Functionals

**CFDFEAGURU** · Jul4-09, 02:55 PM

Hello all,

I have been trying to fill in the gaps in the example of a functional given in chapter 3 of Hartle's book "Gravity" and I am not having much luck. I exhausted wikipedia for help to no avail. Does anyone know of or can provide a good simple example of a functional or just the Lagrangian?

Thanks

**jambaugh** · Jul4-09, 04:14 PM

A functional is simply a function whose domain is a set of functions. i.e. a functional is a mapping from a function to a number.

Example: [The value of the function at x=0]
maps cos to 1, maps sin to 0, and maps a polynomial to the value of its constant term.

Example: Given x as a function of t representing an arbitrary (smooth) particle trajectory,
$LaTeX Code: x \\mapsto S= \\int_{t_1}^{t_2} \\frac{m}{2} \\dot{x}^2(t) - V(x(t)) dt$

maps the whole trajectory of the particle to a single number, S we call the action. V is a potential function defining the potential energy for the particle at any position x.

This is the canonical action for a particle with mass m.
The action integral is the functional. The lagrangian is the integrand we use to define the functional. Generically we have:

$LaTeX Code: x \\mapsto S= \\int_{t_1}^{t_2} L(x,\\dot{x},\\ddot{x},...)dx$

Since a functional is also a type of function we can consider its differentials which are variations in the value of the functional $LaTeX Code: \\delta S$ which depend on variations in the function itself. You will then see that $LaTeX Code: \\delta S$ will be a functional depending on the function x(t) and its variation (functional differential)

.

If you imagine an analytic function as defined by its power series expansion:
$LaTeX Code: x(t) = x_0 + x_1 t + x_2 t^2 + \\cdots$
where the

coefficients can then be though of as coordinates then you can think of the function itself as an infinite dimensional vector.

Thus a functional is a mapping (possibly non-linear) from this infinite dimensional vector space to a one dimensional vector space.

Note that the first example: [the value of the function at 0]
can be re-expressed also in integral form:

$LaTeX Code: f \\mapsto \\int_{-\\infty}^\\infty f(x)\\delta(x)dx$
where we use the Dirac delta function.

More generally we might consider an operator which maps a vector to a vector or a function to a function. Example $LaTeX Code: \\frac{d}{dx}$ .

Most operators on functions can also be expressed in integral form but with an extra variable:

$LaTeX Code: f\\mapsto g: g(x) = \\int_{y_1}^{y_2} f(y)W(x,y)$

Some other examples of operators are e.g. Fourier and Laplace transforms.

In summary:
(ordinary) function maps value to value
functional maps function to value
operator maps function to function

**George Jones** · Jul4-09, 04:15 PM

A general function

is a mapping

$LaTeX Code: f : A \\rightarrow B,$

where

and

are sets. Consequently, $LaTeX Code: f \\left( a \\right) = b$ for

in

and

in

.

A functional is a special type of function where the elements of

are themselves functions and

is either the set of real numbers or the set of complex numbers.

Can you see why the Hartle's Lagrangian is a functional?

I don't have my copy of Hartle home with me, but if you want to ask specific questions about what Hartle does, I can have a look on Monday.

[edit]Obviously, James gave a much more detailed answer.[/edit]

**Bob_for_short** · Jul4-09, 04:28 PM

functional: maps function to value

It is not really so. A functional is an integral where the integrand is considered as a variable, not a fixed function. Otherwise it is indeed a number.

**George Jones** · Jul4-09, 04:51 PM

Originally Posted by Bob_for_short

It is not really so.

The definition of "functional" which James and I gave is the standard definition that is found in most functional analysis books.

Originally Posted by Bob_for_short

A functional is an integral where the integrand is considered as a variable, not a fixed function.

This (with maybe slightly different wording) is a special (i.e., not general) type of functional.

Originally Posted by Bob_for_short

Otherwise it is indeed a number.

No one said that a functional is a number.

**HallsofIvy** · Jul4-09, 05:05 PM

Originally Posted by Bob_for_short

It is not really so. A functional is an integral where the integrand is considered as a variable, not a fixed function. Otherwise it is indeed a number.

Yes, really so. What you give, an integegral which maps a function to a number by taking a specific integral is an example of a functional, not a general functional.

