GR,WM,QM
Ken S. Tucker
density in QM, (also Wave Mechanics) is fairly straightforward.
I think the problem teachers have is relating GR density using
boundary conditions, the same is required for QM, that being
a finite (bounded) volume.
Sometimes something conceptually simple can get involved using
math, like a poker game.
Let me try an example, as I understand, starting with AE's law,
G_uv = T_uv or abbreviated to G = T ,
and we'll skip the constants, and stay conceptual.
The density T at a boundary r from the Sun is
T = M/r^3 == Mass/Volume.
Curvature G is the rate of change of acceleration,
G = M/r^3 , all's well.
De Broglie introduces WM (Wave Mechanics) and gives us the
concept of a "wave group" by employing E=hf so now we have,
T = G = f/r^3 , (M = f),
as a sort of probability density of where the Sun is.
Let me set a wave group frequency to,
F = f/r^3
then that frequency F sets forth the Probability density
of the Sun at any r.
If I have to draw a line demarcating QM from WM I suppose
it would be the Dirac Matrices found QM distinct from WM,
comments welcome of course.
The Dirac Matrices are 4D orthogonal, do we want to express
them in general coordinates?
Regards
Ken S. Tucker
Tom Roberts
> [...]
> G_uv = T_uv or abbreviated to G = T ,
That's not an "abbreviation", that's the Einstein field equation written
first in components and then in tensors. At least that's what everybody
else means with this notation.
[No need to "skip the constants", as one simply
chooses units in which the constant is 1. At least
that's what everybody else does. Other common choices
of units make the constant be 8*pi or 8*pi*G (Newton's
gravitational constant, not the curvature tensor).]
> [...]
> Curvature G is the rate of change of acceleration,
No, it isn't. The curvature G is a TENSOR, and is far more complicated.
[Or if you insist on your "abbreviation" above, then
G is standing for a set of 16 components, not just one,
and "rate of change of acceleration" is at most one
of them.]
> G = M/r^3 , all's well.
All is NOT well -- you just attempted to equate a tensor to a real
number. That's nonsense.
You repeatedly make this sort of mistake. You MUST define your symbols,
and check that every equation equates quantities of similar types.
Mathematics, like Java, is a strongly-typed language (but there is no
compiler to enforce it). You need to be MUCH more careful -- your
carelessness repeatedly leads you off into never-never land....
> [... goes from wrong to worse]
Tom Roberts
Ken S. Tucker
> Ken S. Tucker wrote:
> > [...]
> > G_uv = T_uv or abbreviated to G = T ,
>
> That's not an "abbreviation", that's the Einstein field equation written
> first in components and then in tensors. At least that's what everybody
> else means with this notation.
Good, then I presume you have no problem understanding the notation,
that's what's important, I would further suggest you begin a new
thread
advocating your views on how tensors should be expressed in ascii.
[...]
> > Curvature G is the rate of change of acceleration,
>
> No, it isn't. The curvature G is a TENSOR, and is far more complicated.
Perhaps you could enlighten us with your definition of the rate of
change of acceleration in GR, then we could move forward, following
your views.
Ken
[...]
Oh No
>The relation of an energy-mass density in GR to a probability
> density in QM, (also Wave Mechanics) is fairly straightforward.
I'm afraid it is not straightforward at all, as shown by the Page-
Geilker experiment, and the Eppley-Hannah thought experiment. In fact
many of the issues of compantibility between gtr and qm are centred upon
this issue, and on what happens when the quantum wave function
collapses. If there were a straightforward relationship, then the
collapse of the wave function would instigate an immediate, faster than
light, change in the structure of spacetime.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Ken S. Tucker
> Thus spake Ken S. Tucker <dynam...@vianet.on.ca>
>
> >The relation of an energy-mass density in GR to a probability
> > density in QM, (also Wave Mechanics) is fairly straightforward.
>
> I'm afraid it is not straightforward at all, as shown by the Page-
> Geilker experiment, and the Eppley-Hannah thought experiment. In fact
> many of the issues of compantibility between gtr and qm are centred upon
> this issue, and on what happens when the quantum wave function
> collapses. If there were a straightforward relationship, then the
> collapse of the wave function would instigate an immediate, faster than
> light, change in the structure of spacetime.
Sounds like a subject, here's a couple of ref's if needed,
The description of a probablity density herein is ok with me,
http://en.wikipedia.org/wiki/Wave_function
to start.
