Parity and the spin vector.
Starblade Darksquall
direction as momentum? It would seem to me that it shouldn't. Yet, at
the same time, light obviously has this capability, at least in a
sense.
It would seem to me that if the spin vector was always in the exact
same or exact opposite direciton as momentum, that you couldn't even
find a referenece frame where it switched, since by moving to a
reference frame close to its speed, then turning and going to one at
an angle, passing it, then turning again and then going faster than
it, it wouldn't change at all. So by analogy, you can't just go faster
than it to change its parity. Its parity would remain exactly the same
reguardless of your relative motion.
Spin could be physical, in which case the spin vector would be
independant of its motion, but could be described by taking the dot
product of the momentum vector and the spin vector and determining
whether it was a positive or negative number.
But if it's nonphysical, and ALWAYS points either in the same
direction of the momentum vector or in the opposite direction of the
momentum vector, then it would have to be an intrinsic quality, and
you wouldn't be able to apply any sort of boosts to change that aspect
of it.
(...Starblade Riven Darksquall...)
Old Man
Starblade Darksquall <Starb...@Yahoo.com> wrote in message
news:4aa861fb.03081...@posting.google.com...
> Does the spin vector point in the exact same or exact opposite
> direction as momentum? It would seem to me that it shouldn't. Yet, at
> the same time, light obviously has this capability, at least in a
> sense.
photons have S = 1 and have two possible helicity states: one with
S*k= +1 in the direction of the propagation vector, k,and one with
S*k =- 1 in the opposed direction. The photon is it's own anti-
particle, and the electromagnetic interaction conserves parity. On
the other hand, an electron neutrino possesses only one possible
helicity state wherein a neutrino of the opposite helicity is an anti-
electron neutrino. This is the bases for the observed violation of
parity conservation in weak interactions. [Old Man]
> (...Starblade Riven Darksquall...)
Starblade Darksquall
What I'm asking, though, is how spin is measured. Is spin a quantom
quality or a physical quality? If I shoot a beam of electrons forward,
and then turn around and rotate in an arbitrary angle, will I see the
same spin? What will the relation between momentum and spin be? Does
spin always point in the same or opposite direction of motion or can
it point at some oblique angle to the direction of motion but in a
similar or dissimilar direction?
(...Starblade Riven Darksquall...)
Constantine
Spin is intrinsic angular momentum. It is measured in the same way you
measure angular momentum. When are you going to open a physics book?
Kostas.
Igor
Darksquall) wrote:
Spin can point in the same direction as linear momentum or in the
opposite direction. For most particles, both variations are valid and
represent opposite helicity states. For a particle with finite mass,
the particle can always be stopped and made to reverse its direction.
Conservation of spin angular momentum then says that helicity must
change. Thus, electrons tend to come in right handed and left handed
varieties, easily converted back and forth. Not so with neutrinos.
Nature, for some reason, seems to prefer left handed neutrinos and
right handed antineutrinos only. But this seems to be a quirk of the
weak interaction. Opposite handed neutrinos might yet exist, but we
just haven't seen any.
Starblade Darksquall
But all measures of angular momentum are vectors. You're the one who
needs to open a physics book, not me.
If spin is simply the magnitude of angular momentum, then you'd be
right. But I've never heard of any scalar of angular momentum.
There is a good reason for this. Namely, that there are multiple
possibilities for the choice of axis about which the object is to be
rotated.
(...Starblade Riven Darksquall...)
Starblade Darksquall
What is meant by same and opposite direction? Momentum can be in any
direction, but spin apparantly doesn't do much changing. Wouldn't a
rotational or a lorentz transform push a system with a momentum vector
and spin vector that are parallel into being non-parallel?
(...Starblade Riven Darksquall...)
Richard Herring
Darksquall <Starb...@Yahoo.com> writes
>"Constantine" <Konstantin...@durham.ac.uk> wrote in message
>news:<bhnve2$66n$1...@sirius.dur.ac.uk>...
>> "Starblade Darksquall" <Starb...@Yahoo.com> wrote in message
>> news:4aa861fb.03081...@posting.google.com...
>> > "Old Man" <nom...@nomail.net> wrote in message
>> news:<3f3d2509_4@newsfeed>...
>> > > Starblade Darksquall <Starb...@Yahoo.com> wrote in message
>> > > news:4aa861fb.03081...@posting.google.com...
>> > > > Does the spin vector point in the exact same or exact opposite
>> > > > direction as momentum? It would seem to me that it shouldn't. Yet, at
>> > > > the same time, light obviously has this capability, at least in a
>> > > > sense.
