equations? If it is the latter then I must say it is one of the most convoluted ways of deriving the equations I've ever seen; the standard method of taking a first variation is a lot more intuitive and straightforward.

Astrum
#4
May19-13, 12:37 AM
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Calculus of Variation - Classical Mechanics

Quote Quote by jedishrfu View Post
This writeup on the Brachistochrone problem may help:

http://www.hep.caltech.edu/~fcp/math...alCalculus.pdf
Thanks, checking it out now.


Quote Quote by WannabeNewton View Post
What part of Taylor is this exactly? Is it related to a problem? Or is it the derivation of the Euler-Lagrange equations? If it is the latter then I must say it is one of the most convoluted ways of deriving the equations I've ever seen; the standard method of taking a first variation is a lot more intuitive and straightforward.
This is the derivation of the Euler-Lagrange equations. I'm über confused about it.

Starts on page 218

Here's your virtual cookie, by the way
WannabeNewton
#5
May19-13, 12:56 AM
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See here instead: http://www.colorado.edu/engineering/...20Equation.pdf
Astrum
#6
May19-13, 01:02 AM
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I'm looking into the link first posted. I'm befuddled by line 3 on page 4. But other than that, It it explained it much better than in Taylor's book.

Although I'm still confused by why the function is f(y,y', x). y = this is the curve, of which the constraints are on. y' is the rate of change of this curve, and x is what all of this is dependent on.

[tex]\int^{x_{2}}_{x_{1}} f(y(x), y'(x), x)dx[/tex] this gives us the length of the shortest path? What exactly is f(y,y', x)?
jedishrfu
#7
May19-13, 01:04 AM
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Quote Quote by WannabeNewton View Post
That looks like a really good link too.
jedishrfu
#8
May19-13, 01:06 AM
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Quote Quote by Astrum View Post
Thanks, checking it out now.




This is the derivation of the Euler-Lagrange equations. I'm über confused about it.

Starts on page 218

Here's your virtual cookie, by the way
Oh nooo! WannaBeNewton's gonna get to the cookie first. Are there seconds?
WannabeNewton
#9
May19-13, 01:07 AM
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Quote Quote by jedishrfu View Post
That looks like a really good link too.
Apologies if you thought I meant to take a look at my link as opposed to yours. I meant to take a look at my link as opposed to the explanation in Taylor's text, which I found to be atrocious.
jedishrfu
#10
May19-13, 01:09 AM
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Quote Quote by WannabeNewton View Post
Apologies if you thought I meant to take a look at my link as opposed to yours. I meant to take a look at my link as opposed to the explanation in Taylor's text, which I found to be atrocious.
No apology needed. Can we split the cookie? I liked your link too. It was more to the point.
WannabeNewton
#11
May19-13, 01:12 AM
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Quote Quote by Astrum View Post
What exactly is f(y,y', x)?
It is a ##C^{1}## real valued function of ##y(x)##, ##y'(x)## and ##x##. It can be various things depending on what problem you are working on. For example, it could be the arc-length ##f(y,y',x) = \sqrt{1 + y'^{2}}## of a curve ##y(x)## or, more pertinent to mechanics, the Lagrangian ##L(q(t),\dot{q}(t),t) = \frac{1}{2}m\dot{q}^{2} - U## of a particle of mass ##m## and trajectory ##q(t)## interacting with a potential ##U##.
WannabeNewton
#12
May19-13, 01:12 AM
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Quote Quote by jedishrfu View Post
No apology needed. Can we split the cookie? I liked your link too. It was more to the point.
I'd rather you take the whole thing because I'm eating chips ahoy as we speak xDD
jedishrfu
#13
May19-13, 01:22 AM
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Quote Quote by WannabeNewton View Post
I'd rather you take the whole thing because I'm eating chips ahoy as we speak xDD
Okay, thanks. I just finished my peanut butter sandwich (midnight snack).
Astrum
#14
May19-13, 01:24 AM
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Many thanks for those links, they really cleared things up for me!

As for cookies, have some chemistry cookies:

I need to go and work some problems out, if I have any other questions, I'll be sure to post.
WannabeNewton
#15
May19-13, 01:27 AM
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Chemistry cookies? I feel cheated :[
Astrum
#16
May19-13, 01:34 AM
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Quote Quote by WannabeNewton View Post
Chemistry cookies? I feel cheated :[
http://www.math.umn.edu/~olver/am_/cvz.pdf this link is what I tried to use originally, was kinda confusing.

Alright alright. Have some Einstein chocolate

Never happy!

So, what kinda of class is this taught in, anyway? In a rigorous mathy style
WannabeNewton
#17
May19-13, 01:40 AM
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In a class called calculus of variations, believe it or not lol.


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