Expansion and balloon analogy
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
answer provided as to whether *we* are expanding too.
The question is: are we "painted on" (like magic marker) or are
we objects "taped on" (like pennies).
I have found (in my own head) a way through this. If we were
expanding, so would our eyes. If our eyes were expanding
proportionally wih the Universe, the 1/r^2 intensity change with
distance, would be countered by the r^2 change in light gathering
area of our "eyes". If we were expanding, then distance measures
based on red shift, subtended size, and intensity would give
"two" very different results. Observations with type Ia
supernovae, yield very similar results for the distance measures
obtained.
This line of reasoning is not one I recall having encountered on
my search...
David A. Smith
and...@attglobal.net
We are not expanding the same way the universe on the whole
is. We are held together by electromagnetic forces that
are much stronger than the gravitational effects that govern
universal expansion.
John Anderson
Joe Fischer
>When I first heard of this analogy, I was unsatisfied with the
>answer provided as to whether *we* are expanding too.
>
>The question is: are we "painted on" (like magic marker) or are
>we objects "taped on" (like pennies).
Are you talking about "us"? This is a classical question
about galaxies I think.
Frankly, even with the great respect and admiration I have
for methods and accuracy of measurements made, the galaxies
move so slowly it would seem difficult to say anything definite.
>I have found (in my own head) a way through this. If we were
>expanding, so would our eyes. If our eyes were expanding
>proportionally wih the Universe, the 1/r^2 intensity change with
>distance, would be countered by the r^2 change in light gathering
>area of our "eyes". If we were expanding, then distance measures
>based on red shift, subtended size, and intensity would give
>"two" very different results. Observations with type Ia
>supernovae, yield very similar results for the distance measures
>obtained.
>This line of reasoning is not one I recall having encountered on
>my search...
>David A. Smith
If you are talking about everything expanding, then
the "distances" do not change, the sun is always at a mean
distance of 93 million miles (according to a meter stick which
is also expanding).
There is quite a bit of speculation in any galactic
distance, and even in distant star distances, using the
intensity is one of the ways distances are estimated,
and I think there was a big problem with the original
estimates of some stars, which I should have predicted
as it seemed obvious that there were two possibilities.
There are subtle differences in prediction between
an expanding matter model and Newton, somebody on
the west coast wrote me about 15 or 20 years ago telling
me he was having some heavy devices made to test one
of the differences.
I suggested that there could be a problem doing
the esperiment on Earth, because of friction and using
a pivot, which might cause an object "moving" do only
to gravity to gain momentum, and that would mask any
possible effect.
If there is a general expansion, it has to be at
a very low level, and it must have been involved in
the establishment of density, because surface gravity
has to result from density layers summed and radius.
The sun contains 330000 times as much mass
as the earth, even though it's mean density is close to 1.
But any general expansion would require that
objects all double in radius in the same time period, and
the outward velocity of the surface is the expansion,
the surface gravity is just the change in velocity.
A general expansion is extremely difficult to
visualize and to sort out the processes involved.
But that doesn't matter, it will be useful as
a literal implementation of Einstein's Principle of
Equivalence. Even the event horizon of a
hypothetical black hole is often considered by
some to be accelerating outward, just as the
elevator accelerating upward was the origin
of the concepts that lead to General Relativity
and the inertial motion freefall.
Theoretical Physics is not at all as cut
and dried as some would have us believe,
even the black hole is in question by some
in the same schools where black holes are
popular;
"There has never been direct evidence of a black hole,
said Chapline, while acknowledging there are objects
that general relativity would predict are black holes
at the centers of galaxies. Ironically, Einstein
also didn’t believe in black holes even though he
created general relativity."
http://www.llnl.gov/pao/news/news_releases/2005/SF-05-04-03.html
Joe Fischer
Tom Roberts
> answer provided as to whether *we* are expanding too.
>
> The question is: are we "painted on" (like magic marker) or are
> we objects "taped on" (like pennies).
Classical physics has the problem that it is scale independent. That is,
there is nothing in the equations to determine any length or time scale
uniquely. In a purely classical world you could not tell if we are
expanding or not.
But we do not live in a purely classical world. In our world, quantum
phenomena provide unique scales. In particular, the size of an atom is
determined by various intrinsic properties of its component particles
and by the Lagrangian for electromagnetism. This results in a definite
length scale (e.g. determined by a meterstick in Paris). Similarly, EM
emissions from atomic transitions provide definitive frequency and time
scales.
So while the universe is exapnding on a grand scale, individual objects
within it do not share in the expansion, as long as the following
conditions hold:
* the components of the object are bound together
* the binding can be described by a local Lagrangian
* the length and time scales used in the Lagrangian are related to
length and time intervals obtained from quantized atomic lengths
and energy levels.
(I ignore the trivial case of a pointlike object.)
This applies to objects up to galaxies, and even groups of galaxies. But
that's approaching a region where the above conditions start to break down.
Tom Roberts tjro...@lucent.com
Bill Hobba
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:dl0fq7$c...@netnews.net.lucent.com...
Just a note for David and others who may be interested - I am sure Tom
knows this. The first chapter of Dirac - Principles of Quantum Mechanics
looks at this issue from a more general perspective and explains that if we
are ever to arrive at the ultimate explanation of things then big and small
need to be more than relative concepts - which is one reason QM, while
weird, is to be preferred.
Thanks
Bill
>
>
> Tom Roberts tjro...@lucent.com
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:dl0fq7$c...@netnews.net.lucent.com...
> N:dlzc D:aol T:com (dlzc) wrote:
>> When I first heard of this analogy, I was unsatisfied
>> with the answer provided as to whether *we* are
>> expanding too.
>>
>> The question is: are we "painted on" (like magic
>> marker) or are we objects "taped on" (like pennies).
>
> Classical physics has the problem that it is scale
> independent. That is, there is nothing in the
> equations to determine any length or time scale
> uniquely. In a purely classical world you could not
> tell if we are expanding or not.
True. One might infer a relative change, but since the past is
"writ" then changes must describe the present and its
relationship to the past, no?
> But we do not live in a purely classical world. In our
> world, quantum phenomena provide unique scales.
> In particular, the size of an atom is determined by
> various intrinsic properties of its component particles and by
> the Lagrangian for electromagnetism. This
> results in a definite length scale (e.g. determined by
> a meterstick in Paris). Similarly, EM emissions from
> atomic transitions provide definitive frequency and time
> scales.
Better not the obsolete length standard (which absorbed hydrogen
and secularly lengthened), but the original, the circumference of
the Earth.
> So while the universe is exapnding on a grand scale,
> individual objects within it do not share in the
> expansion, as long as the following conditions hold:
> * the components of the object are bound together
> * the binding can be described by a local Lagrangian
> * the length and time scales used in the Lagrangian
> are related to length and time intervals obtained
> from quantized atomic lengths and energy levels.
>
> (I ignore the trivial case of a pointlike object.)
>
> This applies to objects up to galaxies, and even
> groups of galaxies. But that's approaching a
> region where the above conditions start to break
> down.
Is any system other than the contents of a BH actually bound by
gravity, Mr. Roberts? I'm not trying to heckle, but to make
sense of "ancient" statements by Chris Hillman.
David A. Smith
Tom Roberts
> Is any system other than the contents of a BH actually bound by
Sure. The moon is bound to the earth, and the earth is bound to the sun,
both by gravity. While gravity is indubitably geometrical in GR, these
can be well described by a local Lagrangian, and thus do not share in
the cosmic expansion (to at least the accuracy of the model of their
motions).
Tom Roberts tjro...@lucent.com
Koobee Wublee
>
> Sure. The moon is bound to the earth, and the earth is bound to the sun,
> both by gravity. While gravity is indubitably geometrical in GR, these can
> be well described by a local Lagrangian, and thus do not share in the
> cosmic expansion (to at least the accuracy of the model of their motions).
So, how do you arrive at a certain Lagrangian without defining what it is
through whatever works in the past associated with something that fits the
role of a Lagrangian without understanding why it is a Lagrangian in the
first place? If my question is very confusing, I have made my point. The
Lagrangian as a density of an event taking place does not fit anything you
have in mind. Better check your conclusion with the basic mathematics that
brings you the Euler-Lagrange equation(s). I am sure you know what I am
saying, but you can pretend you don't by attacking my usage of vocabularies.
It is because the Lagrangian is a very abstract measurement that best
described in purely mathematical applications. So, please don't polute and
manipulate the work of Euler/Lagrange.
Speaking of more manipulations of mathematical methods, Lorentz Transforms
as written down first by Larmor become very suspicious. Recall the observed
length reduction along the direction of travel is derived through a two-way
light travel. Why does the spacetime equation (Minkowski) and thus the
Lorentz Transforms derived through just one-way of light travel. In real
world, the one-way light travel just does not compute. As an experimental
physicist, do you not agree?
If you allow the concept of spacetime established with two-way travel of
light, you will find the result not as simply presented by the Minkowski
spacetime equation. The speed of the frames pop out in the spacetime
equation. I am sure you are still confused from what I am saying. So,
allow me to rephrase. The concept of spacetime as a single coordinate is
established through just one-way travel of light. My argument is that it
should be in consistent with how FitzGerold/Lorentz concluded with a length
reduction through two-way travel of light.
Ben Rudiak-Gould
> Classical physics has the problem that it is scale independent. That is,
> there is nothing in the equations to determine any length or time scale
> uniquely. In a purely classical world you could not tell if we are
> expanding or not.
I don't understand this statement. Of course in classical physics there's no
explanation of why, say, the earth doesn't gravitationally collapse beyond a
particular size. But the universe isn't classical. If the universe were
classical, then there would perforce be some classical explanation for the
earth's size, and that explanation, whatever it was, would set a length
scale. Arguing that classical physics can't be correct because it doesn't
set a length scale is begging the question.
-- Ben
Joe Fischer
>Tom Roberts wrote:
>> Classical physics has the problem that it is scale independent. That is,
>> there is nothing in the equations to determine any length or time scale
>> uniquely. In a purely classical world you could not tell if we are
>> expanding or not.
>
>I don't understand this statement.
I don't understand it either. :-) I thought General Relativity was
considered "classical" today.
>Of course in classical physics there's no
>explanation of why, say, the earth doesn't gravitationally collapse beyond a
>particular size.
I thought there was a precise explanation with math that
gives the maximum compressiion of normal matter, red swarfs, white dwarfs,
and neutron stars.
Neutron stars were predicted before they were discovered,
weren't they?
>But the universe isn't classical.
It seems everybody thinks it is, with gravity totally based
on "attraction". It is puzzling to me how Tom and others say
that there is no force acting on freefalling objects, then say
gravity is attractive.
>If the universe were
>classical, then there would perforce be some classical explanation for the
>earth's size, and that explanation, whatever it was, would set a length
>scale.
But would it have to be a fixed length, always the same
as time passes? The matter at the center of the Earth is
greatly compressed, having a density approaching 18 or 20.
The maximum acceleration of gravity is aboyut 1700
miles below the surface, about 1.05 -1,07 g.
This is the type of thing that makes gravity theory
difficult, but nothing is impossible.
>Arguing that classical physics can't be correct because it doesn't
>set a length scale is begging the question.
>-- Ben
Newtonian/Euclidean physics with Mach's "all the other
matter in the universe creates inertia here" can't be correct,
experiment has proven that many times.
The disillusionment is so bad I can't get anybody
to even consider the derivatives of the formula
M1 * M2
F = G -----------------------
d^2
with M1 (gravitational mass of source) removed from
both sides of the equation.
I thought mathematicians usually remove the same
quantity from both sides, but Newton put M1 there.
All that can be observed if there is no contact beteeen
objects is relative acceleration. So that relative acceleration
should be the basic factor in a study of gravity, not a formula
with a fictitious force stuck in it forever.
Joe Fischer
Black Knight
"Ben Rudiak-Gould" <br276d...@cam.ac.uk> wrote in message
news:dl7qgq$5kk$1...@gemini.csx.cam.ac.uk...
What Tom is saying is that there is nothing in an equation that
allows you to determine 1 inch = 2.54 centimetres, and you can't tell
if he is expanding or not. He has a waistline problem, but I can't tell
without tape measure. (Perhaps he missed the "equals" sign.)
In Tom's own little world, he can OBSERVE the accretion disks near black
holes.
The rest of us call that "hallucination".
Tom calls himself a "scientist". Scientists call him a phuckwit.
Androcles.
Black Knight
"Joe Fischer" <efis...@ig1ou.com> wrote in message
news:tmren1988br78ird5...@4ax.com...
Yeah, there's this annoying problem of have one distance between
two masses. They each should have their own distance then Newton
could have written
M1 * M2
F = G -----------------------
d1 * d2
instead, and you could eliminate that irritating Force.
Androcles.
oriel36
In order to provide a ballistics explanation for heliocentric planetary
motion,Newton isolated the solar system from the rest of the Universe
and treated Keplerian orbital motion as strictly isolated within the
solar system.
"Cor. 2. And since these stars are liable to no sensible parallax from
the annual motion of the earth, they can have no force, because of
their immense distance, to produce any sensible effect in our system.
Not to mention that the fixed stars, every where promiscuously
dispersed in the heavens, by their contrary actions destroy their
mutual actions, by Prop. LXX, Book I." Newton
By right,contemporaries should adopt the principle that the Earth has
compound orbital motions,one around the Sun and the other moving in one
direction along with the solar system around the galactic axis,spending
half the time moving in the direction of galactic orbital motion and
half the year against.
If galactic orbital motion of the solar system influences the
variations in speed as a planet moves around the Sun it will also
influence cyclical oscillations in planetary orbits such as seen in
Mercury.
Standard analogy of a boat moving around a buoy in a lake will produce
circular motion but introduce a current and the boat will display
something akin to Keplerian motion.
As for Mach,well the poor guy was not good enough to recognise that
Newton shifted the value for axial rotation from the noon Equation of
Time format to Flamsteed's calendrically based sidereal format and it
was no wonder those unfortunate guys got caught up looking for rotation
of the Earth to inertail space when there is no justification for
it,unless you are compltely indoctrinated.This is what binds
Newtons,aetherists and relativists together in one ugly and overheated
spectacle.
Work with me Joe and we will find compromises.
Tom Roberts
> Tom Roberts wrote:
>> Classical physics has the problem that it is scale independent. That
>> is, there is nothing in the equations to determine any length or time
>> scale uniquely. In a purely classical world you could not tell if we
>> are expanding or not.
>
> I don't understand this statement.
Stated differently: all classical theories are invariant under a uniform
change of scale. This is why we can use different sets of units without
changing the equations of the theory (historical units like CGS and MKS
do change the equations [values of constants in them], because their
historical roots do not mesh perfectly with the equations). For
instance, using geometrical units in GR the only free unit is length,
but there's nothing in the theory itself to determine whether that is
cm, meters, or furlongs. To apply the equations you must pick one, and
that prevents you from ever knowing if the unit you picked actually
remains constant or not -- it is a DEFINITION, and you have no way
WITHIN THE THEORY to determine if the definition is changing over time.
So you define lengths to be measured in cm and forget about the
(classically) unmeasurable possibility that the unit itself might change
-- such unmeasurable things are not part of physics.
Quantum mechanics is different, and the dynamics of the theory
determines the sizes of electromagnetic bound states (e.g. atoms). So
(in principle at least) a meter can be defined as "N Pt atoms lined up
in a row". In QM you _know_ those atoms don't change size (because any
change would violate the theory); in classical physics you don't, and
all you can do is _define_ them to not change size.
[This pushes the problem down a level, so that one must define
the constants of QFT (masses, coupling constants, etc.) to be
fixed. But that is a situation at a different level, because
if they were not constant then the postulated symmetries of
the Lagrangian would not hold, and you would have a completely
different theory.]
So in practice, when applying a classical theory we use units determined
quantum mechanically. This is justified because it works.
[Newton et al implicitly did this without knowing about QM.]
It is misunderstandings of the subtleties here that make some people
confused about the nature of cosmological expansion. And related
misunderstandings have generated the "Divergent matter" nonsense.
> Of course in classical physics
> there's no explanation of why, say, the earth doesn't gravitationally
> collapse beyond a particular size.
Sure there is -- there is pressure inside that prevents the collapse. Of
course the "size" is really a ratio to a _defined_ unit (see above).
> But the universe isn't classical. If
> the universe were classical, then there would perforce be some classical
> explanation for the earth's size, and that explanation, whatever it was,
> would set a length scale.
But the classical theory is mute on whether or not that length scale is
changing -- all lengths are ratios to a _defined_ unit of length, and
say nothing at all about the unit itself.
> Arguing that classical physics can't be
> correct because it doesn't set a length scale is begging the question.
I'm not sure what question you think I am "begging". But it is quite
clear that classical physics is at best an approximation to a better
theory that incorporates quantum concepts.
Tom Roberts tjro...@lucent.com
Joe Fischer
>Ben Rudiak-Gould wrote:
>> Tom Roberts wrote:
>>> Classical physics has the problem that it is scale independent. That
>>> is, there is nothing in the equations to determine any length or time
>>> scale uniquely. In a purely classical world you could not tell if we
>>> are expanding or not.
>>
>> I don't understand this statement.
>
>Stated differently: all classical theories are invariant under a uniform
>change of scale. This is why we can use different sets of units without
>changing the equations of the theory (historical units like CGS and MKS
>do change the equations [values of constants in them], because their
>historical roots do not mesh perfectly with the equations).
Does that mean that classical theories need a Euclidean space
with a separate time ledger, as there is no provision in the models to
have any units change?
> For instance, using geometrical units in GR the only free unit is length,
>but there's nothing in the theory itself to determine whether that is
>cm, meters, or furlongs. To apply the equations you must pick one, and
>that prevents you from ever knowing if the unit you picked actually
>remains constant or not -- it is a DEFINITION, and you have no way
>WITHIN THE THEORY to determine if the definition is changing over time.
>So you define lengths to be measured in cm and forget about the
>(classically) unmeasurable possibility that the unit itself might change
>-- such unmeasurable things are not part of physics.
This does not seem to carry logic from the other statement,
it sounds as if "size" could change in either case, or that size
cannot change in classical theories but could in GR.
>Quantum mechanics is different, and the dynamics of the theory
>determines the sizes of electromagnetic bound states (e.g. atoms). So
>(in principle at least) a meter can be defined as "N Pt atoms lined up
>in a row". In QM you _know_ those atoms don't change size (because any
>change would violate the theory); in classical physics you don't, and
>all you can do is _define_ them to not change size.
I don't understand how you can know that atoms don't change
size over time, isn't the meter standard made out of real atoms?
> [This pushes the problem down a level, so that one must define
> the constants of QFT (masses, coupling constants, etc.) to be
> fixed. But that is a situation at a different level, because
> if they were not constant then the postulated symmetries of
> the Lagrangian would not hold, and you would have a completely
> different theory.]
I haven't interpreted anything you have said that prevents
atoms and molecules from _possibly_ changing size over time in a
way that the change was only detected by surface gravity, and that
only because the change in size over time is not linear.
>So in practice, when applying a classical theory we use units determined
>quantum mechanically. This is justified because it works.
>
> [Newton et al implicitly did this without knowing about QM.]
I think I need a new brain. :-)
>It is misunderstandings of the subtleties here that make some people
>confused about the nature of cosmological expansion. And related
>misunderstandings have generated the "Divergent matter" nonsense.
And how do you explain all the things posted here which are
absurd and do not involve matter expanding.
Or is it only expanding that bothers you?
>> Of course in classical physics
>> there's no explanation of why, say, the earth doesn't gravitationally
>> collapse beyond a particular size.
>
>Sure there is -- there is pressure inside that prevents the collapse. Of
>course the "size" is really a ratio to a _defined_ unit (see above).
There has to be a certain amount of collapse (compression) for
the density at the center to reach 17 or more.
And since the sun is growing in radius as it loses mass, I am
not sure your premise is valid.
Gas bodies are the norm in the visible universe, solids and
liquids are an exception that is a _very_ small percentage.
So for my part, please construct the models to work with
gas bodies first, then think about the solid and liquid.
>> But the universe isn't classical. If
>> the universe were classical, then there would perforce be some classical
>> explanation for the earth's size, and that explanation, whatever it was,
>> would set a length scale.
>
>But the classical theory is mute on whether or not that length scale is
>changing -- all lengths are ratios to a _defined_ unit of length, and
>say nothing at all about the unit itself.
Does that argue that size cannot change over time (as long
as the relative size of objects remains the same?
>> Arguing that classical physics can't be
>> correct because it doesn't set a length scale is begging the question.
>
>I'm not sure what question you think I am "begging". But it is quite
>clear that classical physics is at best an approximation to a better
>theory that incorporates quantum concepts.
>Tom Roberts tjro...@lucent.com
Isn't classical chemistry accountable by the fact that a
liter of any gas at constant pressure and temperature contains
the same number of molecule units?
......and isn't that also an accurate way to measure mass
being atomic and molecular weights are well measured?
Is anything "quite clear" regarding size now to size
yesterday in General Relativity?
To be clear, I need to add, if matter is expanding in
a way that could cause surface gravity to be what it is,
that expansion would have to be many times the rate
of the measured cosmic expansion.
So I am not so "nonsense-ical" to think that the
cosmic expansion occurs down to planets or people.
But if it did, I don't think it would be measurable,
as I would only double in size in a few Billion years.
Joe Fischer
Black Knight
> Ben Rudiak-Gould wrote:
>> Tom Roberts wrote:
>>> Classical physics has the problem that it is scale independent. That is,
>>> there is nothing in the equations to determine any length or time scale
>>> uniquely. In a purely classical world you could not tell if we are
>>> expanding or not.
>>
>> I don't understand this statement.
>
> Stated differently:
Observations of accretion disks near black holes by Tom Roberts
is hallucination.
Androcles
Themistocles
when yoyu are right you are right bleck knight, he never saw
a single black hole anfd he talk about their agregations
its like quantume mecahnings and string theoryes, no body
saw them and they wrote books of them
brian a m stuckless
GR can't get around GUESS adopted Fischer Expansion.
(Albeit, on a simple positive look at what he says):
Fischer Expansion explains not ONLY surface gravity,
but WHY the groundstate electron can't (don't) fall!
[BOTH, which can't be (is NOT) seen mathematically.]
$ brian a m stuckless
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Joe Fischer" <efis...@ig1ou.com> wrote in message
news:8achn1dmj24vm404s...@4ax.com...
> On Mon, 14 Nov 2005 09:31:19 -0600, Tom Roberts
> <tjro...@lucent.com> wrote:
...
> I don't understand how you can know that atoms
> don't change size over time,
If they all follow the same "laws of physics", how would you
know?
> isn't the meter standard made out of real atoms?
No. The meter had not been defined by a physical object since
1960.
http://nvl.nist.gov/pub/nistpubs/jres/104/3/html/j43bee.htm
... and this is why.
> So I am not so "nonsense-ical" to think that the
> cosmic expansion occurs down to planets or people.
> But if it did, I don't think it would be measurable,
> as I would only double in size in a few Billion years.
http://arxiv.org/abs/astro-ph/0203151
... I think the rate of expansion is *much* slower than that. 1
part in 10^11 per year requires 100 billion years to double in
size (unless you want to define size as volume).
David A. Smith
Joe Fischer
>Dear Joe Fischer:
>> isn't the meter standard made out of real atoms?
>
>No. The meter had not been defined by a physical object since
>1960.
>http://nvl.nist.gov/pub/nistpubs/jres/104/3/html/j43bee.htm
Sure, but see Fig. 18 at
http://nvl.nist.gov/pub/nistpubs/jres/104/3/html/j43bee.htm#c3pz7
See, it's getting longer all the time. :-)
Joe Fischer
Joe Fischer
>Dear Joe Fischer:
>Joe Fischer wrote:
>> So I am not so "nonsense-ical" to think that the
>> cosmic expansion occurs down to planets or people.
>> But if it did, I don't think it would be measurable,
>> as I would only double in size in a few Billion years.
>
>http://arxiv.org/abs/astro-ph/0203151
>
>... I think the rate of expansion is *much* slower than that. 1
>part in 10^11 per year requires 100 billion years to double in
>size (unless you want to define size as volume).
>David A. Smith
Somebody's math isn't so good, if the Hubble expanxion
traced back to 13 Billion years, then the doubling rate must be
less than that.
Joe Fischer
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Joe Fischer" <efis...@ig1ou.com> wrote in message
news:f3din15ufgdmi066c...@4ax.com...
The steel one is getting shorter. The Earth's circumference (the
original definition of the meter) isn't changing at all (as far
as we can tell). *That* is (at least one reason) why you don't
use matter for a length standard. Which matter?
David A. Smith
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Joe Fischer" <efis...@ig1ou.com> wrote in message
news:l7din1te65np3u0vs...@4ax.com...
> On Mon, dlzc1 D:cox T:n...@nospam.com> wrote:
>
>>Dear Joe Fischer:
>>Joe Fischer wrote:
>>> So I am not so "nonsense-ical" to think that the
>>> cosmic expansion occurs down to planets or people.
>>> But if it did, I don't think it would be measurable,
>>> as I would only double in size in a few Billion years.
>>
>>http://arxiv.org/abs/astro-ph/0203151
>>
>>... I think the rate of expansion is *much* slower than that.
>> 1 part in 10^11 per year requires 100 billion years to
>> double in size (unless you want to define size as volume).
>
> Somebody's math isn't so good, if the Hubble
> expanxion traced back to 13 Billion years, then the
> doubling rate must be less than that.
Hubble expansion isn't single valued. It was very fast in the
Early Universe, decreased almost to zero, and has "recently"
increased.
Also note that "expansion" and "clock rate in curved spacetime"
are inextricably bound. If space is expanding, global curvature
is decreasing, and ancient clocks (in a more dense past) appear
slowed. It is a hydra with at least two heads.
David A. Smith
Tom Roberts
> On Mon, 14 Nov 2005 09:31:19 -0600, Tom Roberts <tjro...@lucent.com> wrote:
>>Stated differently: all classical theories are invariant under a uniform
>>change of scale.
>
> with a separate time ledger, as there is no provision in the models to
> have any units change?
This does not necessarily mean that. Simple counterexample: GR is
invariant under a uniform change of scale, but needs no "Euclidean space
with a separate time ledger".
Independently, though, Newtonian mechanics does indeed require a
"Euclidean space with a separate time ledger" -- that's one of its
(implicit) assumptions.
>>For instance, using geometrical units in GR the only free unit is length,
>>but there's nothing in the theory itself to determine whether that is
>>cm, meters, or furlongs. To apply the equations you must pick one, and
>>that prevents you from ever knowing if the unit you picked actually
>>remains constant or not -- it is a DEFINITION, and you have no way
>>WITHIN THE THEORY to determine if the definition is changing over time.
