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Variational calculus in particle dynamics

  1. Apr 23, 2014 #1
    I'm reading about the Principle of Least Action.
    As a prelude to it, we look at the functional

    J(x)=[itex]\int f(y(x),y'(x);x) dx[/itex]
    where the limits of integration are x_1 and x_2.
    We want to find the function y(x) that gives the functional an extremum.
    Now, to do this, we write any possible function y(x) as the sum of the sought function y_0(x) and another function η(x) times a constant α, that is:
    y(x)=y_0(x)+αη(x)
    so our aim is then to minimize the functional J as a functional of α,
    J(α)=[itex]\int f(y(α,x),y'(α,x);x) dx[/itex]
    and the book says "The condition that the integral have a stationary value, (i.e. that an extremum results) is that J be independent of α in the first order along the path giving the extremum (α=0), or, equivalenty, that
    [itex]\frac{\partial J}{\partial α}[/itex]=0 at α=0".

    Now I have two questions:
    1. Why not just take the derivative of the functional and put it equal to zero, like we've done when maximizing/minimizing things in Calculus class? and
    2. If we know how to make α=0, why do we have to go to the trouble of differentiating the function at all? Isn't the conclusion just stating an obvious fact and not bringing any solution to the table?

    I hope my reasoning is understandable, I haven't really completely grasped what I'm doing yet...!

    p.s. the text is taken from the book Classical Dynamics of Particles and Systems of Marion and Thornton
     
  2. jcsd
  3. Apr 23, 2014 #2

    Philip Wood

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    Q1: You're not trying to find an extreme value for J. You're trying to find the function y(x) which minimises it. Finding y(x) means determining a recipe for y values for all x values in the specified interval. That's why a special procedure is needed.

    Q2: I humbly suggest that you should try now to get to grips with the actual derivation of the E-L equation, and at least one example of its use. You may well find that Q2 then loses its cogency.
     
  4. Apr 23, 2014 #3

    Matterwave

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    1. That is exactly what you are doing. The functional derivative is defined as:

    $$\int_{x_1}^{x^2}\frac{\delta J[y(x)]}{\delta y(x)}dx\equiv\left.\frac{dJ[y(x)+\alpha h(x)]}{d\alpha}\right|_{\alpha=0}$$

    I don't exactly understand what you mean by this question...

    2. We don't know y_0(x). That's exactly what we are trying to find out! If we knew it, obviously we wouldn't waste our time doing this dance. What we are doing with this method of solutions is using the fundamental lemma of the calculus of variations to turn a functional integral equation (very hard to solve otherwise!) into a much more analysis friendly differential equation (the Euler-Lagrange equation).
     
  5. Apr 24, 2014 #4
    Thank you!
    Looks like the problem is lack of knowledge of multivariable calculus from my part. Thank you for pointing me in the right direction - sometimes asking the right questions can be as hard as finding the right answer!
     
  6. Apr 24, 2014 #5

    Philip Wood

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    I'm currently reading the classic by Cornelius Lanczos; 'The Variational Principles of Mechanics'. It has a good section on the E-L equation,, and is especially helpful ob boundary conditions.It's an old book, but Dover now publish it, so it doesn't cost the earth.
     
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