>> > >
>> > > photons have S = 1 and have two possible helicity states: one with
>> > > S*k= +1 in the direction of the propagation vector, k,and one with
>> > > S*k =- 1 in the opposed direction. The photon is it's own anti-
>> > > particle, and the electromagnetic interaction conserves parity. On
>> > > the other hand, an electron neutrino possesses only one possible
>> > > helicity state wherein a neutrino of the opposite helicity is an anti-
>> > > electron neutrino. This is the bases for the observed violation of
>> > > parity conservation in weak interactions. [Old Man]
>> > >
>> >
>> > quality or a physical quality? If I shoot a beam of electrons forward,
>> > and then turn around and rotate in an arbitrary angle, will I see the
>> > same spin? What will the relation between momentum and spin be? Does
>> > spin always point in the same or opposite direction of motion or can
>> > it point at some oblique angle to the direction of motion but in a
>> > similar or dissimilar direction?
>>
>> measure angular momentum. When are you going to open a physics book?
>>
>needs to open a physics book, not me.
PKB. Though, on reflection, possibly so - if by that you mean that
you're not ready to start on physics until you have learned to formulate
your questions unambiguously. For instance:
>> >Is spin a quantom
>> > quality or a physical quality?
What's that supposed to mean?
>
>If spin is simply the magnitude of angular momentum, then you'd be
>right.
Spin is intrinsic angular momentum, as he said.
>But I've never heard of any scalar of angular momentum.
So you've never heard anyone say that electrons are "spin one-half"
particles, photons are "spin one", and so on? Those are statements that
the magnitude of spin is a constant scalar for those kinds of particle.
>There is a good reason for this. Namely, that there are multiple
>possibilities for the choice of axis about which the object is to be
>rotated.
If you ever learn anything about quantum mechanics (with a 'u' !) you'll
find that the choice is made by the experimenter, not the object. Pick
any axis and measure the spin component of an electron along that axis,
and the result will always be +1/2 or -1/2. There _is_ no uniquely
defined axis "about which the object is to be rotated".
--
Richard Herring
Richard Herring
Darksquall <Starb...@Yahoo.com> writes
>
>What is meant by same and opposite direction? Momentum can be in any
>direction, but spin apparantly doesn't do much changing. Wouldn't a
>rotational or a lorentz transform push a system with a momentum vector
>and spin vector that are parallel into being non-parallel?
Why don't you show us an example? Apply a Lorentz transformation to a
couple of parallel vectors, and show how they become non-parallel.
--
Richard Herring
Starblade Darksquall
The momentum vector and the spin vector don't transform the same under
lorentz transformations because the spin vector is is a pseudovector.
Pseudovectors don't transform the same as 'regular' vectors.
Now obviously the spin vector is a pseudovector because if you reverse
the coordinate axis you reverse the relation between the momentum
vector and the spin vector.
My proof, then, is simply that if you simply apply any regular
transform, perpendicular to motion, the momentum vector is skewed, but
the angular momentum is not affected. Any physics book involving
rotational dynamics will tell you this. The angular momentum is not
affected by translational motion.
(...Starblade Riven Darksquall...)
Igor
Darksquall) wrote:
Spin is what you measure it to be. Same with linear momentum.
Measure either one first, and the only compatible value the other one
can have is along the same axis. This is well known in QM. Angular
and linear momentum cannot have simultaneous eigenvalues along
different axes. So that pretty much narrows down the field to
parallel and anti-parallel spin and linear momentum.
Richard Herring
>Richard Herring <junk@[127.0.0.1]> wrote in message
>news:<H2kda8gCMiQ$Ew...@baesystems.com>...
>> In message <4aa861fb.0308...@posting.google.com>, Starblade
>> Darksquall <Starb...@Yahoo.com> writes
>> >
>> >What is meant by same and opposite direction? Momentum can be in any
>> >direction, but spin apparantly doesn't do much changing. Wouldn't a
>> >rotational or a lorentz transform push a system with a momentum vector
>> >and spin vector that are parallel into being non-parallel?
>>
>> Why don't you show us an example? Apply a Lorentz transformation to a
>> couple of parallel vectors, and show how they become non-parallel.
>
>The momentum vector and the spin vector don't transform the same under
>lorentz transformations because the spin vector is is a pseudovector.
>Pseudovectors don't transform the same as 'regular' vectors.
... under coordinate inversion. If your claim is true, you should be
able to show us an example of a Lorentz transformation which inverts the
axes. (It might save some time if you refresh your memory about the
first of the postulates which lead to the LT.)
>
>Now obviously the spin vector is a pseudovector because if you reverse
>the coordinate axis you reverse the relation between the momentum
>vector and the spin vector.
That sentence is correct.
