Necessity of least action principle
Mike
from physical requirements. And I'm trying to make contact with
physics. It would help if there were a requirement that the variation
of the action be zero. Then Euler-Lagrange equations of motion would
procede from that. So does the evaluation of the path integral require
that there be a classical path, a zero to the functional derivative of
the action?
We are told that as we manually integrate the path integral in the
region of far flung paths, contributions of the integral cancel out
with other parts of far flung paths, leaving only the classical path
that contributes most to the final result. My question is if the
functional derivative of the Action integral does not have a zero for
any path, no classical path, then is it still possible to evaluate the
path integral at all? Or will everything cancel out? I wonder if there
is any mathematical proof that the solvability of the path integral
requires a zero variation of the action. Thanks.
Mike
> We are told that as we manually integrate the path integral in the
> region of far flung paths, contributions of the integral cancel out
> with other parts of far flung paths, leaving only the classical path
> that contributes most to the final result. My question is if the
> functional derivative of the Action integral does not have a zero for
> any path, no classical path, then is it still possible to evaluate the
> path integral at all? Or will everything cancel out? I wonder if there
> is any mathematical proof that the solvability of the path integral
> requires a zero variation of the action. Thanks.
To this end I consider that in QM typically the path integral is
developed by inserting the identity operator an infinite number of
time. i.e.:
<x'|x"> = S dx1 <x'|x1><x1|x">
where S dx1 is the integral with respect to the variable x1, and where
S dx1|x1><x1| is the identity which can be applied an infinite number
of times:
<x'|x"> = S dx1dx2...dxn <x'|x1><x1|x2><x2|...|xn><xn|x">
Then each of the <xi|xj> is recognized to be a Dirac delta function,
<xi|xj> = D(xi-xj).
When this is substituted into the integral above,
<x'|x"> = D(x'-x") = S dx1dx2...dxn D(x'-x1)D(x1-x2)...D(xn-x")
This interesting because the delta function has the same recursive
property as the identity above so that an infinite dimensional
integral can be formed as is done with the identity.
So if we choose the gaussian form of the Dirac delta function that has
a complex exponent, the path integral for a free particle can be shown
to result. See
http://hook.sirus.com/users/mjake/delta_physics.htm
So if I generalize the exponent to be a functional (an action integral
in this case) and ask where might the requirement come from that the
functional derivative of the action be zero, I consider what happens
if I take the functional derivative of the entire path integral. Is
this passed on to taking the functional derivative of the exponent
(the action) in the process? Since the path integral is equal to a
Dirac delta, my question is what is the functional derivative of a
Dirac delta function? Thanks.
Mike
> > I wonder if there
> > is any mathematical proof that the solvability of the path integral
> > requires a zero variation of the action. Thanks.
>
> So if I generalize the exponent to be a functional (an action integral
> in this case) and ask where might the requirement come from that the
> functional derivative of the action be zero, I consider what happens
> if I take the functional derivative of the entire path integral. Is
> this passed on to taking the functional derivative of the exponent
> (the action) in the process? Since the path integral is equal to a
> Dirac delta, my question is what is the functional derivative of a
> Dirac delta function? Thanks.
OK, how does this sound? If we integrate the path integral one more
time, it will equal 1 since it is a delta function. Then if we take
the functional derivative of that, it will equal zero since the
functional derivative of a constant is zero. This much is necessarily
true. But does this sound like a legitimate procedure to follow?
And so when taking the functional derivative of the integrated path
integral, the functional derivative will commute is all the infinite
number of integrations in the path integral and be passed on to the
exponential. I assume the functional derivative of the exponential
will be the exponential times the functional derivative of the
exponent, in this case the action integral. So here is a factor
consisting of the variation of the action. And when this variation of
the integrated path integral is set to zero, the only factor that
could be identically zero is the variation of the action. This may be
where the least action principle comes from. Any comments or
corrections? Thanks.
Mike
>
> OK, how does this sound? If we integrate the path integral one more
> time, it will equal 1 since it is a delta function. Then if we take
> the functional derivative of that, it will equal zero since the
> functional derivative of a constant is zero. This much is necessarily
> true. But does this sound like a legitimate procedure to follow?
>
In the process of thinking about this, I written a brief introduction
of the calculus of variation. And before I continue, I'd like a review
and would appreciate your comments. Thank you. It's at
Mike
OK, how does this sound? If we integrate the path integral one more
time, it should equal 1 since it's a Dirac delta function. Then the
functional derivative of that should be zero, since the functional
derivative of a constant is zero. This much is necessarily True.
So what does it mean if this is not true? If the first variation of
the integration of the path integral is not zero, then what does this
mean? Feel free to offer your thoughts, but I think it means that the
path integral is not a Dirac delta function, not a wavefunction or a
transition function <x|x_0>. And if the path integral as a whole is
not a delta function, then I think this means that one or more of the
infinite number of <xi|xj> that are inside the path integral is not a
delta function. Or at least it means that the infinite multiplication
of the <xi|xj> is not an infinite multiplication of delta functions.
And since the delta functions, <xi|xj> , are expressed in terms of an
exponential whose exponents can be added up to equal the action
integral of one exponential, at least this means the action type
integral is not adding up right. So I can see how this would place
some restrictions on the action integral.
Mike
I think I've completed by derivation of the least action principle
given that the path integral is obtained from principle alone. It's
difficult to write math on these newsgroups, so I've created a webpage
with the derivation. It's at:
http://hook.sirus.com/users/mjake/least_action.htm
I really would appreciated any comments. Thanks.