>>So you define lengths to be measured in cm and forget about the
>>(classically) unmeasurable possibility that the unit itself might change
>>-- such unmeasurable things are not part of physics.
>
> This does not seem to carry logic from the other statement,
> it sounds as if "size" could change in either case, or that size
> cannot change in classical theories but could in GR.
The theory, either GR or NM, cannot determine if the defined unit is
changing.
>>Quantum mechanics is different, and the dynamics of the theory
>>determines the sizes of electromagnetic bound states (e.g. atoms).
>
> I don't understand how you can know that atoms don't change
> size over time, isn't the meter standard made out of real atoms?
All anybody knows is properties of quantities in theories. You are free
to select whatever theory you wish (of course if you choose poorly you
will not make good evaluations of the world around you).
If you select as a theory the standard model, then the masses and
coupling constants (etc.) must all be constant or the symmetries that
determine the standard model won't be valid. The constancy of those
constants implies the constancy of atomic sizes.
That is what I said in the next paragraph:
>> [This pushes the problem down a level, so that one must define
>> the constants of QFT (masses, coupling constants, etc.) to be
>> fixed. But that is a situation at a different level, because
>> if they were not constant then the postulated symmetries of
>> the Lagrangian would not hold, and you would have a completely
>> different theory.]
>
> I haven't interpreted anything you have said that prevents
> atoms and molecules from _possibly_ changing size over time
Go back and re-read what I wrote.
> in a
> way that the change was only detected by surface gravity, and that
> only because the change in size over time is not linear.
"Surface gravity" has nothing whatsoever to do with the symmetries of
the underlying QFT -- they are what constrains atoms to have a definite
and unchaning size.
>>It is misunderstandings of the subtleties here that make some people
>>confused about the nature of cosmological expansion. And related
>>misunderstandings have generated the "Divergent matter" nonsense.
>
> And how do you explain all the things posted here which are
> absurd and do not involve matter expanding.
I make no attempt to explain the absurdities posted around here.
> Or is it only expanding that bothers you?
What bothers me is theories that do not agree with experiment, and
people who advocate them in spite of that.
>>But the classical theory is mute on whether or not that length scale is
>>changing -- all lengths are ratios to a _defined_ unit of length, and
>>say nothing at all about the unit itself.
>
> Does that argue that size cannot change over time (as long
> as the relative size of objects remains the same?
Please re-read what I wrote. The classical theory cannot determine any
changes in its defined unit of length. I said nothing at all about some
abstract "size" and whether or not that "size" might change -- what I
said specifically applies to the UNIT OF LENGTH, and only to the UNIT OF
LENGTH.
> Isn't classical chemistry accountable by the fact that a
> liter of any gas at constant pressure and temperature contains
> the same number of molecule units?
That's a quantum theory. Anything that references atoms or molecules is
a quantum theory.
> Is anything "quite clear" regarding size now to size
> yesterday in General Relativity?
Sure. Using the SI standard of length, in GR that unit does not change,
and measurements made today are directly comparable to measurements made
yesterday. But that unit of length is specified OUTSIDE of GR, and is
ultimately referenced to a quantum theory in which the fundamental
parameters of the QFT are constants, and so are the relevant properties
of atoms.
> To be clear, I need to add, if matter is expanding in
> a way that could cause surface gravity to be what it is,
> that expansion would have to be many times the rate
> of the measured cosmic expansion.
Sure. But how could that possibly explain the orbit of the moon? Or the
constancy of the measured ratio of sizes of earth to moon? Or the
measured constancy of the frequency of zillions of atomic transitions? Or...
Tom Roberts tjro...@lucent.com
TomGee
anything more than that it's getting bigger. That has always explained
the observations of galaxies moving away from each other. The balloon
analogy was born to explain that process.
However, almost everyone argued that we live inside the universe and
not on its surface. The rising raisin bread analogy was born to better
explain the expansion process. It was better than the balloon business
because it has an "inside" to it while the balloon only has a "surface"
to it.
But then it was noted that while galaxies are all moving away from each
other, their coordinates remain the same. I.e., the distance between
them grows but they remain in the same spacial relationship as before.
Now that cannot be explained by balloons or raising breads. It seems
the only explanation possible is that the galaxies are not moving apart
from each other due to their own motions, but due to the expansion of
space!
So, to talk about the expansion of matter in regard to the expansion of
the universe is passe, to say the least. It is not matter that is
expanding, but the distance between it. We are not getting bigger at
all. It is only space that is growing inbetween all discrete matter!
At least, that is the latest guess from the physics community.
For something to do something in our universe requires energy of some
sort, but if space is empty, and unless we overthrow the Principle of
the conservation of energy and E=mc^2, it is not possible for space to
expand. Like so many other things in our universe that seem to be
doing one thing and later we find out they are doing something
different than what we first guessed, there must be a better
explanation than the counter-intuitive idea that space itself is
expanding.
If space was expanding, why only between galaxies and not within them?
The attraction between visible matter is the only reason we can point
to as a possible explanation for that observation. That means either
that the space within galaxies is not expanding or that the bodies
within maintain the same coodinates that make their group a galaxy
while the space around them expands, and they do that due to
gravitational attraction. The latter guess is more likely correct so
long as we cannot think of a way to explain the former.
But now we observe that some bodies within galaxies have speeds beyond
those they should have and some guess that could possibly occur due to
the existence of Dark Matter interactions with visible matter. If that
is so, it could explain how seemingly empty space can expand: It
isn't space that's expanding, but the invisible matter in space that is
causing the galaxies to move apart.
It's my guess that it is Dark Matter that is expanding due to the
impetus of the BB, and that as the impetus wears off some of it becomes
stationary in space. As visible matter collides with DM, interactions
occur that could cause the faster-than-expected motions of galactic
matter. If such interactions do occur, others may also occur, such as
the creation of matter in the birthing of stars.
PD
TomGee wrote:
> Joe, I have never taken the phrase, "the universe is expanding" to mean
> anything more than that it's getting bigger. That has always explained
> the observations of galaxies moving away from each other. The balloon
> analogy was born to explain that process.
>
> However, almost everyone argued that we live inside the universe and
> not on its surface. The rising raisin bread analogy was born to better
> explain the expansion process. It was better than the balloon business
> because it has an "inside" to it while the balloon only has a "surface"
> to it.
>
> But then it was noted that while galaxies are all moving away from each
> other, their coordinates remain the same.
Unsupported pigslop.
> I.e., the distance between
> them grows but they remain in the same spacial relationship as before.
> Now that cannot be explained by balloons or raising breads. It seems
> the only explanation possible is that the galaxies are not moving apart
> from each other due to their own motions, but due to the expansion of
> space!
>
> So, to talk about the expansion of matter in regard to the expansion of
> the universe is passe, to say the least. It is not matter that is
> expanding, but the distance between it. We are not getting bigger at
> all. It is only space that is growing inbetween all discrete matter!
> At least, that is the latest guess from the physics community.
>
> For something to do something in our universe requires energy of some
> sort, but if space is empty, and unless we overthrow the Principle of
> the conservation of energy and E=mc^2, it is not possible for space to
> expand.
Unsupported balderdash.
> Like so many other things in our universe that seem to be
> doing one thing and later we find out they are doing something
> different than what we first guessed, there must be a better
> explanation than the counter-intuitive idea that space itself is
> expanding.
>
> If space was expanding, why only between galaxies and not within them?
> The attraction between visible matter is the only reason we can point
> to as a possible explanation for that observation. That means either
> that the space within galaxies is not expanding or that the bodies
> within maintain the same coodinates that make their group a galaxy
> while the space around them expands, and they do that due to
> gravitational attraction. The latter guess is more likely correct so
> long as we cannot think of a way to explain the former.
>
> But now we observe that some bodies within galaxies have speeds beyond
> those they should have and some guess that could possibly occur due to
> the existence of Dark Matter interactions with visible matter. If that
> is so, it could explain how seemingly empty space can expand: It
> isn't space that's expanding, but the invisible matter in space that is
> causing the galaxies to move apart.
Unsupported dungheaps.
>
> It's my guess that it is Dark Matter that is expanding due to the
> impetus of the BB, and that as the impetus wears off some of it becomes
> stationary in space. As visible matter collides with DM, interactions
> occur that could cause the faster-than-expected motions of galactic
> matter. If such interactions do occur, others may also occur, such as
> the creation of matter in the birthing of stars.
Unsupported birdcrap.
Joe Fischer
>Joe, I have never taken the phrase, "the universe is expanding" to mean
>anything more than that it's getting bigger.
Ok, but that does not precisely describe what _MAY_ be happening,
the original term was "recession of the galaxies. But it is all based on
emperical observations, there was no model before the spectroscopic
data became avaiilable in the late 1920s (unless DeSitter defined a model),
although Einstein definitely had ideas that suggested the possibility.
> That has always explained
>the observations of galaxies moving away from each other. The balloon
>analogy was born to explain that process.
Yes, to explain it to lay people mostly, astronomers and cosmologists
knew what "recession of galaxies" meant.
>However, almost everyone argued that we live inside the universe and
>not on its surface. The rising raisin bread analogy was born to better
>explain the expansion process. It was better than the balloon business
>because it has an "inside" to it while the balloon only has a "surface"
>to it.
Actually neither is very good, bread would be like a medium,
and that is not possible.
>But then it was noted that while galaxies are all moving away from each
>other, their coordinates remain the same. I.e., the distance between
>them grows but they remain in the same spacial relationship as before.
That is a guess-timate, the size and distances are so large
compared to the time it has been observed, and the only data
available is probably spectroscopic with translation mapping and
interpolation.
>Now that cannot be explained by balloons or raising breads. It seems
>the only explanation possible is that the galaxies are not moving apart
>from each other due to their own motions, but due to the expansion of
>space!
Not hardly, that would make space a medium capable of
carrying galaxies, I don't think anybody would realy think that.
They could be moving apart to nothing but thier own
motion with a non-Euclidean component in the motion.
>So, to talk about the expansion of matter in regard to the expansion of
>the universe is passe, to say the least.
Not really, there is much more that needs to be explained and
fitted to a pattern besides spectroscopic data.
> It is not matter that is
>expanding, but the distance between it.
That is not a certainty. The inportant thing is to consider
something that could have caused all the matter in the universe
to be thrown apart, and expanding matter is as good as any
cause of that.
Chances are all the different modern concepts of a
cosmological constant will be wrong, the spherical geometry
of expansion of any kind may not follow Newton or Euclid.
>We are not getting bigger at all.
At least one person feels we may be.
>It is only space that is growing inbetween all discrete matter!
>At least, that is the latest guess from the physics community.
I don't think it is the whole community, guesses are
made by individuals, the rest either concur, object (maybe
silently), or follow like sheep.
>For something to do something in our universe requires energy of some
>sort, but if space is empty, and unless we overthrow the Principle of
>the conservation of energy and E=mc^2, it is not possible for space to
>expand.
I thought you said it is expanding?
>Like so many other things in our universe that seem to be
>doing one thing and later we find out they are doing something
>different than what we first guessed, there must be a better
>explanation than the counter-intuitive idea that space itself is
>expanding.
I agree, in fact it probably is a certainty.
>If space was expanding, why only between galaxies and not within them?
>The attraction between visible matter is the only reason we can point
>to as a possible explanation for that observation. That means either
>that the space within galaxies is not expanding or that the bodies
>within maintain the same coodinates that make their group a galaxy
>while the space around them expands, and they do that due to
>gravitational attraction. The latter guess is more likely correct so
>long as we cannot think of a way to explain the former.
There are other guesses to be made.
>But now we observe that some bodies within galaxies have speeds beyond
>those they should have and some guess that could possibly occur due to
>the existence of Dark Matter interactions with visible matter. If that
>is so, it could explain how seemingly empty space can expand: It
>isn't space that's expanding, but the invisible matter in space that is
>causing the galaxies to move apart.
Not hardly, there is no action at a distance without a mechanism.
>It's my guess that it is Dark Matter that is expanding due to the
>impetus of the BB, and that as the impetus wears off some of it becomes
>stationary in space.
Dark Matter is nothing but matter that is too cool to glow,
and that is a lot different than Dark Energy.
All this points out a clear fact, that no good model exists
that can bring all the bizarre observations together.
>As visible matter collides with DM, interactions
>occur that could cause the faster-than-expected motions of galactic
>matter. If such interactions do occur, others may also occur, such as
>the creation of matter in the birthing of stars.
Actually, a static universe with constant size objects in it
should be much more passive than what is observed.
The many different explanations of all the different
observations has become so speculative and silly they
are now saying that a certain extremely bright and massive
object is a "Black Hole" __FEEDING__ on the matter
around it.
Now black holes have become the brightest objects
in the sky. I prefer to wait for a breakthrough.
Joe Fischer
Tom Roberts
> But then it was noted that while galaxies are all moving away from each
> other, their coordinates remain the same. I.e., the distance between
> them grows but they remain in the same spacial relationship as before.
Hmmm. "Coordinates remain the same" has no meaning at all independent of
the definition of the coordinate system. "Spacial relationship"
generally includes both direction and distance, so your claim here makes
no sense.
> Now that cannot be explained by balloons or raising breads.
Those are _ANALOGIES_, not "explanations".
> It seems
> the only explanation possible is that the galaxies are not moving apart
> from each other due to their own motions, but due to the expansion of
> space!
"expansion of space" has no meaning.
> So, to talk about the expansion of matter in regard to the expansion of
> the universe is passe, to say the least. It is not matter that is
> expanding, but the distance between it. We are not getting bigger at
> all. It is only space that is growing inbetween all discrete matter!
> At least, that is the latest guess from the physics community.
Kinda-sorta.
In the FRW manifolds used for cosmological models, spacetime is
permeated with a dense filling of dust particles. The elapsed proper
time from the big bang for each dust particle defines 3-surfaces of
simultaneity for the cosmologically-preferred time coordinate, and if
one measures the distance between a given pair of dust particles on
successive such surfaces, then this distance is increasing (for the
portion of these manifolds corresponding to our current situation).
> For something to do something in our universe requires energy of some
> sort, but if space is empty, and unless we overthrow the Principle of
> the conservation of energy and E=mc^2, it is not possible for space to
> expand.
Not true in the FRW manifolds. No "energy" is needed for them to expand
in the manner described above.
Attempting to do physics by sound bite is hopeless -- you need to
actually STUDY the relevant theories of physics.
> there must be a better
> explanation than the counter-intuitive idea that space itself is
> expanding.
There is. It's called GR and its application to cosmology via the FRW
manifolds.
> If space was expanding, why only between galaxies and not within them?
See my previous posts in this thread.
>[...]
Tom Roberts tjro...@lucent.com
TomGee
Tom Roberts wrote:
> TomGee wrote:
> > But then it was noted that while galaxies are all moving away from each
> > other, their coordinates remain the same. I.e., the distance between
> > them grows but they remain in the same spacial relationship as before.
>
> Hmmm. "Coordinates remain the same" has no meaning at all independent of
> the definition of the coordinate system. "Spacial relationship"
> generally includes both direction and distance, so your claim here makes
> no sense.
>
>
to understand my plain english. The phrase "spacial relationship"
defines my coordinate system which refers to the direction from each
other of the galaxies observed. I could have worded my concluding
statement better by saying, "but _otherwise_ their spacial relationship
does not change.
>
>
> > Now that cannot be explained by balloons or raising breads.
>
> Those are _ANALOGIES_, not "explanations".
>
>
>
>
> > It seems
> > the only explanation possible is that the galaxies are not moving apart
> > from each other due to their own motions, but due to the expansion of
> > space!
>
> "expansion of space" has no meaning.
>
>
>
>
> > So, to talk about the expansion of matter in regard to the expansion of
> > the universe is passe, to say the least. It is not matter that is
> > expanding, but the distance between it. We are not getting bigger at
> > all. It is only space that is growing inbetween all discrete matter!
> > At least, that is the latest guess from the physics community.
>
> Kinda-sorta.
>
> In the FRW manifolds used for cosmological models, spacetime is
> permeated with a dense filling of dust particles.
>
>
is not a real place - it is a math construct used to calculate the
world lines of objects moving in space. It is an imaginary place that
does not exist in the real universe, only in our minds. Many have come
to believe It is a real place in our universe, but of course that's
nonsense.
>
>
> The elapsed proper
> time from the big bang for each dust particle defines 3-surfaces of
> simultaneity for the cosmologically-preferred time coordinate, and if
> one measures the distance between a given pair of dust particles on
> successive such surfaces, then this distance is increasing (for the
> portion of these manifolds corresponding to our current situation).
>
>
know any English for us poor technobabble-impaired laypersons? Do you
even have a clue as to what the hell you're talking about? You cannot
know the elapsed proper time of a single bit of dust unless you can
track its world line back to the BB, and then it is only a math
construct only remotely related to reality if at all. And why do you
use an undefined "cosmologically-preferred (sic) time coordinate"
system after berating me for not defining mine? And why use all that
nonsense to say that the distance is increasing between dust particles?
I mean, what does that mean, in your language, that is different than
my observation that the distance between matter is increasing?
>
>
> > For something to do something in our universe requires energy of some
> > sort, but if space is empty, and unless we overthrow the Principle of
> > the conservation of energy and E=mc^2, it is not possible for space to
> > expand.
>
> Not true in the FRW manifolds. No "energy" is needed for them to expand
> in the manner described above.
>
>
our universe without the expenditure of some energy either from within
or without an object.
>
>
> Attempting to do physics by sound bite is hopeless -- you need to
> actually STUDY the relevant theories of physics.
>
>
could I talk to you about physics if I studied like you and learned
only nonsense and to make reference to someone's learning when you have
no reasonable retort to their ideas?
>
>
> > there must be a better
> > explanation than the counter-intuitive idea that space itself is
> > expanding.
>
> There is. It's called GR and its application to cosmology via the FRW
> manifolds.
>
>
study of the origin and structure of the universe using topological
spaces or surfaces?
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:4%yef.228$IC3...@newssvr33.news.prodigy.com...
> TomGee wrote:
...
>> For something to do something in our universe requires
>> energy of some sort, but if space is empty, and unless
>> we overthrow the Principle of the conservation of energy
>> and E=mc^2, it is not possible for space to expand.
>
> Not true in the FRW manifolds. No "energy" is needed
> for them to expand in the manner described above.
Not true in General Relativity at all, since gravity is not a
force.
David A. Smith
jem
> Tom Roberts wrote:
>
>>TomGee wrote:
>>>So, to talk about the expansion of matter in regard to the expansion of
>>>the universe is passe, to say the least. It is not matter that is
>>>expanding, but the distance between it. We are not getting bigger at
>>>all. It is only space that is growing inbetween all discrete matter!
>>>At least, that is the latest guess from the physics community.
>>
>>Kinda-sorta.
>>
>>In the FRW manifolds used for cosmological models, spacetime is
>>permeated with a dense filling of dust particles.
>>
>>
>
> No, sorry. You cannot use "spacetime" in this discussion because s-t
> is not a real place - it is a math construct used to calculate the
> world lines of objects moving in space. It is an imaginary place that
> does not exist in the real universe, only in our minds. Many have come
> to believe It is a real place in our universe, but of course that's
> nonsense.
>
The problem isn't his failure to realize that "spacetime" isn't "real",
but your failure to realize that "objects" and "space" aren't any more
real than spacetime; each of these exists only in models of Nature.
TomGee
Joe Fischer wrote:
> On 15 Nov 2005 08:14:01 -0800, "TomGee" <lv...@hotmail.com> wrote:
>
> >Joe, I have never taken the phrase, "the universe is expanding" to mean
> >anything more than that it's getting bigger.
>
> Ok, but that does not precisely describe what _MAY_ be happening,
> the original term was "recession of the galaxies.
>
>
>
>
> But it is all based on
> emperical observations, there was no model before the spectroscopic
> data became avaiilable in the late 1920s (unless DeSitter defined a model),
> although Einstein definitely had ideas that suggested the possibility.
>
>
universe that he fudged on his numbers to invent one. At that time, it
was becoming more and more accepted that the U. is expanding so he must
have known about the possibilities.
>
>
> > That has always explained
> >the observations of galaxies moving away from each other. The balloon
> >analogy was born to explain that process.
>
> Yes, to explain it to lay people mostly, astronomers and cosmologists
> knew what "recession of galaxies" meant.
>
> >However, almost everyone argued that we live inside the universe and
> >not on its surface. The rising raisin bread analogy was born to better
> >explain the expansion process. It was better than the balloon business
> >because it has an "inside" to it while the balloon only has a "surface"
> >to it.
>
> Actually neither is very good, bread would be like a medium,
> and that is not possible.
>
>
filled with DM, then it is DM that is what is expanding and not space,
more likely.
>
>
> >But then it was noted that while galaxies are all moving away from each
> >other, their coordinates remain the same. I.e., the distance between
> >them grows but they remain in the same spacial relationship as before.
>
> That is a guess-timate, the size and distances are so large
> compared to the time it has been observed, and the only data
> available is probably spectroscopic with translation mapping and
> interpolation.
>
>
nature's grand paradoxes for our viewing pleasure....
> >Now that cannot be explained by balloons or raising breads. It seems
> >the only explanation possible is that the galaxies are not moving apart
> >from each other due to their own motions, but due to the expansion of
> >space!
>
> Not hardly, that would make space a medium capable of
> carrying galaxies, I don't think anybody would realy think that.
>
>
You seem to be giving a meaning to the term "medium" different than
commonly accepted. Galaxies do exist in space so space is indeed
capable of carrying galaxies.
>
>
> They could be moving apart to nothing but thier own
> motion with a non-Euclidean component in the motion.
>
>
That's what was thought af first, but if it was due to their own
motion, their coordinates would change.
>
>
> >So, to talk about the expansion of matter in regard to the expansion of
> >the universe is passe, to say the least.
>
> Not really, there is much more that needs to be explained and
> fitted to a pattern besides spectroscopic data.
>
> > It is not matter that is
> >expanding, but the distance between it.
>
> That is not a certainty. The inportant thing is to consider
> something that could have caused all the matter in the universe
> to be thrown apart, and expanding matter is as good as any
> cause of that.
>
>
I disagree. The idea of expanding matter is probably the worst idea to
date about the expansion process of the universe. The BBT is a better
idea because it provides the reason why everything is in motion and we
cannot find a single stationary place in it.
>
>
> Chances are all the different modern concepts of a
> cosmological constant will be wrong, the spherical geometry
> of expansion of any kind may not follow Newton or Euclid.
>
>
Again, I disagree. Chances are that AE's constant will come to have
some validity after all, even though it won't overthrow the expansion
process observations.
>
>
> >We are not getting bigger at all.
>
> At least one person feels we may be.
>
> >It is only space that is growing inbetween all discrete matter!
> >At least, that is the latest guess from the physics community.
>
> I don't think it is the whole community, guesses are
> made by individuals, the rest either concur, object (maybe
> silently), or follow like sheep.
>
> >For something to do something in our universe requires energy of some
> >sort, but if space is empty, and unless we overthrow the Principle of
> >the conservation of energy and E=mc^2, it is not possible for space to
> >expand.
>
> I thought you said it is expanding?
>
>
I said that is the common idea.
>
>
> >Like so many other things in our universe that seem to be
> >doing one thing and later we find out they are doing something
> >different than what we first guessed, there must be a better
> >explanation than the counter-intuitive idea that space itself is
> >expanding.
>
> I agree, in fact it probably is a certainty.
>
> >If space was expanding, why only between galaxies and not within them?
> >The attraction between visible matter is the only reason we can point
> >to as a possible explanation for that observation. That means either
> >that the space within galaxies is not expanding or that the bodies
> >within maintain the same coodinates that make their group a galaxy
> >while the space around them expands, and they do that due to
> >gravitational attraction. The latter guess is more likely correct so
> >long as we cannot think of a way to explain the former.
>
> There are other guesses to be made.
>
>
Yes, I agree.
>
>
> >But now we observe that some bodies within galaxies have speeds beyond
> >those they should have and some guess that could possibly occur due to
> >the existence of Dark Matter interactions with visible matter. If that
> >is so, it could explain how seemingly empty space can expand: It
> >isn't space that's expanding, but the invisible matter in space that is
> >causing the galaxies to move apart.
>
> Not hardly, there is no action at a distance without a mechanism.
>
>
Below, I give the impetus of the BB as the mechanism for DM motions.
>
>
> >It's my guess that it is Dark Matter that is expanding due to the
> >impetus of the BB, and that as the impetus wears off some of it becomes
> >stationary in space.
>
> Dark Matter is nothing but matter that is too cool to glow,
> and that is a lot different than Dark Energy.
>
>
There is no reason to think that because matter too cool to glow should
still reflect light for us to see it. I agree more with Gamow that
invisible matter exists in all the spaces of our universe. Such matter
is invisible because it has no positive energy and mass and our eyes
cannot discern objects having only negative mass/energy.
>
>
> All this points out a clear fact, that no good model exists
> that can bring all the bizarre observations together.
>
>
Well, that's what mine does, and that's why I post here.
>
>
> >As visible matter collides with DM, interactions
> >occur that could cause the faster-than-expected motions of galactic
> >matter. If such interactions do occur, others may also occur, such as
> >the creation of matter in the birthing of stars.
>
> Actually, a static universe with constant size objects in it
> should be much more passive than what is observed.
>
>
Pardon?
>
>
> The many different explanations of all the different
> observations has become so speculative and silly they
> are now saying that a certain extremely bright and massive
> object is a "Black Hole" __FEEDING__ on the matter
> around it.
>
>
Not exactly. It would be the matter surrounding the bh that is bright
but not the bh itself, since light cannot escape it.
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:4%yef.228$IC3...@newssvr33.news.prodigy.com...
>>TomGee wrote:
>>>For something to do something in our universe requires
>>>energy of some sort, but if space is empty, and unless
>>>we overthrow the Principle of the conservation of energy
>>>and E=mc^2, it is not possible for space to expand.
>>
>>Not true in the FRW manifolds. No "energy" is needed
>>for them to expand in the manner described above.
I should have added that TomGee's "Principle of the conservation of
energy" is only valid in the presence of a timelike Killing vector, and
only for coordinates using it as the time coordinate. The FRW manifolds
have no such Killing vector, and that "principle" simply does not apply.
This is one of the remarkable results of Noether's theorem.
Translation from technical jargon into looser english:
Noether's theorem implies that energy is conserved only for situations
in which there is time translation symmetry of the Lagrangian[@] -- i.e.
that no interactions change over time. In GR the Lagrangian includes the
curvature of spacetime, and that means that for energy conservation to
apply there must be essentially no changes whatsoever over time --
nothing can move in any way[#]. Clearly such a situation is not very
useful, and CERTAINLY does not apply to the world we inhabit. And it
does not apply to the FRW manifolds used as cosmological models.
[@] The lagrangian of a system describes all interactions of
the system, both internal and external.
[#] The presence of a timelike Killing vector is the technical
way of saying this.