>
>My proof, then, is simply that if you simply apply any regular
>transform, perpendicular to motion, the momentum vector is skewed, but
>the angular momentum is not affected. Any physics book involving
>rotational dynamics will tell you this.
So cite one. Don't be afraid to quote actual words and equations.
>The angular momentum is not
>affected by translational motion.
--
Richard Herring
Starblade Darksquall
> In message <4aa861fb.03082...@posting.google.com>, Starblade
> Darksquall <Starb...@Yahoo.com> writes
> >Richard Herring <junk@[127.0.0.1]> wrote in message
> >news:<H2kda8gCMiQ$Ew...@baesystems.com>...
> >> In message <4aa861fb.0308...@posting.google.com>, Starblade
> >> Darksquall <Starb...@Yahoo.com> writes
> >> >
> >> >What is meant by same and opposite direction? Momentum can be in any
> >> >direction, but spin apparantly doesn't do much changing. Wouldn't a
> >> >rotational or a lorentz transform push a system with a momentum vector
> >> >and spin vector that are parallel into being non-parallel?
> >>
> >> Why don't you show us an example? Apply a Lorentz transformation to a
> >> couple of parallel vectors, and show how they become non-parallel.
> >
> >The momentum vector and the spin vector don't transform the same under
> >lorentz transformations because the spin vector is is a pseudovector.
> >Pseudovectors don't transform the same as 'regular' vectors.
>
> ... under coordinate inversion. If your claim is true, you should be
> able to show us an example of a Lorentz transformation which inverts the
> axes. (It might save some time if you refresh your memory about the
> first of the postulates which lead to the LT.)
The Lorentz Transformation does not invert the axes. However, it can
invert the momentum vector with a proper boost.
> >
> >Now obviously the spin vector is a pseudovector because if you reverse
> >the coordinate axis you reverse the relation between the momentum
> >vector and the spin vector.
>
> That sentence is correct.
> >
> >My proof, then, is simply that if you simply apply any regular
> >transform, perpendicular to motion, the momentum vector is skewed, but
> >the angular momentum is not affected. Any physics book involving
> >rotational dynamics will tell you this.
>
> So cite one. Don't be afraid to quote actual words and equations.
>
In specific, I mean if spin and translational momentum are parallel,
then if you change the reference frame so that there is movement
orthogonal to the original motion spin is unaffected.
This is because in the equation L = r x p both r and p are measured
with respect to the origin. Since we can take the center of mass to be
the origin without changing the result of the equation, then if origin
is in motion both r and p remain invariant.
> >The angular momentum is not
> >affected by translational motion.
(...Starblade Riven Darksquall...)
Lester Zick
>On 19 Aug 2003 05:47:11 -0700, Starb...@Yahoo.com (Starblade
>Darksquall) wrote:
>
[. . .]
>>
>>What is meant by same and opposite direction? Momentum can be in any
>>direction, but spin apparantly doesn't do much changing. Wouldn't a
>>rotational or a lorentz transform push a system with a momentum vector
>>and spin vector that are parallel into being non-parallel?
>>
>>(...Starblade Riven Darksquall...)
>
>Spin is what you measure it to be. Same with linear momentum.
>Measure either one first, and the only compatible value the other one
>can have is along the same axis. This is well known in QM. Angular
>and linear momentum cannot have simultaneous eigenvalues along
>different axes. So that pretty much narrows down the field to
>parallel and anti-parallel spin and linear momentum.
Can you supply an online citation for this? It sounds like you're
saying that some physical or mechanical consideration restricts spin
and p to the same or opposite directions.
>
Regards - Lester
remove DEL in address for email
Richard Herring
Well, of course. If I'm walking faster than you, you appear to be
receding, but that isn't an "inversion".
What does that same transformation do to the angular momentum?
Don't just wave your hands, do the calculation.
>> >
>> >Now obviously the spin vector is a pseudovector because if you reverse
>> >the coordinate axis you reverse the relation between the momentum
>> >vector and the spin vector.
>>
>> That sentence is correct.
>> >
>> >My proof, then, is simply that if you simply apply any regular
>> >transform, perpendicular to motion, the momentum vector is skewed, but
>> >the angular momentum is not affected. Any physics book involving
>> >rotational dynamics will tell you this.
>>
>> So cite one. Don't be afraid to quote actual words and equations.
>>
>
>In specific, I mean if spin and translational momentum are parallel,
>then if you change the reference frame so that there is movement
>orthogonal to the original motion spin is unaffected.
No, that's just as vague. Write out the equations. Do the calculation.
>
>This is because in the equation L = r x p both r and p are measured
>with respect to the origin. Since we can take the center of mass to be
>the origin without changing the result of the equation,
Only in the centre of mass frame. In any other frame it is necessarily
moving. Do the calculation.
