| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
13 May 2006 04:58:08 PM |
| Object: |
Understanding the Action Integral |
Please help me understand why the Action Integral is so important.
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| User: "Sam Wormley" |
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| Title: Re: Understanding the Action Integral |
13 May 2006 06:41:20 PM |
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wrote:
Please help me understand why the Action Integral is so important.
See: http://en.wikipedia.org/wiki/Lagrangian
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| User: "" |
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| Title: Re: Understanding the Action Integral |
13 May 2006 07:52:56 PM |
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So it is merely a formalism, with no new content to offer?
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| User: "" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 09:23:26 AM |
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The idea of the action integral appears to spring from:
x = 1/2 a t^2
But you need to multiply by a to get anywhere:
ax = 1/2 at * at = 1/2 v^2
But the usefulness of that escapes me. I don't see any new content
there.
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| User: "Ken S. Tucker" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 10:57:20 PM |
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wrote:
Please help me understand why the Action Integral is so important.
It's the total cost.
Energy is defined as Rate of Action, and energy
is the same as Work.
I hire alice and bob to hammer in nails.
alice does 10 per hour and bob 20 per hour,
I pay pay alice $10/hr and bob $20/hr.
bob works twice as hard as alice.
The Action is hammering one nail.
bob does twice the rate of action as alice.
My project needs 100 nails hammered in,
so the Action Integral (aka sum) is $100
for the project.
Including gratutities, your owe me $110.
Now hows that for integrated action??
Best Regards
Ken S. Tucker
PS: I posted that way to provide a feel
for action...ouch!
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| User: "" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 01:04:06 PM |
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Ken S. Tucker wrote:
actionintegral@yahoo.com wrote:
Please help me understand why the Action Integral is so important.
It's the total cost.
Energy is defined as Rate of Action, and energy
is the same as Work.
I hire alice and bob to hammer in nails.
alice does 10 per hour and bob 20 per hour,
I pay pay alice $10/hr and bob $20/hr.
bob works twice as hard as alice.
The Action is hammering one nail.
bob does twice the rate of action as alice.
My project needs 100 nails hammered in,
so the Action Integral (aka sum) is $100
for the project.
Including gratutities, your owe me $110.
Now hows that for integrated action??
Best Regards
Ken S. Tucker
Thank you for your analogy. But as you have it set up, nails per hour
corresponds to
a velocity. nails per hour per hour would correspond to an
acceleration.
(nails * nails)/(hour*hour) would correspond to energy. Hence action in
your analogy
would correspond to (nails*nails)/hour
And changing units from nails/hour to dollars/hour has no significance!
PS: I posted that way to provide a feel
for action...ouch!
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| User: "Mike" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 01:52:28 PM |
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wrote:
Ken S. Tucker wrote:
wrote:
Please help me understand why the Action Integral is so important.
It's the total cost.
Energy is defined as Rate of Action, and energy
is the same as Work.
I hire alice and bob to hammer in nails.
alice does 10 per hour and bob 20 per hour,
I pay pay alice $10/hr and bob $20/hr.
bob works twice as hard as alice.
The Action is hammering one nail.
bob does twice the rate of action as alice.
My project needs 100 nails hammered in,
so the Action Integral (aka sum) is $100
for the project.
Including gratutities, your owe me $110.
Now hows that for integrated action??
Best Regards
Ken S. Tucker
Thank you for your analogy. But as you have it set up, nails per hour
corresponds to
a velocity. nails per hour per hour would correspond to an
acceleration.
(nails * nails)/(hour*hour) would correspond to energy. Hence action in
your analogy
would correspond to (nails*nails)/hour
And changing units from nails/hour to dollars/hour has no significance!
Do not pay attention to cranks. Most people in these ngs are serious
case cranks and hide behind orthodox relativity views. That was a
crank's masterpiece.
Mike
PS: I posted that way to provide a feel
for action...ouch!
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| User: "Ken S. Tucker" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 02:42:07 PM |
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wrote:
Ken S. Tucker wrote:
wrote:
Please help me understand why the Action Integral is so important.
It's the total cost.
Energy is defined as Rate of Action, and energy
is the same as Work.
I hire alice and bob to hammer in nails.
alice does 10 per hour and bob 20 per hour,
I pay pay alice $10/hr and bob $20/hr.
bob works twice as hard as alice.
The Action is hammering one nail.
bob does twice the rate of action as alice.
My project needs 100 nails hammered in,
so the Action Integral (aka sum) is $100
for the project.
Including gratutities, your owe me $110.
Now hows that for integrated action??
Best Regards
Ken S. Tucker
If you goto the Quantum Theory, E = h*f
and note bobs frequency = 2 * alices frequency,
(given they both strike equally hard), you may
get a sense of QT in familiar terms to start.
