Is Action trickery?
Ken S. Tucker
I removed this from the "Causality" thread.
On Aug 25, 12:39 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
...
> >Let's consider Action. System S1 passes an action h
> >to system S2,
> >S1-h ---->h----> h+S2
> >using any of the 4 known conventional forces, we
> >term, gravity, EM, weak and strong. The Action h
> >is held to be a constant and invariant, apart from
> >the nature of the causal force.
> >The concept of Action also has angular momentum
> >(spin) and also the product of two charges "a" and "b",
> >each is fundamental.
Charles writes,
> In my view, action is no more than a mathematical trick. It is not
> fundamental.
((Tucker grabbed his chest and swallowed a
1/2 bottle of glycerine capsules, I'm ok now))
Let Action = Energy*Time.
(h=6.625*10^-27 ergs*seconds).
Tucker does carpentry and is the boss, and
watches Alice hammer in 100 nails/hour and
Bob 200 nails/ hour.
They both use the same hammer strike.
Bob strikes twice as fast as Alice, thus
has twice the frequency, I can hear.
I pay for energy, such as Kilowatt*hours,
(Power*Time) in electricity.
Each nail hammered in takes a fixed
amount of energy.
Bob expends twice the energy that Alice
does per hour.
Energy(Bob) = 200 Actions/hour
Energy(Alice) = 100 Actions/hour
I pay Bob $200/hour and Alice $100/hour.
Next: another Guy' is flying by at 8/10 c,
where gamma=1/sqrt(1-v^2/c^2) = 10/6.
To Guy' the Energy' = Energy*10/6 of what
Tucker measures because of Guy's kinetic
energy, and Guy' sees Tuckers clock run
slower by an amount, Time' = Time*6/10.
Guy and Tucker agree that,
Action = Energy' * Time' = Energy * Time
and then Tucker and Guy agree that $1/action
is the relativistic rate of payment.
======================
I think, prior to conclusions, we might consider
Weinbergs "Grav&Cosmo", chp 12, "THE ACTION
PRINCIPLE", since he worked hard to write it.
Eq.(12.3.2), begins an interest followed by,
"*Thus the energy-momentum tensor... if and only
if the matter action is a scalar*".
===========================
> >We theoreticians seem bound by rules that respect
> >the indivisability of "h", so an ordinary derivative
> >CANNOT be employed to "h", there is no such
> >thing as "dh", empirically, (that I ever read).
> Actually there is such a concept. The principle of least action makes
> dh=0. We only have empirical law when action is minimised.
Thanks for the correction.
> Regards
> Charles Francis
> http://www.teleconnection.info/rqg/MainIndex
Cheers
Ken S. Tucker
Igor
Except that action is the integral of the mechanical Lagrangian over
time. And a mechanical Lagrangian, while having the same units as
energy, is not energy.
Chalky
> Except that action is the integral of the mechanical Lagrangian over
> time. And a mechanical Lagrangian, while having the same units as
> energy, is not energy.
So, given that Newton died before Lagrange was born, what was the
definition of action in Newton's third law of motion?
And why is this usually interpreted as meaning that the forces match,
not that the energies match?
======================================= MODERATOR'S COMMENT:
Law III was discussed some times earlier in this group, please have a look there, too
Chalky
OK but from a quick perusal of this I could not work out how this
question was actually resolved (if it ever was).
1) Action = force?
2) Action = force x time?
3) Action = energy x time?
The only definition (3) which seems to be explicitly ruled out is the
one now being used (since no energy is expended when a weight rests on
a table)
Oh No
the concept of inertial force, in the sense of centrifugal was also
developed later). The concept of energy was developed over the next few
generations. Action in the sense of Lagrange is a different concept
again.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Chalky
Tucker could agree equally simply to pay a 'piece rate' of $1/nail,
irrespective of time taken, and Guy would have no difficulty agreeing
with that relativistic rate too. This has very little, if anything, to
do with energy expended in Kilowatt*hours
[If Tucker had a modicum of common sense he could also invest in a
nail gun and thus slash the total labor cost per nail]
: )
======================================= MODERATOR'S COMMENT:
Please snip. Especially so since this post is really just a joke.
Ken S. Tucker
On Aug 27, 8:40 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Aug 27, 12:26 am, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> > On Aug 25, 12:39 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
> > ...
>
> > > >Let's consider Action. System S1 passes an action h
> > > >to system S2,
> > > >S1-h ---->h----> h+S2
> > > >using any of the 4 known conventional forces, we
> > > >term, gravity, EM, weak and strong. The Action h
> > > >is held to be a constant and invariant, apart from
> > > >the nature of the causal force.
> > > >The concept of Action also has angular momentum
> > > >(spin) and also the product of two charges "a" and "b",
> > > >each is fundamental.
>
> > Charles writes,
> > > In my view, action is no more than a mathematical trick. It is not
> > > fundamental.
>
The challenge is too take Action, Energy, Power,
frequency, Planck's constant "h", money and put
those into more familiar terms like hammering nails
paying electrical bills and wages then going relativistic
to provide an example of how all those things basically
relate.
For me it is quite interesting to see that I pay for
action (not energy) in my electric bill, it only looks
like I pay for energy in my rest frame, because I
use the second as a fixed invariant constant, but
relative to Guy, (flying by at .8c) we agree on paying
$1/action, both Action and money being invariant.
That seems reasonable, because a $100 bill on my
desk is still a $100 bill relative to Guy, (because the
denomination of the bill is invariant).
So now we can consider this exchange
S1-h ---->h----> h+S2
as an exchange of Action or a monetary payment.
> ======================================= MODERATOR'S COMMENT:
> Please snip. Especially so since this post is really just a joke.
I promise to next time.
Regards
Ken S. Tucker
Chalky
> For me it is quite interesting to see that I pay for
> action (not energy) in my electric bill,
If that were true, your bills would be 9 times as high if you paid
quarterly than if you paid monthly.
In practice you pay for the total energy consumed in an agreed time
period (time typically measured as a proportion of a revolution of the
Earth round the Sun), not the total energy consumed multiplied by that
time period.
> both Action and money being invariant.
Money is so NOT invariant that it buys substantially less energy now
than it did 6 months ago (and I didn't even need to move to discover
that).
However, despite your garbled analogy in places, you have succeeded in
convincing me that action is Lorentz invariant (as is any given number
of nails)
What I find potentially interesting is why this would be (for action).
The boring answer would be that the relativistic energy transform is
the inverse of the relativistic time transform. A more potentially
intriguing explanation might be that, with units of force x distance x
time, it provides a more intimate fusion of the concepts of energy
(force x distance) and momentum (force x time) , than the energy-
momentum 4 vector achieves?
Ken S. Tucker
On Aug 28, 4:25 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On Aug 27, 7:28 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
>
> > For me it is quite interesting to see that I pay for
> > action (not energy) in my electric bill,
>
> If that were true, your bills would be 9 times as high if you paid
> quarterly than if you paid monthly.
> In practice you pay for the total energy consumed in an agreed time
> period (time typically measured as a proportion of a revolution of the
> Earth round the Sun), not the total energy consumed multiplied by that
> time period.
>
> > both Action and money being invariant.
>
> Money is so NOT invariant that it buys substantially less energy now
> than it did 6 months ago and I didn't even need to move to discover
> that).