--
Richard Herring
Igor
You'll find it in any elementary QM textbook. It's a fairly well
known principle.
Starblade Darksquall
Is the whether something is spin up or spin down in any way affected
by which reference frame you are in? I would think not. If it is only
the measured property which is important, then perhaps if one measures
the angular momentum in a 'boosted' reference frame one will also get
the same measurement of spin up or spin down.
(...Starblade Riven Darksquall...)
Starblade Darksquall
I did do it. Look below.
> >> >
> >> >Now obviously the spin vector is a pseudovector because if you reverse
> >> >the coordinate axis you reverse the relation between the momentum
> >> >vector and the spin vector.
> >>
> >> That sentence is correct.
> >> >
> >> >My proof, then, is simply that if you simply apply any regular
> >> >transform, perpendicular to motion, the momentum vector is skewed, but
> >> >the angular momentum is not affected. Any physics book involving
> >> >rotational dynamics will tell you this.
> >>
> >> So cite one. Don't be afraid to quote actual words and equations.
> >>
> >
> >In specific, I mean if spin and translational momentum are parallel,
> >then if you change the reference frame so that there is movement
> >orthogonal to the original motion spin is unaffected.
>
> No, that's just as vague. Write out the equations. Do the calculation.
L = I * w, where I represents the moment of inertia, and must be taken
with respect to the axis or axes about which the object is rotating,
and w is the angular velocity, which is also with respect to the axis
or axes about which the object is rotating. As you can see, this
equation is independent of reference frame, at least with respect to
the galilean transformation.
> >
> >This is because in the equation L = r x p both r and p are measured
> >with respect to the origin. Since we can take the center of mass to be
> >the origin without changing the result of the equation,
>
> Only in the centre of mass frame. In any other frame it is necessarily
> moving. Do the calculation.
Of course it's moving! And I did do the calculation. The angular
momentum is invariant with respect to which reference frame is used.
There is no such thing as 'moving' angular momentum. Only spinning
affects angular momentum.
(...Starblade Riven Darksquall...)
Richard Herring
I looked. I don't see a Lorentz transformation.
[snip]
--
Richard Herring
Lester Zick
>On Thu, 21 Aug 2003 14:08:37 GMT, lester...@worldnet.att.net
>(Lester Zick) wrote:
>
>>On Thu, 21 Aug 2003 03:55:43 GMT, Igor <bx...@bfn.org> wrote:
>>
>>>On 19 Aug 2003 05:47:11 -0700, Starb...@Yahoo.com (Starblade
>>>Darksquall) wrote:
>>>
>>[. . .]
>>>>
>>>>What is meant by same and opposite direction? Momentum can be in any
>>>>direction, but spin apparantly doesn't do much changing. Wouldn't a
>>>>rotational or a lorentz transform push a system with a momentum vector
>>>>and spin vector that are parallel into being non-parallel?
>>>>
>>>>(...Starblade Riven Darksquall...)
>>>
>>>Spin is what you measure it to be. Same with linear momentum.
>>>Measure either one first, and the only compatible value the other one
>>>can have is along the same axis. This is well known in QM. Angular
>>>and linear momentum cannot have simultaneous eigenvalues along
>>>different axes. So that pretty much narrows down the field to
>>>parallel and anti-parallel spin and linear momentum.
>>
>>Can you supply an online citation for this? It sounds like you're
>>saying that some physical or mechanical consideration restricts spin
>>and p to the same or opposite directions.
>
>
>You'll find it in any elementary QM textbook. It's a fairly well
>known principle.
>
>
Well then let me ask whether the classical planetary model for the
atom conforms to this idea? My impression is that the spin of orbital
electron lies normal to the plane of the orbit and thus normal to
orbital p.
Wolf Kirchmeir
>Well then let me ask whether the classical planetary model for the
>atom conforms to this idea? My impression is that the spin of orbital
>electron lies normal to the plane of the orbit and thus normal to
>orbital p.
AFAIK, the classical planetary model of the atom is irrelevant to this issue.
--
Best Wishes,
Wolf Kirchmeir, Blind River ON
"Not that brains are everything --
you'll also need a skull to put them in." (Nancy Franklin, 1997)
<just one w and plain ca for correct address>
Lester Zick
<wwol...@sympatico.can> wrote:
>On Fri, 22 Aug 2003 14:47:10 GMT, Lester Zick wrote:
>
>>Well then let me ask whether the classical planetary model for the
>>atom conforms to this idea? My impression is that the spin of orbital
>>electron lies normal to the plane of the orbit and thus normal to
>>orbital p.