This article is also ok,
http://en.wikipedia.org/wiki/Wave_function_collapse
Suppose we're in a large windowless rocket, in freefall but
with some free floating sphere's within it, (Baez uses coffee
grains) then the relative acceleration of the sphere's provides
us with the g-field curvature, and then the proximity to the
gravitating mass. In turn the proximity to the mass can be
defined as a density. The actual location of the gravitating
mass is unknown, but once the atmosphere is encountered
the actualization begins based on a measurable EM action,
as the rocket reacts with the atmoshpere.
That 're-entry' I'd describe as a "collapsing wave function"
since we're uncertain whether or not we're falling into a huge
gas giant like Jupiter or into Earth, but the collapse requires
(IMHO) a photon exchange.
Regards
Ken S. Tucker
> Regards
>
> --
> Charles Francis
> moderator sci.physics.foundations.
> charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
> braces)
>
> http://www.rqgravity.net
======================================= MODERATOR'S COMMENT:
Please trim (delete) uncommented quoted text. -fd
Rich L.
...
> Suppose we're in a large windowless rocket, in freefall but
> with some free floating sphere's within it, (Baez uses coffee
> grains) then the relative acceleration of the sphere's provides
> us with the g-field curvature, and then the proximity to the
> gravitating mass. In turn the proximity to the mass can be
> defined as a density. The actual location of the gravitating
> mass is unknown, but once the atmosphere is encountered
> the actualization begins based on a measurable EM action,
> as the rocket reacts with the atmoshpere.
> �That 're-entry' I'd describe as a "collapsing wave function"
> since we're uncertain whether or not we're falling into a huge
> gas giant like Jupiter or into Earth, but the collapse requires
> (IMHO) a photon exchange.
> Regards
> Ken S. Tucker
>
Ken,
The problem with using the QM probability density as the stress-energy
tensor in GR is that there are situations where the probability
density may extend in some directions for quite some distance/time
before it "collapses" to what finally happens. Consider, for example,
two correlated (entangled) particles emitted billions of years ago and
radiating out into space. The probability density in QM will be a
cloud that expands roughly spherically in all directions. If neither
of the particles is "measured" (whatever that means...) for several
billion years, then the probability density cloud will extend over
billions of cubic light years. According to QM, when one of these
particles is finally "measured", then this huge cloud
"instantaneously" collapses to a form that represents only the paths
the particles "actually" traveled. If the probability cloud was the
input to the stress-energy tensor, this implies that the stress-energy
tensor also abruptly changes, as does the geometry of space.
One post recently suggested that the geometry of space is a
superposition of all the states corresponding to all the possible QM
wave function paths. I still have severe problems with this view
because the "instantaneous collapse" of the wave function is not
something that is relativistically invariant. If one observer sees it
as instantaneous, there must be other observers who would see it as
propagating in one direction or another, at much faster than the speed
of light. Also, what does it mean for the geometry of space to
suddenly crystallize into a particular state when the wave function
collapses? Does that mean that some particle somewhere suddenly
experiences a gravitational "force" that wasn't there a moment
before? Wouldn't this be a rather spooky way for the universe to
work? (By "spooky" here I mean that this is not our common
experience.)
I know I'm swimming up stream a bit, but I'm inclined to believe that
there is something fundamentally real about the location of a particle
(or energy of any form) at any moment. I'm not sure how to reconcile
this view with quantum mechanics, but I have great difficulty
understanding how the universe can maintain conservation of energy-
momentum unless there is some sort of reality beyond what QM seems to
allow. I think we need to re-think our concepts of time and causality
and events before we will resolve this issue, but I certainly haven't
managed all that yet myself.
Rich L.
Tom Roberts
As usual, your excessively loose way of talking is almost useless. There is no
general or generic "rate of change of acceleration" in GR. So let me enumerate
some possibilities.
The "rate of change of acceleration" of an object could refer to its coordinate
acceleration or to its 4-acceleration. The former is essentially useless in
general discussions, as the choice of coordinates is often the major
contribution to coordinate acceleration (and its derivatives) -- coordinates are
completely arbitrary so I won't attempt to discuss it.
The rate of change of 4-acceleration of a small pointlike object is 1/m times
the rate of change of the object's 4-momentum P (for constant mass m of the
object, which is implied by its being pointlike). We know
DP/d\tau = F
where \tau is its proper time, and F is the total applied force on the object
(F=0 for gravity, which appears inside the connection of the covariant
derivative). So the rate of change of such an object's 4-acceleration is clearly
D(F/m)/d\tau. I doubt that is very useful.