Have fun...
Ken
Thank you for your analogy. But as you have it set up, nails per hour
corresponds to
a velocity. nails per hour per hour would correspond to an
acceleration.
(nails * nails)/(hour*hour) would correspond to energy. Hence action in
your analogy
would correspond to (nails*nails)/hour
And changing units from nails/hour to dollars/hour has no significance!
PS: I posted that way to provide a feel
for action...ouch!
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| User: "Mike" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 02:06:57 PM |
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Ken S. Tucker wrote:
actionintegral@yahoo.com wrote:
Please help me understand why the Action Integral is so important.
It's the total cost.
Energy is defined as Rate of Action, and energy
is the same as Work.
Energy cannot be defined as rate of action because action is a scalar
magnitude measured along a path and the rate of such magnitude dioes
not exist.
Mike
I hire alice and bob to hammer in nails.
alice does 10 per hour and bob 20 per hour,
I pay pay alice $10/hr and bob $20/hr.
bob works twice as hard as alice.
The Action is hammering one nail.
bob does twice the rate of action as alice.
My project needs 100 nails hammered in,
so the Action Integral (aka sum) is $100
for the project.
Including gratutities, your owe me $110.
Now hows that for integrated action??
Best Regards
Ken S. Tucker
PS: I posted that way to provide a feel
for action...ouch!
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| User: "" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 02:13:23 PM |
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Well, if you start with Action = Energy * Time, then I can see how
Energy = Action / Time
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| User: "Mike" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 03:49:05 PM |
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wrote:
Well, if you start with Action = Energy * Time, then I can see how
Energy = Action / Time
That is NOT how action is defined in modern physics,
Mike
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| User: "" |
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| Title: Re: Understanding the Action Integral |
20 May 2006 11:43:38 AM |
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I understand what you are saying. If we start with the definition that
Action is the integral of
the langrangian over a path, then the endpoints of that path correspond
to different times.
As the object moves over the path, the value of that integral will
change. The time rate of change of the value of the integral will have
the dimensions of Energy. Hence the time rate
of change of Action is Energy.
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| User: "Ken S. Tucker" |
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| Title: Re: Understanding the Action Integral |
20 May 2006 12:38:38 PM |
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wrote:
I understand what you are saying. If we start with the definition that
Action is the integral of
the langrangian over a path, then the endpoints of that path correspond
to different times.
Yup, see below...
As the object moves over the path, the value of that integral will
change. The time rate of change of the value of the integral will have
the dimensions of Energy. Hence the time rate
of change of Action is Energy.
Yes, I think so. In my simple post about alice
and bob, if we use bob, the time to integrate is
1/2 that of using alice, because alices rate of
action is 1/2 of bob's.
I regard myself as a student of "action" so I really
like your handle!
Cheers
Ken
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| User: "Mike" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 01:53:44 PM |
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wrote:
Please help me understand why the Action Integral is so important.
See the following:
http://www.eftaylor.com/leastaction.html#ogborn
Mike
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| User: "" |
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| Title: Re: Understanding the Action Integral |
19 May 2006 02:01:37 PM |
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Actually, that's what got me started on this. I think I am getting
close to seeing the usefulness of the action approach
but not yet.
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| User: "PD" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 12:48:52 PM |
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wrote:
Please help me understand why the Action Integral is so important.
There are certain quantities that seem to stand out in physics, either
because for one reason or another they appear to be invariant (stay the
same no matter what happens) or appear to take extremal values (a
minimum for some quantities or a maximum for some other properties).
Momentum, for example, appears to be a quantity that is invariant in a
closed system, no matter what happens inside the system. This property
is what makes momentum stand out as a physically interesting quantity.
The spacetime interval appears to be a quantity that is maximized for
objects in free-fall. The fact that this quantity is also invariant, no
matter which local inertial observer is looking at it, also makes it a
physically interesting and useful quantity.
Likewise, the action appears to be a quantity whose integral is
stationary for true trajectories (behaviors) of objects in a physical
system. This fact is what makes it a physically interesting and useful
quantity.
It allows you to make computations about complex systems that would be
difficult to do any other way, just like knowing that energy and
momentum are invariant make computations about complex systems easier.
PD
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| User: "" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 02:04:56 PM |
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PD wrote:
....
Likewise, the action appears to be a quantity whose integral is
stationary for true trajectories (behaviors) of objects in a physical
system. This fact is what makes it a physically interesting and useful
quantity.
It allows you to make computations about complex systems that would be
difficult to do any other way, just like knowing that energy and
momentum are invariant make computations about complex systems easier.