>
>AFAIK, the classical planetary model of the atom is irrelevant to this issue.
>
I wonder if you would explain why. As far as I know electrons have
always been described with quantum spins of 1/2 h/2pi even in the
classical model. And those spins were always paired in opposing
directions normal to the plane of the orbit.
Wolf Kirchmeir
>>AFAIK, the classical planetary model of the atom is irrelevant to this issue.
>>
>I wonder if you would explain why. As far as I know electrons have
>always been described with quantum spins of 1/2 h/2pi even in the
>classical model. And those spins were always paired in opposing
>directions normal to the plane of the orbit.
>
>
>Regards - Lester
The planetary model doesn't entail electron spin. That comes from QM. So the
properties of the planetary model don't matter one way or the other. Besides,
the planetary model is highly misleading, eg, there is "plane of the orbit"
in actual fact.
Lester Zick
<wwol...@sympatico.can> wrote:
>On Sat, 23 Aug 2003 14:29:18 GMT, Lester Zick wrote:
>
>>>AFAIK, the classical planetary model of the atom is irrelevant to this issue.
>>>
>>I wonder if you would explain why. As far as I know electrons have
>>always been described with quantum spins of 1/2 h/2pi even in the
>>classical model. And those spins were always paired in opposing
>>directions normal to the plane of the orbit.
>>
>>
>>Regards - Lester
>
>The planetary model doesn't entail electron spin. That comes from QM. So the
>properties of the planetary model don't matter one way or the other. Besides,
>the planetary model is highly misleading, eg, there is "plane of the orbit"
>in actual fact.
>
>
I think electrons must have been thought to spin even in the classical
model of atomic structure. I'm not suggesting the model itself has
anything to do with electron spin. But I think the electron has to
spin regardless and I just wonder how this is consistent with Igor's
observation.
In fact I've been wondering if he wasn't referring to
particle-antiparticle pair creation. Otherwise I just don't see any
reason that spin and linear p should lie parallel or anti parallel to
one another.
Joseph Legris
Do I detect the faint rumblings of another endless, pointless thread
developing? Before things get too silly, let's first establish the
meanings of the words spin, parallel and linear p.
Go ahead Lester. By the way, if your definitions include any words other
than "not", or "all", please define them as well. On second thought,
you'd better define those too.
--
Joe Legris
Starblade Darksquall
> On Sat, 23 Aug 2003 12:55:44 -0400 (EDT), "Wolf Kirchmeir"
> <wwol...@sympatico.can> wrote:
>
> >On Sat, 23 Aug 2003 14:29:18 GMT, Lester Zick wrote:
> >
> >>>AFAIK, the classical planetary model of the atom is irrelevant to this issue.
> >>>
> >>I wonder if you would explain why. As far as I know electrons have
> >>always been described with quantum spins of 1/2 h/2pi even in the
> >>classical model. And those spins were always paired in opposing
> >>directions normal to the plane of the orbit.
> >>
> >>
> >>Regards - Lester
> >
> >The planetary model doesn't entail electron spin. That comes from QM. So the
> >properties of the planetary model don't matter one way or the other. Besides,
> >the planetary model is highly misleading, eg, there is "plane of the orbit"
> >in actual fact.
> >
> >
> I think electrons must have been thought to spin even in the classical
> model of atomic structure. I'm not suggesting the model itself has
> anything to do with electron spin. But I think the electron has to
> spin regardless and I just wonder how this is consistent with Igor's
> observation.
>
The classical models DO indicate a spin. There are spin 1/2 and spin
-1/2 electrons even in the classical orbit. At least during the
transition phases between classical and quantum. But I may be wrong.
In High School chemistry they showed me the electron orbitals, but
they didn't introduce the concept of uncertainty. Or is it that my
High School chemistry textbook was just really up to date, perhaps?
> In fact I've been wondering if he wasn't referring to
> particle-antiparticle pair creation. Otherwise I just don't see any
> reason that spin and linear p should lie parallel or anti parallel to
> one another.
>
I think I understand this better. Spin can only be MEASURED along one
axis. But what gets me confused is why it should change with reference
frame. If it were a quantom effect, and you can only measure the spin
once, then why wouldn't it always be spin up or spin down reguardless
of reference frame?
I would think that quantom physics would be different than classical
physics in this reguard, so that things do not change spin from one
reference frame to another. Not only that, but the magnitude of the
spin ought to be the same reguardless of which way it's pointing since
angular momentum is quantized.
>
> Regards - Lester
>
> remove DEL in address for email
(...Starblade Riven Darksquall...)
Lester Zick
Well, Joe, this isn't my thread. I'm just trying to clarify someone
else's comment. It seems strange. The terms are his.