What is truly amazing is that energy conservation works so well in
Newtonian physics -- its pervasiveness there is what made people
erroneously think that conservation of energy is a general "principle".
Noether's theorem (1918) delineated the limits of such conservation
"laws" by relating them to symmetries of the Lagrangian. At base this
has to do with the fact that gravitation is so incredibly small, and the
terms in the Lagrangian that generate violations of energy conservation
are completely negligible in Newtonian situations.
[Similar remarks apply to momentum conservation.]
> Not true in General Relativity at all, since gravity is not a
> force.
It goes much deeper than that. See above.
Tom Roberts tjro...@lucent.com
TomGee
Tom Roberts wrote:
> N:dlzc D:aol T:com (dlzc) wrote:
> > "Tom Roberts" <tjro...@lucent.com> wrote in message
> > news:4%yef.228$IC3...@newssvr33.news.prodigy.com...
> >>TomGee wrote:
> >>>For something to do something in our universe requires
> >>>energy of some sort, but if space is empty, and unless
> >>>we overthrow the Principle of the conservation of energy
> >>>and E=mc^2, it is not possible for space to expand.
> >>
> >>Not true in the FRW manifolds. No "energy" is needed
> >>for them to expand in the manner described above.
>
> I should have added that TomGee's "Principle of the conservation of
> energy" is only valid in the presence of a timelike Killing vector, and
> only for coordinates using it as the time coordinate.
>
>
the Principle of Conservation of Mass and Energy that supports the
inferred equivalence of mass and energy shown in E=mc^2. This
principle unifies mass and energy. Sorry for all the inconvenience my
error has obviously caused you....
contortions over the non-principle of conservation of energy.
Tom Roberts
> Tom Roberts wrote:
>>permeated with a dense filling of dust particles.
>
> No, sorry. You cannot use "spacetime" in this discussion because s-t
> is not a real place - it is a math construct used to calculate the
> world lines of objects moving in space.
Sure I can use the concept of spacetime -- I am discussing the MODEL.
Remember that in cosmology (and physics in general), all we really can
discuss are models....
>>The elapsed proper
>>time from the big bang for each dust particle defines 3-surfaces of
>>simultaneity for the cosmologically-preferred time coordinate, and if
>>one measures the distance between a given pair of dust particles on
>>successive such surfaces, then this distance is increasing (for the
>>portion of these manifolds corresponding to our current situation).
>
> Do you
> know any English for us poor technobabble-impaired laypersons?
If you are interested in cosmology, you MUST learn the language, or you
will be limited to the role of an infant.
> Do you
> even have a clue as to what the hell you're talking about?
Yes.
> You cannot
> know the elapsed proper time of a single bit of dust unless you can
> track its world line back to the BB, and then it is only a math
> construct only remotely related to reality if at all.
In the MODEL one can do that.
> And why do you
> use an undefined "cosmologically-preferred (sic) time coordinate"
> system after berating me for not defining mine?
I defined it in that paragraph. Perhaps you should go back and read it
again. <shrug>
> And why use all that
> nonsense to say that the distance is increasing between dust particles?
It's not nonsense -- without the background description, "distance" has
no meaning.
>>[in the FRW manifolds] No "energy" is needed for them to expand
>>in the manner described above.
>
> No, sorry. There can be no free lunch here. Motion is not possible in
> our universe without the expenditure of some energy either from within
> or without an object.
Just because a sound bite you happen to like applies in Newtonian
mechanics does not mean it applies everywhere. You are wrong. Though I
suppose it is debatable whether or not the concept "motion" applies to
such a cosmological expansion.... In any case, in the FRW manifolds the
constant-cosmological-time distance between a given pair of dust
particles changes over cosmological time, and no energy is involved. In
one class of FRW manifolds, that distance can change from increasing to
decreasing, and still no energy is involved.
>>>there must be a better
>>>explanation than the counter-intuitive idea that space itself is
>>>expanding.
>>
>>There is. It's called GR and its application to cosmology via the FRW
>>manifolds.
>
> GR talks about the expansion of space?
No. Re-read what I wrote. I said the GR is indeed the "better
explanation" you claimed there must be. But it is not amenable to sound
bites, so you'll either have to STUDY or remain mystified.
Tom Roberts tjro...@lucent.com
Joe Fischer
>Joe Fischer wrote:
>> On 15 Nov 2005 08:14:01 -0800, "TomGee" <lv...@hotmail.com> wrote:
>> >Joe, I have never taken the phrase, "the universe is expanding" to mean
>> >anything more than that it's getting bigger.
>>
>> Ok, but that does not precisely describe what _MAY_ be happening,
>> the original term was "recession of the galaxies.
>>
>That's correct. I agree.
A few years ago I posted a couple of articles that described a
cyclic BB with a long period of coasting galaxies and then an expansion
of all the stars filling in all the space between them, with another BB,
then the whole process starting again.
Not that anybody should worry.
>> But it is all based on
>> emperical observations, there was no model before the spectroscopic
>> data became avaiilable in the late 1920s (unless DeSitter defined a model),
>> although Einstein definitely had ideas that suggested the possibility.
>>
>It is commonly thought that Einstein wanted so badly to prove a static
>universe that he fudged on his numbers to invent one.
No, I have read the papers, and he was using all the information
available at the time. The stars of the Milky Way (the local galaxy)
were the only stars thought to exist until about 1927 or so.
I am not sure if the Magellanic Clouds were thought to be of
the local galaxy, or just more nebulous clouds called nebulae.
>At that time, it
>was becoming more and more accepted that the U. is expanding so he must
>have known about the possibilities.
Not when he proposed the cosmological factor around 1916.
I did not find anything about anything expanding before an
address to the Prussion Academy in 1921, but even that did not
refer to galaxies.
>> > That has always explained
>> >the observations of galaxies moving away from each other. The balloon
>> >analogy was born to explain that process.
>>
>> Yes, to explain it to lay people mostly, astronomers and cosmologists
>> knew what "recession of galaxies" meant.
>>
>> >However, almost everyone argued that we live inside the universe and
>> >not on its surface. The rising raisin bread analogy was born to better
>> >explain the expansion process. It was better than the balloon business
>> >because it has an "inside" to it while the balloon only has a "surface"
>> >to it.
>>
>> Actually neither is very good, bread would be like a medium,
>> and that is not possible.
>
>But of course it is. Space is a medium for matter and if space is
>filled with DM, then it is DM that is what is expanding and not space,
>more likely.
You have accepted the wrong definition of Dark Matter,
which is only planets, dust and burned out or dim stars.
It is possible that black holes are considered to be dark
matter, I don't know.
Are you thinking of Dark Energy/
>> >But then it was noted that while galaxies are all moving away from each
>> >other, their coordinates remain the same. I.e., the distance between
>> >them grows but they remain in the same spacial relationship as before.
>>
>> That is a guess-timate, the size and distances are so large
>> compared to the time it has been observed, and the only data
>> available is probably spectroscopic with translation mapping and
>> interpolation.
>>
>No, it has been shown to be so through observation. Another of
>nature's grand paradoxes for our viewing pleasure....
All objects in free space move in inertial motion, so the spacial
relationhip would not change, and especially would not change in
a mere 80 years of long tube spectrographs.
I think it takes 250 million years for the Milky Way to rotate
once, and I don't think Pluto has completed one full orbit since
being discoveref.
>> >Now that cannot be explained by balloons or raising breads. It seems
>> >the only explanation possible is that the galaxies are not moving apart
>> >from each other due to their own motions, but due to the expansion of
>> >space!
>>
>> Not hardly, that would make space a medium capable of
>> carrying galaxies, I don't think anybody would realy think that.
>
>You seem to be giving a meaning to the term "medium" different than
>commonly accepted. Galaxies do exist in space so space is indeed
>capable of carrying galaxies.
Rocket flight exists in air, but the air isn't needed,
there is less drag in space and the engines work much better.
>> They could be moving apart to nothing but thier own
>> motion with a non-Euclidean component in the motion.
>That's what was thought af first, but if it was due to their own
>motion, their coordinates would change.
I don't think so, the idea that anything is capable of altering
the path of a single star is absurd, let alone a galaxy.
>> >So, to talk about the expansion of matter in regard to the expansion of
>> >the universe is passe, to say the least.
>>
>> Not really, there is much more that needs to be explained and
>> fitted to a pattern besides spectroscopic data.
>>
>> > It is not matter that is
>> >expanding, but the distance between it.
>>
>> That is not a certainty. The inportant thing is to consider
>> something that could have caused all the matter in the universe
>> to be thrown apart, and expanding matter is as good as any
>> cause of that.
>I disagree. The idea of expanding matter is probably the worst idea to
>date about the expansion process of the universe.
I didn't say there was any connection, if gravity is connected
with any expansion of matter, it would be much faster than the
observed recession of the galaxies.
>The BBT is a better
>idea because it provides the reason why everything is in motion and we
>cannot find a single stationary place in it.
An expansion of matter seems to me to be more compatible
with an expansion of the universe than static matter.
There are lots of things to consider, like the temperature
of the upper atmosphere being extremely hot, and an expansion
of matter might fit a pattern of observations better than static
matter.
>> Chances are all the different modern concepts of a
>> cosmological constant will be wrong, the spherical geometry
>> of expansion of any kind may not follow Newton or Euclid.
>Again, I disagree. Chances are that AE's constant will come to have
>some validity after all, even though it won't overthrow the expansion
>process observations.
Not from galaxies speeding up or slowing down, but there
could be an observational effect that might make something
other than inertial motion appear.
>> >For something to do something in our universe requires energy of some
>> >sort, but if space is empty, and unless we overthrow the Principle of
>> >the conservation of energy and E=mc^2, it is not possible for space to
>> >expand.
>>
>> I thought you said it is expanding?
>>
>I said that is the common idea.
There are a lot of ideas, many not very logical.
Nature is logical when all the facts are known.
>> Not hardly, there is no action at a distance without a mechanism.
>Below, I give the impetus of the BB as the mechanism for DM motions.
I don't know anything about Dark Matter moving.
>> >It's my guess that it is Dark Matter that is expanding due to the
>> >impetus of the BB, and that as the impetus wears off some of it becomes
>> >stationary in space.
>>
>> Dark Matter is nothing but matter that is too cool to glow,
>> and that is a lot different than Dark Energy.
>There is no reason to think that because matter too cool to glow should
>still reflect light for us to see it.
Then you don't realize how difficult it is to see even small bright
stars, may I suggest a search of google for "intrinsic brightness",
and "visual magnitude" etc.
> I agree more with Gamow that
>invisible matter exists in all the spaces of our universe. Such matter
>is invisible because it has no positive energy and mass and our eyes
>cannot discern objects having only negative mass/energy.
>> All this points out a clear fact, that no good model exists
>> that can bring all the bizarre observations together.
>Well, that's what mine does, and that's why I post here.
And now is it time to ask what yours is/does? :-)
>> >As visible matter collides with DM, interactions
>> >occur that could cause the faster-than-expected motions of galactic
>> >matter. If such interactions do occur, others may also occur, such as
>> >the creation of matter in the birthing of stars.
>>
>> Actually, a static universe with constant size objects in it
>> should be much more passive than what is observed.
>Pardon?
Even a spherical galaxy has to have every star moving
outward to prevent all of them coming together.
At least that is what Einstein said, and I think that
means the same thing gave them the original velocity
that gave the galaxies their original velocity of recession.
>> The many different explanations of all the different
>> observations has become so speculative and silly they
>> are now saying that a certain extremely bright and massive
>> object is a "Black Hole" __FEEDING__ on the matter
>> around it.
>Not exactly. It would be the matter surrounding the bh that is bright
>but not the bh itself, since light cannot escape it.
But I think there are only very few of these puzzling objects
close enough to get a good look at.
Chances are that more money is being spent on observation
of distant objects is more than spent on the space program
And it takes a lot of time, with a lot of speculating going on.
Joe Fischer
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:dlfjf6$s...@netnews.net.lucent.com...
> N:dlzc D:aol T:com (dlzc) wrote:
...
Hopefully covered in more (basic, fundamental) detail in MTW's
"Gravitation" chapters 5 and 25 (myself having not even read the
introduction yet...) ;>)
Thanks for the clarification, broadening.
David A. Smith
Tom Roberts
> Tom Roberts wrote:
>>energy" is only valid in the presence of a timelike Killing vector, and
>>only for coordinates using it as the time coordinate.
>
> Whoa! Whoa! I left out a term of the principle I referred to. It is
> the Principle of Conservation of Mass and Energy that supports the
> inferred equivalence of mass and energy shown in E=mc^2. This
> principle unifies mass and energy.
What you call it does not matter. It still does not hold in the FRW
manifolds, nor, in general, in GR.
Tom Roberts tjro...@lucent.com
Koobee Wublee
>
> Noether's theorem implies that energy is conserved only for situations in
> which there is time translation symmetry of the Lagrangian[@] -- i.e. that
> no interactions change over time. In GR the Lagrangian includes the
> curvature of spacetime, and that means that for energy conservation to
> apply there must be essentially no changes whatsoever over time --
> nothing can move in any way[#]. Clearly such a situation is not very
> useful, and CERTAINLY does not apply to the world we inhabit. And it does
> not apply to the FRW manifolds used as cosmological models.
Speaking of uniformity of language, the following equation should be
universally accepted by all. It describes a segment of observed (thus
observable and subject to be observed with curvature) spacetime.
(ds)^2 = g^ij dx_i dx_j
What is the Lagrangian for this extremely general case?
From the Lagrangian Method of minimizing an action where the action is the
time elapsed where a particle moves in a distance as described by the above
generic equation, the Lagrangian can be derived by dividing the equation
with (g^00 dx_0 dx_0) which yields as follows.
Lagrangian = (ds/dx_0)^2 / g^00 + g^ij (dx_i/dx_0) (dx_j/dx_0) / g^00 = 1
The rest is just purely mathematical methods. The partial derivative of the
Lagrangian as defined above is always zero with respect to ds. Therefore,
according to Noether's Theorem, there is always a conserved quantity. This
quantity is none other than the energy. Conservation of energy(an observed
quantity) is ubiquitously a universal phenomenon even in GR and even in
binary stars. The anomaly observed from the binary stars is nothing more
than the failed universally observed conserved conservation of angular
momentum where it is not generally conserved. It is all in the math where
in this case is very simple. You can invent more fancy phrases and
vocabularies to erroneously justify your position on the lack of a universal
conservation of energy, but the mathematical method represents a much
stronger case contrary to your wordly manipulations of phrases where any
good lawyers should have no problems performing.
> [@] The lagrangian of a system describes all interactions of
> the system, both internal and external.
Again, the Lagrangian is a very abstract mathematical entity representing a
density of an action over the parameter if minimized would result in a
minimal effort of this action. It is a density of an action. Its concept
is so abstract that it cannot be safely regarded as a candidate to describe
all interactions of a given system.
> [#] The presence of a timelike Killing vector is the technical
> way of saying this.
You are bringing up more vocabularies to describe the same thing.
> What is truly amazing is that energy conservation works so well in
> Newtonian physics -- its pervasiveness there is what made people
> erroneously think that conservation of energy is a general "principle".
> Noether's theorem (1918) delineated the limits of such conservation "laws"
> by relating them to symmetries of the Lagrangian. At base this has to do
> with the fact that gravitation is so incredibly small, and the terms in
> the Lagrangian that generate violations of energy conservation are
> completely negligible in Newtonian situations.
You are wrong on this. The reason why NM demands the conservation of energy
to be universal is because the conservation of energy can only be observed
to be universally so. It is all in the math. How can you argue against the
math?
> [Similar remarks apply to momentum conservation.]
Amen!
Koobee Wublee
>
> This does not necessarily mean that. Simple counterexample: GR is
> invariant under a uniform change of scale, but needs no "Euclidean space
> with a separate time ledger".
>
>
> The theory, either GR or NM, cannot determine if the defined unit is
> changing.
Are you serious? Energy is speed squared. If you define a certain distance
as traveled by a speeding object in a certain time, you will find the
observed energy is so out-of-the-ordinary, and the expected distance of
travel according to this observed quantity of energy does not match your
definition of distance.
>>>Quantum mechanics is different, and the dynamics of the theory determines
>>>the sizes of electromagnetic bound states (e.g. atoms).
More exceotions in your [plural] world of what the concept of physics
represents?
>> I don't understand how you can know that atoms don't change
>> size over time, isn't the meter standard made out of real atoms?
Are you referring to the Mercury's orbital anomaly under the concept of GR
where it is fudged to fit the observation through manipulations on the
hopelessly unobservable accuracies of several abstract quantities (such as
the observed energy and the observed angular momentum) involved?
> All anybody knows is properties of quantities in theories. You are free to
> select whatever theory you wish (of course if you choose poorly you will
> not make good evaluations of the world around you).
So, Grossmann/Einstein/Hilbert presented GR where both Grossmann and Hilbert
walked away from afterwards.
> If you select as a theory the standard model, then the masses and coupling
> constants (etc.) must all be constant or the symmetries that determine the
> standard model won't be valid. The constancy of those constants implies
> the constancy of atomic sizes.
Yes, indeed. Remember the speed of light squared!
> I make no attempt to explain the absurdities posted around here.
Me, too.
> What bothers me is theories that do not agree with experiment, and people
> who advocate them in spite of that.
Yes, indeed. This also goes for a theory that is fudged to fit the
experimental result.
> The classical theory cannot determine any changes in its defined unit of
> length. I said nothing at all about some abstract "size" and whether or
> not that "size" might change -- what I said specifically applies to the
> UNIT OF LENGTH, and only to the UNIT OF LENGTH.
Even in classical theory, one should be able to tell if the speed of light
in vacuum has suddenly reduced by half! Hint: c^2.
Joe Fischer
>
>"Tom Roberts" <tjro...@lucent.com> wrote in message
>news:ugeef.113$4o7...@newssvr24.news.prodigy.net...
>>
>> This does not necessarily mean that. Simple counterexample: GR is
>> invariant under a uniform change of scale, but needs no "Euclidean space
>> with a separate time ledger".
>>
>> [...]
>>
>> The theory, either GR or NM, cannot determine if the defined unit is
>> changing.
>
>Are you serious? Energy is speed squared. If you define a certain distance
>as traveled by a speeding object in a certain time, you will find the
>observed energy is so out-of-the-ordinary, and the expected distance of
>travel according to this observed quantity of energy does not match your
>definition of distance.
I apologize for the subject confusion, this thread was originally
about the recession of the galaxies, and injected some statements and
questions about a model where there is rapid global expansion all the
way down to atoms or maybe even atomic particles.
It is not a delusion or an insanity, just a 60 year pasttime thinking
about gravity.
>>>>Quantum mechanics is different, and the dynamics of the theory determines
>>>>the sizes of electromagnetic bound states (e.g. atoms).
>
>More exceotions in your [plural] world of what the concept of physics
>represents?
>
>>insert Joe Fischer wrote:
>>> I don't understand how you can know that atoms don't change
>>> size over time, isn't the meter standard made out of real atoms?
>
>Are you referring to the Mercury's orbital anomaly under the concept of GR
>where it is fudged to fit the observation through manipulations on the
>hopelessly unobservable accuracies of several abstract quantities (such as
>the observed energy and the observed angular momentum) involved?
Sorry somebody did not carry through with attributes, this question
was about a global general expansion where matter expands, and time
slows, so that c is always c, in fact many constants would arise in this
particular model.
>Even in classical theory, one should be able to tell if the speed of light
>in vacuum has suddenly reduced by half! Hint: c^2.
Sorry for the mixup, I will try to stay within the subject line topic.
Joe Fischer
brian a m stuckless
message > news:ugeef.113$4o7...@newssvr24.news.prodigy.net...
> > This does not necessarily mean that. Simple counterexample: GR is
> > invariant under a uniform change of scale, but needs no "Euclidean
> > space with a separate time ledger".
GR TEST-mass had NO size, NO shape, NO iNside, NO OUTside and NO mass.
GR TEST-mass had been HOPED to have mass, but G_uv can't RELATE to it.
> > [...]
> > The theory, either GR or NM, cannot determine if the defined unit
> > is changing.
>
> Are you serious? Energy is speed squared. --
..This is TOTAL dis-information THEORY.!!
> -- If you define a certain
> distance as traveled by a speeding object in a certain time, you
> will find the observed energy is so out-of-the-ordinary, and the
> expected distance of travel according to this observed quantity of
> energy does not match your definition of distance.
The increased iNPUT *REQUiRED* is due to neglected AMBiENT media.!!
( ..NOT due to any *ACTUAL* or iNTRiNSiC *particle* ENERGY ..duh. )
Which, AMOUNTs to DELiBERATE *massive* Ph.Tivity dis-iNFORMATiON.!!
> >>>Quantum mechanics is different, and the dynamics of the theory
> >>>determines the sizes of electromagnetic bound states (e.g. atoms).
>
> More exceotions in your [plural] world of what the concept of
> physics represents?
>
> >> I don't understand how you can know that atoms don't
> >> change size over time,
Avagadro's NUMBER = Ra*k = no*vm*k
..where vm is the STP&g PARTiCLE volume.
..where no is the particle NUMBER per cubic meter.
..where Ra is the Molar ENERGY ..per Mol ..per degree Kelvin.
..where k is Boltzmann's ENERGY ..per (mol part) ..per degree Kelvin.
> >> isn't the meter standard made out of real atoms?
>
> Are you referring to the Mercury's orbital anomaly under the
> concept of GR where it is fudged to fit the observation through
> manipulations --
Mercury's GR TEST-mass does NOT attract the Sun (..no G-field).!!
Note, this, is the ONLY reason WHY GR TEST-mass HAS no G-field.!!
Note, this, is the ONLY historical GR TEST-mass FOR no G-field.!!
Note that the Sun, for the Mercury GR TEST, DiD HAVE a G-field.!!
No CONSiDERATiON was given to kinematic-viscosity-LiKE effects.!!
brian a m stuckless
>><> >><> >><> >><> >><>
> -- on the hopelessly unobservable accuracies of several abstract
> quantities (such as the observed energy and the observed angular
> momentum) involved?
>
> > All anybody knows is properties of quantities in theories. You
> > are free to select whatever theory you wish (of course if you
> > choose poorly you will not make good evaluations of the world
> > around you).
>
> So, Grossmann/Einstein/Hilbert presented GR where both Grossmann
> and Hilbert walked away from afterwards.
>
> > If you select as a theory the standard model, then the masses
> > and coupling constants (etc.) must all be constant or the
> > symmetries that determine the standard model won't be valid.
> > The constancy of those constants implies the constancy of atomic
> > sizes.
Atomic SiZEs are NOT constant, but VARY with the LOCAL STP&g.!!
> Yes, indeed. Remember the speed of light squared!
Yes ..c^2 = 1 / Uo*Eo.!! Go go Google GROUP SEARCH < BiGGER.wpd >.
> > I make no attempt to explain the absurdities posted around here.
Diligently "..no attempt".!!
> Me, too.
>
> > What bothers me is theories that do not agree with experiment,
> > and people who advocate them in spite of that.
>
> Yes, indeed. This also goes for a theory that is fudged to fit
> the experimental result.
>
> > The classical theory cannot determine any changes in its defined
> > unit of length. I said nothing at all about some abstract "size"
> > and whether or not that "size" might change -- what I said --
GR TEST-mass had NO size, NO shape, NO iNside, NO OUTside and NO mass.
GR TEST-mass had been HOPED to have mass, but G_uv can't RELATE to it.
Hope this helps.
brian a m stuckless
>><> >><> >><> >><> >><>
> > specifically applies to the UNIT OF LENGTH, and only to the UNIT
> > OF LENGTH.
>
> Even in classical theory, one should be able to tell if the speed
> of light in vacuum has suddenly reduced by half! Hint: c^2.
TomGee
Joe Fischer wrote:
> On 16 Nov 2005 07:12:40 -0800, "TomGee" <lv...@hotmail.com> wrote:
>
> >Joe Fischer wrote:
> >> On 15 Nov 2005 08:14:01 -0800, "TomGee" <lv...@hotmail.com> wrote:
> >> >Joe, I have never taken the phrase, "the universe is expanding" to mean
> >> >anything more than that it's getting bigger.
> >>
> >> Ok, but that does not precisely describe what _MAY_ be happening,
> >> the original term was "recession of the galaxies.
> >>
> >That's correct. I agree.
>
> >>
> >> >However, almost everyone argued that we live inside the universe and
> >> >not on its surface. The rising raisin bread analogy was born to better
> >> >explain the expansion process. It was better than the balloon business
> >> >because it has an "inside" to it while the balloon only has a "surface"
> >> >to it.
> >>
> >> Actually neither is very good, bread would be like a medium,
> >> and that is not possible.
> >
> >But of course it is. Space is a medium for matter and if space is
> >filled with DM, then it is DM that is what is expanding and not space,
> >more likely.
>
> You have accepted the wrong definition of Dark Matter,
> which is only planets, dust and burned out or dim stars.
>
>
>
>
> It is possible that black holes are considered to be dark
> matter, I don't know.
>
>
>
>
> Are you thinking of Dark Energy/
>
>
model of Dark Matter and Dark Energy. Like with real matter (RM),
i.e., visible matter, neg. mass/energy go together. The distinction
between the two is that one has the property of gravitational
attraction and the other more likely has a repulsive property toward
RM.
>
>
> >> >But then it was noted that while galaxies are all moving away from each
> >> >other, their coordinates remain the same. I.e., the distance between
> >> >them grows but they remain in the same spacial relationship as before.
> >>
> >> That is a guess-timate, the size and distances are so large
> >> compared to the time it has been observed, and the only data
> >> available is probably spectroscopic with translation mapping and
> >> interpolation.
> >>
> >No, it has been shown to be so through observation. Another of
> >nature's grand paradoxes for our viewing pleasure....
>
> All objects in free space move in inertial motion, so the spacial
> relationhip would not change, and especially would not change in
> a mere 80 years of long tube spectrographs.
>
>
group were all at constant velocity wrt each other, but what is
observed is each galaxy moving away from the others. Put 4 marbles on
a carpet as a square having equal sides and then move each marble away
from the others an equal amount and note that the distances between
them have increased while their positions wrt to each other remain the
same as before.
>
>
> I think it takes 250 million years for the Milky Way to rotate
> once, and I don't think Pluto has completed one full orbit since
> being discoveref.
>
> >> >Now that cannot be explained by balloons or raising breads. It seems
> >> >the only explanation possible is that the galaxies are not moving apart
> >> >from each other due to their own motions, but due to the expansion of
> >> >space!
> >>
> >> Not hardly, that would make space a medium capable of
> >> carrying galaxies, I don't think anybody would realy think that.
> >
> >You seem to be giving a meaning to the term "medium" different than
> >commonly accepted. Galaxies do exist in space so space is indeed
> >capable of carrying galaxies.