The money is invariant, what you can buy with it
is shrinking :-).
> However, despite your garbled analogy in places,
Thanks, I'll try to improve on my garbling.
> you have succeeded in
> convincing me that action is Lorentz invariant (as is any given number
> of nails)
Good, (ok $1/nail).
> What I find potentially interesting is why this would be (for action).
> The boring answer would be that the relativistic energy transform is
> the inverse of the relativistic time transform.
Agreed, I find GR is easier to understand using the
invariance of action.
> A more potentially
> intriguing explanation might be that, with units of force x distance x
> time, it provides a more intimate fusion of the concepts of energy
> (force x distance) and momentum (force x time) , than the energy-
> momentum 4 vector achieves?
I suggest NOT using force in GR, the force vanishes
by definition, even Lorentz Force.
I've been following the thread on 4 momentum.
Here's a paragraph on the subject,
http://physics.trak4.com/MST_PlancksConstant.pdf
Regards
Ken S. Tucker
Chalky
> The money is invariant, what you can buy with it
> is shrinking :-).
No. Money is merely a mechanism for exchange. Since the abandonment of
the 'gold standard' it is the value of money which decreases with
time.
Interestingly, "One Pound Sterling" may well have originally meant
"One Pound Weight of Sterling Silver"
In this context, it is obviously the value of money (expressed in GBP)
that has depreciated subsequently, not the value of everything else
that has escalated.
> > you have succeeded in
> > convincing me that action is Lorentz invariant (as is any given number
> > of nails)
>
> Good, (ok $1/nail).
No. A 4" nail remains a 4" nail (in its rest frame)
What it costs (in monetary terms) varies with both location and time.
> > What I find potentially interesting is why this would be (for action).
> > The boring answer would be that the relativistic energy transform is
> > the inverse of the relativistic time transform.
>
> Agreed, I find GR is easier to understand using the
> invariance of action.
>
> > A more potentially
> > intriguing explanation might be that, with units of force x distance x
> > time, it provides a more intimate fusion of the concepts of energy
> > (force x distance) and momentum (force x time) , than the energy-
> > momentum 4 vector achieves?
>
> I suggest NOT using force in GR, the force vanishes
> by definition, even Lorentz Force.
If something I experience and use fruitfully every day vanishes within
the context of a particular theoretical model, I am inclined to
conclude that this represents a weakness of the theoretical model, not
a weakness in my own experiences, actions and observations.
Oh No
>> I suggest NOT using force in GR, the force vanishes
>> by definition, even Lorentz Force.
>
>If something I experience and use fruitfully every day vanishes within
>the context of a particular theoretical model, I am inclined to
>conclude that this represents a weakness of the theoretical model, not
>a weakness in my own experiences, actions and observations.
May I just clarify that the Lorentz force does not vanish in GR. Only
the gravitational force vanish in a local inertial reference frames.
Cl.Massé
>> > In my view, action is no more than a mathematical trick. It is not
>> > fundamental.
In the derivation of the Schrödinger equation from the Hamilton-Jacobi
equation, S becomes the phase of the wave. It is especially clear for
light, were S is the time taken from a point to another, and then the phase
difference between those points since its speed is constant. In first
quantization, the least action principle is equivalent to the Fermat
theorem, that gives the points of constructive interference, where the phase
is stationary as a function of the path. Feynman's paths integrals are but
a rehashing of those ideas.
It is even interesting to note that the least action fulfills all the
properties of a distance. That's why Einstein used it to define the
geometry of space-time, through light rays. That is surely useful for those
who are investigating theories without prior space-time.
I would say, action more fundamental than anything else, and still more so
in gauge theories.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
Cl.Massé
>> Except that action is the integral of the mechanical Lagrangian over
>> time. And a mechanical Lagrangian, while having the same units as
>> energy, is not energy.
In the case of light, it has the dimension of time.
"Chalky" <chalk...@bleachboys.co.uk> a écrit dans le message de
news:49c4f334-a0c1-4760...@m45g2000hsb.googlegroups.com...
> So, given that Newton died before Lagrange was born, what was the
> definition of action in Newton's third law of motion?
>
> And why is this usually interpreted as meaning that the forces match,
> not that the energies match?
Because that conserves the momentum, while the *kinetic* energy is not
conserved in general. Cf. F = dp/dt.
Ken S. Tucker
On Aug 29, 7:29 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>
> >> I suggest NOT using force in GR, the force vanishes
> >> by definition, even Lorentz Force.
>
> >If something I experience and use fruitfully every day vanishes within
> >the context of a particular theoretical model, I am inclined to
> >conclude that this represents a weakness of the theoretical model, not
> >a weakness in my own experiences, actions and observations.
It's a matter of understanding the meaning of
Lorentz force mathematically (#). While you are
sitting on your chair, can you regard yourself
as being at rest?
> May I just clarify that the Lorentz force does not vanish in GR. Only
> the gravitational force vanish in a local inertial reference frames.
Charles, Lorentz force does vanish in GR.(#)
Acceleration is relative in GR, as uniform motion
is relative in SR.
See the dichotomy, if all uniform motion is relative
then how does a change in uniform motion create
an absolute acceleration? So you now circulate
back to absolute uniform motion if the Lorentz
force is an absolute force and acceleration.
(#) This thread has a more complete explanation,
"Vanishing of Lorentz Force?" , Feb,12 2008.
Perhaps we should consider the equation
U^u;w U^w =0 ,
where U^u and U^w are 4 velocities ?
I think it's vital to RQG.
Oh No
>Hi Charles and Chalky.
>
>On Aug 29, 7:29 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>>
>> >> I suggest NOT using force in GR, the force vanishes
>> >> by definition, even Lorentz Force.
>>
>> >If something I experience and use fruitfully every day vanishes within
>> >the context of a particular theoretical model, I am inclined to
>> >conclude that this represents a weakness of the theoretical model, not
>> >a weakness in my own experiences, actions and observations.
>
>It's a matter of understanding the meaning of
>Lorentz force mathematically (#). While you are
>sitting on your chair, can you regard yourself
>as being at rest?
>
You can regard yourself as being at rest, but not as being inertial.
There is an active force exerted by the chair.
>> May I just clarify that the Lorentz force does not vanish in GR. Only
>> the gravitational force vanish in a local inertial reference frames.
>
>Charles, Lorentz force does vanish in GR.(#)
>Acceleration is relative in GR, as uniform motion
>is relative in SR.
>See the dichotomy, if all uniform motion is relative
>then how does a change in uniform motion create
>an absolute acceleration? So you now circulate
>back to absolute uniform motion if the Lorentz
>force is an absolute force and acceleration.
"absolute" or, rather *proper* acceleration is the acceleration away
from inertial motion. Do not confuse inertial motion (meaning no active
forces) with uniform motion, which is relative.
Chalky
> Hi Charles and Chalky.
> On Aug 29, 7:29 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> > Thus spake Chalky <chalkys...@bleachboys.co.uk>
> > >> by definition, even Lorentz Force.
>
> > >If something I experience and use fruitfully every day vanishes within
> > >the context of a particular theoretical model, I am inclined to
> > >conclude that this represents a weakness of the theoretical model, not
> > >a weakness in my own experiences, actions and observations.