Lester Zick
Darksquall) wrote:
This is consistent with my own experience except I don't really see
what uncertainty has to do with it.
>
>> In fact I've been wondering if he wasn't referring to
>> particle-antiparticle pair creation. Otherwise I just don't see any
>> reason that spin and linear p should lie parallel or anti parallel to
>> one another.
>>
>
>I think I understand this better. Spin can only be MEASURED along one
>axis. But what gets me confused is why it should change with reference
>frame. If it were a quantom effect, and you can only measure the spin
>once, then why wouldn't it always be spin up or spin down reguardless
>of reference frame?
>
>I would think that quantom physics would be different than classical
>physics in this reguard, so that things do not change spin from one
>reference frame to another. Not only that, but the magnitude of the
>spin ought to be the same reguardless of which way it's pointing since
>angular momentum is quantized.
>
It's not clear what measurement has to do with the comment. I'm just
not sure exactly what Igor is talking about. As I understand it he
maintains - and he says QM maintains as well - that spin and p are
necessarily colinear properties of particles. I'm just asking whether
this would hold for the classical model of the atom as well where p
rotates and I believe electron spin is characterized as lying normal
to the plane of electron rotation.
Wolf Kirchmeir
>>The classical models DO indicate a spin. There are spin 1/2 and spin
>>-1/2 electrons even in the classical orbit. At least during the
>>transition phases between classical and quantum. But I may be wrong.
>>In High School chemistry they showed me the electron orbitals, but
>>they didn't introduce the concept of uncertainty. Or is it that my
>>High School chemistry textbook was just really up to date, perhaps?
>
>This is consistent with my own experience except I don't really see
>what uncertainty has to do with it.
Measuring spin destroys information.
Igor
The problem is that there never was a classical theory of particle
spin. It's a concept that has its roots entirely in QM.
Traditionally, effects of spin were observed as very fine splitting of
lines in atomic emission spectra. Pauli was the first person to show
how this could come about if we think about electrons as having
intrinsic angular momenta, i.e. behaving like tiny spinning tops. The
analogy works up to a point, but can have some nonsensical
consequences if taken too seriously. It's known that mathematically
spin has the same properties as classical angular momentum in terms of
operators and wave functions. And that's really all that can be said
about it.
Igor
It has nothing to do with pair production, per se. Spin and linear
momentum can point in any direction, independent of each other.
However QM deals with the mathematics of the outcomes of measurements,
and it says that only spin and linear momentum components in the same
direction can have simultaneous eigenvalues in any given system. That
means that they are compatible dynamical variables and we can measure
both at the same time to as large a precision as necessary.
Measurements of any other pairs of components will always interfere
with each other in much the same way as measurements of particle
location and momentum do.
Since spin obeys the same mathematical rules as orbital angular
momentum, the same rules apply to orbital angular momentum and linear
momentum as well.
Igor
Darksquall) wrote:
The electron orbitals you deal with in Chemistry are not at all
classical physics. They were first derived from solutions to the
Schrodinger wave equation back in the 1920's and 30's. It's all
quantum based.
Igor
No spin in classical physics. Only orbital angular momentum.
Igor
<wwol...@sympatico.can> wrote:
>On Sun, 24 Aug 2003 00:03:49 GMT, Lester Zick wrote:
>
>>>The classical models DO indicate a spin. There are spin 1/2 and spin
>>>-1/2 electrons even in the classical orbit. At least during the
>>>transition phases between classical and quantum. But I may be wrong.
>>>In High School chemistry they showed me the electron orbitals, but
>>>they didn't introduce the concept of uncertainty. Or is it that my
>>>High School chemistry textbook was just really up to date, perhaps?
>>
>>This is consistent with my own experience except I don't really see
>>what uncertainty has to do with it.
>
>Measuring spin destroys information.
Not by itself. In QM, I can measure any one thing to arbitrary
accuracy. When we try to make simultaneous measurements of pairs of
quantities is where we sometimes run into problems. That's where the
uncertainty principle comes in. Some pairs of measurements are
compatible and some are not. Half the battle in QM is knowing which
ones are which.
ZZBunker
It's not really a problem though.
It was only a problem for Feynmann, since he was
somewhat arbitrarily clueless what "uncertainity" means.
Joseph Legris
So who cross-posted it to sci.cognitive? It's your baby as far as I can
tell.
The source of the terms is irrelevant - they are all grist for your
knowledge mill. Information goes in and nonsense comes out.
--
Joe Legris
Lester Zick
Whatever.