>
> Rocket flight exists in air, but the air isn't needed,
> there is less drag in space and the engines work much better.
>
>
>
>
> >> They could be moving apart to nothing but thier own
> >> motion with a non-Euclidean component in the motion.
>
> >That's what was thought af first, but if it was due to their own
> >motion, their coordinates would change.
>
> I don't think so, the idea that anything is capable of altering
> the path of a single star is absurd, let alone a galaxy.
>
>
they alter each other's paths according to the laws of physics, right?
That is not absurd; it's an accepted observation.
>
>
> >> >So, to talk about the expansion of matter in regard to the expansion of
> >> >the universe is passe, to say the least.
> >>
> >> Not really, there is much more that needs to be explained and
> >> fitted to a pattern besides spectroscopic data.
> >>
> >> > It is not matter that is
> >> >expanding, but the distance between it.
> >>
> >> That is not a certainty.
>
>
>
>
> The inportant thing is to consider
> >> something that could have caused all the matter in the universe
> >> to be thrown apart, and expanding matter is as good as any
> >> cause of that.
>
> >I disagree. The idea of expanding matter is probably the worst idea to
> >date about the expansion process of the universe.
>
> I didn't say there was any connection, if gravity is connected
> with any expansion of matter, it would be much faster than the
> observed recession of the galaxies.
>
> >The BBT is a better
> >idea because it provides the reason why everything is in motion and we
> >cannot find a single stationary place in it.
>
> An expansion of matter seems to me to be more compatible
> with an expansion of the universe than static matter.
>
>
>
>
> There are lots of things to consider, like the temperature
> of the upper atmosphere being extremely hot, and an expansion
> of matter might fit a pattern of observations better than static
> matter.
>
> >> Chances are all the different modern concepts of a
> >> cosmological constant will be wrong, the spherical geometry
> >> of expansion of any kind may not follow Newton or Euclid.
>
> >Again, I disagree. Chances are that AE's constant will come to have
> >some validity after all, even though it won't overthrow the expansion
> >process observations.
>
> Not from galaxies speeding up or slowing down, but there
> could be an observational effect that might make something
> other than inertial motion appear.
>
>
what we see now.
>
>
> >> >For something to do something in our universe requires energy of some
> >> >sort, but if space is empty, and unless we overthrow the Principle of
> >> >the conservation of energy and E=mc^2, it is not possible for space to
> >> >expand.
> >>
> >> I thought you said it is expanding?
> >>
> >I said that is the common idea.
>
> There are a lot of ideas, many not very logical.
>
> Nature is logical when all the facts are known.
>
> >> Not hardly, there is no action at a distance without a mechanism.
>
> >Below, I give the impetus of the BB as the mechanism for DM motions.
>
> I don't know anything about Dark Matter moving.
>
> >> >It's my guess that it is Dark Matter that is expanding due to the
> >> >impetus of the BB, and that as the impetus wears off some of it becomes
> >> >stationary in space.
> >>
> >> Dark Matter is nothing but matter that is too cool to glow,
> >> and that is a lot different than Dark Energy.
>
>
>
>
> >There is no reason to think that because matter too cool to glow should
> >still reflect light for us to see it.
>
> Then you don't realize how difficult it is to see even small bright
> stars, may I suggest a search of google for "intrinsic brightness",
> and "visual magnitude" etc.
>
>
sensitive only to positive matter because all visible matter has
temperature and if we are close enough to it we can see it. We cannot
see the furniture in a closed room from the outside of it either
because our vision is blocked. But from inside the room we can. DM is
matter that exists but is invisible to our eyes because it has no
positive mass/energy and thus it exists without any temperature.
Tom Roberts
> Speaking of uniformity of language, the following equation should be
> universally accepted by all. It describes a segment of observed (thus
> observable and subject to be observed with curvature) spacetime.
> (ds)^2 = g^ij dx_i dx_j
It's more naturally written:
(ds)^2 = g_ij dx^i dx^j
because thise are the more natural definitions of the metric components
{g_ij} and the tangent vector components {dx^i}.
> What is the Lagrangian for this extremely general case?
> From the Lagrangian Method of minimizing an action where the action is the
> time elapsed where a particle moves in a distance as described by the above
> generic equation, the Lagrangian can be derived by dividing the equation
> with (g^00 dx_0 dx_0) which yields as follows.
> Lagrangian = (ds/dx_0)^2 / g^00 + g^ij (dx_i/dx_0) (dx_j/dx_0) / g^00 = 1
NO! NO! NO!
That is an attempt to minimize wrt COORDINATE TIME, and it happens to be
INCORRECT (x_0 is your time coordinate, not "g^00 dx_0 dx_0"). Note also
that this is supposed to be a general argument, and for badly chosen
coordinates such attempts to use coordinate time can yield nonsense.
Relativity is a theory of invariants, and you started with an invariant
above, and you should stick with invariants. In this case, the correct
time coordinate to use is clearly the object's own proper time, which
I'll label \tau. (ds)^2 above is just the (timelike) path length along
the trajectory of the object, and leads directly to a suitable Lagrangian:
L = \integral g_ij (dx^i/d\tau) (dx^j/d\tau) d\tau
Where the {x^i} are considered functions of the proper time \tau (they
define the object's trajectory), and the {g_ij} are considered functions
of the {x^i(\tau)} (the metric is a function on the manifold). The
integration is done between specified points on the trajectory (values
of \tau).
When you perform the usual variational calculus on this Lagrangian, you
obtain the usual geodesic equation for the metric g.
Another suitable Lagrangian uses sqrt(|ds)^2|). The result of
the variation is the same geodesic equation, but the
parameterization of the path turns out to be different.
> The rest is just purely mathematical methods. The partial derivative of the
> Lagrangian as defined above is always zero with respect to ds. Therefore,
> according to Noether's Theorem, there is always a conserved quantity. This
> quantity is none other than the energy.
ds is not part of the parameterization of your (incorrect) Lagrangian,
and Noether's theorem does not apply to it. The "partial derivative with
respect to ds" makes no sense -- ds does not appear on the RHS of your
Lagrangian equation above (and it should be s, not ds).
My L (defined above) is invariant over translations in \tau, and the
Noether current corresponding to this symmetry is indeed the energy of
the particle.
Exercise for the reader: As we know, energy is coordinate
dependent; in what coordinates is that energy expressed?
What is the common name for this energy? Hint: for which
locally-inertial frame is \tau the time coordinate?
> Conservation of energy(an observed
> quantity) is ubiquitously a universal phenomenon even in GR
Not true. For instance, the L defined above is valied ONLY for a "test
particle", meaning its influence on the rest of the world is ignored.
But if one takes that into account, energy is not conserved, because
miniscule gravitational waves will in general be emitted, and they will
destroy the \tau translation symmetry.
Exercise for the reader: relate this lack of symmetry
to the previous exercise. What is wrong? Hint: I alluded to
the error in the previous paragraph. (this probably makes
no sense until you know the correct answer for the first
exercise.)
> the Lagrangian is a very abstract mathematical entity [...] Its concept
> is so abstract that it cannot be safely regarded as a candidate to describe
> all interactions of a given system.
Nonsense. Every fundamental theory of physics we have today is derived
from a Lagrangian.
>>[#] The presence of a timelike Killing vector is the technical
>> way of saying this.
> You are bringing up more vocabularies to describe the same thing.
Yes. That's what I said. <shrug>
>>What is truly amazing is that energy conservation works so well in
>>Newtonian physics -- its pervasiveness there is what made people
>>erroneously think that conservation of energy is a general "principle".
>>Noether's theorem (1918) delineated the limits of such conservation "laws"
>>by relating them to symmetries of the Lagrangian. At base this has to do
>>with the fact that gravitation is so incredibly small, and the terms in
>>the Lagrangian that generate violations of energy conservation are
>>completely negligible in Newtonian situations.
>
> You are wrong on this. The reason why NM demands the conservation of energy
> to be universal is because the conservation of energy can only be observed
> to be universally so. It is all in the math. How can you argue against the
> math?
Clearly you do not understand the math. I am not "arguing against the
math", I am describing what the math actually says: energy is conserved
only for systems having a Lagrangian with a time translation symmetry.
_THAT_ is what Noether's theorem says.
You claim "the conservation of energy to be universal". If so, how is it
that it is clearly NOT conserved for a subsystem interacting with
another subsystem? -- Not very "universal", is it? Especially since the
largest "system" we humans could construct is clearly a tiny SUBsystem
of the universe.
Noether's theorem explains this: in NM the total system Lagrangian has
time translation symmetry, but for interacting subsystems, the
Lagrangians for the individual subsystems do not have time translation
symmetry.
Before Noether's theorem, the non-conservation of energy
for subsystems was a puzzle. Ad hoc rules had to be created
to avoid embarrassment, but no underlying explanation for
them had been given until Noether. Most authors simply
ignored this puzzle and treated those rules as "self
evident".
Tom Roberts tjro...@lucent.com
Tom Roberts
Tom Roberts wrote:
> My L (defined above) is invariant over translations in \tau, and the
> Noether current corresponding to this symmetry is indeed the energy of
> the particle.
That's only true for suitable metrics (i.e. ones with a timelike Killing
vector; see other recent posts in this newsgroup).
Tom Roberts tjro...@lucent.com
TomGee
separate concepts of mass and energy. Whatever it is you're going on
about is not relevant to this discussion unless you can explain the
relationship to it.
Tom Roberts
> Tom Roberts wrote:
>>>I left out a term of the principle I referred to. It is
>>>the Principle of Conservation of Mass and Energy that supports the
>>>inferred equivalence of mass and energy shown in E=mc^2. This
>>>principle unifies mass and energy.
>>
>>What you call it does not matter. It still does not hold in the FRW
>>manifolds, nor, in general, in GR.
>
> Oh, it matters, alright.
Not for the question of whether or not this "conservation principle" is
valid.
> There's a biiigg difference in the two
> separate concepts of mass and energy.
Yes. Mass is in general not conserved, but energy is conserved IN
SYSTEMS WITH A LAGRANGIAN THAT IS INVARIANT OVER TIME TRANSLATIONS.
NOTE: when I discuss the mass of a system, I mean the sum of the masses
of its components. In the context of "conservation of mass" that is what
was meant in Newtonian mechanics, and is the relevant meaning here.
Mass is often approximately conserved to high accuracy in low-energy
systems (e.g. our everyday lives; Newtonian mechanics). Here
"low-energy" means per elementary particle.
Before the advent of multi-MeV particle accelerators and mass
spectrometers (before ~1930), one might have thought that mass was
conserved. Today we know for sure it is not.
> Whatever it is you're going on
> about is not relevant to this discussion unless you can explain the
> relationship to it.
Plain and simple, mass is not conserved (again, by "mass of a system" I
mean the sum of the masses of all components of a system). So, for
instance, an electron plus a positron total 1.022 MeV/c^2 of mass, but
when they annihilate they yield two gamma rays each with zero mass. In
principle, even things like light bulbs and running horses do not
conserve their masses, but the violations are too small to measure in
practice.
Plain and simple, energy is not conserved, except in systems with a
Lagrangian that is invariant over time translations.
Tom Roberts tjro...@Lucent.com
TomGee
Tom Roberts wrote:
> TomGee wrote:
> > Tom Roberts wrote:
> >>TomGee wrote:
> >>>I left out a term of the principle I referred to. It is
> >>>the Principle of Conservation of Mass and Energy that supports the
> >>>inferred equivalence of mass and energy shown in E=mc^2. This
> >>>principle unifies mass and energy.
> >>
> >>What you call it does not matter. It still does not hold in the FRW
> >>manifolds, nor, in general, in GR.
> >
> > Oh, it matters, alright.
>
> Not for the question of whether or not this "conservation principle" is
> valid.
>
>
energy do not disappear from existence, only from our eyes.
>
>
> > There's a biiigg difference in the two
> > separate concepts of mass and energy.
>
> Yes. Mass is in general not conserved, but energy is conserved IN
> SYSTEMS WITH A LAGRANGIAN THAT IS INVARIANT OVER TIME TRANSLATIONS.
>
>
talking about empirical evidence, not math constructs. We see some
effects that are attributed to DM as an explanation, a theory, an idea.
Your objection is based on imaginary math constructs and specifically
on one system type that is at best a rule of such math constructs.
>
>
> NOTE: when I discuss the mass of a system, I mean the sum of the masses
> of its components. In the context of "conservation of mass" that is what
> was meant in Newtonian mechanics, and is the relevant meaning here.
>
> Mass is often approximately conserved to high accuracy in low-energy
> systems (e.g. our everyday lives; Newtonian mechanics). Here
> "low-energy" means per elementary particle.
>
> Before the advent of multi-MeV particle accelerators and mass
> spectrometers (before ~1930), one might have thought that mass was
> conserved. Today we know for sure it is not.
>
>
> > Whatever it is you're going on
> > about is not relevant to this discussion unless you can explain the
> > relationship to it.
>
> Plain and simple, mass is not conserved (again, by "mass of a system" I
> mean the sum of the masses of all components of a system). So, for
> instance, an electron plus a positron total 1.022 MeV/c^2 of mass, but
> when they annihilate they yield two gamma rays each with zero mass. In
> principle, even things like light bulbs and running horses do not
> conserve their masses, but the violations are too small to measure in
> practice.
>
>
visible particles disappear leaving tracks in their wake, but they only
disappear from our view as they become negative mass and energy, in my
opinion.
>
>
> Plain and simple, energy is not conserved, except in systems with a
> Lagrangian that is invariant over time translations.
>
>
Energy has been overthrown not just in specific and rare situations but
as a general rule. Perhaps if you explained why you believe energy is
conserved only in systems having points of equilibrium of its bodies
that never deteriorate, it may help to explain why you think it is
germane to this discussion.
PD
TomGee wrote:
> That would be true if the motions of each galaxy within the observed
> group were all at constant velocity wrt each other, but what is
> observed is each galaxy moving away from the others. Put 4 marbles on
> a carpet as a square having equal sides and then move each marble away
> from the others an equal amount and note that the distances between
> them have increased while their positions wrt to each other remain the
> same as before.
An interestingly unconventional use of the physics term "position". I
wonder what other physics terms TomGee has interestingly unconventional
uses for.
PD
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:ugeef.113$4o7...@newssvr24.news.prodigy.net...
>>changing.
>
> Are you serious?
Yes.
> Energy is speed squared.
Not in SR/GR. This is _NOT_ Newtonian mechanics. But no matter....
> If you define a certain distance
> as traveled by a speeding object in a certain time, you will find the
> observed energy is so out-of-the-ordinary, and the expected distance of
> travel according to this observed quantity of energy does not match your
> definition of distance.
Not true. You would always measure the "expected" values, because
_everything_ available to you would change with any underlying variation
in the _defined_ unit of length.
This is a simple consequence of the fact that the equations of classical
mechanics are invariant under a scale change. At base, that is why we
can define so many different systems of units.
Example: the Maxwell's equations are invariant under a
15-parameter group sometimes called the conformal group
(not to be confused with a geometrical group with the
same name). A subgroup of it is the Poincare' group,
which is projected out of the complete invariance group
if you require that the length and time scales remain
unchanged.
>>The classical theory cannot determine any changes in its defined unit of
>>length. I said nothing at all about some abstract "size" and whether or
>>not that "size" might change -- what I said specifically applies to the
>>UNIT OF LENGTH, and only to the UNIT OF LENGTH.
>
> Even in classical theory, one should be able to tell if the speed of light
> in vacuum has suddenly reduced by half! Hint: c^2.
I was using geometrical units, in which c=1. So the length scale is
identical to the time scale, and a change in either is a change in both.
Tom Roberts tjro...@lucent.com
Androcles
> I was using geometrical units, in which c=1. So the length scale is
> identical to the time scale, and a change in either is a change in both.
>
c = 0, Roberts.
(AB +BA)/(t'A-tA) = c = 0, Roberts. There is no 2AB
when light goes from A to B and back to A, Roberts.
I said that in 1999, Roberts.
I am continually unamazed at how some people are incompetent in everything
they attempt. Tom Roberts seems to get something important wrong in
everything he writes, regardless of subject. <shrug>
Androcles.
Koobee Wublee
>> What is the Lagrangian for this extremely general case?
>> From the Lagrangian Method of minimizing an action where the action is
>> the time elapsed where a particle moves in a distance as described by the
>> above generic equation, the Lagrangian can be derived by dividing the
>> equation with (g^00 dx_0 dx_0) which yields as follows.
>> Lagrangian = (ds/dx_0)^2 / g^00 + g^ij (dx_i/dx_0) (dx_j/dx_0) / g^00 = 1
>
> NO! NO! NO!
>
> That is an attempt to minimize wrt COORDINATE TIME, and it happens to be
> INCORRECT (x_0 is your time coordinate, not "g^00 dx_0 dx_0"). Note also
> that this is supposed to be a general argument, and for badly chosen
> coordinates such attempts to use coordinate time can yield nonsense.
>
> Relativity is a theory of invariants, and you started with an invariant
> above, and you should stick with invariants. In this case, the correct
> time coordinate to use is clearly the object's own proper time, which I'll
> label \tau. (ds)^2 above is just the (timelike) path length along the
> trajectory of the object, and leads directly to a suitable Lagrangian:
>
> L = \integral g_ij (dx^i/d\tau) (dx^j/d\tau) d\tau
I am afraid your Lagrangian does not satisfy the condition that leads to the
Lagrangian Method. Has it been that long since you have learned it which I
assume you must have. So, you need to review that subject. For a quick
review,
Action = integral of Lagrangian
Where the Lagrangian is a density to this action and in this case the action
is the elapsed time whether it is the observed time or the proper time. You
can write an equation for the Lagrangian that results in an action of the
elapsed proper time and also another equation for the Lagrangian that
results in an action of the elapsed observed time. Other than these two
Lagrangians, we can also write down another equation for the Lagrangian that
results in an action of the elapsed segment of proper spacetime. That last
Lagrangian is not surprising to exist because the motion of an observed
particle is not subject to the minimum action of space itself but only
either the proper time or the observed time. If you have done your
dilligence, you will find that all these three Lagrangians give the same
geodesic (trying to speak your language) equations. So, the Lagrangians are
** Based on the observed time, dx^0
sqrt[(ds/dx^0)^2 / g_00 + |g_ij| (dx^i/dx^0) (dx^j/dx^0) / g_00]
i, j = 0, 1, 2, 3
i + j != 0
** Based on the proper time, dT, where ds = c dT
sqrt[g_ij (dx^i/dT) (dx^j/dT)] / c
** Based on the proper spacetime, ds
sqrt[g_ij (dx^i/ds) (dx^j/ds)]
Which is only valid for non-photons
ds for photons = 0
You can also remove the square root because all these Lagrangians equal to
1. They are very simply derived from
(ds)^2 = g_ij dx^i dx^j
> [... snipped due to faulty memory of what Lagrangian is]
> Exercise for the reader: relate this lack of symmetry
> to the previous exercise. What is wrong? Hint: I alluded to
> the error in the previous paragraph. (this probably makes
> no sense until you know the correct answer for the first
> exercise.)
And the action item for Dr. Roberts is to review the Lagrangian Method. So,
we can discuss logically and clearly how Noether's Theorem applies to
indicate a universal phenomenon to the conservation of observed energy.
>> the Lagrangian is a very abstract mathematical entity [...] Its concept
>> is so abstract that it cannot be safely regarded as a candidate to
>> describe all interactions of a given system.
>
> Nonsense. Every fundamental theory of physics we have today is derived
> from a Lagrangian.
If you find an equation that suit your line of thought without justifying
its validity does fall in the realm of fundamental theory of physics. In
this case, you can pull out an equation out of I-don't-want-to-know-where
and call it the Lagrangian is totally bogus. <shrug>
>> You are wrong on this. The reason why NM demands the conservation of
>> energy to be universal is because the conservation of energy can only be
>> observed to be universally so. It is all in the math. How can you argue
>> against the math?
>
> Clearly you do not understand the math. I am not "arguing against the
> math", I am describing what the math actually says: energy is conserved
> only for systems having a Lagrangian with a time translation symmetry.
> _THAT_ is what Noether's theorem says.
You have only discover one Lagrangian. There are more.
> You claim "the conservation of energy to be universal". If so, how is it
> that it is clearly NOT conserved for a subsystem interacting with another
> subsystem? -- Not very "universal", is it? Especially since the largest
> "system" we humans could construct is clearly a tiny SUBsystem of the
> universe.
It is very surprising for an experimental physicist to ask me this question.
Let me use the example I gave in the past. Say you want to measure and thus
observe the speed of a car driving by. You can initially utilize your nose
to do so. For most of us, we detect coherently a Doppler effect in odor, we
have to conclude the observation as a failure and move on to incorporate
more sophisticated observational methodologies. Clearly, any of the
Lagrangians I described above does not indicate a universal conservation of
angular momentum this should be a start to find out how the conservation of
energy seems to fail in a binary system instead of creating gravitons. YOU
ARE NO GOD!
> Noether's theorem explains this: in NM the total system Lagrangian has
> time translation symmetry, but for interacting subsystems, the Lagrangians
> for the individual subsystems do not have time translation symmetry.
Again, all three Langrangians I described above converge on the same
geodesic equations. This means you are wrong in your statement above.
> Before Noether's theorem, the non-conservation of energy
> for subsystems was a puzzle. Ad hoc rules had to be created
> to avoid embarrassment, but no underlying explanation for
> them had been given until Noether. Most authors simply
> ignored this puzzle and treated those rules as "self
> evident".
You seem not to be ashamed by pulling out a Lagrangian in a ad hoc matter.
On the contrary to your statements above, Noether's Theorem strengthens the
principle where the conservation of observed energy is universally
fundamental.
> My L (defined above) is invariant over translations in \tau, and the
> Noether current corresponding to this symmetry is indeed the energy of
> the particle.
>
> That's only true for suitable metrics (i.e. ones with a timelike Killing
> vector; see other recent posts in this newsgroup).
In this case, the Lagrangian is equal to 1 to satisfy the requirement
leading to the Lagrangian Method. Your instint is correct. You need to
trust yourself many years ago before you forget about the Lagrangian Method
due to the inevitable aging. You should not blame yourself for getting old.
If you don't, <shrug>
Koobee Wublee
>> Energy is speed squared.
>
> Not in SR/GR. This is _NOT_ Newtonian mechanics. But no matter....
So, you don't call the equation, E = m c^2, having a speed term squared. I
am afraid you are losing it.
> [... rest of nonsense snipped]
Dirk Van de moortel
"Koobee Wublee" <kub...@cox.net> wrote in message news:QZyff.1192$pF.18@fed1read04...
I am afraid you are behaving like a little kid again:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LosingIt.html
Dirk Vdm
Dirk Van de moortel
"Koobee Wublee" <kub...@cox.net> wrote in message news:VQyff.1191$pF.698@fed1read04...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:dlidra$c...@netnews.net.lucent.com...
>
> >> What is the Lagrangian for this extremely general case?
> >> From the Lagrangian Method of minimizing an action where the action is
> >> the time elapsed where a particle moves in a distance as described by the
> >> above generic equation, the Lagrangian can be derived by dividing the
> >> equation with (g^00 dx_0 dx_0) which yields as follows.
> >> Lagrangian = (ds/dx_0)^2 / g^00 + g^ij (dx_i/dx_0) (dx_j/dx_0) / g^00 = 1
> >
> > NO! NO! NO!
> >
> > That is an attempt to minimize wrt COORDINATE TIME, and it happens to be
> > INCORRECT (x_0 is your time coordinate, not "g^00 dx_0 dx_0"). Note also
> > that this is supposed to be a general argument, and for badly chosen
> > coordinates such attempts to use coordinate time can yield nonsense.
> >
> > Relativity is a theory of invariants, and you started with an invariant
> > above, and you should stick with invariants. In this case, the correct
> > time coordinate to use is clearly the object's own proper time, which I'll
> > label \tau. (ds)^2 above is just the (timelike) path length along the
> > trajectory of the object, and leads directly to a suitable Lagrangian:
> >
> > L = \integral g_ij (dx^i/d\tau) (dx^j/d\tau) d\tau
>
> I am afraid your Lagrangian does not satisfy the condition that leads to the
... original, but removed from the archives:
http://groups.google.co.uk/groups?selm=bdq09.28353$Fq6.2...@news2.west.cox.net
But we still have the reply:
http://groups.google.co.uk/groups?&threadm=V1r09.661180$352.138570@sccrnsc02
| "Scholarly Fungi" <scholar...@yahoo.com> wrote in message
| news:bdq09.28353$Fq6.2...@news2.west.cox.net...
| > It is also unfortunate that most of the folks blindly embracing this
| > holohaux come from the white supremacists. I don't see what this would gain
| > for them other than trying to antagonize the Jews. However, this is
| > history. When I was in my early high school years, I independently came up
| > with what Butz was saying without knowing his existence. Hey, I am very
| > proud of my humble analytical skills.
... or to the original, but removed from the archives:
http://groups.google.co.uk/groups?&threadm=NnI09.31007$Fq6.3...@news2.west.cox.net
But we still have the reply:
http://groups.google.co.uk/groups?&threadm=a1777f85.0207...@posting.google.com
| "Scholarly Fungi" <scholar...@yahoo.com> wrote in message
| > news:<NnI09.31007$Fq6.3...@news2.west.cox.net>...
| > All history is written upon congruency among the historians but except one.
| > The Holocaust was born in the court rooms of Nueremberg. It is a complete
| > hoax.
| >
| > I did not know of Arthur Butz, but I independently came up with that
| > hypothesis noticing the tremendous amount of inconsistencies while studying
| > holohoax in high school.
... or to the original, but removed from the archives:
http://groups.google.co.uk/groups?&threadm=fytJa.78487%24%2542.6441%40fed1read06
But we still have the reply:
http://groups.google.co.uk/groups?&threadm=XvKJa.100908$hd6.25327@fed1read05
| "Australopithecus Afarensis" <lu...@olduvaigorge.net> wrote in message
| news:fytJa.78487$%42.6441@fed1read06...
| > Thanks for posting all that and your own comments at the end. There are so
| > many lies after lies conjured up against the Nazis. I guess I'd better read
| > "Mein Kampf" to get it from the horse's mouth. It will be on my
| > things-to-do list for the near future.