The rate of change of 4-acceleration of an object that is not small or is not
pointlike is problematical, as such an object might not have a definite
4-momentum or 4-velocity. In any case, there is no simple equation for the
motion of such an object, and the full Einstein field equation must be used. No
general statements can be made.
One might attempt to look for a field that corresponds to "acceleration", in
analogy to the gravitational acceleration in Newtonian mechanics. This search
fails in GR, as there is no such field; consequently attempting to discuss "rate
of change" of such a quantity also fails.
One might perform that last search in an approximation to GR. It also fails for
the approximation known as linearized GR, as the "gravitational force" on a
small pointlike object is not a field and depends on the object's position AND
velocity. In the approximation known as Newtonian Gravitation one can indeed
find such a field -- but one has completely abandoned GR.
Note that NONE of these are related to curvature, as in your original claim.
> then we could move forward, following
> your views.
You would have much more success in discussing such topics, and would be able to
"move forward", if you actually STUDIED General Relativity. Whatever it is you
are doing does not work.
Tom Roberts
Oz
> Also, what does it mean for the geometry of space to suddenly
>crystallize into a particular state when the wave function collapses? Does that
>mean that some particle somewhere suddenly experiences a gravitational "force"
>that wasn't there a moment before? Wouldn't this be a rather spooky way for the
>universe to work? (By "spooky" here I mean that this is not our common
>experience.)
>
>I know I'm swimming up stream a bit, but I'm inclined to believe that there is
>something fundamentally real about the location of a particle (or energy of any
>form) at any moment. I'm not sure how to reconcile this view with quantum
>mechanics, but I have great difficulty understanding how the universe can
>maintain conservation of energy- momentum unless there is some sort of reality
>beyond what QM seems to allow. I think we need to re-think our concepts of time
>and causality and events before we will resolve this issue, but I certainly
>haven't managed all that yet myself.
I'm afraid I am an infrequent reader and even more infrequent poster,
but no matter. Like pretty well everyone here I consider this quite
often. Having oscillated over some 30 years between the particle and
wave representation and indeed pretty well every mix in between I came
down quite strongly on the side of the hugely unpopular wave
description. There are a number of reasons for this which I may briefly
sketch over later.
So I DON'T see a proton as existing at all. That's right, no such thing.
Its one way (very useful but unphysical) to describe an EM wave. I see
quantum effects ONLY on emitters and detectors which are ALL (notice)
inherently quantised devices. That is they display a change of state
from one state to another.
A 'typical' detector is an array of resonant aerials (atoms, usually)
exposed to an incident EM wave. These react to the incoming wave but not
with each one taking an average figure but with occasional 'spikes', and
its these that trigger the change of state. The maths for this has been
done, but I no longer have the reference.
Its common to consider some photon being emitted by some distant galaxy
and being adsorbed by a particular atom on earth, as if it really worked
that way. Of course it doesn't. The probability of one individual event
pair like that is absolutely infinitesimal. Instead the star emits
untold billions of 'photons' or as I would say, an oscillating em field
of a certain size and strength. On earth we set out our detectors and
they eventually accumulate enough energy (resonance, see) to trigger one
change of state in one atom of the detector. That's not actually one
photon being emitted in that galaxy and travelling all the way here to
'hit' our detector. Its am EM field transmitting for enough amplitude
and time to trigger a state change in our detector.
You can use the same argument for the photoelectric effect and pretty
well all (perhaps all) the mechanisms that claim to detect 'a photon'.
Now QED is perfectly reasonable. Its just how an em wave would work when
surrounded by such quantised detectors. Of course they seem to go every
which way, its because they DO go every which way.
Particle physicists tend to forget that their complex statistical
techniques for handling "a particle interaction" is really not one
interaction but one chosen out of quadrillions that were happening, and
they forget about the energy dumped elsewhere by the 'particle' emitter
that is their source.
I rather doubt inherently quantised 'particles' like electrons, which
can be perfectly well described as waves (more or less) behave
significantly differently. The key difference is really one of 'typical
wavelength. However material scientists are perfectly happy to have
electrons smeared out over the whole VOLUME of a crystal (metal) and can
make accurate quantitative analyses on this assumption. The real
question is why do we have quantised particles? I don't know. However
the idea of small dimensions (which may have a size that depends on the
energy) is particularly attractive.
This post is long enough.
--
Oz
Rich L.
You bring up a number of issues that I've also thought about off and
on over the last 30 years. I've come to some different opinions and
would like to argue a couple of your points:
On Sep 30, 6:44�am, Oz <O...@mailcatch.com> wrote:
...