>
> It's a matter of understanding the meaning of
> Lorentz force mathematically (#).
The Lorentz force describes the force on a charge which is moving
relative to a (stationary) magnetic field.
If you choose a reference frame in which that charge is not moving,
the magnetic field would be, presumably.
If you choose a reference frame in which a positive charge is not
accelerating, then a negative charge would be, presumably.
Consequently, I can't see any physically meaningful way in which you
can claim that the Lorentz force vanishes, relativistically.
Peter
> >> > In my view, action is no more than a mathematical trick. It is not
> >> > fundamental.
> "Cl\.Massé" <a...@c.com> writes:
> In the derivation of the Schrödinger equation from the Hamilton-Jacobi
> equation, S becomes the phase of the wave.
In the 2nd footnote of the 2nd Communication 1926, S. expressis verbis
refrains from this result of his 1st Communication
> It is especially clear for
monochromatic
> light, were S is the time taken from a point to another, and then the phase
> difference between those points since its speed is constant. In first
> quantization, the least action principle is equivalent to the Fermat
> theorem,
which is the role of "first quantization" here?
> that gives the points of constructive interference, where the phase
> is stationary as a function of the path.
is constructive interference tied to stationarity of phase?
> Feynman's paths integrals are but a rehashing of those ideas.
> It is even interesting to note that the least action fulfills all the
> properties of a distance. That's why Einstein used it to define the
> geometry of space-time, through light rays. That is surely useful for
> those who are investigating theories without prior space-time.
I think so, too
Let's recall Leibniz's formula
action = mass x path length x velocity
When velocity=const (light), action ~ path length, but what, if not?
> I would say, action more fundamental than anything else, and still more so
> in gauge theories.
I'm wondering about the intimate relationship between Lagrangian, L, and
symmetry, where L itself has no immediate physical meaning (Helmholtz called
L the 'kinetic potential', but this applies not generally), do you know an
explanation of that?
Thank you,
Peter
Ken S. Tucker
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
So are you suggesting that "absolute" acceleration
exists in GR?
Ken S. Tucker
If a particle (or a charge) has zero acceleration is there
a force on that particle?
Regards
Ken S. Tucker
Oh No
>If a particle (or a charge) has zero acceleration is there
>a force on that particle?
This is a different definition of force, using Newton's second law and
applied to force treated as a 3-vector. We can choose a non-inertial
frame in which the impressed force balances the inertial force (as we do
when we consider our own local frame, in which gravity balance the force
exerted by the chair we are sitting on), but usually we describe this as
a state of equilibrium between forces, not a state in which there are no
forces.
Oh No
>> "absolute" or, rather *proper* acceleration is the acceleration away
>> from inertial motion. Do not confuse inertial motion (meaning no active
>> forces) with uniform motion, which is relative.
>
>So are you suggesting that "absolute" acceleration exists in GR?
I do not like that terminology, but iirc you pointed out that Weinberg
uses "absolute acceleration" to mean the derivative of 4-velocity with
respect to proper time. I would prefer to call it "proper acceleration".
This quantity is a 4-vector and, multiplied by (scalar) mass, gives the
Force 4-vector, as used in the Lorentz force law
(Force)^i = F^ij J_j
This is a covariant equation, holding in all coordinate systems
(inertial or not).
http://www.teleconnection.info/rqg/IntroductionToTensors#Electromagnetic
Force
Ken S. Tucker
On Aug 30, 6:20 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
> >If a particle (or a charge) has zero acceleration is there
> >a force on that particle?
>
> This is a different definition of force, using Newton's second law and
> applied to force treated as a 3-vector. We can choose a non-inertial
> frame in which the impressed force balances the inertial force (as we do
> when we consider our own local frame, in which gravity balance the force
> exerted by the chair we are sitting on), but usually we describe this as
> a state of equilibrium between forces, not a state in which there are no
> forces.
Agreeable for Newton, although imv GR treats
that sitting on a chair, using the stress energy
tensor, where the detail is in how the cushion
is deformed, it's like storing energy in a spring,
that will increase it's g-field via E=mc^2.
Ken S. Tucker
On Aug 30, 6:20 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
> >> "absolute" or, rather *proper* acceleration is the acceleration away
> >> from inertial motion. Do not confuse inertial motion (meaning no active
> >> forces) with uniform motion, which is relative.
>
> >So are you suggesting that "absolute" acceleration exists in GR?
>
> I do not like that terminology, but iirc you pointed out that Weinberg
> uses "absolute acceleration" to mean the derivative of 4-velocity with
> respect to proper time.
Yes in Grav& Cosmo, DU^u/Ds = f^u/m , Eq.(5.1.11).
However note you can consider your motion as zero
relatively to yourself so that DU^u/Ds=0 for any FoR.
Most importantly, the Laws of Physics must be true
in your FoR.
> I would prefer to call it "proper acceleration".
> This quantity is a 4-vector and, multiplied by (scalar) mass, gives the
> Force 4-vector, as used in the Lorentz force law
>
> (Force)^i = F^ij J_j
>
> This is a covariant equation, holding in all coordinate systems
> (inertial or not).
True, however see GR1916 Eq.(65a) and onward,
and find that to vanish in GR, (This link might work)...
http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200...
((Simple Simon posted one that does)) I covered that in
the thread "Vanishing of Lorentz Force?" Feb 12, 2008.
You may interested to find that AE's GR1916 Eq.(66)
requires the Lorentz force to vanish as does Weinberg's
equivalent Eq.(5.3.7).
> http://www.teleconnection.info/rqg/IntroductionToTensors#Electromagnetic
> Force
Does your treatment prevent an electron from spiralling
into the nucleus?
We find f_u =0 will create a "quantum geodesic" to
prevent the classically predicted spiralling, and that is
posted here, "Vanishing of Lorentz Force?" Feb 12, 2008.
Oh No
>Hi Charles
>
>On Aug 30, 6:20 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>>
>> >> "absolute" or, rather *proper* acceleration is the acceleration away
>> >> from inertial motion. Do not confuse inertial motion (meaning no active
>> >> forces) with uniform motion, which is relative.
>>
>> >So are you suggesting that "absolute" acceleration exists in GR?
>>
>> I do not like that terminology, but iirc you pointed out that Weinberg
>> uses "absolute acceleration" to mean the derivative of 4-velocity with
>> respect to proper time.
>
>Yes in Grav& Cosmo, DU^u/Ds = f^u/m , Eq.(5.1.11).
>
>However note you can consider your motion as zero
>relatively to yourself so that DU^u/Ds=0 for any FoR.
No. This is not true. You only have DU^u/Ds=0 if you are in inertial
motion.
>Most importantly, the Laws of Physics must be true
>in your FoR.
>
>> I would prefer to call it "proper acceleration".
>> This quantity is a 4-vector and, multiplied by (scalar) mass, gives the
>> Force 4-vector, as used in the Lorentz force law
>>
>> (Force)^i = F^ij J_j
>>
>> This is a covariant equation, holding in all coordinate systems
>> (inertial or not).
>
>True, however see GR1916 Eq.(65a) and onward,
>and find that to vanish in GR, (This link might work)...
>http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200...
http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pd
f
>
>You may interested to find that AE's GR1916 Eq.(66)
>requires the Lorentz force to vanish as does Weinberg's
>equivalent Eq.(5.3.7).