PD
I followed the demonstration that the action integral is stationary
along the true path. But I noticed that the integrand was L-V. And what
is V? The integral of the acceleration over the true path!
It seems that the intelligence lies in V, and not in the action
integral. I don't see how the action integral adds anything new to the
picture.
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| User: "PD" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 02:14:34 PM |
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wrote:
PD wrote:
...
Likewise, the action appears to be a quantity whose integral is
stationary for true trajectories (behaviors) of objects in a physical
system. This fact is what makes it a physically interesting and useful
quantity.
It allows you to make computations about complex systems that would be
difficult to do any other way, just like knowing that energy and
momentum are invariant make computations about complex systems easier.
PD
I followed the demonstration that the action integral is stationary
along the true path. But I noticed that the integrand was L-V.
Are you sure? Look again.
And what
is V? The integral of the acceleration over the true path!
No, V is not the integral of the acceleration over the true path. Check
your units.
It seems that the intelligence lies in V, and not in the action
integral.
I don't know what you mean by "intelligence". The intelligence is
knowing how to *use* the action, not what's in it.
I don't see how the action integral adds anything new to the
picture.
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| User: "" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 02:46:06 PM |
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Thank you for your reply.
By "intelligence" I mean that V is precisely that quantity that makes
the action integral work.
V appears to me to be what is special.
The fact that T-V integrates to 0 over the true path seems incidental.
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| User: "PD" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 02:54:36 PM |
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wrote:
Thank you for your reply.
By "intelligence" I mean that V is precisely that quantity that makes
the action integral work.
V appears to me to be what is special.
The fact that T-V integrates to 0 over the true path seems incidental.
Unfortunately, the integral of V is not stationary.
Perhaps what you mean by "intelligence" is that V is where all the
forces are, where all the interactions between the parts of the system
are. That is true.
The *motion*, the kinematics if you will, live in the T.
But it's the interplay between the T and the V that make the integral
of the action stationary.
PD
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| User: "Mike" |
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| Title: Re: Understanding the Action Integral |
18 May 2006 03:22:40 PM |
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PD wrote:
actionintegral@yahoo.com wrote:
Please help me understand why the Action Integral is so important.
There are certain quantities that seem to stand out in physics, either
because for one reason or another they appear to be invariant (stay the
same no matter what happens) or appear to take extremal values (a
minimum for some quantities or a maximum for some other properties).
Momentum, for example, appears to be a quantity that is invariant in a
closed system, no matter what happens inside the system. This property
is what makes momentum stand out as a physically interesting quantity.
The spacetime interval appears to be a quantity that is maximized for
objects in free-fall. The fact that this quantity is also invariant, no
matter which local inertial observer is looking at it, also makes it a
physically interesting and useful quantity.
Mathematically interesting yes. But there is nothing physical about it.
Unless you know something I do not. By the way, inertial abservers are
also mathematical abstractions. In reality, there are no such observer.
That is, unless you know something I do not.
Likewise, the action appears to be a quantity whose integral is
stationary for true trajectories (behaviors) of objects in a physical
system. This fact is what makes it a physically interesting and useful
quantity.
Mathematically interesting only. Physically, nobody knows what actions
are present even in the simplest kind of systems.
It allows you to make computations about complex systems that would be
difficult to do any other way, just like knowing that energy and
momentum are invariant make computations about complex systems easier.
Just avoid boundary conditions. Keep the legacy os simplicity.
Mike
PD
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| User: "Edward Green" |
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| Title: On the Physical -- Was: Understanding the Action Integral |
04 Jun 2006 10:32:22 AM |
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Mike wrote:
PD wrote:
The spacetime interval appears to be a quantity that is maximized for
objects in free-fall. The fact that this quantity is also invariant, no
matter which local inertial observer is looking at it, also makes it a
physically interesting and useful quantity.
Mathematically interesting yes. But there is nothing physical about it.
Curious, and I think, erroneous point of view. Can you give an
operational definition of "physical" which supports your claim?
Physical systems are described by mathematical models. You could make
the fruitless claim that nothing in the model is "physical", only the
system. I don't think that's what you mean.
You could claim that the model contains arbitrary elements, like the
origin of coordinates. You could say that the arbitrary choice among a
class of equivalent models which we have not yet learned to cast as a
single entity is an unphysical choice.
I don't think that that's what you mean either. I have a feeling what
you have in mind is that given some collection of variables -- assume
for the sake of argument unambiguous up to units -- describing a
physical system, that some combinations of the variables are more
physical than others.
I do not subscribe to this brand of semantics. If height and albedo of
the skin are physical variables, then height/albedo is also a physical
variable. There may be some basis for a distinction along the lines
that height and albedo are more primitive than their quotient, but it
will take additional work to operationalize this.