... or to the original, but removed from the archives:
http://groups.google.co.uk/groups?&threadm=uMeDa.59118%24%2542.39687%40fed1read06
Reply:
http://groups.google.co.uk/groups?&threadm=v3jsdv43ch9d4nnd1...@4ax.com
| On Tue, 3 Jun 2003 21:42:04 -0700, "Australopithecus Afarensis"
| <lu...@olduvaigorge.net> wrote:
|
| >Thanks for answering these questions fair and square.
| >
| >Although I don't speak for all other Australopithecine, I certainly want to
| >be as less nationalistic as possible. I am an individual just trying to
| >learn as much as I can before my short life expires on this earth.
| >
| >OK, now the media and "media"-controlled educational history have painted
| >the Nazis as the most fiendish group of people ever lived through out the
| >entire history of mankind. When I was growing up, I was constantly reminded
| >that the Nazis were so genocidal, they will kill any non-Germans in a heart
| >beat. After getting constantly bombarded with Nazi atrocities, I was very
| >much like the rest. Well, until one clip of film showing mountains of hair
| >inside a giant oven, the purpose was to show how many people murdered and
| >cremated. As a young scientist-to-be, it just hit me that the whole sh*t
| >was a lie. As far as I knew, the human hair would burn first. After
| >meticulous research and reasoning, I have concluded the WWII Nazis were no
| >more atrocious than any other governments in the 20th century or beyond.
| >Many of these information mostly came out after the explosion of the
| >internet where all skeletons in the closets finally have a chance to tell
| >their side of the story. Now, what is your plan to the public to shed these
| >negative sentiments accused against your political group?
| >
Bilge
>"Tom Roberts" <tjro...@lucent.com> wrote in message
>news:dlidra$c...@netnews.net.lucent.com...
>
>>> What is the Lagrangian for this extremely general case?
>>> From the Lagrangian Method of minimizing an action where the action is
>>> the time elapsed where a particle moves in a distance as described by the
>>> above generic equation, the Lagrangian can be derived by dividing the
>>> equation with (g^00 dx_0 dx_0) which yields as follows.
>>> Lagrangian = (ds/dx_0)^2 / g^00 + g^ij (dx_i/dx_0) (dx_j/dx_0) / g^00 = 1
>>
>> NO! NO! NO!
>>
>> That is an attempt to minimize wrt COORDINATE TIME, and it happens to be
>> INCORRECT (x_0 is your time coordinate, not "g^00 dx_0 dx_0"). Note also
>> that this is supposed to be a general argument, and for badly chosen
>> coordinates such attempts to use coordinate time can yield nonsense.
>>
>> Relativity is a theory of invariants, and you started with an invariant
>> above, and you should stick with invariants. In this case, the correct
>> time coordinate to use is clearly the object's own proper time, which I'll
>> label \tau. (ds)^2 above is just the (timelike) path length along the
>> trajectory of the object, and leads directly to a suitable Lagrangian:
>>
>> L = \integral g_ij (dx^i/d\tau) (dx^j/d\tau) d\tau
>
>I am afraid your Lagrangian does not satisfy the condition that leads to the
>Lagrangian Method.
I'm afraid you have no idea what you are talking about. The idea is
have a lagrangian which has no explicit coordinate dependence, since
the lagrangian is supposed to be a scalar. For example, the solution
to your perennial objections to the precession of mercuries lies in
the lagrangian,
L = (m/4)(s_dot)^2
where m is the mass of mercury and s_dot is obtained from the line
element in the schwarzchild geometry,
s^2 = (1-2M/r)dt^2 - dr^2/(1-2M/r) - r^2 d\Omega^2.
by taking the derivative with respect to the proper time, d\tau. That
works for _any_ choice of coordinates you want to use in writing the
line element. The only reason for choosing (t, r, \theta, \phi) is because
the symmetry is evident in terms which are useful from the perspective of
measuring numbers here on earth and the coordinates are a good description
of the geometry for this application.
>Has it been that long since you have learned it which I assume you
>must have. So, you need to review that subject. For a quick review,
>
>Action = integral of Lagrangian
>
>Where the Lagrangian is a density to this action and in this case the
Where the lagrangian density is _functional_ (not a function).
For example, the lagrangian I gave above, (m/4)(s_dot)^2, s_dot
can be a function of any coordinates you want to use. A simple
choice only has the advantage of reducing the amount of effort
involved in extracting a useful result.
>action is the elapsed time whether it is the observed time or the
>proper time.
If you want to make life hard, you are free to do so. Since it
can't matter how you obtain the answer, the only freedom you are
exercising is the freedom to eschew physical intuition for sake
of additional work and an increase in the number of opportunities
to go wrong. Most physicists call that a lose-lose proposition,
and prefer to use physical intuition to build in as much symmetry
as possible right off the bat.
>You can write an equation for the Lagrangian that results in an
>action of the elapsed proper time and also another equation for
>the Lagrangian that results in an action of the elapsed observed time.
You can also solve for the field of a point charge in terms of
elliptical cylinder functions, but the question is, why? Doing so
only shows contempt for the physics with the reward of a solution
that is hard to obtain and illustrates a poor choice of coordinates
rather than any physics that would otherwise be apparent by thinking
before calculating.
>** Based on the observed time, dx^0
To use a familiar phrase, that is not even wrong.
[...]
>> Clearly you do not understand the math. I am not "arguing against the
>> math", I am describing what the math actually says: energy is conserved
>> only for systems having a Lagrangian with a time translation symmetry.
>> _THAT_ is what Noether's theorem says.
>
>You have only discover one Lagrangian. There are more.
Why does it matter? Any lagrangian will work. The only difference
is the amount of effort required to eliminate the irrelevant consequences
of not choosing the one which most closely resembles the physics, in
advance.
>> You claim "the conservation of energy to be universal". If so, how is it
>> that it is clearly NOT conserved for a subsystem interacting with another
>> subsystem? -- Not very "universal", is it? Especially since the largest
>> "system" we humans could construct is clearly a tiny SUBsystem of the
>> universe.
>
>It is very surprising for an experimental physicist to ask me this question.
Why is that surprising? The job description of physicist includes
a desire to find out what nature does, not make nature conform to
the ANSI standard for a universe. What is surprising is that you
claim to know a lot about lagrangian formalism, yet fail to understand
the entire connection between symmetry and conservation laws (apart
from repeating the soundbites occasionally). It should be obvious
that a lagrangian which depends explictly on the time, cannot conserve
energy.
[...]
>more sophisticated observational methodologies. Clearly, any of the
>Lagrangians I described above does not indicate a universal conservation of
>angular momentum this should be a start to find out how the conservation of
Clearly, you are completely confused. It is unnecessary to know the
details of any particular lagrangian to determine a general conservation
law. All you need to know is whether it's possible to eliminate the
dependence of the lagrangian on a coordinate. So, if you are looking
for a conserved angular momentum, the fisrt order of business is to
eliminate from the lagrangian, any explicit dependence on the angular
coordinates. If the lagrangian has spherical symmetry, then obviously,
a dependence on the angular coordinates is inherent in the choice of
coordinates, not the actual physics.
If additional effort is your thing, why not choose geocentric
coordinates to describe the planetary orbits in the solar system,
write down the lagrangian and try to prove angular momentum
is conserved by the epicycles and retrograde motion in those
coordinates?
>> Noether's theorem explains this: in NM the total system Lagrangian has
>> time translation symmetry, but for interacting subsystems, the Lagrangians
>> for the individual subsystems do not have time translation symmetry.
>
>Again, all three Langrangians I described above converge on the same
>geodesic equations. This means you are wrong in your statement above.
>
>> Before Noether's theorem, the non-conservation of energy
>> for subsystems was a puzzle. Ad hoc rules had to be created
>> to avoid embarrassment, but no underlying explanation for
>> them had been given until Noether. Most authors simply
>> ignored this puzzle and treated those rules as "self
>> evident".
>
>You seem not to be ashamed by pulling out a Lagrangian in a ad hoc matter.
Since there is no procedure for constructing a lagrangian, the only
means of obtaining one is to use physical intuition to make an intelligent
guess. If such a procedure existed, we would have a theory of everything.
The utility of the lagrangian formalism lies in what one can say about
the physics for an any lagrangian based only on the properties of the
lagrangian. The utility of noether's theorems is that one can obtain
a lot of physics based on eliminating the dependence of lagrangain
on some quantity with a symmtery argument.
>On the contrary to your statements above, Noether's Theorem strengthens
>the principle where the conservation of observed energy is universally
>fundamental.
Conservation of energy is not a principle. It's a conservation
law which hinges on the lagrangian not having any explicit time
dependence. In general relativity, that means being able to
eliminate any explicit time dependence from the metric.
>> My L (defined above) is invariant over translations in \tau, and the
>> Noether current corresponding to this symmetry is indeed the energy of
>> the particle.
>>
>> That's only true for suitable metrics (i.e. ones with a timelike Killing
>> vector; see other recent posts in this newsgroup).
>
>In this case, the Lagrangian is equal to 1 to satisfy the requirement
>leading to the Lagrangian Method.
What the lagrangian is ``equal to'' is not only irrelevant, but
doesn't even make sense. What matters is that the variation
vanishes.
>Your instint is correct. You need to trust yourself many years ago
>before you forget about the Lagrangian Method due to the inevitable
>aging. You should not blame yourself for getting old. If you don't,
><shrug>
You ought to start with old fashioned classical mechanics to gain
some understanding of the lagrangian before pontificating about
how it's used in more complex theories.
Koobee Wublee
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
news:slrndo32jc....@radioactivex.lebesque-al.net...
> I'm afraid you have no idea what you are talking about. The idea is
> have a lagrangian which has no explicit coordinate dependence, since
> the lagrangian is supposed to be a scalar. For example, the solution
> to your perennial objections to the precession of mercuries lies in
> the lagrangian,
>
> L = (m/4)(s_dot)^2
>
> where m is the mass of mercury and s_dot is obtained from the line
> element in the schwarzchild geometry,
>
> s^2 = (1-2M/r)dt^2 - dr^2/(1-2M/r) - r^2 d\Omega^2.
>
> by taking the derivative with respect to the proper time, d\tau. That
> works for _any_ choice of coordinates you want to use in writing the
> line element. The only reason for choosing (t, r, \theta, \phi) is because
> the symmetry is evident in terms which are useful from the perspective of
> measuring numbers here on earth and the coordinates are a good description
> of the geometry for this application.
The Lagrangian I have described all are scalar in total aggreement with you.
Furthermore, all the Lagrangians I have described being a constant also
satisfy the conditions imposed by the Lagrangian Method that qualify them to
be Lagrangians. One of the Lagrangians I described is just a constant times
the Lagrangian you have above. Recall that I had (reference provided at the
end of this post)
> ** Based on the proper time, dT, where ds = c dT
>
> sqrt[g_ij (dx^i/dT) (dx^j/dT)] / c
This is only true if there is no proper spatial component. The reason why I
have a sqrt() that you don't is that the Langrangian is a constant. So, now
you have presented a special case of the Lagrangian where there is no proper
spatial component that is exactly what I have described. How can you accuse
me of bening wrong without implicating yourself?
Stop hiding under Dr. Robert's skirt and take cheap shots at scholars like
myself. This advice also goes for moortel and Hobba. The trio
Bilge-moortel-Hobba makes up a team of three stooges, each of whose only
existence seems to hide under a respected poster and takes cheap shots at
scholars who question the sanity of modern physics.
> >Has it been that long since you have learned it which I assume you
> >must have. So, you need to review that subject. For a quick review,
> >
> >Action = integral of Lagrangian
> >
> >Where the Lagrangian is a density to this action and in this case the
>
> Where the lagrangian density is _functional_ (not a function).
> For example, the lagrangian I gave above, (m/4)(s_dot)^2, s_dot
> can be a function of any coordinates you want to use. A simple
> choice only has the advantage of reducing the amount of effort
> involved in extracting a useful result.
Your Lagrangian is only a constant. I have not seen anyone so capable of
describing a constant like that. To some, it may seem indicate great
scholarship in you. However, to true scholars, you are just BSing.
> >action is the elapsed time whether it is the observed time or the
> >proper time.
>
> If you want to make life hard, you are free to do so. Since it
> can't matter how you obtain the answer, the only freedom you are
> exercising is the freedom to eschew physical intuition for sake
> of additional work and an increase in the number of opportunities
> to go wrong. Most physicists call that a lose-lose proposition,
> and prefer to use physical intuition to build in as much symmetry
> as possible right off the bat.
You are the one who try very hard to confuse anyone by calling a constant
_functional_ and not a function. The Lagrangian associated with the study
of spacetime is only a constant, and it can be derived from the equation
(ds)^2 = g_ij dx^i dx^j
> >You can write an equation for the Lagrangian that results in an
> >action of the elapsed proper time and also another equation for
> >the Lagrangian that results in an action of the elapsed observed time.
>
> You can also solve for the field of a point charge in terms of
> elliptical cylinder functions, but the question is, why? Doing so
> only shows contempt for the physics with the reward of a solution
> that is hard to obtain and illustrates a poor choice of coordinates
> rather than any physics that would otherwise be apparent by thinking
> before calculating.
Since when you observe an event, it is your observed displacement divided by
your own elapsed time and not your observed displacement divided by some one
else's elapsed time, it becomes very important to describe any motion as
observed by you to be your observed displacement divided by your elapased
time. Luckily, the Lagrangian I described in the previous post which is
referenced at the end of this post addresses this issue and is qualified to
the condition imposed by the Lagrangian Method. After all, since the
Lagrangian is just a constant, all the Lagrangians I described result in the
same geodesic equations. So, recall another Lagrangian I described which is
more useful because it addresses one's observed speed on the event of
observation.
> ** Based on the observed time, dx^0
>
> sqrt[(ds/dx^0)^2 / g_00 + |g_ij| (dx^i/dx^0) (dx^j/dx^0) / g_00]
> i, j = 0, 1, 2, 3
> i + j != 0
dx^0 is your elapsed time.
> >> Clearly you do not understand the math. I am not "arguing against the
> >> math", I am describing what the math actually says: energy is conserved
> >> only for systems having a Lagrangian with a time translation symmetry.
> >> _THAT_ is what Noether's theorem says.
> >
> >You have only discover one Lagrangian. There are more.
>
> Why does it matter? Any lagrangian will work. The only difference
> is the amount of effort required to eliminate the irrelevant consequences
> of not choosing the one which most closely resembles the physics, in
> advance.
No, again, a Lagrangian must satisfies the condition imposed by the
Lagrangian Method to be a Lagrangian. All the Lagranigians I have described
do just that, and they all arrive at the same geodesic equations.
> >> You claim "the conservation of energy to be universal". If so, how is
> >> it
> >> that it is clearly NOT conserved for a subsystem interacting with
> >> another
> >> subsystem? -- Not very "universal", is it? Especially since the largest
> >> "system" we humans could construct is clearly a tiny SUBsystem of the
> >> universe.
> >
> >It is very surprising for an experimental physicist to ask me this
> >question.
>
> Why is that surprising? The job description of physicist includes
> a desire to find out what nature does, not make nature conform to
> the ANSI standard for a universe. What is surprising is that you
> claim to know a lot about lagrangian formalism, yet fail to understand
> the entire connection between symmetry and conservation laws (apart
> from repeating the soundbites occasionally). It should be obvious
> that a lagrangian which depends explictly on the time, cannot conserve
> energy.
According to Noether's Theorem, there is one condition that allows the
conservation of a conserved quantity. This condition is when the partial
derivative of the Lagrangian with respect to the quantity that is conserved
vanishes. For the Lagrangian you presented, although it is a function of
(ds/dTau), it is independent of s. Thus, the Lagrangian with respect to s
even in the very general case is always null. The case of conserved quanity
in the very general case indicates a universal principle of the conservation
of observed energy where all enegy described is an observed phenomenon
without any exceptions. Thus, the conservation of observed energy becomes a
fundamental principle.
> [...]
> >more sophisticated observational methodologies. Clearly, any of the
> >Lagrangians I described above does not indicate a universal conservation
> >of
> >angular momentum this should be a start to find out how the conservation
> >of
>
> Clearly, you are completely confused. It is unnecessary to know the
> details of any particular lagrangian to determine a general conservation
> law. All you need to know is whether it's possible to eliminate the
> dependence of the lagrangian on a coordinate. So, if you are looking
> for a conserved angular momentum, the fisrt order of business is to
> eliminate from the lagrangian, any explicit dependence on the angular
> coordinates. If the lagrangian has spherical symmetry, then obviously,
> a dependence on the angular coordinates is inherent in the choice of
> coordinates, not the actual physics.
As I said before, all the Lagrangians I described all indicated a universal
conservation of energy, and they all arrive at the same geodesic equations.
> If additional effort is your thing, why not choose geocentric
> coordinates to describe the planetary orbits in the solar system,
> write down the lagrangian and try to prove angular momentum
> is conserved by the epicycles and retrograde motion in those
> coordinates?
The Lagrangian in general does not indicate a conserved angular momentum. I
will leave it as a home work exercise for you. In a nutshell, energy is
always conserved but not angular momentum.
> Since there is no procedure for constructing a lagrangian, the only
> means of obtaining one is to use physical intuition to make an intelligent
> guess. If such a procedure existed, we would have a theory of everything.
> The utility of the lagrangian formalism lies in what one can say about
> the physics for an any lagrangian based only on the properties of the
> lagrangian. The utility of noether's theorems is that one can obtain
> a lot of physics based on eliminating the dependence of lagrangain
> on some quantity with a symmtery argument.
Stop BSing again. A Lagrangian must satisfy the conditions described by the
Lagrangian Method. The Lagrangian I described all are described from the
generalized spacetime equation
(ds)^2 = g_ij dx^i dx^j
> [... sniiped due to repetition of the same topic]
> You ought to start with old fashioned classical mechanics to gain
> some understanding of the lagrangian before pontificating about
> how it's used in more complex theories.
OK, how do you derive the classical Lagrangian of the kinetic energy plus
the potential energy? Does it satisfy the conditions of the Lagrangian
Method? Just because the classical quantity which is credited as the
Lagrangian works so well solving classical problems, it does not indicate
the classical Lagrangian as being a true Lagrangian in the general case.
Without any mathematical support to indicate it being so, the classical
Lagrangian is just a mathematical wonder that mother nature occasionally
indulged us with.
> Lagrangian Method. Has it been that long since you have learned it which
> I
> assume you must have. So, you need to review that subject. For a quick
> review,
>
> Action = integral of Lagrangian
>
> Where the Lagrangian is a density to this action and in this case the
> action
> is the elapsed time whether it is the observed time or the proper time.
> You
> can write an equation for the Lagrangian that results in an action of the
> elapsed proper time and also another equation for the Lagrangian that
> results in an action of the elapsed observed time. Other than these two
> Lagrangians, we can also write down another equation for the Lagrangian
> that
> results in an action of the elapsed segment of proper spacetime. That
> last
> Lagrangian is not surprising to exist because the motion of an observed
> particle is not subject to the minimum action of space itself but only
> either the proper time or the observed time. If you have done your
> dilligence, you will find that all these three Lagrangians give the same
> geodesic (trying to speak your language) equations. So, the Lagrangians
> are
>
> ** Based on the observed time, dx^0
>
>> Clearly you do not understand the math. I am not "arguing against the
>> math", I am describing what the math actually says: energy is conserved
>> only for systems having a Lagrangian with a time translation symmetry.
>> _THAT_ is what Noether's theorem says.
>
> You have only discover one Lagrangian. There are more.
>
>> You claim "the conservation of energy to be universal". If so, how is it
>> that it is clearly NOT conserved for a subsystem interacting with another
>> subsystem? -- Not very "universal", is it? Especially since the largest
>> "system" we humans could construct is clearly a tiny SUBsystem of the
>> universe.
>
> It is very surprising for an experimental physicist to ask me this
> question.
> Let me use the example I gave in the past. Say you want to measure and
> thus
> observe the speed of a car driving by. You can initially utilize your
> nose
> to do so. For most of us, we detect coherently a Doppler effect in odor,
> we
> have to conclude the observation as a failure and move on to incorporate
> more sophisticated observational methodologies. Clearly, any of the
> Lagrangians I described above does not indicate a universal conservation
> of
> angular momentum this should be a start to find out how the conservation
> of
> energy seems to fail in a binary system instead of creating gravitons.
> YOU
> ARE NO GOD!
>
>> Noether's theorem explains this: in NM the total system Lagrangian has
>> time translation symmetry, but for interacting subsystems, the
>> Lagrangians
>> for the individual subsystems do not have time translation symmetry.
>
> Again, all three Langrangians I described above converge on the same
> geodesic equations. This means you are wrong in your statement above.
>
>> Before Noether's theorem, the non-conservation of energy
>> for subsystems was a puzzle. Ad hoc rules had to be created
>> to avoid embarrassment, but no underlying explanation for
>> them had been given until Noether. Most authors simply
>> ignored this puzzle and treated those rules as "self
>> evident".
>
> You seem not to be ashamed by pulling out a Lagrangian in a ad hoc matter.
> On the contrary to your statements above, Noether's Theorem strengthens
> the
> principle where the conservation of observed energy is universally
> fundamental.
>
>> My L (defined above) is invariant over translations in \tau, and the
>> Noether current corresponding to this symmetry is indeed the energy of
>> the particle.
>>
>> That's only true for suitable metrics (i.e. ones with a timelike Killing
>> vector; see other recent posts in this newsgroup).
>
> In this case, the Lagrangian is equal to 1 to satisfy the requirement
> leading to the Lagrangian Method. Your instint is correct. You need to
Bilge
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
>news:slrndo32jc....@radioactivex.lebesque-al.net...
>
>> I'm afraid you have no idea what you are talking about. The idea is
>> have a lagrangian which has no explicit coordinate dependence, since
>> the lagrangian is supposed to be a scalar. For example, the solution
>> to your perennial objections to the precession of mercuries lies in
>> the lagrangian,
>>
>> L = (m/4)(s_dot)^2
>>
>> where m is the mass of mercury and s_dot is obtained from the line
>> element in the schwarzchild geometry,
>>
>> s^2 = (1-2M/r)dt^2 - dr^2/(1-2M/r) - r^2 d\Omega^2.
>>
>> by taking the derivative with respect to the proper time, d\tau. That
>> works for _any_ choice of coordinates you want to use in writing the
>> line element. The only reason for choosing (t, r, \theta, \phi) is because
>> the symmetry is evident in terms which are useful from the perspective of
>> measuring numbers here on earth and the coordinates are a good description
>> of the geometry for this application.
>
>The Lagrangian I have described all are scalar in total aggreement with you.
>Furthermore, all the Lagrangians I have described being a constant also
>satisfy the conditions imposed by the Lagrangian Method that qualify them to
>be Lagrangians.
i.e., another one of your own inventions which you've deliberately
associated with something that vaguely sounds like it came from physics?
[...]
>This is only true if there is no proper spatial component. The reason why I
>have a sqrt() that you don't is that the Langrangian is a constant. So, now
>you have presented a special case of the Lagrangian where there is no proper
>spatial component that is exactly what I have described. How can you accuse
>me of bening wrong without implicating yourself?
Because you're wrong and I'm not and the most obvious justification
for my claim is that your derived result doesn't match the experiemental
data and mine does. End of story.
>Stop hiding under Dr. Robert's skirt and take cheap shots at scholars like
>myself.
Let's see... You've spent months working on this problem without getting
a result that matches the data or anyone else's calculation and you've
consistently argued with _everyone_ who has attempted to point out your
errors on the basis of your self-described expertise, even though us idiots
don't seem to have the difficulty you're having. You'll excuse me if I
treat your idea of sholarship the same way I'd treat it when considering
having surgery performed by a surgeon who believes his success rate of zero
is due to being a gifted and scholarly surgeon.
[...]
>> Where the lagrangian density is _functional_ (not a function).
>> For example, the lagrangian I gave above, (m/4)(s_dot)^2, s_dot
>> can be a function of any coordinates you want to use. A simple
>> choice only has the advantage of reducing the amount of effort
>> involved in extracting a useful result.
>
>Your Lagrangian is only a constant.
If you believe that, despite the fact that I just told you what
it was and wrote the equations for you, your mind holds the derivative
of a constant.
>I have not seen anyone so capable of describing a constant like that.
Then perhaps you should take a detour through the variational calculus
from whence the lagrangian emerged. The lagrangian is a functional that
depends on some parameters and first derivatives of those parameters
and which vanishes at the endpoints.
>To some, it may seem indicate great scholarship in you. However, to
>true scholars, you are just BSing.
Don't go into medicine. That argument will never convince a jury
that cutting off someone's left arm to treat a hangnail on his
right hand iss evidence of scholarship, not malpractice.
[...]
>> to go wrong. Most physicists call that a lose-lose proposition,
>> and prefer to use physical intuition to build in as much symmetry
>> as possible right off the bat.
>
>You are the one who try very hard to confuse anyone by calling a constant
>_functional_ and not a function.
Look it up.
[...]
>> Why does it matter? Any lagrangian will work. The only difference
>> is the amount of effort required to eliminate the irrelevant consequences
>> of not choosing the one which most closely resembles the physics, in
>> advance.
>
>No, again, a Lagrangian must satisfies the condition imposed by the
>Lagrangian Method to be a Lagrangian.
I can see all of that scholarship paid off. I must admit, that I never
considered the possibilty that I used the word lagrangian to mean something
that doesn't satisfy its own definition. Nice going, slick.
[...]
>
>According to Noether's Theorem, there is one condition that allows the
According to the evidence in your own posts, your knowledge of noether's
theorem extends to spelling ``noether's'' correctly, although I don't
dismiss the possibility that you use a spell checker.
[...]
>As I said before, all the Lagrangians I described all indicated a universal
>conservation of energy, and they all arrive at the same geodesic equations.
Can you estimate a date for when you'll finish (to within a calendar
quarter)?
>> If additional effort is your thing, why not choose geocentric
>> coordinates to describe the planetary orbits in the solar system,
>> write down the lagrangian and try to prove angular momentum
>> is conserved by the epicycles and retrograde motion in those
>> coordinates?
>
>The Lagrangian in general does not indicate a conserved angular momentum.
If it has spherical symmetry it does. Really. It's not just an
urban legend.
Koobee Wublee
> Koobee Wublee:
> >
> >"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
> >news:slrndo32jc....@radioactivex.lebesque-al.net...
> >
> >> I'm afraid you have no idea what you are talking about. The idea is
> >> have a lagrangian which has no explicit coordinate dependence, since
> >> the lagrangian is supposed to be a scalar. For example, the solution
> >> to your perennial objections to the precession of mercuries lies in
> >> the lagrangian,
> >>
> >> L = (m/4)(s_dot)^2
> >>
> >> where m is the mass of mercury and s_dot is obtained from the line
> >> element in the schwarzchild geometry,
> >>
> >> s^2 = (1-2M/r)dt^2 - dr^2/(1-2M/r) - r^2 d\Omega^2.
> >>
> >> by taking the derivative with respect to the proper time, d\tau. That
> >> works for _any_ choice of coordinates you want to use in writing the
> >> line element. The only reason for choosing (t, r, \theta, \phi) is
> >> because
> >> the symmetry is evident in terms which are useful from the perspective
> >> of
> >> measuring numbers here on earth and the coordinates are a good
> >> description
> >> of the geometry for this application.