That isn't true. The vanishing of the Lorentz force is required for
inertial motion, not in general.
>
>> http://www.teleconnection.info/rqg/IntroductionToTensors#Electromagnetic
>> Force
>
>Does your treatment prevent an electron from spiralling
>into the nucleus?
For that one must deal with qed, not the classical Lorentz force.
Ken S. Tucker
At the outset, let me explain I explore many different
ways to solve a problem.
On Aug 30, 11:15 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
> >Hi Charles
> >On Aug 30, 6:20 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> >> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
> >> >> "absolute" or, rather *proper* acceleration is the acceleration away
> >> >> from inertial motion. Do not confuse inertial motion (meaning no active
> >> >> forces) with uniform motion, which is relative.
>
> >> >So are you suggesting that "absolute" acceleration exists in GR?
>
> >> I do not like that terminology, but iirc you pointed out that Weinberg
> >> uses "absolute acceleration" to mean the derivative of 4-velocity with
> >> respect to proper time.
>
> >Yes in Grav& Cosmo, DU^u/Ds = f^u/m , Eq.(5.1.11).
>
> >However note you can consider your motion as zero
> >relatively to yourself so that DU^u/Ds=0 for any FoR.
>
> No. This is not true. You only have DU^u/Ds=0 if you are in inertial
> motion.
Hmm, if you have a "proper change" in your acceleration
given by DU^u/Ds as you allude to, and a fella on the other
side of the Earth (Australia) claims the same, why are
you both at relative rest?
> >Most importantly, the Laws of Physics must be true
> >in your FoR.
>
> >> I would prefer to call it "proper acceleration".
> >> This quantity is a 4-vector and, multiplied by (scalar) mass, gives the
> >> Force 4-vector, as used in the Lorentz force law
>
> >> (Force)^i = F^ij J_j
>
> >> This is a covariant equation, holding in all coordinate systems
> >> (inertial or not).
>
> >True, however see GR1916 Eq.(65a) and onward,
> >and find that to vanish in GR, (This link might work)...
> >http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200...
>
> http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf
Thanks buddy.
> >You may interested to find that AE's GR1916 Eq.(66)
> >requires the Lorentz force to vanish as does Weinberg's
> >equivalent Eq.(5.3.7).
>
> That isn't true. The vanishing of the Lorentz force is required for
> inertial motion, not in general.
Charles, I've cross referenced to two Nobel Prize
winners, provided their Eqs, and explained the
meaning of those equations using analogies and
mathematics. I/we went further to check the impact
of those equations on electron orbitals, and found
empirical agreement, so I think the study is worthy.
> >>http://www.teleconnection.info/rqg/IntroductionToTensors#Electromagnetic
> >> Force
>
> >Does your treatment prevent an electron from spiralling
> >into the nucleus?
>
> For that one must deal with qed, not the classical Lorentz force.
Charles, you imply classical Lorentz force does not
apply to that problem...(what ever classical means).
However the GR application (f_u=0) of the Lorentz
force does indeed prevent the "classical" idea of
electrons from spiralling into the nucleus without
needing qed.
That is good empirical evidence for f_u=0, and as I've
showed at this link,
http://groups.google.com/group/sci.physics.foundations/browse_thread/thread/a3453bd8b85313a4/2c63bc0e5d42d357?hl=en&lnk=gst&q=Vanishing+of+Lorentz+Force%3F#2c63bc0e5d42d357
(Vanishing of Lorentz Force?)
is entirely in accord with EM and moreover EMR,
so I'm keeping an open mind.
Cl.Massé
"Peter" <end...@dekasges.de> a écrit dans le message de
news:guest.20080830081313$04...@news.killfile.org...
> monochromatic
>> light, were S is the time taken from a point to another, and then the
>> phase difference between those points since its speed is constant. In
>> first quantization, the least action principle is equivalent to the
>> Fermat theorem,
> which is the role of "first quantization" here?
In the Lagrangian, the position and the velocity correspond to a point
particle, and not to the field strength like in second quantization. Then
there is an exact equivalence with the least action principle for a point
particle.
>> that gives the points of constructive interference, where the phase
>> is stationary as a function of the path.
> is constructive interference tied to stationarity of phase?
Yes, but the converse is not true in general. If the phase as a function of
the path is at a maximum or a minimum, that is stationary, it is obvious the
interference is constructive. If not, there are enough paths to cancel the
interference.
>> Feynman's paths integrals are but a rehashing of those ideas.
>>
>> It is even interesting to note that the least action fulfills all the
>> properties of a distance. That's why Einstein used it to define the
>> geometry of space-time, through light rays. That is surely useful for
>> those who are investigating theories without prior space-time.
> I think so, too
>
> Let's recall Leibniz's formula
>
> action = mass x path length x velocity
>
> When velocity=const (light), action ~ path length, but what, if not?
>> I would say, action more fundamental than anything else, and still more
>> so in gauge theories.
> I'm wondering about the intimate relationship between Lagrangian, L, and
> symmetry, where L itself has no immediate physical meaning (Helmholtz
> called L the 'kinetic potential', but this applies not generally), do you
> know an explanation of that?
L is defined up to an additive constant, which correspond to a gauge
transformation for the wave. Any symmetry of L can so be linked to a gauge
symmetry through the action. L has no more physical meaning than the phase.
Peter
> >> ...
> >> In first quantization, the least action principle is equivalent to the
> >> Fermat theorem,
> > which is the role of "first quantization" here?
> In the Lagrangian, the position and the velocity correspond to a point
> particle, and not to the field strength like in second quantization. Then
> there is an exact equivalence with the least action principle for a point
> particle.
Hm
> >> that gives the points of constructive interference, where the phase
> >> is stationary as a function of the path.
> > is constructive interference tied to stationarity of phase?
> Yes, but the converse is not true in general. If the phase as a function
> of the path is at a maximum or a minimum, that is stationary, it is obvious
> the interference is constructive. If not, there are enough paths to cancel
> the interference.
have to rethink ;-)
> >> Feynman's paths integrals are but a rehashing of those ideas.
only that?
> >> It is even interesting to note that the least action fulfills all the
> >> properties of a distance. That's why Einstein used it to define the
> >> geometry of space-time, through light rays. That is surely useful for
> >> those who are investigating theories without prior space-time.
> > I think so, too
> >
> > Let's recall Leibniz's formula
> >
> > action = mass x path length x velocity
> >
> > When velocity=const (light), action ~ path length, but what, if not?
what, if not?
> >> I would say, action more fundamental than anything else, and still more
> >> so in gauge theories.
I agree that the gauge theories exhibit a great power of unification. I'm
unhappy, however, about the manner how it deals classically with electrical
charge conservation. The gauge freedom in Maxwell's theory can be removed,
when Maxwell's original set of equations (1864) is freed from its
incompleteness and redundancy.
> > I'm wondering about the intimate relationship between Lagrangian, L, and
> > symmetry, where L itself has no immediate physical meaning (Helmholtz
> > called L the 'kinetic potential', but this applies not generally), do you
> > know an explanation of that?
> L is defined up to an additive constant, which correspond to a gauge
> transformation for the wave. Any symmetry of L can so be linked to a gauge
> symmetry through the action.
How this looks like for the case momentum=const.?