**CFDFEAGURU** · Jul4-09, 05:19 PM

The example given in Hartle is

(V(x)=0) defines a free particle moving between points $LaTeX Code: _{}x$ A at $LaTeX Code: _{}t$ A and $LaTeX Code: _{}x$ B at $LaTeX Code: _{}t$ B

In Newtonian physics with a constant velocity of ( $LaTeX Code: _{}x$ B- $LaTeX Code: _{}x$ A)/T, where T= $LaTeX Code: _{}t$ B- $LaTeX Code: _{}t$ A When half of the time, T, has elasped the particle is at the position ( $LaTeX Code: _{}x$ B+ $LaTeX Code: _{}x$ A)/2

Now we compare the action of this path satisfying Newton's laws with paths that move from $LaTeX Code: _{}x$ A with a constant velocity to some different position X in total time T/2 and then with a different constant velocity to get to $LaTeX Code: _{}x$ B in time T. The action S(X) for these paths is a function of X, which is easy to calculate from S[x(t)]= $LaTeX Code: \\int$ dt L((x $LaTeX Code: ^{.}$ (t),x(t)) (The limits of integration are $LaTeX Code: _{}t$ A to $LaTeX Code: _{}t$ B) because the velocity is constant on each leg, namely, (X- $LaTeX Code: _{}x$ A)/(T/2) on the first leg and ( $LaTeX Code: _{}x$ B-X)/(T/2) on the second. The action along any leg with a constant velocity, V, for a time, t, is mV $LaTeX Code: ^{}2$ t/2

The sum of both legs is

S(X)=m[( $LaTeX Code: _{}x$ B-X)^2+(X- $LaTeX Code: _{}x$ A)^2]/T

The extremal paths are defined where dS/dX = 0.

The subsequent answer is given, but is that all this is required is to take the derivative of S(X) and then set it equal to zero and then solve for X?

**Fredrik** · Jul4-09, 06:17 PM

Originally Posted by CFDFEAGURU

...is that all this is required is to take the derivative of S(X) and then set it equal to zero and then solve for X?

Your notation makes that a pain to read. As for your last question, I'm not sure if you mean "is that all that's required...", "is all this required...", or something else entirely.

One thing you can do to find the function that minimizes the action functional is to consider a one-parameter family of functions $LaTeX Code: x_\\epsilon$ such that

is the function that minimizes the action:

$LaTeX Code: 0=\\frac{d}{d\\epsilon}\\bigg|_0 S[x_\\epsilon]=\\int_{t_1}^{t_2}dt \\frac{d}{d\\epsilon}\\bigg|_0 L(x_\\epsilon(t),\\dot x_\\epsilon(t))=\\cdots$

This condition gives you the Euler-Lagrange equations, which is what you use instead of Newton's second in this formulation of classical mechanics.

Note that the Lagrangian L is just a function of several variables (in this case just two). For example $LaTeX Code: (a,b)\\mapsto b^2/2+V(a)$ is the Lagrangian of a single point particle in one dimension. If we instead consider the map $LaTeX Code: x\\mapsto L_t[x]=L(x(t),\\dot x(t))$ , L_t is a functional. Actually, it's a different functional for each value of t.

Originally Posted by George Jones

The definition of "functional" which James and I gave is the standard definition that is found in most functional analysis books.

What if the domain of definition is some unspecified vector space. Wouldn't you still call it a functional?

**Bob_for_short** · Jul4-09, 06:19 PM

Originally Posted by HallsofIvy

Yes, really so. What you give, an integegral which maps a function to a number by taking a specific integral is an example of a functional, not a general functional.

If we speak of action S, then the integrand is not only variable function but also an unknown function. So S depends strongly on the integrand. Than is why it is called "functional" (nearly function). A simple integral with a known (fixed) function is a number, not a functional.

The Lagrange equations serve to find the unknown variables. No one injects them, when found, into the action expression. Nobody cares of the action value on the real trajectory.

**Fredrik** · Jul4-09, 06:34 PM

I see what you're saying Bob. It's like when people call f(x) a "function". I always find that annoying. f(x) is a number. f is a function. f is the map that takes x to f(x). What you're saying is the corresponding statement for functionals. S[f] isn't a functional. It's just a number. S is a functional. It's the map that takes f (any f in the domain of definition) to S[f].

That's a valid point, but I think it's clear that both George and James understand that. When Jambaugh said "is a mapping from a function to a number" it would have been better to say something like "is a mapping from a set of functions to a set of numbers", but I think it's clear from the rest of what he said that that's what he meant.