Lester Zick
<wwol...@sympatico.can> wrote:
>On Sun, 24 Aug 2003 00:03:49 GMT, Lester Zick wrote:
>
>>>The classical models DO indicate a spin. There are spin 1/2 and spin
>>>-1/2 electrons even in the classical orbit. At least during the
>>>transition phases between classical and quantum. But I may be wrong.
>>>In High School chemistry they showed me the electron orbitals, but
>>>they didn't introduce the concept of uncertainty. Or is it that my
>>>High School chemistry textbook was just really up to date, perhaps?
>>
>>This is consistent with my own experience except I don't really see
>>what uncertainty has to do with it.
>
>Measuring spin destroys information.
>
I understand. But as far as I can tell this has no bearing on whether
spin and linear p must be colinear.
Lester Zick
You're sure particle spin in classical physics was never taken to be
1/2 h/2pi?
Lester Zick
This is more or less what I thought. But it seems at odds with your
original statement. However it's probably not worth pursuing further.
But thanks for the clarification.
>However QM deals with the mathematics of the outcomes of measurements,
>and it says that only spin and linear momentum components in the same
>direction can have simultaneous eigenvalues in any given system. That
>means that they are compatible dynamical variables and we can measure
>both at the same time to as large a precision as necessary.
>Measurements of any other pairs of components will always interfere
>with each other in much the same way as measurements of particle
>location and momentum do.
>
>Since spin obeys the same mathematical rules as orbital angular
>momentum, the same rules apply to orbital angular momentum and linear
>momentum as well.
>
Regards - Lester
Lester Zick
Just a point of historical curiosity. Were there not measurements of
particle oscillations in magnetic fields interpreted as evidence of
spin prior to the advent of QM?
Wolf Kirchmeir
>>Measuring spin destroys information.
>
>Not by itself. In QM, I can measure any one thing to arbitrary
>accuracy. When we try to make simultaneous measurements of pairs of
>quantities is where we sometimes run into problems. That's where the
>uncertainty principle comes in. Some pairs of measurements are
>compatible and some are not. Half the battle in QM is knowing which
>ones are which.
I thought that's what I said... just with fewer words, is all... :-)
Igor
(Lester Zick) wrote:
Yes, I'm sure that's not true, since there's no Planck's constant in
classical physics either. Planck's discovery of that constant about a
hundred years ago led to the QM revolution.
Igor
(Lester Zick) wrote:
Well perhaps someone may have noticed line splitting in the 18th
century. But you must understand that back then nobody even
understood what the lines really meant. They knew that each element
had its own emission signature but that was about it. QM eventually
cleared up several decades worth of questions.
Lester Zick
Sure, I understand. However what I meant was that the discovery of
Planck's constant happened in the conceptual context of classical
physics and was experimentally measured and defined in that context.
We could note analogous considerations with respect to Newtonian
concepts being defined in the context of Galilean and Keplerian
celestial mechanics even though it gave rise to a whole new branch of
celestial mechanics.
Lester Zick
You know I'll go out on a non contentious limb here. But I suspect
that magnetic resonance experiments were conducted on atoms and
particles in the early twentieth century specifically to measure spin.
In other words I think the experimenters knew what they were measuring
and were quite surprised by the constant measure of spin between
protons and electrons and by opposing orbital spins.
This all happened as a result of the planetary model of the atom once
the atomic and particle nature of the atom had become apparent and
before quantum ideas were applied to explain why spin and orbital
energy appeared quantized in terms of Planck's constant.
Prior to those measures I don't think anyone really suspected a causal
link between the two and I think it was these classical experimental
measures that really defined the arena of QM. Granted that there were
puzzling insights prior to this with respect to radiation vis-a-vis
Planck's experimental measures. I just doubt anyone really expected
atomic and particulate matter to exhibit such spin properties as they
do.
However there is no need to beat a dead horse. I appreciate your
replies and consider that you have explained what you originally meant
with the observation regarding the colinearity of particle spin and p.
Igor
(Lester Zick) wrote:
Well, I guess you could say that. Alot of the issues addressed by
Planck came from classical physics. The traditional view of radiation
was in serious need of an overhaul at the dawn of the twentieth
century, and Planck provided just the right insight at just the right
time. One of the interesting things is that Planck and Einstein,
within just a few years of each other, essentially marked the
separation of the era that is usually defined as classical and modern
physics, with their far-reaching discoveries.
Igor
(Lester Zick) wrote:
You might want to look up the experimental efforts of a fellow by the
name of George Uhlenbeck . I forget exactly in what decade his
experiments were performed, but he's usually credited with first
discovering spin effects by examining the spectral line splitting. It
was Wolfgang Pauli who eventually made theoretical sense out of these
results and gave us the modern theory of particle spin.
As you can tell, there is a lot of physical history that goes into all
of this. Frankly, I think it's a fascinating subject. As you can
also probably tell, I enjoy getting into the history probably as much
as the math. If you're interested, there are several good books on
this topic. The one I remember enjoying the most is something called
"The Second Creation : Makers of the Revolution in Twentieth-Century
Physics " by Robert P. Crease. See if you can locate it. I recommend
it tremendously.
Lester Zick
>On Mon, 25 Aug 2003 15:10:02 GMT, lester...@worldnet.att.net
>(Lester Zick) wrote:
>
[. . .]
>>
>>You know I'll go out on a non contentious limb here. But I suspect
>>that magnetic resonance experiments were conducted on atoms and
>>particles in the early twentieth century specifically to measure spin.
>>In other words I think the experimenters knew what they were measuring
>>and were quite surprised by the constant measure of spin between
>>protons and electrons and by opposing orbital spins.
>>
>>This all happened as a result of the planetary model of the atom once
>>the atomic and particle nature of the atom had become apparent and
>>before quantum ideas were applied to explain why spin and orbital
>>energy appeared quantized in terms of Planck's constant.
>>
>>Prior to those measures I don't think anyone really suspected a causal
>>link between the two and I think it was these classical experimental
>>measures that really defined the arena of QM. Granted that there were
>>puzzling insights prior to this with respect to radiation vis-a-vis
>>Planck's experimental measures. I just doubt anyone really expected
>>atomic and particulate matter to exhibit such spin properties as they
>>do.
>>
>>However there is no need to beat a dead horse. I appreciate your
>>replies and consider that you have explained what you originally meant
>>with the observation regarding the colinearity of particle spin and p.
>
>You might want to look up the experimental efforts of a fellow by the
>name of George Uhlenbeck . I forget exactly in what decade his
>experiments were performed, but he's usually credited with first
>discovering spin effects by examining the spectral line splitting. It
>was Wolfgang Pauli who eventually made theoretical sense out of these
>results and gave us the modern theory of particle spin.
>
>As you can tell, there is a lot of physical history that goes into all
>of this. Frankly, I think it's a fascinating subject. As you can
>also probably tell, I enjoy getting into the history probably as much
>as the math. If you're interested, there are several good books on
>this topic. The one I remember enjoying the most is something called
>"The Second Creation : Makers of the Revolution in Twentieth-Century
>Physics " by Robert P. Crease. See if you can locate it. I recommend
>it tremendously.
I think you're correct. The history of these events is important. And
along similar lines I would include the history of computers and
computing in a similar vein.
It's always amused me that the principal players in Hollywood enjoy so
much adulation and so many public awards when those who have made
modern society what it is today enjoy so little. Maybe there's a moral
in there somewhere.
Lester Zick
>On Mon, 25 Aug 2003 14:58:24 GMT, lester...@worldnet.att.net
>(Lester Zick) wrote:
>
>>On Mon, 25 Aug 2003 03:27:20 GMT, Igor <bx...@bfn.org> wrote:
>>
>>>On Sun, 24 Aug 2003 15:16:07 GMT, lester...@worldnet.att.net
>>>(Lester Zick) wrote:
>>>
>>>>On Sun, 24 Aug 2003 04:21:58 GMT, Igor <bx...@bfn.org> wrote:
[. . .]
>>>>
>>>Yes, I'm sure that's not true, since there's no Planck's constant in
>>>classical physics either. Planck's discovery of that constant about a
>>>hundred years ago led to the QM revolution.
>>
>>Sure, I understand. However what I meant was that the discovery of
>>Planck's constant happened in the conceptual context of classical
>>physics and was experimentally measured and defined in that context.
>>We could note analogous considerations with respect to Newtonian
>>concepts being defined in the context of Galilean and Keplerian
>>celestial mechanics even though it gave rise to a whole new branch of
>>celestial mechanics.
>>>
>
>Well, I guess you could say that. Alot of the issues addressed by
>Planck came from classical physics. The traditional view of radiation
>was in serious need of an overhaul at the dawn of the twentieth
>century, and Planck provided just the right insight at just the right
>time. One of the interesting things is that Planck and Einstein,
>within just a few years of each other, essentially marked the
>separation of the era that is usually defined as classical and modern
>physics, with their far-reaching discoveries.
It's an interesting parallel. Of course every discovery in science
marks the end of an era and the beginning of a new era in some
respect. But these were especially portentous.
However I should advise you that I have serious problems with both
general concepts. Not so much with the facts as with their application
and interpretation. I'm doing this not to advance my own arguments in
the present context but to alert you to the potential pitfalls of
further discussion as I'm considered one of the more significant
trolls on sci.physics.