> L has no more physical meaning than the phase.
For this, you need h, correct? If yes, how to treat this classically?
Thank you,
Peter
Oh No
>Hi Charles.
>At the outset, let me explain I explore many different
>ways to solve a problem.
>
>On Aug 30, 11:15 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
>> >Hi Charles
>> >On Aug 30, 6:20 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> >> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>>
>> >> >> "absolute" or, rather *proper* acceleration is the acceleration away
>> >> >> from inertial motion. Do not confuse inertial motion (meaning no active
>> >> >> forces) with uniform motion, which is relative.
>>
>> >> >So are you suggesting that "absolute" acceleration exists in GR?
>>
>> >> I do not like that terminology, but iirc you pointed out that Weinberg
>> >> uses "absolute acceleration" to mean the derivative of 4-velocity with
>> >> respect to proper time.
>>
>> >Yes in Grav& Cosmo, DU^u/Ds = f^u/m , Eq.(5.1.11).
>>
>> >However note you can consider your motion as zero
>> >relatively to yourself so that DU^u/Ds=0 for any FoR.
>>
>> No. This is not true. You only have DU^u/Ds=0 if you are in inertial
>> motion.
>
>Hmm, if you have a "proper change" in your acceleration
>given by DU^u/Ds as you allude to, and a fella on the other
>side of the Earth (Australia) claims the same, why are
>you both at relative rest?
Because of the geometry of spacetime surrounding the planet. Expressions
like DU^u/Ds apply locally.
>
>> >Most importantly, the Laws of Physics must be true
>> >in your FoR.
>>
>> >> I would prefer to call it "proper acceleration".
>> >> This quantity is a 4-vector and, multiplied by (scalar) mass, gives the
>> >> Force 4-vector, as used in the Lorentz force law
>>
>> >> (Force)^i = F^ij J_j
>>
>> >> This is a covariant equation, holding in all coordinate systems
>> >> (inertial or not).
>>
>> >True, however see GR1916 Eq.(65a) and onward,
>> >and find that to vanish in GR, (This link might work)...
>> >http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200...
>>
>> http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf
>
>Thanks buddy.
>
>> >You may interested to find that AE's GR1916 Eq.(66)
>> >requires the Lorentz force to vanish as does Weinberg's
>> >equivalent Eq.(5.3.7).
>>
>> That isn't true. The vanishing of the Lorentz force is required for
>> inertial motion, not in general.
>
>Charles, I've cross referenced to two Nobel Prize
>winners, provided their Eqs, and explained the
>meaning of those equations using analogies and
>mathematics. I/we went further to check the impact
>of those equations on electron orbitals, and found
>empirical agreement, so I think the study is worthy.
In gr we are mostly concerned with the gravitational force, and oftern
study situations in which the Lorentz force is zero. That is very
different from saying that the Lorentz force vanishes, which would mean
that there is never an e.m. force.
>
>> >>http://www.teleconnection.info/rqg/IntroductionToTensors#Electromagnetic
>> >> Force
>>
>> >Does your treatment prevent an electron from spiralling
>> >into the nucleus?
>>
>> For that one must deal with qed, not the classical Lorentz force.
>
>Charles, you imply classical Lorentz force does not
>apply to that problem...(what ever classical means).
>However the GR application (f_u=0) of the Lorentz
>force
The Lorentz force in GR is given by
(Force)^i = F^ij J_j
>does indeed prevent the "classical" idea of
>electrons from spiralling into the nucleus without
>needing qed.
>That is good empirical evidence for f_u=0, and as I've
>showed at this link,
>http://groups.google.com/group/sci.physics.foundations/browse_thread/th
>read/a3453bd8b85313a4/2c63bc0e5d42d357?hl=en&lnk=gst&q=Vanishing+of+Lor
>entz+Force%3F#2c63bc0e5d42d357
>(Vanishing of Lorentz Force?)
>is entirely in accord with EM and moreover EMR,
>so I'm keeping an open mind.
>
That would imply that electrons are in geodesic motion, and so
gravitationally bound to the nucleus. The em force is much stronger.
harry
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:8d81cf0c-4787-4311...@l33g2000pri.googlegroups.com...
That deformation corresponds to Hooke's force law. This is an obvious case
of force equilibrium.
Regards,
Harald
Ken S. Tucker
I also agree.
> Let's recall Leibniz's formula
>
> action = mass x path length x velocity
>
> When velocity=const (light), action ~ path length, but what, if not?
I find, mass=energy/c^2 , "path length" = 1 wavelength = L,
energy= h*f , f*L = c , (f=frequency) then,
action = (hf/c^2)Lc = hfL/c = h which looks good.
but the so-called "path-length" uses L.
Fred & Andy (IIRC) developed a photon theory that required
2L, (2 cycles) which is advanced.
I find an exact relation between the classical cycle
and the action "h" is still debateable, but I lean to
Fred and Andy's model.
> > I would say, action more fundamental than anything else, and still more so
> > in gauge theories.
>
> I'm wondering about the intimate relationship between Lagrangian, L, and
> symmetry, where L itself has no immediate physical meaning (Helmholtz called
> L the 'kinetic potential', but this applies not generally), do you know an
> explanation of that?
I'll try. A ball bouncing inelastically on a hard
surface repeats its trajectory cyclically with
a constant Lagrangian. If that ball where to hit
a place on the surface that is soft, that surface
would heat up and the ball would loose kinetic
energy at that bounce. Hence the Lagrangian
puts heat (kinetic energy) in that soft surface
and the Lagrange reduces.
> Thank you,
> Peter
Best again to you.
Ken S. Tucker
Ken S. Tucker
> "Ken S. Tucker" <dynam...@vianet.on.ca> wrote in messagenews:8d81cf0c-4787-4311...@l33g2000pri.googlegroups.com...
True in Newtonian thinking, however we are in GR,
using the stress tensor.
Newtons ideas compared to GR is like comparing
the game of Checkers to Chess, a different set of
rules and thinking.
Regards
Ken S. Tucker
harry
I'm not "in GR", in the sense that I don't share the orginal philosophy
behind it. Nevertheless, when talking about "force equilibrium" one does not
refer to "stress tensors" but to forces, and Hooke's force law is applied in
the 21 century in nearly every scale.
> Newtons ideas compared to GR is like comparing
> the game of Checkers to Chess, a different set of
> rules and thinking.
Yes indeed!
Regards,
Harald
Ken S. Tucker
I didn't know that Harald, sometime when you get a
chance I like to read more about that.
> Nevertheless, when talking about "force equilibrium" one does not
> refer to "stress tensors" but to forces, and Hooke's force law is applied in
> the 21 century in nearly every scale.
Not necessarily, force cannot be directly measured,
it is an extrapolation from a change in stress such as,
http://en.wikipedia.org/wiki/Piezoelectricity
Consider getting on a bathroom scale, the meter spins
to output the change in force. If I *cheat* I can change
the meter to read less, all the way to zero. Therefore
I have done a transformation to a CS wherein the force
vanishes.
Now when I step off the scale, I get a change in force
per change in time, and the scale reads negative.
You can pich up how Hooke's Law relates to GR here,
http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
> > Newtons ideas compared to GR is like comparing
> > the game of Checkers to Chess, a different set of
> > rules and thinking.
>
> Yes indeed!