**Bob_for_short** · Jul4-09, 06:42 PM

Originally Posted by Fredrik

I see what you're saying Bob. It's like when people call f(x) a "function". I always find that annoying. f(x) is a number. f is a function. f is the map that takes x to f(x). What you're saying is the corresponding statement for functionals. S[f] isn't a functional. It's just a number. S is a functional. It's the map that takes f (any f in the domain of definition) to S[f].

That's a valid point, but I think it's clear that both George and James understand that. When Jambaugh said "is a mapping from a function to a number" it would have been better to say something like "is a mapping from a set of functions to a set of numbers", but I think it's clear from the rest of what he said that that's what he meant.

Yes, I agree, and I believe they understand it right. I just wanted to explain to the author of OP that there is difference between X(t) before finding it and after that. For example, Hamiltonian H is a form containing unknown variables x and p. When found and injected into H, they give the system energy E which is a number. Unfortunately they often denote unknown variables and the equation solutions with the same symbols.

**jambaugh** · Jul4-09, 09:52 PM

Just some followup comments...
One of the virtues of natural language in literature is a detriment in mathematics and that is ambiguity. I like to distinguish: "f is a mapping from (set) A to (set) b" from "f maps a (in A) to b (in B)."
Symbolically we use distinct arrows to make this clear:
$LaTeX Code: f:A \\to B$
$LaTeX Code: f:a\\mapsto b$
this second being equivalent to saying f(a)=b.

Another minor distinction, remember that the Lagrangian is not itself the functional. It together with the domain of integration uniquely defines the action functional so for practical purposes we identify the choice of Lagrangian with the choice of action functional.

**CFDFEAGURU** · Jul6-09, 02:18 PM

Fredrik,

Sorry for the poor use/abuse of the Latex for the illustration of the mathematics. What I meant to say was that once you obtain this equation S(X)=m[(B-X)^2+(X-A)^2]/T. Is the solution to this equation found by calculating the derivative, setting the derivative equal to zero and then solving for X?

**Fredrik** · Jul6-09, 06:21 PM

That example is pretty weird. The goal is to find the function x (out of all the continuous real-valued functions of one real variable) that makes S[x] as small or as large as possible, and you're only considering functions that have a graph that consists of two straight lines.

**CFDFEAGURU** · Aug16-09, 12:11 PM

Hello all,

Thanks for the advice and definitions above. Here is the problem given in Hartle's book. This is problem 5 in chapter 3.

Consider the functional

S[x(t)] = $LaTeX Code: \\int[(dx(t)/dt)^2+x^2(t)]dt$

The integral is from zero to T. (Can someone show me how to create an integral with the limits of integration in LaTex?)

Find the curve x(t) satisfying the conditions x(0)=0, x(T)=1. which makes S[x(t)] an extremum. What is the extremum value of S[x(t)]? Is it a maximum or a minimum?

Note: This is not a homework problem. I am teaching myself GR and I just need some help here.

Any help on this would be greatly appreciated.

Thanks
Matt

**Fredrik** · Aug16-09, 12:51 PM

Originally Posted by CFDFEAGURU

(Can someone show me how to create an integral with the limits of integration in LaTex?)

Option 1: $LaTeX Code: \\int_a^b$ Option 2: $LaTeX Code: \\int\\limits_a^b$

Originally Posted by CFDFEAGURU

Consider the functional

S[x(t)] = $LaTeX Code: \\int[(dx(t)/dt)^2+x^2(t)]dt$

The integral is from zero to T.

First note that S is the functional and that S[x] is a number. The expression S[x(t)] doesn't make sense since it's a functional acting on a number. (See #10). Also note that the integral is of the form

$LaTeX Code: \\int_0^T L(xsingle-quote(t),x(t)) dt$

and that L is just a polynomial in two variables. There are three options here: a) use the Euler-Lagrange equations, b) derive the E-L equations for an arbitrary L, and then use them for this specific L, c) derive the E-L euqations for this specific integral.

The way I like to derive the E-L equations starts like this: Let x be the function that maximizes or minimizes the integral, and let $LaTeX Code: \\{x_\\epsilon\\}$ be a one-parameter family of functions with

. Now we must have

$LaTeX Code: 0=\\frac{d}{d\\epsilon}\\bigg|_0 S[x_\\epsilon]$

If you figure out the appropriate way to rewrite the right-hand side, you have found the Euler-Lagrange equations. Post #3 in this thread does this calculation for for a scalar field theory, so you can check that out if you get stuck, but I recommend that you try it yourself first. (Oops, now I see that I omitted some details in the other thread, but it should still be useful).