> So I DON'T see a proton as existing at all. That's right, no such thing.
> Its one way (very useful but unphysical) to describe an EM wave. I see
> quantum effects ONLY on emitters and detectors which are ALL (notice)
> inherently quantised devices. That is they display a change of state
> from one state to another.
I'm tending towards a contrary opinion. What appears to be
fundamental is the emission and absorption of energy quanta (photons,
electrons, whatever). What we attribute to being wave-like character
I am suspecting is a "rest mass frequency" observed in the local
observers frame. It is no longer clear to me that we need to describe
the propagation of the quanta explicitly because it is impossible to
observe, therefore a meaningless question to ask about. All we know
is that something was emitted here and then appears over there.
Everything else in our theories is mathematical constructions that
enable us to calculate probabilities, but I am suspecting that much of
these constructions are not real, or are like shadows of what is
really going on.
> A 'typical' detector is an array of resonant aerials (atoms, usually)
> exposed to an incident EM wave. These react to the incoming wave but not
> with each one taking an average figure but with occasional 'spikes', and
> its these that trigger the change of state. The maths for this has been
> done, but I no longer have the reference.
>
> Its common to consider some photon being emitted by some distant galaxy
> and being adsorbed by a particular atom on earth, as if it really worked
> that way. Of course it doesn't. The probability of one individual event
> pair like that is absolutely infinitesimal. Instead the star emits
> untold billions of 'photons' or as I would say, an oscillating em field
> of a certain size and strength. On earth we set out our detectors and
> they eventually accumulate enough energy (resonance, see) to trigger one
> change of state in one atom of the detector. That's not actually one
> photon being emitted in that galaxy and travelling all the way here to
> 'hit' our detector. Its am EM field transmitting for enough amplitude
> and time to trigger a state change in our detector.
This is an idea I've considered as well, but ended up rejecting. My
problem with this has to do with conservation of energy-momentum. I
am inclined to believe that conservation of energy-momentum is not a
statistical thing, but must always conform with the uncertainty
principle (delta_E * delta_T is about equal to h_bar, always). If the
source is radiating a wave and absorbers are responding to this wave
locally and without any coordination with other absorbers, then it is
easy for conservation of energy-momentum to be true statistically, on
average, but it would not be true in detail for every emission-
absorbtion. Sometimes the photon would be emitted and nothing would
be absorbed anywhere, and sometimes one photon would be emitted and
two absorbed. I think the fact that we never see the latter in high
energy particle experiments argues that conservation of energy-
momentum must be strictly true, not just statistically true.
...
> Particle physicists tend to forget that their complex statistical
> techniques for handling "a particle interaction" is really not one
> interaction but one chosen out of quadrillions that were happening, and
> they forget about the energy dumped elsewhere by the 'particle' emitter
> that is their source.
I don't buy this argument. I believe the particle physicists would
have noticed, on bubble chamber tracks for example, if there was a
significant probability for one emission event to result in multiple
detection events. You'd occasionally see a particle decay and instead
of getting the expected combination of decay products, they'd see some
products that were doubled. The whole idea of conservation laws
(parity, charge, strangeness, etc.) is to explain the relationships
that are observed rigorously, not just statistically.
Rich L.
Oz
>Oz,
>
>You bring up a number of issues that I've also thought about off and
>on over the last 30 years. I've come to some different opinions and
>would like to argue a couple of your points:
>
>On Sep 30, 6:44�am, Oz <O...@mailcatch.com> wrote:
> ...
>> So I DON'T see a proton
>as existing at all. That's right, no such thing.
>> Its one way (very useful but unphysical) to describe an EM wave. I see
>> quantum effects ONLY on emitters and detectors which are ALL (notice)
>> inherently quantised devices. That is they display a change of state
>> from one state to another.
>
>I'm tending towards a contrary opinion. What appears to be
>fundamental is the emission and absorption of energy quanta (photons,
>electrons, whatever).
That's what I just said.
>What we attribute to being wave-like character
>I am suspecting is a "rest mass frequency" observed in the local
>observers frame.
Yes, but that's a wave, right? 'frequency' is a wave thingy.
> It is no longer clear to me that we need to describe
>the propagation of the quanta explicitly because it is impossible to
>observe, therefore a meaningless question to ask about.
Yes, except you have to describe things like diffraction, and that does
rather seem to be an interaction of propagating things, doesn't it?
>All we know
>is that something was emitted here and then appears over there.