When we stepped onto the scale, we changed the stress
in a quantized way, either by generating electricity, or by
compressing (and thus heating) a spring. What GR can
do is explain that quantized output.
> Regards,
> Harald
Same to you Harald.
Peter
> You can pich up how Hooke's Law relates to GR here,
> http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
I have not found Hooke's law there, please help
J.E is not the mechanical energy density, u_mech, as stated there
Best wishes,
Peter
Ken S. Tucker
On Sep 3, 2:24 pm, Peter <end...@dekasges.de> wrote:
> > You can pich up how Hooke's Law relates to GR here,
> >http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
>
> I have not found Hooke's law there, please help
Yes, here's Hooke's "Law" in this ref,
http://en.wikipedia.org/wiki/Piezoelectricity#Mathematical_description
> J.E is not the mechanical energy density, u_mech, as stated there.
J.E =0 , always, so there is no such thing as J.E,
however that equation is important as a constraint,
since it always vanishes. IOW's if it doesn't vanish
we're in trouble, because J.E=0 is proven empirically
and forms the basis of QT.
>From the standpoint of a technician (like myself)
who needs to understand current flow, I'll rewrite
J.E as,
J./\E = (J=qV) . (/\E + /\E') = 0
and solve to,
qV./\E = - qV./\E' .
which are in Power units.
In the case of a conducting wire the current "qV" is
constant, but there is a *quantized* power loss
" -qV./\E " via heating the extension cord to the
electric lawn mower.
Electricians call that a voltage drop through the
conductor due to resistance.
Regards
Ken S. Tucker
PS: Did I just demo GR and QT using a lawn mower
and extension cord ???
harry
That chance pops up immediately here below. :-)
>> Nevertheless, when talking about "force equilibrium" one does not
>> refer to "stress tensors" but to forces, and Hooke's force law is applied
>> in
>> the 21 century in nearly every scale.
>
> Not necessarily, force cannot be directly measured,
> it is an extrapolation from a change in stress such as,
> http://en.wikipedia.org/wiki/Piezoelectricity
It depends on what you mean with "directly measured". But we may take
Hooke's law as a definition of force. :-)
> Consider getting on a bathroom scale, the meter spins
> to output the change in force. If I *cheat* I can change
> the meter to read less, all the way to zero. Therefore
> I have done a transformation to a CS wherein the force
> vanishes.
That "vanishing" *is* cheating: the spring is not in equilibrium, and that
can even be easily verified. In other words, that force vanishing is a
simple example of hiding what is physically truly there.
Note: that happens to be exactly what makes me disagree with the original GR
philosophy, according to which gravitational fields can be made to "really
vanish" so that a body's gravitational field "does not exist", eventhough in
fact it's only apparently so.
> Now when I step off the scale, I get a change in force
> per change in time, and the scale reads negative.
>
> You can pich up how Hooke's Law relates to GR here,
> http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
I will quickly admit that there is a connection. :-)
>> > Newtons ideas compared to GR is like comparing
>> > the game of Checkers to Chess, a different set of
>> > rules and thinking.
>>
>> Yes indeed!
>
> When we stepped onto the scale, we changed the stress
> in a quantized way, either by generating electricity, or by
> compressing (and thus heating) a spring. What GR can
> do is explain that quantized output.
The word "quantized" confused me for a second. :-)
Of course GR can explain that output, but I don't think that GR does a
better job at that than classical (pre-GR) mechanics.
Cheers,
Harald
Ken S. Tucker
> "Ken S. Tucker" <dynam...@vianet.on.ca> wrote in messagenews:6eb92307-8207-4e10...@r15g2000prh.googlegroups.com...
Not really, it's 3D, see
http://en.wikipedia.org/wiki/Hooke's_law
> > Consider getting on a bathroom scale, the meter spins
> > to output the change in force. If I *cheat* I can change
> > the meter to read less, all the way to zero. Therefore
> > I have done a transformation to a CS wherein the force
> > vanishes.
>
> That "vanishing" *is* cheating: the spring is not in equilibrium, and that
> can even be easily verified. In other words, that force vanishing is a
> simple example of hiding what is physically truly there.
How do you find that physical truth?
(I think you'll need to use the force to output
energy/time = Power for measurement)
> Note: that happens to be exactly what makes me disagree with the original GR
> philosophy, according to which gravitational fields can be made to "really
> vanish" so that a body's gravitational field "does not exist", eventhough in
> fact it's only apparently so.
That slides from the PoE to Tidal effects.
> > Now when I step off the scale, I get a change in force
> > per change in time, and the scale reads negative.
>
> > You can pich up how Hooke's Law relates to GR here,
> >http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
>
> I will quickly admit that there is a connection. :-)
>
> >> > Newtons ideas compared to GR is like comparing
> >> > the game of Checkers to Chess, a different set of
> >> > rules and thinking.
>
> >> Yes indeed!
>
> > When we stepped onto the scale, we changed the stress
> > in a quantized way, either by generating electricity, or by
> > compressing (and thus heating) a spring. What GR can
> > do is explain that quantized output.
>
> The word "quantized" confused me for a second. :-)
> Of course GR can explain that output, but I don't think that GR does a
> better job at that than classical (pre-GR) mechanics.
I can demo my understanding using vanilla light
tensors with a helping of physics on the side.
GR puts many concepts in the same pot.
An "energy density" (aka stress tensor) , T^uv has
a Divergence (T^uv;v=0) of zero, meaning energy
in = energy out of a small (infinitesmal) volume,
to stand for the Conservation of Energy and as a
sideline, force f^u = T^uv;v =0.
OTOH, T^uv is NOT constant, it can be changed
*incrementally* "/\", by (T^uv + /\T^uv);v =0.
where /\T^uv is an incremental change in the
energy density, IOW's, a finite change /\T^uv,
where /\T^uv;v=0.
It's at that point we become mathematically
illiterate, as we accept an incrementally change
"/\T^uv" but NOT a continous variation of that
incremental change which is /\T^uv;v = 0.
> Cheers,
> Harald
Cheers back :-)
Ken S. Tucker
Cl.Massé
news:1220516...@sicinfo3.epfl.ch...
> Note: that happens to be exactly what makes me disagree with the original
> GR philosophy, according to which gravitational fields can be made to
> "really vanish" so that a body's gravitational field "does not exist",
> eventhough in fact it's only apparently so.
The equivalence principle is only local, so that you can't make a central
field of force to globally vanish. That's because space-time is curved.
Ken S. Tucker
Suppose we have a photon coincident with a
point, figuratively,
point
~~~~~O~~~~~>
At point "O", the photon is either a {1,0}, ie. it's there
or NOT there.
We can represent that by the Kronecker delta,
that has a *covariant* derivative of zero, however
/\T^uv =/=0, when the photon is coincident with the
point "O", because the photon *instanteously* changes
the energy density at that point "O" and then changes
back to zero, instanteously.
I think that is describing: what is the derivative of
a digital variation?
Regards
Ken S. Tucker
======================================= MODERATOR'S COMMENT:
/\T^uv;v is not an "incremental change"; your photon represents an infinite energy density - this and other flaws let me reject this posting
Ken S. Tucker
> On Sep 4, 11:50 am, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> > On Sep 4, 2:55 am, "harry" <harald.vanlintelButNotT...@epfl.ch> wrote:
Hi Mr. Moderator.