When a feature of a physical model is glibly dismissed as unphysical,
an unspecified combination of arbitrariness and unprimitiveness is
lurking behind an affective classification. The wave function is often
denigrated this way. What we might say instead is that we have a vague
idea that there are some ways of looking at the model which are closer
in an unknown sense to the working of the modeled, and also that there
may be arbitrary elements in the model.
<...>
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| User: "noshellswill" |
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| Title: Re: On the Physical -- Was: Understanding the Action Integral |
04 Jun 2006 11:42:59 PM |
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On Sun, 04 Jun 2006 08:32:22 -0700, Edward Green wrote:
Mike wrote:
PD wrote:
The spacetime interval appears to be a quantity that is maximized for
objects in free-fall. The fact that this quantity is also invariant, no
matter which local inertial observer is looking at it, also makes it a
physically interesting and useful quantity.
Mathematically interesting yes. But there is nothing physical about it.
Curious, and I think, erroneous point of view. Can you give an
operational definition of "physical" which supports your claim?
Physical systems are described by mathematical models. You could make
the fruitless claim that nothing in the model is "physical", only the
system. I don't think that's what you mean.
You could claim that the model contains arbitrary elements, like the
origin of coordinates. You could say that the arbitrary choice among a
class of equivalent models which we have not yet learned to cast as a
single entity is an unphysical choice.
I don't think that that's what you mean either. I have a feeling what
you have in mind is that given some collection of variables -- assume
for the sake of argument unambiguous up to units -- describing a
physical system, that some combinations of the variables are more
physical than others.
I do not subscribe to this brand of semantics. If height and albedo of
the skin are physical variables, then height/albedo is also a physical
variable. There may be some basis for a distinction along the lines
that height and albedo are more primitive than their quotient, but it
will take additional work to operationalize this.
When a feature of a physical model is glibly dismissed as unphysical,
an unspecified combination of arbitrariness and unprimitiveness is
lurking behind an affective classification. The wave function is often
denigrated this way. What we might say instead is that we have a vague
idea that there are some ways of looking at the model which are closer
in an unknown sense to the working of the modeled, and also that there
may be arbitrary elements in the model.
<...>
EG:
Unphysical = can't be directly measured. (More)-unphysical = can't be
measured and leading to the assertion of further unmeasurables.
nss
********
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| User: "" |
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| Title: Re: Understanding the Action Integral |
24 May 2006 12:45:28 PM |
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Imagine an object about to move along a path. We will calculate the
action integral along that path. Now if we look only at the region very
close to the object, where the acceleration can be considered constant,
then of course the object will take the path where T-V = 0, since this
is just ax = .5 v v or x = .5 a t t.
Along the path of the object, we are simply taking the cumulative
effect of these path "choices" and calling it the action integral.
Once a student understands that objects interact through acceleration,
isn't that the whole story? Doesn't the action integral just say "ok,
now do that over and over again until you reach the end point."
If the action integral is just a cumulative effect , isn't it LESS
useful than conservation of energy, which at least gives us a relation
between a, x and v?
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| User: "" |
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| Title: Re: Understanding the Action Integral |
24 May 2006 12:34:23 PM |
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Imagine an object about to move along a path. We will calculate the
action integral along that path. Now if we look only at the region very
close to the object, where the acceleration can be considered constant,
then of course the object will take the path where T-V = 0, since this
is just ax = .5 v v or x = .5 a t t.
Along the path of the object, we are simply taking the cumulative
effect of these path "choices" and calling it the action integral.
Once a student understands that objects interact through acceleration,
isn't that the whole story? Doesn't the action integral just say "ok,
now do that over and over again until you reach the end point."
If the action integral is just a cumulative effect , isn't it LESS
useful that conservation of energy, which at least shows us a relation
between a, x and v?
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| User: "" |
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| Title: Re: Understanding the Action Integral |
24 May 2006 09:14:07 PM |
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In article <1148492063.855103.311560@g10g2000cwb.googlegroups.com>, writes:
Imagine an object about to move along a path. We will calculate the
action integral along that path. Now if we look only at the region very
close to the object, where the acceleration can be considered constant,
then of course the object will take the path where T-V = 0, since this
is just ax = .5 v v or x = .5 a t t.
Along the path of the object, we are simply taking the cumulative
effect of these path "choices" and calling it the action integral.
Once a student understands that objects interact through acceleration,
isn't that the whole story? Doesn't the action integral just say "ok,
now do that over and over again until you reach the end point."
If the action integral is just a cumulative effect , isn't it LESS
useful that conservation of energy, which at least shows us a relation
between a, x and v?
Actually, it is more useful.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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