>Everything else in our theories is mathematical constructions that
>enable us to calculate probabilities, but I am suspecting that much of
>these constructions are not real, or are like shadows of what is
>really going on.
I just said that too. They are means of arriving at the correct answer
in an easy way, no doubt about that. Unfortunately they tend to lead to
some strange devices. Using statistical point like objects to represent
a wave is just fine, but one doesn't really think of a sonata as being
made of a load of square pulses, even if that works as a mathematical
convenience.
>
>> A 'typical' detector is an array of resonant aerials (atoms, usually)
>> exposed to an incident EM wave. These react to the incoming wave but not
>> with each one taking an average figure but with occasional 'spikes', and
>> its these that trigger the change of state. The maths for this has been
>> done, but I no longer have the reference.
>>
>> Its common to consider some photon being emitted by some distant galaxy
>> and being adsorbed by a particular atom on earth, as if it really worked
>> that way. Of course it doesn't. The probability of one individual event
>> pair like that is absolutely infinitesimal. Instead the star emits
>> untold billions of 'photons' or as I would say, an oscillating em field
>> of a certain size and strength. On earth we set out our detectors and
>> they eventually accumulate enough energy (resonance, see) to trigger one
>> change of state in one atom of the detector. That's not actually one
>> photon being emitted in that galaxy and travelling all the way here to
>> 'hit' our detector. Its am EM field transmitting for enough amplitude
>> and time to trigger a state change in our detector.
>
>This is an idea I've considered as well, but ended up rejecting. My
>problem with this has to do with conservation of energy-momentum.
Why? Clearly a detector that detects 1 ue-v will detect 1 ue-v. What
else? We know classical EM waves transmit momentum. We can only confirm
a really accurate energy-momentum transfer when we have plenty of
absorption of the em wave to beat the noise, so unsurprisingly we get
apparent (and actual) e-momentum transfer. However I would bet that if
you took any random detection of a 'photon' (is an energy jump) it would
NOT be *precisely* as expected, because there would be noise and other
confounding effects. So that individual photon would NOT show accurate
e-momentum transfer (but h is your friend here in explaining). Clearly
as energy goes up, noise goes down, but the concept remains the same.
>I
>am inclined to believe that conservation of energy-momentum is not a
>statistical thing, but must always conform with the uncertainty
>principle (delta_E * delta_T is about equal to h_bar, always).
Ie its statistical, ok?
>If the
>source is radiating a wave and absorbers are responding to this wave
>locally and without any coordination with other absorbers,
Quite clearly they can NOT be responding without co-ordination with
other local absorbers. They are (effectively) all tuned to the same
frequency and as such will be emitting and absorbing em radiation, in
essence a multiply coupled excited pendulum system.
>then it is
>easy for conservation of energy-momentum to be true statistically, on
>average, but it would not be true in detail for every emission-
>absorbtion.
As I said.
>Sometimes the photon would be emitted and nothing would
>be absorbed anywhere,
Only if the energy density in the device (which would have to be closed
or you would never know if it had escaped) went up (ie the em radiation
was just bouncing around). Remember, I never said energy was not
conserved, of course it is.
>and sometimes one photon would be emitted and
>two absorbed.
How would you know? But I jest, energy is conserved, there is energy in
any em wave.
>I think the fact that we never see the latter in high
>energy particle experiments argues that conservation of energy-
>momentum must be strictly true, not just statistically true.
I don't see where I suggested energy momentum was NOT conserved. What I
said was you can only see it statistically. That's a noise thing. Low
energy events (radio waves, even light) are hard to detect and have very
low e-momentum and are noisy, high energy the reverse.
>> Particle physicists tend to forget that their complex statistical
>> techniques for handling "a particle interaction" is really not one
>> interaction but one chosen out of quadrillions that were happening, and
>> they forget about the energy dumped elsewhere by the 'particle' emitter
>> that is their source.
>
>I don't buy this argument. I believe the particle physicists would
>have noticed, on bubble chamber tracks for example, if there was a
>significant probability for one emission event to result in multiple
>detection events.
Its called noise, and of course it happens. It happens a LOT in low
energy events, which are generally thus avoided.
>You'd occasionally see a particle decay and instead
>of getting the expected combination of decay products, they'd see some
>products that were doubled.
Why?
>The whole idea of conservation laws
>(parity, charge, strangeness, etc.) is to explain the relationships
>that are observed rigorously, not just statistically.
In what way does that affect a purely wavelike interpretation?
--
Oz