This is why I dislike using tensor analysis in SPF unless
Jay Yablon is available, because it's unpopular, though I
can see my "style" can be improved.
You can pick up the T^uv;v = 0 here,
http://en.wikipedia.org/wiki/Stress-energy_tensor#In_general_relativity
I can improve my delivery using (T^uv);v =0.
Then with a finite increment " /\T^uv " inserted in the
brackets, we form,
(T^uv + /\T^uv);v = 0 = (T^uv);v + (/\T^uv);v .
It follows that (/\T^uv);v = 0, although (/\T^uv) =/=0.
Is that better?
If that's ok, (and this gets posted) I'll try to address
the issue of "infinite energy density" you mentioned.
Regards
Ken S. Tucker
Cl.Massé
>> so in gauge theories.
"Peter" <end...@dekasges.de> a écrit dans le message de
news:guest.20080830081313$04...@news.killfile.org...
> I'm wondering about the intimate relationship between Lagrangian, L, and
> symmetry, where L itself has no immediate physical meaning (Helmholtz
> called
> L the 'kinetic potential', but this applies not generally), do you know an
> explanation of that?
If the action is to be associated with something like a distance, the least
action principle simply states that a system evolves along a straight line
in some generalized space, that reduces to the ordinary space for a free
particle. If the starting point and the ending point are given, a straight
line that passes through them is the trajectory of the system. There is
nothing mysterious in that, even in the point of view of causality.
Alternatively, if the starting point and the initial direction are given,
there is also a trajectory, and that is the equivalent to the Hamilton
formalism. It is the same difference as for limiting conditions and initial
conditions.
In general relativity, gravitation is explicitly described like that, the
trajectories are geodesics of the curved ordinary space. It is actually a
gauge theory where the local gauge transformation is an element of SO(3,1).
The Poincaré group is an isometry one, which conserve the distances.
For a charged point particle in an electromagnetic field, the Lagrangian
depends on the scalar and the vector potentials. These can be locally
changed for the same electromagnetic field. The action then changes also,
and that is analogous to the local phase change in quantum gauge theory.
The trajectory, on the other hand, is the same. It is like a change of
coordinates in the generalized space.
The action, as well as the phase of the wave function, have an analogous
physical meaning to the coordinates. But this meaning is up to a change of
coordinate system, that is, up to a gauge transformation. The Lagrangian
then has the role of the metric.
Oh No
>Charles writes,
>
>>> > In my view, action is no more than a mathematical trick. It is not
>>> > fundamental.
>
>light, were S is the time taken from a point to another, and then the phase
>difference between those points since its speed is constant. In first
>quantization, the least action principle is equivalent to the Fermat
>is stationary as a function of the path. Feynman's paths integrals are but
>a rehashing of those ideas.
>
>It is even interesting to note that the least action fulfills all the
>properties of a distance. That's why Einstein used it to define the
>geometry of space-time, through light rays. That is surely useful for those
>who are investigating theories without prior space-time.
>
>in gauge theories.
>
While the path integral gives a reason for the working of the classical
action principle, it does not make the action fundamental. Rather it
reduces it to a property of wave mechanics. As I think of wave mechanics
as also being a clever mathematical device rather than a fundamental
description of matter I am a long way from seeing action as fundamental.
Peter
> While the path integral gives a reason for the working of the classical
> action principle, it does not make the action fundamental. Rather it
> reduces it to a property of wave mechanics. As I think of wave mechanics
> as also being a clever mathematical device rather than a fundamental
> description of matter I am a long way from seeing action as fundamental.
I agree as, together with the very derivation (!), Schrödinger himself has
named 4 (afaik) deficiencies of his derivation, in particular, that there
should be a proper maths for quantum systems rather than the eigenvalue
methods for classical systems.
However, these deficiencies can be removed, so that Schrödinger's theory
becomes much more than merely "a clever mathematical device", eg, an
explanation for the so-called tunnel effect.
Best wishes,
Peter
clm...@online.fr
On 31 aoūt, 23:13, Peter <end...@dekasges.de> wrote:
> only that?
No, still much less. I'm not a Feynman idolizer.
>> > Let's recall Leibniz's formula
>> >
>> > action = mass x path length x velocity
>> >
>> > When velocity=const (light), action ~ path length, but what, if not?
It isn't the distance of the ordinary space, but it is still a
distance in some convenient space.
> How this looks like for the case momentum=const.?
The momentum isn't defined in the Lagrangian formalism. To go into
the Hamiltonian formalism, the generalized momentum p = @L/@q' is
used. Then p' = @H/@q. Let's make a point transformation that singles
out the coordinate of the center of mass Q. The total momentum P is
conserved iff the Hamiltonian doesn't depend exlicitely on Q. As H =
PQ' + pq' - L, where Q' is expressed as a function of P, that amounts
to @L/@Q = 0, that is, the Lagrangian doesn't explicitely depend on Q,
which excludes the case of a potential depending on Q. There is
nothing new, the conservation of momentum is associated to the
translation symmetry.
>> L has no more physical meaning than the phase.
> For this, you need h, correct? If yes, how to treat this classically?
h is only a constant to go from a unit to another. The Lagrangian is
also defined up to a multiplicative constant.
Peter
> >> Feynman's paths integrals are but a rehashing of those ideas.
> > only that?
> No, still much less. I'm not a Feynman idolizer.
Me not, too (see, eg, my papers on Huygens' principle in Eur. J. Phys. 1996
and in Icfai Univ. J. Optics 2008)
> >> > Let's recall Leibniz's formula
> >> >
> >> > action = mass x path length x velocity
> >> >
> >> > When velocity=const (light), action ~ path length, but what, if not?
> It isn't the distance of the ordinary space, but it is still a
> distance in some convenient space.
Can you provide an exemple?
> > How this looks like for the case momentum=const.?
> The momentum isn't defined in the Lagrangian formalism. To go into
> the Hamiltonian formalism, the generalized momentum p = @L/@q' is
> used. Then p' = @H/@q. Let's make a point transformation that singles
> out the coordinate of the center of mass Q. The total momentum P is
> conserved iff the Hamiltonian doesn't depend exlicitely on Q. As H =
> PQ' + pq' - L, where Q' is expressed as a function of P, that amounts
> to @L/@Q = 0, that is, the Lagrangian doesn't explicitely depend on Q,
> which excludes the case of a potential depending on Q. There is
> nothing new, the conservation of momentum is associated to the
> translation symmetry.
how this, if "the momentum isn't defined in the Lagrangian formalism"
> >> L has no more physical meaning than the phase.
> > For this, you need h, correct? If yes, how to treat this classically?
> h is only a constant to go from a unit to another.
poor Planck ;-)
> The Lagrangian is also defined up to a multiplicative constant.
this depends on the very definition - it seems that you are a supporter of
the approach 'the equation of motion is all - the state is nothing' :-(
> Liberty, Equality, Profitability.
poor science
P.
Cl.Massé
>> distance in some convenient space.
"Peter" <end...@dekasges.de> a écrit dans le message de
news:guest.20080909223621$66...@news.killfile.org...
> Can you provide an exemple?
