Integrating a Functional

Question

the so called "Feynmann path integral", which, as far as I understand, means "integrating" a functional (action) on some infinite-dimentional space of configurations (fields) of a system.

leaves me wondering, how does one set up the integral of a functional such as $J[y] = \int_0^1 f(x,y,y')dx = \int_0^1 (y'^2 + xy)dx \ , \ y(0) = 0, y(1) = 1$ with E-L equation $x-2y''=0$ and solution $y(x) = \tfrac{x^3}{12} + \frac{11x}{12}$ and evaluate it explicitly? Furthermore, what does it mean to do such a thing?

You just integrate it. Without any kind of physical meaning, it's hard to give a meaning to the value of the integral. The principle of least action would mean this is the path $y(t)$ which minimizes the difference between the kinetic energy ($\dot{y}^2$) and the potential energy ($2xy$)--the $2$ coming from the fact that kinetic energy is really $\frac{1}{2}\dot{y}^2$. If $x$ really is a distance, then I would have to think harder about how this fits into the principle of least action. — Jared, Apr 22 '14 at 06:03
Do you mean integrate as in $\int J[y] dy$ or as in just evaluate $J[y] = \int_a^b Ldx$ for the minimum? Does integrative over an infinite-dimensional space of configurations just mean we're integrating a multi-dimensional lagrangian, i.e. $L = \int_V\mathcal{L}dV$? — bolbteppa, Apr 22 '14 at 07:13

Integrating a Functional

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