> >
> >The Lagrangian I have described all are scalar in total aggreement with
> >you.
>
> Then why do you keep getting the wrong result?
No, I don't get the wrong result. You have not understood what you have.
> >Furthermore, all the Lagrangians I have described being a constant also
> >satisfy the conditions imposed by the Lagrangian Method that qualify them
> >to
> >be Lagrangians.
>
> Is the ``Lagrangian Method'' like the ``Minkowski Spacetime Equation,''
> i.e., another one of your own inventions which you've deliberately
> associated with something that vaguely sounds like it came from physics?
No. Try chapter 5.2 of the following.
http://www.courses.fas.harvard.edu/~phys16/Textbook/ch5.pdf
> [...]
> >This is only true if there is no proper spatial component. The reason
> >why I
> >have a sqrt() that you don't is that the Langrangian is a constant. So,
> >now
> >you have presented a special case of the Lagrangian where there is no
> >proper
> >spatial component that is exactly what I have described. How can you
> >accuse
> >me of bening wrong without implicating yourself?
>
> Because you're wrong and I'm not and the most obvious justification
> for my claim is that your derived result doesn't match the experiemental
> data and mine does. End of story.
What type of observaton are you referring to on the Lagrangian?
> >Stop hiding under Dr. Robert's skirt and take cheap shots at scholars
> >like
> >myself.
>
> Let's see... You've spent months working on this problem without getting
> a result that matches the data or anyone else's calculation and you've
> consistently argued with _everyone_ who has attempted to point out your
> errors on the basis of your self-described expertise, even though us
> idiots
> don't seem to have the difficulty you're having. You'll excuse me if I
> treat your idea of sholarship the same way I'd treat it when considering
> having surgery performed by a surgeon who believes his success rate of
> zero
> is due to being a gifted and scholarly surgeon.
No, only a few minutes.
> [...]
> >> Where the lagrangian density is _functional_ (not a function).
> >> For example, the lagrangian I gave above, (m/4)(s_dot)^2, s_dot
> >> can be a function of any coordinates you want to use. A simple
> >> choice only has the advantage of reducing the amount of effort
> >> involved in extracting a useful result.
> >
> >Your Lagrangian is only a constant.
>
> If you believe that, despite the fact that I just told you what
> it was and wrote the equations for you, your mind holds the derivative
> of a constant.
You need to understand the difference between total derivative and partial
derivative. The Lagrangian Method involves the partial derivatives.
> >I have not seen anyone so capable of describing a constant like that.
>
> Then perhaps you should take a detour through the variational calculus
> from whence the lagrangian emerged. The lagrangian is a functional that
> depends on some parameters and first derivatives of those parameters
> and which vanishes at the endpoints.
Variational calculus is nothing more than the Lagrangian Method.
> >To some, it may seem indicate great scholarship in you. However, to
> >true scholars, you are just BSing.
>
> Don't go into medicine. That argument will never convince a jury
> that cutting off someone's left arm to treat a hangnail on his
> right hand iss evidence of scholarship, not malpractice.
Trying to change the subject?
> [...]
> >> to go wrong. Most physicists call that a lose-lose proposition,
> >> and prefer to use physical intuition to build in as much symmetry
> >> as possible right off the bat.
> >
> >You are the one who try very hard to confuse anyone by calling a constant
> >_functional_ and not a function.
>
> Look it up.
I just did. The Lagrangian in the study of GR is just a constant that is
derived through the following equation.
(ds)^2 = g_ij dx^i dx^j
> [...]
>
> >> Why does it matter? Any lagrangian will work. The only difference
> >> is the amount of effort required to eliminate the irrelevant
> >> consequences
> >> of not choosing the one which most closely resembles the physics, in
> >> advance.
> >
> >No, again, a Lagrangian must satisfies the condition imposed by the
> >Lagrangian Method to be a Lagrangian.
>
> I can see all of that scholarship paid off. I must admit, that I never
> considered the possibilty that I used the word lagrangian to mean
> something
> that doesn't satisfy its own definition. Nice going, slick.
> [...]
What is your point?
> >According to Noether's Theorem, there is one condition that allows the
>
> According to the evidence in your own posts, your knowledge of noether's
> theorem extends to spelling ``noether's'' correctly, although I don't
> dismiss the possibility that you use a spell checker.
We are talking about Noether's Theorem and not my spelling capability. You
need to stay on the topic.
> [...]
> >As I said before, all the Lagrangians I described all indicated a
> >universal
> >conservation of energy, and they all arrive at the same geodesic
> >equations.
>
> Can you estimate a date for when you'll finish (to within a calendar
> quarter)?
Not at all, I have already show that given any Lagrangian that satisfies the
condition imposed by the Lagrangian Method this Lagrangian is independent of
(s). Thus, the parital derivative of this Lagrangian with respect to s or
Tau or x^0 has to be null. According to Noether's Theorem, this indicates
always indicates a conserved parameter which is none other than the observed
energy. Period.
> >> If additional effort is your thing, why not choose geocentric
> >> coordinates to describe the planetary orbits in the solar system,
> >> write down the lagrangian and try to prove angular momentum
> >> is conserved by the epicycles and retrograde motion in those
> >> coordinates?
> >
> >The Lagrangian in general does not indicate a conserved angular momentum.
>
> If it has spherical symmetry it does. Really. It's not just an
> urban legend.
Given 2 equal-massed object circulating each other, there is no spherical
symmetry. Thus, the Langrangian Method deos not indicate an conserved
quantity in the approppriate angular displacement. Therefore, angular
momentum is not generally conserved.
I take it that you believe by the farce of nature that angular momentum is
always conserved but not energy. This is so ludicrous on your part, but I
don't believe you can understand this. So, go back to hide under Dr.
Robert's skirt. He does not need any spokesperson to speak for himself.
You have shown yourself to be a total failure.
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:dlidra$c...@netnews.net.lucent.com...
Yes, the Lagrangian is buried inside there; the integral itself is
called the action.
> we can discuss logically and clearly how Noether's Theorem applies to
> indicate a universal phenomenon to the conservation of observed energy.
Except, of course, that Noether's theorem states quite explicitly that
energy is conserved ONLY for systems with a Lagrangian that is invariant
under time translations. <shrug>
And you never responded to the fact that for two interacting subsystems,
if conservation of energy is "universal", why isn't energy conserved
in each subsystem? The answer is, of course, that as long as the
interactions between a subsystem and its surroundings are independent of
time then energy will be conserved in the subsystem; but if the
interactions vary over time then the subsystem's Lagrangian won't be
time independent and energy in the subsystem won't be conserved. That
_IS_ what Noether's theorem says about such a case. <shrug>
As I said before, before Noether this was a puzzle[#], and
ad hoc rules had to be established to save energy
conservation. So "... for an isolated system that does
not interact with its surroundings" had to be added to
the "principle of conservation of energy".
Not very "universal" is it?
[#] Admitedly one that not many people worried about. It
was considered self-evident that energy won't be conserved
in a subsystem that interacts with its surroundings.
> [...]
Tom Roberts tjro...@lucent.com
Koobee Wublee
> Koobee Wublee wrote:
>> "Tom Roberts" <tjro...@lucent.com> wrote in message
>> news:dlidra$c...@netnews.net.lucent.com...
>>> L = \integral g_ij (dx^i/d\tau) (dx^j/d\tau) d\tau
>
> Yes, the Lagrangian is buried inside there; the integral itself is called
> the action.
Thanks for verifying. The Lagrangian as you have described above should be
L = sqrt[g_ij (dx^i/d\tau) (dx^j/d\tau)] = g_ij (dx^i/d\tau) (dx^j/d\tau)
Where
** ds = c dTau
** c = 1
>> we can discuss logically and clearly how Noether's Theorem applies to
>> indicate a universal phenomenon to the conservation of observed energy.
>
> Except, of course, that Noether's theorem states quite explicitly that
> energy is conserved ONLY for systems with a Lagrangian that is invariant
> under time translations. <shrug>
Yes, in this case with the Lagrangian of your choice that satisfies the
conditions imposed by the Lagrangian Method, the partial derivative of this
Lagrangian with respect to x^0 (observed time) is always zero because you
have chosen to describe x^i as functions of Tau not x^0. You can also write
down an equally valid Lagrangian based on x^0 and not Tau. Since you have
in this case describe the Lagrangian as a function of x^0, it is independent
of Tau. So, the partial derivative of this new Lagrangian with respect to
Tau is always zero. Thus, according to Noether's Theorem, there is always a
conserved quantity in any system. Since it is time related, this quantity
is the observed energy. This indicates the conservation of energy is indeed
a univeral principle.
> And you never responded to the fact that for two interacting subsystems,
> if conservation of energy is "universal", why isn't energy conserved in
> each subsystem? The answer is, of course, that as long as the interactions
> between a subsystem and its surroundings are independent of time then
> energy will be conserved in the subsystem; but if the interactions vary
> over time then the subsystem's Lagrangian won't be time independent and
> energy in the subsystem won't be conserved. That _IS_ what Noether's
> theorem says about such a case. <shrug>
Yes, I have. Any general Lagrangian does not indicate a universal
conservation in angular momentum. This is all in the very simple math which
you have now reviewed.
> As I said before, before Noether this was a puzzle[#], and
> ad hoc rules had to be established to save energy
> conservation. So "... for an isolated system that does
> not interact with its surroundings" had to be added to
> the "principle of conservation of energy".
> Not very "universal" is it?
Yes, all thanks to Emma Noether. Now, we understand why we observe a
conservation of angular momentum in our solar system where the mass of the
sun is way larger than any planet. However, in general, the conservation of
angular momentum is not true. Unlike angular momentum, Noether's Theorem
always indicates a conservation of energy.
> [#] Admitedly one that not many people worried about. It
> was considered self-evident that energy won't be conserved
> in a subsystem that interacts with its surroundings.
>
> > [...]
Thanks for going back to review the Lagrangian Method. I know it has been a
long time for you. Now, we can speak logically in the same language.
brian a m stuckless
Sub-SYSTEMs are "submerged" in SYSTEM "working fluid" AMBiENT.
Sub-SYSTEMs ONLY EXCHANGE energy with "working fluid" AMBiENT.
Go-go Google GROUP SEARCH: < My BiGGER bang > ..My REAL STUFF.
TOTAL LaGrangian h*f Dis-iNFORMATiON follows with Tom Roberts.
brian a m stuckless
donsto...@hotmail.com
at random (well, plus your content leaves a little to be desired).
Have a nice day.
Bilge
>> Then why do you keep getting the wrong result?
>
>No, I don't get the wrong result. You have not understood what you have.
When did your calculation of the precession of mercury start agreeing
with the correct result?
>> Is the ``Lagrangian Method'' like the ``Minkowski Spacetime Equation,''
>> i.e., another one of your own inventions which you've deliberately
>> associated with something that vaguely sounds like it came from physics?
>
>No. Try chapter 5.2 of the following.
>
>http://www.courses.fas.harvard.edu/~phys16/Textbook/ch5.pdf
Let me repeat the question, since you didn't answer it.
Is the ``Lagrangian Method'' like the ``Minkowski Spacetime Equation,''
i.e., another one of your own inventions which you've deliberately
associated with something that vaguely sounds like it came from physics?
>> [...]
>> Because you're wrong and I'm not and the most obvious justification
>> for my claim is that your derived result doesn't match the experiemental
>> data and mine does. End of story.
>
>What type of observaton are you referring to on the Lagrangian?
Try rereading what I wrote.
[...]
>> If you believe that, despite the fact that I just told you what
>> it was and wrote the equations for you, your mind holds the derivative
>> of a constant.
>
>You need to understand the difference between total derivative and partial
>derivative. The Lagrangian Method involves the partial derivatives.
No shit, sherlock. The sooner you figure out what that means,
the sooner you can stop babbling.
[...]
>> Then perhaps you should take a detour through the variational calculus
>> from whence the lagrangian emerged. The lagrangian is a functional that
>> depends on some parameters and first derivatives of those parameters
>> and which vanishes at the endpoints.
>
>Variational calculus is nothing more than the Lagrangian Method.
Hence what I said about functionals.
[...]
>> Look it up.
>
>I just did. The Lagrangian in the study of GR is just a constant that is
>derived through the following equation.
>
>(ds)^2 = g_ij dx^i dx^j
No, it isn't. That is called a line element. That is not the same thing
as a particle moving in that metric unless you construct the line element
for both masses to get a single metric and line element containing _both_
masses. If you want to impress me, by all means, find a metric for two
masses moving in their own metric with no approximations. That would
reduce the list of unsolved problems is general relativity by 1.
[...]
>> According to the evidence in your own posts, your knowledge of noether's
>> theorem extends to spelling ``noether's'' correctly, although I don't
>> dismiss the possibility that you use a spell checker.
>
>We are talking about Noether's Theorem and not my spelling capability. You
>need to stay on the topic.
No, you refuse to discuss anything but your misconceptions about noether's
theorem. Before you reply with a slew of bullshit, claiming otherwise,
demonstrate that you can derive conservation of energy using nother's
theorem for something simple, like special relativity (without copying
the derivations I've posted previously or someone else's work from a
quote mining expedition).
If you can't do that, then don't bother trying to tell me I don't
know what I'm talking about. I can reference my own posts in which
I've done exactly that.
[...]
>> Can you estimate a date for when you'll finish (to within a calendar
>> quarter)?
>
>Not at all, I have already show that given any Lagrangian that satisfies the
>condition imposed by the Lagrangian Method this Lagrangian is independent of
>(s). Thus, the parital derivative of this Lagrangian with respect to s or
>Tau or x^0 has to be null. According to Noether's Theorem, this indicates
>always indicates a conserved parameter which is none other than the observed
>energy. Period.
No. You haven't. However, if you pulled your head out, you would
discover that you don't know what you are talking about.
[...]
>> If it has spherical symmetry it does. Really. It's not just an
>> urban legend.
>
>Given 2 equal-massed object circulating each other, there is no spherical
>symmetry.
Don't be an idiot. Wait - nevermind, it's too late for that.
If the symmetry is manifest, noether's theorem would be of almost
no value. You are clearly clueless.
>Thus, the Langrangian Method deos not indicate an conserved quantity
in the approppriate angular displacement.
Ohm then obviously your same theory about what noether's theorem
means, proves that energy and momentum aren't conserved, since
I observe that velocities are not zero and not even constant.
Nice going, slick.
>Therefore, angular momentum is not generally conserved.
Go buy a textbook and study it. Preferably starting at a level
you can understand.
Tom Roberts
> You need to understand the difference between total derivative and partial
> derivative. The Lagrangian Method involves the partial derivatives.
The first sentence is completely correct, and _YOU_ need to take your
own words to heart. The second is not correct (it's only partly correct,
and the omission is the crux of your error); your error is illustrated
in other things you say:
Koobee Wublee wrote:
> in this case with the Lagrangian of your choice that satisfies the
> conditions imposed by the Lagrangian Method, the partial derivative of this
> Lagrangian with respect to x^0 (observed time) is always zero because you
> have chosen to describe x^i as functions of Tau not x^0.
You are confused about the roles of variables and how partial
derivatives work. In particular: x^0 and Tau are _not_ independent
variables.
The beauty and power of the Lagrangian method is that you can express
the Lagrangian in any variables whatsoever, then apply Euler's method to
obtain equations of motion in terms of those variables. The (classical)
Euler-Lagrange equations involve a TOTAL derivative wrt t and partial
derivatives wrt the generalized coordinates and generalized velocities.
In geometry, for the path of an object, the variable t must be a path
parameter of the curve, and it is the TOTAL derivative wrt t which appears.
So your statement above is nonsense -- you must express the Lagrangian
in terms of x^0 for TOTAL derivatives wrt x^0 to make sense; ditto for
tau (separately).
1D Example: let q=f(t), and q.=g(t) [#]. For L=mq.^2-k*q^2, d/dt(@L/@q.)
is not zero, even though t does not appear in the exression for L. Here
d/dt is a _TOTAL_ derivative and @/@q. is a _partial_ derivative --
remember that for this to make sense L = L(q,q.) -- that is, q and q.
are considered to be independent variables here, and in particular: q
and q. are both functions of t. Attempting to discuss this in terms of a
completely irrelevant variable tau would make no sense (tau is
irrelevant BECAUSE it does not appear in the equations).
[#] Normally the dot is placed over the q; in ASCII I just
put it after the q -- q. is the generalized velocity
corresponding to q
> You can also write
> down an equally valid Lagrangian based on x^0 and not Tau. Since you have
> in this case describe the Lagrangian as a function of x^0, it is independent
> of Tau. So, the partial derivative of this new Lagrangian with respect to
> Tau is always zero.
See above with x^0 and tau interchanged.
> Thus, according to Noether's Theorem, there is always a
> conserved quantity in any system.
You are also confused about what Noether's theorem says. It is not
@L/@t=0 that matters, it is the fact that one can perform a transform on
{q,q.} that leaves the Lagrangian unchanged that matters -- energy is
conserved if t->t+k leaves the Lagrangian unchanged for all constants k,
and this is so even if @L/@t is nonzero (of course the {q,q.} must then
change in some miraculous way to cancel the explicit t dependence). The
thing you missed above is: one _MUST_ express _ALL_ of the variables
{q,q.} as functions of t (above that means x^0 and tau, separately for
the two cases).
IOW: there are an infinite number of other variables v for which
@L/@v=0. Do you seriously think there are an infinite number of
"conserved quantities" corresponding to each and every one of them????
-- Noether's theorem applies to transformations of the {q,q.} that leave
the Lagrangian unchanged, not to fictitious or irrelevant variables that
happen not to appear in the Lagrangian.
A cyclical variable q1 corresponds to a conserved quantity
because one can apply the transform q1=q1+k (all others
unchanged) without changing the value of the Lagrangian.
While indeed @L/@q1=0, that is NOT what matters here.
Exercise for the reader: Given that q1 is a cyclical variable
(i.e. it does not appear in the expression for L), clearly
L is unchanged for q1=k*q1 -- why does that not correspond to
another conserved quantity? Hint1: consider q1.. Hint2:
there is no typo in Hint1. Hint3: what would we call a
variable q2 for which neither q2 nor q2. appears in L?
Tom Roberts tjro...@lucent.com
Koobee Wublee
> Koobee Wublee wrote:
>> in this case with the Lagrangian of your choice that satisfies the
>> conditions imposed by the Lagrangian Method, the partial derivative of
>> this Lagrangian with respect to x^0 (observed time) is always zero
>> because you have chosen to describe x^i as functions of Tau not x^0.
>
> Dr. Roberts wrote:
> You are confused about the roles of variables and how partial derivatives
> work. In particular: x^0 and Tau are _not_ independent variables.
I have never claimed x^0 and Tau are indepedent variables.
> Dr. Roberts wrote:
> The beauty and power of the Lagrangian method is that you can express the
> Lagrangian in any variables whatsoever, then apply Euler's method to
> obtain equations of motion in terms of those variables. The (classical)
> Euler-Lagrange equations involve a TOTAL derivative wrt t and partial
> derivatives wrt the generalized coordinates and generalized velocities.
You have reviewed your Lagrangian Method but not very thoroughly. I will
point out how later in this post.
> Dr. Roberts wrote:
> In geometry, for the path of an object, the variable t must be a path
> parameter of the curve, and it is the TOTAL derivative wrt t which
> appears.
So, although mathematically it is correct to write down the velocity as a
derivative of the observed displacement with respect to the proper time, it
is rather absurd to do so. You have just pointed this out, but it is not a
subject of our discussion.
> Dr. Roberts wrote:
> So your statement above is nonsense -- you must express the Lagrangian in
> terms of x^0 for TOTAL derivatives wrt x^0 to make sense; ditto for tau
> (separately).
Have you not read my posts first before you responded? I said
> Koobee Wublee wrote several posts ago:
> Dr. Roberts wrote:
> 1D Example: let q=f(t), and q.=g(t) [#]. For L=mq.^2-k*q^2, d/dt(@L/@q.)
> is not zero, even though t does not appear in the exression for L. Here
> d/dt is a _TOTAL_ derivative and @/@q. is a _partial_ derivative --
> remember that for this to make sense L = L(q,q.) -- that is, q and q. are
> considered to be independent variables here, and in particular: q and q.
> are both functions of t.
For this Lagrangian which is classical in nature, it is yes as you have
described. However, from the **** equation which I call the spacetime
equation and you call it whatever which I still don't know what,
** (ds)^2 = g_ij dx^i dx^j
It does yield several Lagrangians that satisfy the conditions imposed by the
Lagrangian Method which makes them to be true Lagrangians. They are
functions
** L(q(t), dq(t)/dt), where (t = x^9)
** L(q(s), dq(s)/ds)
** L(q(T), dq(T)/dT), where (T = Tau) and (s = c Tau)
> Dr. Roberts wrote:
> Attempting to discuss this in terms of a
> completely irrelevant variable tau would make no sense (tau is irrelevant
> BECAUSE it does not appear in the equations).
No, it is not silly to discuss them. The Lagrangian Method demands you to
discuss them. Thus,
** For L(q(t), dq(t)/dt), @L/@s = 0, where (q = s, x, y, z)
So, d(@L/@(ds/dt))/dt = @L/@s = 0
** For L(q(s), dq(s)/ds), @L/@t = 0, where (q = t, x, y, z)
So, d(@L/@(dt/ds))/ds = @L/@t = 0
** For L(q(T), dq(T)/dT), @L/@t = 0, where (q = t, x, y, z)
So, d(@L/@(dt/dT))/dT = @L/@t = 0
These Lagrangians all indicate a conservation of energy.
> Dr. Roberts wrote:
> You are also confused about what Noether's theorem says. It is not @L/@t=0
> that matters, it is the fact that one can perform a transform on {q,q.}
> that leaves the Lagrangian unchanged that matters -- energy is conserved
> if t->t+k leaves the Lagrangian unchanged for all constants k, and this is
> so even if @L/@t is nonzero (of course the {q,q.} must then change in some
> miraculous way to cancel the explicit t dependence). The thing you missed
> above is: one _MUST_ express _ALL_ of the variables {q,q.} as functions of
> t (above that means x^0 and tau, separately for the two cases).
(@L/@t = 0) is a special case of Noether's Theorem. So, I still don't see
where I am confused over.
> Dr. Roberts wrote:
> IOW: there are an infinite number of other variables v for which @L/@v=0.
> Do you seriously think there are an infinite number of "conserved
> quantities" corresponding to each and every one of them????
All other independent variables are independent of our system, so to our
system, these variables are just as good as being conserved. Yes, I really
seriously think there are an infinite number of 'conserved quantities' in a
given system.
> Dr. Roberts wrote:
> Noether's theorem applies to transformations of the {q,q.} that leave the
> Lagrangian unchanged, not to fictitious or irrelevant variables that
> happen
> not to appear in the Lagrangian.
All these Lagarangians being derived from the **** (spacetime) equation are
constants. So, dL/dt = dL/ds = dL/dT = 0. You can never escape the fact
that the Noether's Theorem indicates a universal principle of the
conservaton of observed energy. You need to go back to review what
Noether's Theorem tells you, and you have not thoroughly reviewed what the
Lagrangian Method really is.
> Dr. Roberts wrote:
> A cyclical variable q1 corresponds to a conserved quantity
> because one can apply the transform q1=q1+k (all others
> unchanged) without changing the value of the Lagrangian.
> While indeed @L/@q1=0, that is NOT what matters here.
>
> Exercise for the reader: Given that q1 is a cyclical variable
> (i.e. it does not appear in the expression for L), clearly
> L is unchanged for q1=k*q1 -- why does that not correspond to
> another conserved quantity? Hint1: consider q1.. Hint2:
> there is no typo in Hint1. Hint3: what would we call a
> variable q2 for which neither q2 nor q2. appears in L?
This means there is more home work for Dr. Roberts. He needs to go back to
thoroughly understand what Lagrangian Method is completely with the
conditions imposed as well as the derivation. He also needs to review what
Noether Theorem really tells him.
Koobee Wublee
> When did your calculation of the precession of mercury start agreeing
> with the correct result?
This is not related to the subject of our conversion. However, in the past,
I have shown unmistakenly that GR does not predict a 42" of advance per
century associated with Mercury's orbital perihelion. It predicts a null
result just like the Newtonian physics does.
> Let me repeat the question, since you didn't answer it.
>
> Is the ``Lagrangian Method'' like the ``Minkowski Spacetime Equation,''
> i.e., another one of your own inventions which you've deliberately
> associated with something that vaguely sounds like it came from physics?
Let me repeat my answer.
No. Try chapter 5.2 of the following.
http://www.courses.fas.harvard.edu/~phys16/Textbook/ch5.pdf
> >> [...]
> >> Because you're wrong and I'm not and the most obvious justification
> >> for my claim is that your derived result doesn't match the
> >> experiemental
> >> data and mine does. End of story.
> >
> >What type of observation are you referring to on the Lagrangian?
>
> Try rereading what I wrote.
I did. You have not described what observation.
> [...]
> >> If you believe that, despite the fact that I just told you what
> >> it was and wrote the equations for you, your mind holds the derivative
> >> of a constant.
> >
> >You need to understand the difference between total derivative and
> >partial
> >derivative. The Lagrangian Method involves the partial derivatives.
>
> No shit, sherlock. The sooner you figure out what that means,
> the sooner you can stop babbling.
When are you going to figure it out?
> [...]
> >> Then perhaps you should take a detour through the variational calculus
> >> from whence the lagrangian emerged. The lagrangian is a functional that
> >> depends on some parameters and first derivatives of those parameters
> >> and which vanishes at the endpoints.
> >
> >Variational calculus is nothing more than the Lagrangian Method.
>
> Hence what I said about functionals.
Hence, you have not understood what the Lagrangian is.
> [...]
> >> Look it up.
> >
> >I just did. The Lagrangian in the study of GR is just a constant that is
> >derived through the following equation.
> >
> >(ds)^2 = g_ij dx^i dx^j
>
> No, it isn't. That is called a line element. That is not the same thing
> as a particle moving in that metric unless you construct the line element
> for both masses to get a single metric and line element containing _both_
> masses. If you want to impress me, by all means, find a metric for two
> masses moving in their own metric with no approximations. That would
> reduce the list of unsolved problems is general relativity by 1.
This is an segment in spacetime. You need to understand that spacetime is
composed of space and time. Is it really that hard to understand this for
you?
> [...]
> >> According to the evidence in your own posts, your knowledge of
> >> noether's
> >> theorem extends to spelling ``noether's'' correctly, although I don't
> >> dismiss the possibility that you use a spell checker.
> >
> >We are talking about Noether's Theorem and not my spelling capability.
> >You
> >need to stay on the topic.
>
> No, you refuse to discuss anything but your misconceptions about
> noether's
> theorem. Before you reply with a slew of bullshit, claiming otherwise,
> demonstrate that you can derive conservation of energy using nother's
> theorem for something simple, like special relativity (without copying
> the derivations I've posted previously or someone else's work from a
> quote mining expedition).
Noether's Theorem is already derived by Emma Noether. Go look it up.
> If you can't do that, then don't bother trying to tell me I don't
> know what I'm talking about. I can reference my own posts in which
> I've done exactly that.
We don't want to reference any of your posts. They are all BS. Just look
up Noether's Theorem.
> [...]
> >> Can you estimate a date for when you'll finish (to within a calendar
> >> quarter)?
> >
> >Not at all, I have already show that given any Lagrangian that satisfies
> >the
> >condition imposed by the Lagrangian Method this Lagrangian is independent
> >of
> >(s). Thus, the parital derivative of this Lagrangian with respect to s
> >or
> >Tau or x^0 has to be null. According to Noether's Theorem, this
> >indicates
> >always indicates a conserved parameter which is none other than the
> >observed
> >energy. Period.
>
> No. You haven't. However, if you pulled your head out, you would
> discover that you don't know what you are talking about.
You need to understand what the Lagrangian Method is.
> [...]
> >> If it has spherical symmetry it does. Really. It's not just an
> >> urban legend.
> >
> >Given 2 equal-massed object circulating each other, there is no spherical
> >symmetry.
>
> Don't be an idiot. Wait - nevermind, it's too late for that.
>
> If the symmetry is manifest, noether's theorem would be of almost
> no value. You are clearly clueless.
If the symmetry is not manifest, Noether's Theorem would be of value. You
are still clearly clueless.
> >Thus, the Langrangian Method deos not indicate an conserved quantity
> in the approppriate angular displacement.
>
> Ohm then obviously your same theory about what noether's theorem
> means, proves that energy and momentum aren't conserved, since
> I observe that velocities are not zero and not even constant.
> Nice going, slick.
No.
> >Therefore, angular momentum is not generally conserved.
>
> Go buy a textbook and study it. Preferably starting at a level
> you can understand.
I have. If you have too, you need to return that textbook and ask for a
complete refund.
Bilge
>For this Lagrangian which is classical in nature, it is yes as you have
>described. However, from the **** equation which I call the spacetime
>equation and you call it whatever which I still don't know what,
>
>** (ds)^2 = g_ij dx^i dx^j
It's called a line element. How could you possibly have read anything
about relativity without knowing that?
>It does yield several Lagrangians that satisfy the conditions imposed by the
>Lagrangian Method which makes them to be true Lagrangians.
A lagrangian is a functional which depends upon (1) a coordinate,
(2) the first derivative of the coordinate. Coordinate means any
generalized coordinate which itself may be a function of the coordinates.
In relativistic theories, the generalized coordinates are called fields.
>They are
>functions
>
>** L(q(t), dq(t)/dt), where (t = x^9)
>** L(q(s), dq(s)/ds)
>** L(q(T), dq(T)/dT), where (T = Tau) and (s = c Tau)
The general form of the lagrangian is L[q, Dq].
>> Dr. Roberts wrote:
>> Attempting to discuss this in terms of a
>> completely irrelevant variable tau would make no sense (tau is irrelevant
>> BECAUSE it does not appear in the equations).
>
>No, it is not silly to discuss them. The Lagrangian Method demands you to
>discuss them. Thus,
>
>** For L(q(t), dq(t)/dt), @L/@s = 0, where (q = s, x, y, z)
>
> So, d(@L/@(ds/dt))/dt = @L/@s = 0
Wrong. q == q(t, x, y, z), d_u q == d_u q(t, x, y, z). The
derivatives are with respect to q and d_u q
\delta L[q, q_dot] = (dL/dq)\delta q + dL/d(d_uq)\delta d_u q
>> Dr. Roberts wrote:
>> You are also confused about what Noether's theorem says. It is not @L/@t=0
>> that matters, it is the fact that one can perform a transform on {q,q.}
>> that leaves the Lagrangian unchanged that matters -- energy is conserved
>> if t->t+k leaves the Lagrangian unchanged for all constants k, and this is
>> so even if @L/@t is nonzero (of course the {q,q.} must then change in some
>> miraculous way to cancel the explicit t dependence). The thing you missed
>> above is: one _MUST_ express _ALL_ of the variables {q,q.} as functions of
>> t (above that means x^0 and tau, separately for the two cases).
>
>(@L/@t = 0) is a special case of Noether's Theorem.
No, it isn't. It's a simple fact that in classical mechanics, the time
is an affine paramater and if the lagrangian doesn't explicitly depend
on t, then,
dH/dt = (d/dt) [p.v - L] = dp.v/dt - dL/dt
dH/dt = - dL/dt (assuming that L satisfies the euler-lagrange eqns).
Since H is the total energy, dH/dt = -dL/dt = 0. That was known and
used extensives long before noether was even born. Saying thats a
special case of noether's theorem is like saying a^2 = c^2 is a special
case of the pythagorean theorem for b = 0.
>So, I still don't see where I am confused over.
Only because you aren't interested in knowing. If you were interested,
you wouldn't keep arguing about it everytime someone tries to explain it.
Mike
Tom Roberts wrote:
[snip senseless conversation]
>
> And you never responded to the fact that for two interacting subsystems,
> if conservation of energy is "universal", why isn't energy conserved
> in each subsystem? The answer is, of course, that as long as the
> interactions between a subsystem and its surroundings are independent of
> time then energy will be conserved in the subsystem; but if the
> interactions vary over time then the subsystem's Lagrangian won't be
> time independent and energy in the subsystem won't be conserved. That
> _IS_ what Noether's theorem says about such a case. <shrug>
>
> As I said before, before Noether this was a puzzle[#], and
> ad hoc rules had to be established to save energy
> conservation. So "... for an isolated system that does
> not interact with its surroundings" had to be added to
> the "principle of conservation of energy".
> Not very "universal" is it?
It was a puzzle only for morons. Some books stated the AXIOM of
conservation of energy correctly but some other books skipped the
"isolated system" condition. It was just a confusion of morons rather
than a confusion in physics that some Noether came to clarify.
>
> [#] Admitedly one that not many people worried about. It
> was considered self-evident that energy won't be conserved
> in a subsystem that interacts with its surroundings.
Yet, although self-evident, the condition was not present in the
statements for the law of conservation. Do you understand why? I guess
you do not...
Noether theorem is no more than a rediscovery of an implicit assumption
of classical physics. It is implicit in the virtual work function used
to derive the Lagrangian originally. So if you all want to understand
what the Lagrangian means, go and find the original derivation. There
can be no Lagragian without local energy conservation and to say that
energy conservation results from the Lagrangian and time symmetry,
although ultimately true, is a SHAM because the premises of a deduction
can never become the conclusion.
For God's shake, where is this world heading to...
Mike
>
>
> > [...]
>
>
> Tom Roberts tjro...@lucent.com
Koobee Wublee
> Koobee Wublee:
>
> >For this Lagrangian which is classical in nature, it is yes as you have
> >described. However, from the **** equation which I call the spacetime
> >equation and you call it whatever which I still don't know what,
> >
> >** (ds)^2 = g_ij dx^i dx^j
>
> It's called a line element. How could you possibly have read anything
> about relativity without knowing that?
Whatever it is called, but I like the phrase 'a segment in spacetime'.
> >It does yield several Lagrangians that satisfy the conditions imposed by
> >the
> >Lagrangian Method which makes them to be true Lagrangians.
>
> A lagrangian is a functional which depends upon (1) a coordinate,
> (2) the first derivative of the coordinate. Coordinate means any
> generalized coordinate which itself may be a function of the coordinates.
> In relativistic theories, the generalized coordinates are called fields.
(1) is true, but mathematically ijn general (2) is not necessarily. For the
study of General Relativity, the Lagrangian becomes the simplest case where
your (2) applies.
> >They are
> >functions
> >
> >** L(q(t), dq(t)/dt), where (t = x^9)
> >** L(q(s), dq(s)/ds)
> >** L(q(T), dq(T)/dT), where (T = Tau) and (s = c Tau)
>
> The general form of the lagrangian is L[q, Dq].
They are as all these text books stated as well as I have stated above.
> >> Dr. Roberts wrote:
> >> Attempting to discuss this in terms of a
> >> completely irrelevant variable tau would make no sense (tau is
> >> irrelevant
> >> BECAUSE it does not appear in the equations).
>
> >No, it is not silly to discuss them. The Lagrangian Method demands you
> >to
> >discuss them. Thus,
> >
> >** For L(q(t), dq(t)/dt), @L/@s = 0, where (q = s, x, y, z)
> >
> > So, d(@L/@(ds/dt))/dt = @L/@s = 0
>
> Wrong. q == q(t, x, y, z), d_u q == d_u q(t, x, y, z). The
> derivatives are with respect to q and d_u q
No, I have already defined q as a function of (s, x, y, z) in order for the
above equations to be true. This is the Lagrangian Method which you need to
study.
> \delta L[q, q_dot] = (dL/dq)\delta q + dL/d(d_uq)\delta d_u q
What does '\delta' mean? Where does this equation lead to?
> >> Dr. Roberts wrote:
> >> You are also confused about what Noether's theorem says. It is not
> >> @L/@t=0
> >> that matters, it is the fact that one can perform a transform on {q,q.}
> >> that leaves the Lagrangian unchanged that matters -- energy is
> >> conserved
> >> if t->t+k leaves the Lagrangian unchanged for all constants k, and this
> >> is
> >> so even if @L/@t is nonzero (of course the {q,q.} must then change in
> >> some
> >> miraculous way to cancel the explicit t dependence). The thing you
> >> missed
> >> above is: one _MUST_ express _ALL_ of the variables {q,q.} as functions
> >> of
> >> t (above that means x^0 and tau, separately for the two cases).
> >
> >(@L/@t = 0) is a special case of Noether's Theorem.
>
> No, it isn't. It's a simple fact that in classical mechanics, the time
> is an affine paramater and if the lagrangian doesn't explicitly depend
> on t, then,
The following article does address Neother's Theorem. It does not agree
with you.
http://www.courses.fas.harvard.edu/~phys16/Textbook/ch5.pdf
> dH/dt = (d/dt) [p.v - L] = dp.v/dt - dL/dt
Where are H? L? What is your justification to state 'H = p.v - L'?
> dH/dt = - dL/dt (assuming that L satisfies the euler-lagrange eqns).
>
> Since H is the total energy, dH/dt = -dL/dt = 0. That was known and
> used extensives long before noether was even born. Saying thats a
> special case of noether's theorem is like saying a^2 = c^2 is a special
> case of the pythagorean theorem for b = 0.
This is total nonsense.
> >So, I still don't see where I am confused over.
>
> Only because you aren't interested in knowing. If you were interested,
> you wouldn't keep arguing about it everytime someone tries to explain it.
As I said before, you have failed to capitalize on any textbooks. It is
your best interest to return all textbooks and demand for a full refund.
Let someone else have a change to learn about physics.
Bilge
>It was a puzzle only for morons. Some books stated the AXIOM of
>conservation of energy correctly but some other books skipped the
>"isolated system" condition.
Oh? Go ahead, prove energy is conserved for any system you define,
without using noether's theorem. First make sure you define
energy unambiguously. Good luck.
Bilge
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
>news:slrndo9ucf....@radioactivex.lebesque-al.net...
>
>> When did your calculation of the precession of mercury start agreeing
>> with the correct result?
>
>This is not related to the subject of our conversion. However, in the past,
>I have shown unmistakenly that GR does not predict a 42" of advance per
understand. You just keep adding additional proof. Lets see you
apply noether's theorem to the example I gave you the last time
you tried to make that silly claim.
Bilge
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
>news:slrndodv8u....@radioactivex.lebesque-al.net...
>> Koobee Wublee:
>>
>> >For this Lagrangian which is classical in nature, it is yes as you have
>> >described. However, from the **** equation which I call the spacetime
>> >equation and you call it whatever which I still don't know what,
>> >
>> >** (ds)^2 = g_ij dx^i dx^j
>>
>> It's called a line element. How could you possibly have read anything
>> about relativity without knowing that?
>
>Whatever it is called, but I like the phrase 'a segment in spacetime'.
[...]
>> >** L(q(t), dq(t)/dt), where (t = x^9)
>> >** L(q(s), dq(s)/ds)
>> >** L(q(T), dq(T)/dT), where (T = Tau) and (s = c Tau)
>>
>> The general form of the lagrangian is L[q, Dq].
>
>They are as all these text books stated as well as I have stated above.
textbooks. Look in wald, for the additional sophistication needed
for more advanced topics like relativity.
[...]
>> \delta L[q, q_dot] = (dL/dq)\delta q + dL/d(d_uq)\delta d_u q
>
>What does '\delta' mean? Where does this equation lead to?
Thank you confirming you have never actually tried to understand
what the lagrangian is, or how to use it properly. It is literally
impossible to know what a lagrangian is without recognizing the
expression for the variation, given that variational calculus
is what you keen calling ``The Lagrangian Method.'' Go buy a real
textbook.
Tom Roberts
Koobee Wublee
> symptoms of inebriation]:
> s is not a valid coordinate.
All the derivations associated with GR will disagree with you. Here is just
one example.
http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter3.pdf
Go to page 157. What do equations 3.5.1, 3.5.2, and 3.5.3 indicate? You
are wrong!
However, you are not entirely wrong. 's' is only a valid coordinate for
non-photons. The professional such as the authors realized this. They
resort to use another parameter 'sigma' which is cleverly presented on page
143.
As I have pointed out, you can also write a Lagrangian as the density of the
action representing of elapsed observed time (x^0) where the observed time
is minimized according to the Lagrangian Method. This Lagrangian works for
photons as well. After poo-poo-ing this Lagrangian based on (x^0), have you
had a change of heart? After all, all the mathematics do indicate this
Lagrangian is indeed a very valid Lagrangian.
brian a m stuckless
YOU too, can EARN $50 a pop ..mis-spelling "LaGrangian".!!
brian a m stuckless
>><> >><> >><> >><> >><>
Bilge
>> symptoms of inebriation]:
>> s is not a valid coordinate.
>
>All the derivations associated with GR will disagree with you. Here is just
>one example.
>
>http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter3.pdf
>
>Go to page 157. What do equations 3.5.1, 3.5.2, and 3.5.3 indicate?
different letters confuses you even more than you're confused about the
difference in what the authors call \sigma and s (which are exactly the
same quantities that I called s and \tau). I didn't see all of what tom
posted, but I'm quite certain he isn't confused by any of this.
So, let's see - you take offense when anyone complains about you
inventing your own terminolgy, yet, you find it really confusing
when I use different letts in equations than whatever you've dug
up from the so-called ``professionals?'' I guess that's self-con-
sistent if your goal is to force people to guess what you mean
in order to reject what they tell you and cite a reference that
used different letters in the same equations. You'll excuse me if
I don't have a lot of sympathy for someone who goes out of his
way to reject something I've written in terms of the most commonly
used symbols in a less compact way than it could be written if
I was talking to someone who understood the subject.
>You are wrong!
Once again, you demonstrate that inherent efficiency in rejecting
anything anyone tells you by not bothering to spend the effort
in trying and understand something you plan to reject anyway.
>However, you are not entirely wrong. 's' is only a valid coordinate for
>non-photons.
It's not a coordinate. It's the affine parameter that paramaterizes
the proper time of an object along a geodesic trajectory. I told
you exactly why you're confused about 3 posts ago. You are confusing
the metric which defines the geodesics with the proper time of an
object moving along those geodesics. (what they call \sigma^2 is
what I called s^2. What they call s is what I called \tau.)
>The professional such as the authors realized this.
Oh, I forgot - a ``professional'' is someone from out of town.
What they realize is quite different from what you think those
authors realize. In particular, the exact same equations are
found in both wald and schutz, but since you won't bother to
look at a textbook, I'm not going to bother hunting down the
page relevant pages.
>They resort to use another parameter 'sigma' which is cleverly
>presented on page 143.
Holy affine paramater, scatman! They used a different letter.
It sounds like the work of that villian, the rindler.
>As I have pointed out, you can also write a Lagrangian as the density of the
>action representing of elapsed observed time (x^0) where the observed time
>is minimized according to the Lagrangian Method.
All you've pointed out is that you're immune to education. So long
as you are determined to remain ignorant, I'm certainly not about to
waste the effort required to post anything which assumes otherwise.
>This Lagrangian works for photons as well. After poo-poo-ing this
>Lagrangian based on (x^0), have you had a change of heart?
What coordinate system does x^0 refer to? If you had bothered to
figure out what x^0 refers to, instead of presuming every observer
uses the same x^0, you could have saved yourself some effort.
Let me know when you decide you're more interested in understanding
physics than you are in being famous for proving that general relativity
is wrong and being credited for the terminology of the new paridigm.
>After all, all the mathematics do indicate this Lagrangian is indeed
>a very valid Lagrangian.
How would you know? You keep arguing that a lagrangian is something
other than what everyone else calls a lagrangian and don't even
seem to know variational calculus is involved. (You might consider
looking up emmy noether to see what her mathematical specialty was
and why hilbert hired her to assist in his pursuit of the field
equations - that is where noether's theorems originated.)
Mike
First, I have asked you to be a man, not a degenerate, and stop
re-directing posts to alt.morons like a little kid who makes a
mischieve but fears to face the consequences. Wear pants and face the
truth about what you are. You direct to alt.morons because it IS THERE
YOU ACTUALLY BELONG.
you go ahead and learn the basics. Learn for instance that AXIOMS are
not provable propositions. It is just something you accept to be true
because otherwise your world would not make sense.
Noether's theorem is just about symmetries and related conservation AND
NOT about just conservation as many want to extend it to. Conservation
is already built into the Lagrangian. Noether just linked it to the
symmetries. However, the key is that even thge "symmetries" are already
assumed implicitely in the derivation of the Lagrangian. So, from a
deduction POV, Noether's theorem carries no new information. It is just
uncovering OF a hidden premise.
Mike
Androcles
"Mike" <ele...@yahoo.gr> wrote in message
news:1133089148.3...@o13g2000cwo.googlegroups.com...
>
> Bilge wrote:
>> Mike:
>> >
>> >It was a puzzle only for morons. Some books stated the AXIOM of
>> >conservation of energy correctly but some other books skipped the
>> >"isolated system" condition.
>>
>> Oh? Go ahead, prove energy is conserved for any system you define,
>> without using noether's theorem. First make sure you define
>> energy unambiguously. Good luck.
>
> First, I have asked you to be a man, not a degenerate, and stop
> re-directing posts to alt.morons like a little kid who makes a
> mischieve but fears to face the consequences. Wear pants and face the
> truth about what you are. You direct to alt.morons because it IS THERE
> YOU ACTUALLY BELONG.
Why do you even bother with the moron? He's done that for years,
he's not about to grow up.
Androcles.
Mike
Androcles wrote:
> "Mike" <ele...@yahoo.gr> wrote in message
> news:1133089148.3...@o13g2000cwo.googlegroups.com...
> >
> > Bilge wrote:
> >> Mike:
> >> >
> >> >It was a puzzle only for morons. Some books stated the AXIOM of
> >> >conservation of energy correctly but some other books skipped the
> >> >"isolated system" condition.
> >>
> >> Oh? Go ahead, prove energy is conserved for any system you define,
> >> without using noether's theorem. First make sure you define
> >> energy unambiguously. Good luck.
> >
> > First, I have asked you to be a man, not a degenerate, and stop
> > re-directing posts to alt.morons like a little kid who makes a
> > mischieve but fears to face the consequences. Wear pants and face the
> > truth about what you are. You direct to alt.morons because it IS THERE
> > YOU ACTUALLY BELONG.
>
> Why do you even bother with the moron? He's done that for years,
> he's not about to grow up.
>
> Androcles.
>
> >
I guess you right. There is no point in doing that. I won't argue with
him any more. Just tell'em he is a moron. :)
Mike
Androcles
>
> Androcles wrote:
>> "Mike" <ele...@yahoo.gr> wrote in message
>> news:1133089148.3...@o13g2000cwo.googlegroups.com...
>> >
>> > Bilge wrote:
>> >> Mike:
>> >> >
>> >> >It was a puzzle only for morons. Some books stated the AXIOM of
>> >> >conservation of energy correctly but some other books skipped the
>> >> >"isolated system" condition.
>> >>
>> >> Oh? Go ahead, prove energy is conserved for any system you define,
>> >> without using noether's theorem. First make sure you define
>> >> energy unambiguously. Good luck.
>> >
>> > First, I have asked you to be a man, not a degenerate, and stop
>> > re-directing posts to alt.morons like a little kid who makes a
>> > mischieve but fears to face the consequences. Wear pants and face the
>> > truth about what you are. You direct to alt.morons because it IS THERE
>> > YOU ACTUALLY BELONG.
>>
>> Why do you even bother with the moron? He's done that for years,
>> he's not about to grow up.
>>
>> Androcles.
>>
>> >
>
> I guess you right. There is no point in doing that. I won't argue with
> him any more. Just tell'em he is a moron. :)
>
> Mike
Has it occured to you that he doesn't want a reply?
He thinks he has a victory by having the last word.
If nobody replies to him, he will have succeeded.
I wish him every success, he'll have nobody to
converse with. Then he'll either fuck off or start
a thread of his own which we can redirect to alt.moron.
It's quite easy, go to Google groups, select "show options",
"Reply", "Add follow up header". I never read his shit,
he's been killfiled for a long time.
A.
Bilge
>Bilge wrote:
>> Mike:
>> >
>> >It was a puzzle only for morons. Some books stated the AXIOM of
>> >conservation of energy correctly but some other books skipped the
>> >"isolated system" condition.
>>
>> Oh? Go ahead, prove energy is conserved for any system you define,
>> without using noether's theorem. First make sure you define
>> energy unambiguously. Good luck.
>
>First, I have asked you to be a man, not a degenerate,
You could save some typing if you waited until you had a reason
to ask.
> and stop
>re-directing posts to alt.morons like a little kid who makes a
Start posting something relevant to this newsgroup and save yourself
the tantrum:
>mischieve but fears to face the consequences. Wear pants and face the
>truth about what you are. You direct to alt.morons because it IS THERE
>YOU ACTUALLY BELONG.
Why do you care what I'm wearing when I post? Some sort of
bizarre newsgroup fetish?
>you go ahead and learn the basics. Learn for instance that AXIOMS are
>not provable propositions.
GADZOOKS! Who would have known?
> It is just something you accept to be true
>because otherwise your world would not make sense.
And one of those axioms is that you never post anything that
makes sense, the world does make sense.
>Noether's theorem is just about symmetries and related conservation AND
>NOT about just conservation as many want to extend it to. Conservation
OK, so what you're telling me is that you refuse to answer the
question and prefer shovelling bullshit. That's ok, since the
question was rhetorical. I really didn't expect you to be able
post anything but bullshit and you lived up to my expectations.
Mike
Bilge wrote:
[snip crap]
>
> Why do you care what I'm wearing when I post? Some sort of
> bizarre newsgroup fetish?
>
I do not. You should care. Pretending to be a man while you behavior
shows otherwise must be of concern to you.
Mike
Koobee Wublee
>
> >http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter3.pdf
> >
> >Go to page 157. What do equations 3.5.1, 3.5.2, and 3.5.3 indicate?
>
> Holy letter transposition, scatman! The equations indicate that using
> different letters confuses you even more than you're confused about the
> difference in what the authors call \sigma and s (which are exactly the
> same quantities that I called s and \tau). I didn't see all of what tom
> posted, but I'm quite certain he isn't confused by any of this.
It is more than what you are claiming. You need to read the whote chapter
more carefully.
> So, let's see - you take offense when anyone complains about you
> inventing your own terminolgy, yet, you find it really confusing
> when I use different letts in equations than whatever you've dug
> up from the so-called ``professionals?'' I guess that's self-con-
> sistent if your goal is to force people to guess what you mean
> in order to reject what they tell you and cite a reference that
> used different letters in the same equations. You'll excuse me if
> I don't have a lot of sympathy for someone who goes out of his
> way to reject something I've written in terms of the most commonly
> used symbols in a less compact way than it could be written if
> I was talking to someone who understood the subject.
So, in one instance you use 's' and the other 'sigma' that is because 'ds =
0'. Yeah, you are full of BS. Proper usage of math is not about BS.
> >You are wrong!
>
> Once again, you demonstrate that inherent efficiency in rejecting
> anything anyone tells you by not bothering to spend the effort
> in trying and understand something you plan to reject anyway.
Your accusation is very faulty.
> >However, you are not entirely wrong. 's' is only a valid coordinate for
> >non-photons.
>
> It's not a coordinate. It's the affine parameter that paramaterizes
> the proper time of an object along a geodesic trajectory.
It is used as a coordinate by almost all authors.
> I told
> you exactly why you're confused about 3 posts ago. You are confusing
> the metric which defines the geodesics with the proper time of an
> object moving along those geodesics. (what they call \sigma^2 is
> what I called s^2. What they call s is what I called \tau.)
What you call s^2 or \tau are manipulated as if they are coordinates. I
still don't see your point of insult.
> >The professional such as the authors realized this.
>
> Oh, I forgot - a ``professional'' is someone from out of town.
Of the topic, but apparently you have a very proprietary dictionary.
> What they realize is quite different from what you think those
> authors realize. In particular, the exact same equations are
> found in both wald and schutz, but since you won't bother to
> look at a textbook, I'm not going to bother hunting down the
> page relevant pages.
I still don't see your objections. Even Wald and Schutz, which I take your
word for, manipulated your so affine parameter as if a coordinate. It is
time to call it a coordinate instead of beating around the bush, and make it
more unusual than what it actually represents.
> >They resort to use another parameter 'sigma' which is cleverly
> >presented on page 143.
>
> Holy affine paramater, scatman! They used a different letter.
> It sounds like the work of that villian, the rindler.
Yes, you wrote that a few sentences ago.
> All you've pointed out is that you're immune to education. So long
> as you are determined to remain ignorant, I'm certainly not about to
> waste the effort required to post anything which assumes otherwise.
If I am immune to education, I would not get where I am today. Education is
a rational effort by the students to learn, study, and rationalize. After
doing so, a learned student with enough education should be able to discern
rights from wrongs. Apparently, you are not there yet.
> >This Lagrangian works for photons as well. After poo-poo-ing this
> >Lagrangian based on (x^0), have you had a change of heart?
>
> What coordinate system does x^0 refer to? If you had bothered to
> figure out what x^0 refers to, instead of presuming every observer
> uses the same x^0, you could have saved yourself some effort.
Every observer would have his own x^0.
> Let me know when you decide you're more interested in understanding
> physics than you are in being famous for proving that general relativity
> is wrong and being credited for the terminology of the new paridigm.
Now.
> >After all, all the mathematics do indicate this Lagrangian is indeed
> >a very valid Lagrangian.
>
> How would you know? You keep arguing that a lagrangian is something
> other than what everyone else calls a lagrangian and don't even
> seem to know variational calculus is involved. (You might consider
> looking up emmy noether to see what her mathematical specialty was
> and why hilbert hired her to assist in his pursuit of the field
> equations - that is where noether's theorems originated.)
A true Lagrangian is a density to an event that satisfies all the
constraints of the Lagrangian Method. When the event represents a minimal
action, the Euler-Lagrangian equations come about.
Tom Roberts
> Tom Roberts wrote:
>> s is not a valid coordinate.
>
> Go to page 157. What do equations 3.5.1, 3.5.2, and 3.5.3 indicate? You
> are wrong!
Like so many other people around here you need to SIGNIFICANTLY improve
the accuracy with which you read. Your claim is wrong.
The text says QUITE CLEARLY that s is the ARC-LENGTH PARAMETER OF THE
PATH, and makes no mention of using it as a coordinate at all.
In the Lagrangian approach to determining a geodesic path in a manifold
(the subject of those equations), the distinction between the path
parameter and the coordinates is ABSOLUTELY CRUCIAL. The fact that you
do not understand this seems to me to be the center of your confusions.
In the Lagrangian approach to Newtonian mechanics, that is
equivalent to confusing time with a generalized coordinate.
Doing so yields complete NONSENSE.
And you never attempted the little exercise I suggested:
[...] what would we call a
variable q2 for which neither q2 nor q2. appears in L?
Until you understand why the answer to this question is "an irrelevant
variable" you don't understand this at all.
Here's another one:
Exercise: What is the conserved Noether current corresponding
to an "irrelevant variable" (as defined above)?
Hint: do not guess, apply Noether's actual theorem.
Tom Roberts tjro...@lucent.com
Koobee Wublee
> Koobee Wublee wrote:
> > Tom Roberts wrote:
>>> s is not a valid coordinate.
>>
>> http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter3.pdf
>> Go to page 157. What do equations 3.5.1, 3.5.2, and 3.5.3 indicate? You
>> are wrong!
> TYPO: page 167.
Point well taken. Originally, I had the actual physical page number of 56
on it. After realizing that I should be using the documented page number, I
somewhat neglected to change the middile digit. Sorry about that, but I did
give the actual equation numbers. So, you had no problem to locate what I
was talking about. At least, youn have to give me credit on that foresight.
> Like so many other people around here you need to SIGNIFICANTLY improve
> the accuracy with which you read. Your claim is wrong.
>
> The text says QUITE CLEARLY that s is the ARC-LENGTH PARAMETER OF THE
> PATH, and makes no mention of using it as a coordinate at all.
Does it matter? If this parameter is manipulated as if it is a coordinate
which it is done so. The context is that 's' is manipulated as if it is a
coordinate. So, if you really insist. Fine, 's' is not a coordinate, but
it can be manipulated as if it is a coordinate. Happier?
> In the Lagrangian approach to determining a geodesic path in a manifold
> (the subject of those equations), the distinction between the path
> parameter and the coordinates is ABSOLUTELY CRUCIAL. The fact that you do
> not understand this seems to me to be the center of your confusions.
>
> In the Lagrangian approach to Newtonian mechanics, that is
> equivalent to confusing time with a generalized coordinate.
> Doing so yields complete NONSENSE.
The Lagrangian Method does not care about whether it is a coordinate or not.
All it cares about is that whether it is a state variable or not. Calling
this variable a coordinate or not is rather silly. It is like arguing
Miller Lite is 'less filling' or 'taste great'.
> And you never attempted the little exercise I suggested:
>
> [...] what would we call a
> variable q2 for which neither q2 nor q2. appears in L?
Yes, I have. My answer remains as what I wrote. Recall
Koobee Wublee wrote several posts ago:
> All other independent variables are independent of our system,
> so to our system, these variables are just as good as being conserved.
> Yes, I really seriously think there are an infinite number of 'conserved
> quantities' in a given system."
> Until you understand why the answer to this question is "an irrelevant
> variable" you don't understand this at all.
>
> Here's another one:
>
> Exercise: What is the conserved Noether current corresponding
> to an "irrelevant variable" (as defined above)?
> Hint: do not guess, apply Noether's actual theorem.
You really do give out tough homework assignments. OK, give me a few days
to chew on that one. This assignment has no impact on what we are
discussing, but it is a good homework assignment never-the-less.
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:dmfhqt$4...@netnews.net.lucent.com...
>>PATH, and makes no mention of using it as a coordinate at all.
>
> Does it matter?
Yes. Very much so.
> If this parameter is manipulated as if it is a coordinate
> which it is done so.
It cannot be done. A coordinate must be a field on the spacetime
manifold, and s cannot be one -- it is PATH DEPENDENT. That is, there is
no unique assignment of values of s to points in the manifold, because
the value you obtain for s is path dependent.
What you suggest is as nonsensical as attempting to use time as a
"spatial coordinate" in Newton/Lagrange mechanics -- it simply DOES NOT
MAKE SENSE.
> The Lagrangian Method does not care about whether it is a coordinate or not.
Not true. Just LOOK at the Euler equations, and note the difference
between the way coordinates appear in them and the way the path-length
parameter appears. In particular: all coordinates (and velocities)
appear as partial derivatives, but the path-length pararmeter appears in
a total derivative. This is DIRECTLY related to the difference between
coordinates and path-length discussed above.
>>And you never attempted the little exercise I suggested:
>> [...] what would we call a
>> variable q2 for which neither q2 nor q2. appears in L?
>
> Yes, I have. My answer remains as what I wrote. Recall
>>All other independent variables are independent of our system,
>>so to our system, these variables are just as good as being conserved.
>>Yes, I really seriously think there are an infinite number of 'conserved
>>quantities' in a given system."
And you are still wrong. <shrug>
>>Exercise: What is the conserved Noether current corresponding
>>to an "irrelevant variable" (as defined above)?
>>Hint: do not guess, apply Noether's actual theorem.
>
> You really do give out tough homework assignments. OK, give me a few days
> to chew on that one. This assignment has no impact on what we are
> discussing, but it is a good homework assignment never-the-less.
Your confusions about this are DIRECTLY related to the discussion. To
move the discussion forward, I'll give the answer: the "conserved"
Noether current corresponding to changes of an irrelevant variable is 0.
Yes, the number 0 [#]. Not very enlightening or useful, is it? And
applying the term "conserved" to it is a PUN at best (the theorem really
says that (d/dt)0=0, a trivial identity).
In particular, the irrelevant variable itself is not
"conserved" as you seem to think -- clearly the Lagrangian
has no power over such variables at all, and cannot possibly
affect or restrict their values.
[#] This assumes the irrelevant variable is real.
Irrelevant variables are indeed irrelevant, and the Lagrangian can say
nothing at all about them. This OUGHT to be obvious.
Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message news:slrndoj9m5....@radioactivex.lebesque-al.net...
> Koobee Wublee:
[snip]
> >After all, all the mathematics do indicate this Lagrangian is indeed
> >a very valid Lagrangian.
>
> How would you know? You keep arguing that a lagrangian is something
> other than what everyone else calls a lagrangian and don't even
> seem to know variational calculus is involved.
Remember this one?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html
;-)
Dirk Vdm
Koobee Wublee
>
>> If this parameter is manipulated as if it is a coordinate which it is
>> done so.
>
> It cannot be done. A coordinate must be a field on the spacetime manifold,
> and s cannot be one -- it is PATH DEPENDENT. That is, there is no unique
> assignment of values of s to points in the manifold, because the value you
> obtain for s is path dependent.
>
> What you suggest is as nonsensical as attempting to use time as a "spatial
> coordinate" in Newton/Lagrange mechanics -- it simply DOES NOT MAKE SENSE.
I do think minimizing the elapsed proper spacetime is rather silly myself.
Since it is mathematically sound to do so, I have to go along with it. If
you really think it is absurd, go complaining it to Ciufolini/Wheeler and
the rest of the GR community. In the meantime, the Lagrangian as a density
of the action of elapsed observed time is the best choice of all. It is
even valid for photons where all others fail.
>> The Lagrangian Method does not care about whether it is a coordinate or
>> not.
>
> Not true. Just LOOK at the Euler equations, and note the difference
> between the way coordinates appear in them and the way the path-length
> parameter appears. In particular: all coordinates (and velocities) appear
> as partial derivatives, but the path-length pararmeter appears in a total
> derivative. This is DIRECTLY related to the difference between coordinates
> and path-length discussed above.
No, I don't notice anything you have raised. Euler-Lagrange equations deal
with variables. Who cares if they are coordinate or something else like you
are so vehimently defending.
>>>And you never attempted the little exercise I suggested:
>>> [...] what would we call a
>>> variable q2 for which neither q2 nor q2. appears in L?
>>
>> Yes, I have. My answer remains as what I wrote. Recall
>>>All other independent variables are independent of our system,
>>>so to our system, these variables are just as good as being conserved.
>>>Yes, I really seriously think there are an infinite number of 'conserved
>>>quantities' in a given system."
>
> And you are still wrong. <shrug>
If you think I am wrong, <shrug>
>>>Exercise: What is the conserved Noether current corresponding
>>>to an "irrelevant variable" (as defined above)?
>>>Hint: do not guess, apply Noether's actual theorem.
>>
>> You really do give out tough homework assignments. OK, give me a few
>> days to chew on that one. This assignment has no impact on what we are
>> discussing, but it is a good homework assignment never-the-less.
>
> Your confusions about this are DIRECTLY related to the discussion. To move
> the discussion forward, I'll give the answer: the "conserved" Noether
> current corresponding to changes of an irrelevant variable is 0. Yes, the
> number 0 [#]. Not very enlightening or useful, is it? And applying the
> term "conserved" to it is a PUN at best (the theorem really says that
> (d/dt)0=0, a trivial identity).
>
> In particular, the irrelevant variable itself is not
> "conserved" as you seem to think -- clearly the Lagrangian
> has no power over such variables at all, and cannot possibly
> affect or restrict their values.
>
You gave me an excellent homework problem to solve which I was looking
forward to do so and then mercifully took it away from me. That is not
fair. In the meantime, I have asked you to review the Lagrangian Method
thoroughly even how it is derived, and yet you have not done so. The
derivation of the Lagrangian involves a total derivative of this Lagrangian
and integration by part. It is rather silly to take the derivative of
something that is not a dependent variable. Please also go back to review
the Leibnitz/Newton's rules of derivative. You cannot get anymore basic
than that.
> [#] This assumes the irrelevant variable is real.
>
> Irrelevant variables are indeed irrelevant, and the Lagrangian can say
> nothing at all about them. This OUGHT to be obvious.
Stop going off on a tangent. What you are saying of (@L/@u =0) not
necessarily indicating a conserved quantity is so absurd. In doing so, you
have to prove Noether's Theorem wrong which I doubt you can. In the
meantime, you just cannot escape the destiny where
** @L/@s = 0, if L is a function of s.
** @L/@x^0 = 0, if L is a function f x^0.
** @L/@Tau = 0, if L is a function of Tau.
All these valid Lagrangians indicate a conserved quantity in energy. Energy
conservation becomes a fundamental principle.
Ilja Schmelzer
"Tom Roberts" <tjro...@lucent.com> schrieb
> What is truly amazing is that energy conservation works so well in
> Newtonian physics -- its pervasiveness there is what made people
> erroneously think that conservation of energy is a general "principle".
The problem is much deeper.
In the canonical QM scheme we have an operator H so that
i d_t psi = H psi
H is a self-adjoint operator, and may be used for measurement.
Eigenstates of H are preserved, that means, the value of H
is conserved. The thing measured by H is named energy.
As a consequence, conservation of energy is a basic
feature of quantum theory.
QFT is not an exception. Even if field theory makes all
this more complex, the wave function becomes a functional
and H an operator acting on spaces of functionals, the
basic scheme remains the same. Special relativity doesn't
change this game too. We have this scheme in every inertial
frame, and have to show that observable effects remain unchanged,
but in every such frame we have the same old picture.
Thus, missing energy conservation is a serious
problem for GR quantization.
Ilja
The thing named the operator which measures energy.
The consequence is the conservation of energy.
> Noether's theorem (1918) delineated the limits of such conservation
> "laws" by relating them to symmetries of the Lagrangian. At base this
> has to do with the fact that gravitation is so incredibly small, and the
> terms in the Lagrangian that generate violations of energy conservation
> are completely negligible in Newtonian situations.
>
> [Similar remarks apply to momentum conservation.]
>
>
> > Not true in General Relativity at all, since gravity is not a
> > force.
>
> It goes much deeper than that. See above.
>
>
> Tom Roberts tjro...@lucent.com
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:dmhvnd$f...@netnews.net.lucent.com...
>>coordinate" in Newton/Lagrange mechanics -- it simply DOES NOT MAKE SENSE.
>
> I do think minimizing the elapsed proper spacetime is rather silly myself.
Your phrase "proper spacetime" is unique to you (and is indeed rather
silly). But you have identified it with what everyone else calls proper
time (for timelike paths). Minimizing it is not "silly", this is the
_definition_ of a (timelike) geodesic in spacetime.
> If
> you really think it is absurd, go complaining it to Ciufolini/Wheeler and
> the rest of the GR community.
_THEY_ don't make this confusion, only YOU do so. <shrug>
>>>The Lagrangian Method does not care about whether it is a coordinate or
>>>not.
>>Not true. Just LOOK at the Euler equations, and note the difference
>>between the way coordinates appear in them and the way the path-length
>>parameter appears. In particular: all coordinates (and velocities) appear
>>as partial derivatives, but the path-length pararmeter appears in a total
>>derivative.
>
> No, I don't notice anything you have raised.
That's just one more illustration of the fact that you need to learn how
to read. <shrug>
> Euler-Lagrange equations deal
> with variables.
No, they deal with (generalized) coordinates and time (in classical
mechanics) or proper time (for a timelike geodesic in SR/GR). These
different TYPES of variables have different ROLES in the theory.
>>Irrelevant variables are indeed irrelevant, and the Lagrangian can say
>>nothing at all about them. This OUGHT to be obvious.
>
> Stop going off on a tangent.
This is not a "tangent", it is directly related to your confusions and
mis-statements.
> What you are saying of (@L/@u =0) not
> necessarily indicating a conserved quantity is so absurd.
I'm pointing out that considering the number 0 to be "conserved" is
silly, which is what Noether's theorem says for your irrelevant variable
u (which also must have @L/@u. = 0).
> In doing so, you
> have to prove Noether's Theorem wrong
I repeat: you need to learn how to READ. Go back and _READ_ what I
wrote. I was USING Noether's theorem, applying it to your irrelevant
variables and concluding that the "conserved current" corresponding to
them is the number 0.
> ** @L/@s = 0, if L is a function of s.
> ** @L/@x^0 = 0, if L is a function f x^0.
> ** @L/@Tau = 0, if L is a function of Tau.
>
> All these valid Lagrangians indicate a conserved quantity in energy.
No, they don't necessarily do so. One needs dL/dTau=0 for mass to be
conserved, where d/dTau is the _TOTAL_ derivative wrt Tau (proper time).
For dL/dx^0=0 then energy is conserved in the inertial frame with time
coordinate x^0. It is ESSENTIAL that the _TOTAL_ derivative be used,
because this conservation requires time translation invariance of L, and
the partial derivatives simply do not ensure that but the total
derivative does.
Remember the sound bite is: time translation invariance of L
implies energy conservation. That is, L(t)=L(t+deltat) for all
values of deltat implies energy conservation.
lim[deltat->0] (L(t+delta-t)-L(t))/deltat is the TOTAL
derivative, not the partial derivative.
To understand this you MUST learn the difference between partial and
total derivatives. And you MUST learn what Noether's theorem actually
says and not GUESS about it.
> Energy
> conservation becomes a fundamental principle.
No, it is not. Energy is only conserved when the Lagrangian has time
translation invariance. Your irrelevant variables. and your errors in
understanding the distinction between partial and total derivatives, and
your errors in understanding the difference between coordinates and the
path-parameter, do not affect this.
[This is going nowhere. I cannot teach you to read.
Unless you avoid repeating yourself, don't expect me to
respond any further.]
Tom Roberts tjro...@lucent.com
Bilge
>"Tom Roberts" <tjro...@lucent.com> schrieb
>> What is truly amazing is that energy conservation works so well in
>> Newtonian physics -- its pervasiveness there is what made people
>> erroneously think that conservation of energy is a general "principle".
>
>The problem is much deeper.
>
>In the canonical QM scheme we have an operator H so that
>
>i d_t psi = H psi
>
>H is a self-adjoint operator, and may be used for measurement.
>Eigenstates of H are preserved, that means, the value of H
>is conserved. The thing measured by H is named energy.
The reason H is the total energy is because the hamiltonian turned
out to be what noether's theorem said was conserved under a time
translation. Actually, what most people measure are the off-diagonal
matrix elements, since those correspond to an interaction between an
initial and final state.
>As a consequence, conservation of energy is a basic
>feature of quantum theory.
>
>QFT is not an exception. Even if field theory makes all
>this more complex, the wave function becomes a functional
>and H an operator acting on spaces of functionals, the
>basic scheme remains the same. Special relativity doesn't
>change this game too. We have this scheme in every inertial
>frame, and have to show that observable effects remain unchanged,
>but in every such frame we have the same old picture.
I think you missed the point. In newtonian theory, conservation
laws were ad hoc assumptions which everyone assumed were correct,
but no one was able to justify (or even give a formal definition
for the conserved quantities). As far as complexity goes, it's
all about the same. The basic difference is that the name changes
from generalized coordinate to field.
>Thus, missing energy conservation is a serious problem for GR quantization.
Why? Ever consider the possibility that it might be a feature that
holds a key rather than something to be avoided? We have a rigorous
definition for conserved energy. General relativity doesn't say that
energy isn't conserved, per se. It says you can't define a conserved
energy in general. That is not quite the same thing, since you can't
fail to conserve something that isn't well defined in the first place.
Genral relativity (and many other theories) don't conserve burfl,
since the burfl symmetry isn't respected by the lagrangian. But,
since invariance under burfl isn't well-defined, I might want to
reconsider my reasons for expecting burfl to be conserved.
Maybe it isn't supposed to be.
Ilja Schmelzer
"Bilge" <dub...@radioactivex.lebesque-al.net> schrieb
> Ilja Schmelzer:
> >"Tom Roberts" <tjro...@lucent.com> schrieb
> >> What is truly amazing is that energy conservation works so well in
> >> Newtonian physics -- its pervasiveness there is what made people
> >> erroneously think that conservation of energy is a general
"principle".
> >
> >The problem is much deeper.
> >In the canonical QM scheme we have an operator H
> >is conserved. The thing measured by H is named energy.
> >feature of quantum theory.
> I think you missed the point.
The purpose of my posting was to make a point, not to understand
one of your points. My point was:
>> Thus, missing energy conservation is a serious problem for GR
>> quantization.
>
> Why? Ever consider the possibility that it might be a feature that
> holds a key rather than something to be avoided? We have a rigorous
> definition for conserved energy. General relativity doesn't say that
> energy isn't conserved, per se. It says you can't define a conserved
> energy in general.
Thus, it says that some essential feature of quantum theory, the operator
H, cannot be defined too. Because its eigenvalues define energy.
That's the answer to your "why".
It means, or you have to modify GR to allow the definition of a
conserved energy, or you have to modify QM.
> That is not quite the same thing, since you can't
> fail to conserve something that isn't well defined in the first place.
> Genral relativity (and many other theories) don't conserve burfl,
> since the burfl symmetry isn't respected by the lagrangian. But,
> since invariance under burfl isn't well-defined, I might want to
> reconsider my reasons for expecting burfl to be conserved.
> Maybe it isn't supposed to be.
Maybe. But if you have to unify your theory with another
theory which uses that burfl in a central place, this is at least
a hint that, maybe, you should be better able to define burfl.
Ilja
Tom Roberts
> eigenvalues [of H] define energy.
No. Say rather that the eigenvalues of H define measurable values of the
observable corresponding to the operator H. For the case when the
Lagrangian is invariant under time translations, we can identigy that
observable with energy; for other cases I believe we cannot.
The _definition_ of energy is the Noether current corresponding to time
translation. AFAIK that is related to H only under certain circumstances.
> It means, or you have to modify GR to allow the definition of a
> conserved energy, or you have to modify QM.
I don't think that is necessarily required. It might be part of the
solution, it might not.
> [Bilge's burfl]
> if you have to unify your theory with another
> theory which uses that burfl in a central place, this is at least
> a hint that, maybe, you should be better able to define burfl.
A hint, yes. A requirement, no. Because it could be that the other
theory using burfl is not necessary. For QM and energy that seems to be
the case.
Note also that if physics is indeed local, then the lack of a global
definition energy in GR is not a problem -- local energy conservation is
exact in GR.
Tom Roberts tjro...@lucent.com
zzbu...@netscape.net
N:dlzc D:aol T:com (dlzc) wrote:
> When I first heard of this analogy, I was unsatisfied with the
> answer provided as to whether *we* are expanding too.
No we are are not expanding, since the only
thing that is possible to expand is physics idiots
and their associate in religously stupid
and equivalently very dead -- Pythagorus.
Since just like moderrn physics idiots he
reasoned by analogy with Doric Columns
and secret matrices. And the only result
he came up the absurb result
that sqrt(2) is both equal to pi and isn't equal to pi.
With the equally idiotic mathematical advice:
"It just depends on how you look at the problem".
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
<zzbu...@netscape.net> wrote in message
news:1133482208.9...@g14g2000cwa.googlegroups.com...
>
> N:dlzc D:aol T:com (dlzc) wrote:
>> When I first heard of this analogy, I was unsatisfied
>> with the answer provided as to whether *we* are
>> expanding too.
>
> No we are are not expanding, since the only
> thing that is possible to expand is physics idiots
> and their associate in religously stupid
> and equivalently very dead -- Pythagorus.
"Pythagoras". Don't know if you knew this, but he was peddling
himself as divine before the end...
David A. Smith
zzbu...@netscape.net
N:dlzc D:aol T:com (dlzc) wrote:
> Dear zzbunker:
>
> <zzbu...@netscape.net> wrote in message
> news:1133482208.9...@g14g2000cwa.googlegroups.com...
> >
> > N:dlzc D:aol T:com (dlzc) wrote:
> >> When I first heard of this analogy, I was unsatisfied
> >> with the answer provided as to whether *we* are
> >> expanding too.
> >
> > No we are are not expanding, since the only
> > thing that is possible to expand is physics idiots
> > and their associate in religously stupid
> > and equivalently very dead -- Pythagorus.
>
> "Pythagoras". Don't know if you knew this, but he was peddling
> himself as divine before the end...
But so did Alexander The Great. But thery'tre still both dead.
>
> David A. Smith
Ilja Schmelzer
"Tom Roberts" <tjro...@lucent.com> schrieb
> Ilja Schmelzer wrote:
> > eigenvalues [of H] define energy.
>
> No. Say rather that the eigenvalues of H define measurable values of the
> observable corresponding to the operator H. For the case when the
> Lagrangian is invariant under time translations, we can identigy that
> observable with energy; for other cases I believe we cannot.
Indeed, I have implicitly assumed that H does not depend on t.
Or do you think the fundamental theory of everything will work with
a time-dependent operator H(t)?
> The _definition_ of energy is the Noether current corresponding to time
> translation. AFAIK that is related to H only under certain circumstances.
In the Hamilton formalism H defines the energy. Always.
But that's not important. Usually these definitions coinside.
> > It means, or you have to modify GR to allow the definition of a
> > conserved energy, or you have to modify QM.
>
> I don't think that is necessarily required. It might be part of the
> solution, it might not.
I see no other way. If we leave QM unmodified, we have an energy
operator. (There was the attempt of canonical quantization with an
energy operator with eigenvalues only 0 (Wheeler-deWitt). It means
d_t Psi = 0. Thus, no evolution in time. This is, from my point of view,
already an essential modification of QM, even if we have formally
written down a Schrödinger equation.)
> > [Bilge's burfl]
> > if you have to unify your theory with another
> > theory which uses that burfl in a central place, this is at least
> > a hint that, maybe, you should be better able to define burfl.
>
> A hint, yes. A requirement, no.
There is, indeed, the other possibility to modify QM in such a way
that we no longer have the Schrödinger equation.
> Because it could be that the other
> theory using burfl is not necessary. For QM and energy that seems to be
> the case.
There is not a single reason to suspect that QM is unnecessary.
> Note also that if physics is indeed local, then the lack of a global
> definition energy in GR is not a problem -- local energy conservation is
> exact in GR.
First, H is global. As global as possible.
Second, to name the equation nabla_m T^mn = 0 "local energy conservation"
is only an unfortunate naming convention. The only justification is that
in the flat space limit it becomes a conservation law. But it is not a
conservation
law. Because it does not have the form partial_m something^m = 0.
Ilja
Bilge
>
[...]
>> Why? Ever consider the possibility that it might be a feature that
>> holds a key rather than something to be avoided? We have a rigorous
>> definition for conserved energy. General relativity doesn't say that
>> energy isn't conserved, per se. It says you can't define a conserved
>> energy in general.
>
>Thus, it says that some essential feature of quantum theory, the operator
>H, cannot be defined too. Because its eigenvalues define energy.
Transformations between hamiltonians are unitary transformations,
not lorentz boosts.
>That's the answer to your "why".
since different hamiltonians may not even describe the same
particles. For example, the kobayshi-moskawa matrix rotates
weak eigenstates into strong eigenstates and vice verse.
>It means, or you have to modify GR to allow the definition of a
>conserved energy, or you have to modify QM.
Why is that? An observer who defines a hamiltonian H, defines
the measurement of some energy _right_ now, _right_ here and
since right now and right here are completely indeterminate,
there is no problem defining the energy and momentum. It doesn't
depend upon what any other observer (except on the light cone)
defines as right here and right now, because those observers
cannot be reached by a light signal.
>> That is not quite the same thing, since you can't
>> fail to conserve something that isn't well defined in the first place.
>> Genral relativity (and many other theories) don't conserve burfl,
>> since the burfl symmetry isn't respected by the lagrangian. But,
>> since invariance under burfl isn't well-defined, I might want to
>> reconsider my reasons for expecting burfl to be conserved.
>> Maybe it isn't supposed to be.
>
>Maybe. But if you have to unify your theory with another
>theory which uses that burfl in a central place, this is at least
>a hint that, maybe, you should be better able to define burfl.
``My theory'' in this case is relativity, and space and time have no
unique central place in the theory. The entire point is that space and
time disappear from the equations containing any physics. I'm not so hung
up on classical reality that I'm going to force nature to be something it
isn't. You might feel the need to define burfl, but I'm willing to let
nature tell me if defining it is inconsistent with the physics that
describes this universe.