Let's take a free massive particle. If the starting and ending points
(x1, t1) and (x2, t2) are given, the trajectory is just the straight line
joining them, and the (least) action is some distance, not the Cartesian
one, between the points, in some unit.
We can also take the Lagrangian corresponding to the Minkowskian
pseudo-distance:
L = - m c^2 sqrt(1 - (x'/c)^2)
It gives the same equation of motion: x" = 0, and reduces to the usual
Lagrangian (up to an additive constant) for x' << c.
>> The momentum isn't defined in the Lagrangian formalism. To go into
>> the Hamiltonian formalism, the generalized momentum p = @L/@q' is
>> used. Then p' = @H/@q. Let's make a point transformation that singles
>> out the coordinate of the center of mass Q. The total momentum P is
>> conserved iff the Hamiltonian doesn't depend explicitly on Q. As H =
>> PQ' + pq' - L, where Q' is expressed as a function of P, that amounts
>> to @L/@Q = 0, that is, the Lagrangian doesn't explicitly depend on Q,
>> which excludes the case of a potential depending on Q. There is
>> nothing new, the conservation of momentum is associated to the
>> translation symmetry.
> how this, if "the momentum isn't defined in the Lagrangian formalism"
@L/@Q' is conserved, even if P doesn't appear explicitly in the Lagrangian.
The equation of motion gives directly @L/@Q = 0.
>> h is only a constant to go from a unit to another.
> poor Planck ;-)
Or poor Einstein who did exactly the same with c? When something like this
is discovered, it is a big breakthrough because it is the unification of two
quantities that was thought different.
>> The Lagrangian is also defined up to a multiplicative constant.
> this depends on the very definition - it seems that you are a supporter of
> the approach 'the equation of motion is all - the state is nothing' :-(
No, the very definition is a function of q, q' and t that gives the correct
equations of motion through the least action principle. It is written
1/2m q'^2 only because it is more elegant. That's exactly the same for some
gauges. Note in addition that the physics isn't changed if other units are
taken.
The equations of motion describe the state evolution, what's the problem?
The Lagrangian doesn't describe the state, but the motion.
Peter
> >> It isn't the distance of the ordinary space, but it is still a
> >> distance in some convenient space.
> > Can you provide an exemple?
> Let's take a free massive particle. If the starting and ending points
> (x1, t1) and (x2, t2) are given, the trajectory is just the straight line
> joining them, and the (least) action is some distance, not the Cartesian
> one, between the points, in some unit.
>
> We can also take the Lagrangian corresponding to the Minkowskian
> pseudo-distance:
>
> L = - m c^2 sqrt(1 - (x'/c)^2)
>
> It gives the same equation of motion: x" = 0, and reduces to the usual
> Lagrangian (up to an additive constant) for x' << c.
ok - this points to the issue I have discussed some time ago in this group,
whether the potential energy induces a certain metrics for the motion of
bodies
> >> The momentum isn't defined in the Lagrangian formalism. To go into
> >> the Hamiltonian formalism, the generalized momentum p = @L/@q' is
> >> used. Then p' = @H/@q. Let's make a point transformation that singles
> >> out the coordinate of the center of mass Q. The total momentum P is
> >> conserved iff the Hamiltonian doesn't depend explicitly on Q. As H =
> >> PQ' + pq' - L, where Q' is expressed as a function of P, that amounts
> >> to @L/@Q = 0, that is, the Lagrangian doesn't explicitly depend on Q,
> >> which excludes the case of a potential depending on Q. There is
> >> nothing new, the conservation of momentum is associated to the
> >> translation symmetry.
> > how this, if "the momentum isn't defined in the Lagrangian formalism"
> @L/@Q' is conserved, even if P doesn't appear explicitly in the Lagrangian.
>
> The equation of motion gives directly @L/@Q = 0.
yes, I know
> >> h is only a constant to go from a unit to another.
> > poor Planck ;-)
> Or poor Einstein who did exactly the same with c? When something like this
> is discovered, it is a big breakthrough because it is the unification of
> two quantities that was thought different.
thus, not just a unit converter ;-)
> >> The Lagrangian is also defined up to a multiplicative constant.
> > this depends on the very definition - it seems that you are a supporter of
> > the approach 'the equation of motion is all - the state is nothing' :-(
> No, the very definition is a function of q, q' and t that gives the correct
> equations of motion through the least action principle. It is written
> 1/2m q'^2 only because it is more elegant. That's exactly the same for
> some gauges. Note in addition that the physics isn't changed if other
> units are taken.
>
> The equations of motion describe the state evolution, what's the problem?
> The Lagrangian doesn't describe the state, but the motion.
all Lagrangians I know are of dimension energy (it's not a matter of units,
but of dimension), and this is necessary for reproducing the Newtonian
equation of motion and for connecting to a Hamiltonian which can represent
the total energy of a system
Best wishes,
Peter
Ken S. Tucker
On Sep 11, 2:46 pm, Peter <end...@dekasges.de> wrote:
...
> All Lagrangians I know are of dimension energy (it's not a matter of units,
> but of dimension), and this is necessary for reproducing the Newtonian
> equation of motion and for connecting to a Hamiltonian which can represent
> the total energy of a system
If anyone has Weinberg's "Grav&Cosmo", the
Lagrangian is well explained beginning on
pg.220. An interesting equation is simply,
dtau/dt = 1 - L , Eq.(9.2.2),
Let me call dtau = ds, then with a bit of
calculus find, (a good approximation)...
(using Eq.(9.2.2) above),
(dt/ds = 1+L , when L is small),
d^2(t) = dL ds , d^2(s)= -dL dt , Eq.(1)
and see how the time paradox developes using
the Lagrangian "L", in those.
The "ds" can be thought of as the uniform rate
at which a clocks ticks in CS k and likewise
for dt in CS K, and the d^2(t) and d^2(s) are
the accelerations of those clock *rates* as
"defined" by the change in the Lagrangian "dL".
Correct me if I'm wrong: In Newton Gravity, "L"
is constant so that dL=0 and dt^2 =0 because
both dt and ds are sync'd constant...tick, tick.
If I understand correctly, there is a subtle
variation in the Lagrange "L" , "dL" transforming
from Newton's theory to GR, wherein GR the L
is NOT constant.
A clearer notation, going forward, uses dt'=ds,
to provide Eq.(1) rewritten in a relativistic format,
d^2(t) = dL dt' , d^2(t')= -dL dt , Eq.(2).
As Peter pointed out, the L is in the dimension
of energy, so I'll rewrite Eq.(2) in terms of
increments to highlight Planck's constant "h",
as we evolve,
/\^2(t') = /\E' dt' , /\^2(t)= -/\E /\t , Eq.(3)
where /\E' = - /\E and
h = /\E' /\t' = /\E /\t is an invariant to yield,
/\^2(t') = h, /\^2(t) = -h , /\^2(t') = -/\^2(t) , Eq.(4) .
That find's Eq.(4) *quantizes* the "twin paradox".
Observers stationary in CS's K and K' will find the
variation of the Lagrangian results in a transfer of
time, such as the "second" via /\^2(t') = -/\^2(t),
with the time deduction applied to K, transferred
to K' , in relation to the excange of action.
Regards
Ken S. Tucker
kxsxt8
Peter
> but of dimension), and this is necessary for reproducing the Newtonian
> equation of motion
please ignore that and take only this: