Uncertainty and Relativity
Todd Desiato
in its Energy, position, momentum and frequency. The expressions for
Heisenbergs Uncertainty principle show the minimum uncertainty obtainable by
an observation of a free particle/wave. They are;
(delta_E)(delta_t) > or = h/2
which reads the "probability that a particle has energy within the interval
delta_E" times the "probability that a particle is measured within the
interval delta_t" is greater than or equal to a constant which is the
"minimum" action of the particle, i.e. 1/2 cycle of its oscillation.
(delta_p)(delta_r) > or = h/2
which reads the "probability that a particle has momentum within the
interval delta_p" times the "probability that a particle is measured within
the interval delta_r" (radial to the most probable location) is greater than
or equal to the minimum action.
For a free particle its Energy is just mc^2, but its momentum is not zero
as it would classically be considered. Because of zero point motion,
"zitterbewegung" a particle is never at rest, it always has an intrinsic
motion. That makes (delta_p) a finite interval (>0) and then delta_r is also
a finite interval, even for a point-like particle like an electron. So
basically delta - _E,p,t,r_ are never zero if a particle actually exists.
Now here's the good part. Take the ratio of the two uncertainties, we get;
(delta_E)
----------- (delta_t) > or = (delta_r)
(delta_p)
Now delta_r was chosen here for a reason. In Cartesian coordinates we get;
(delta_r) = sqrt[(delta_x)^2 + (delta_y)^2 + (delta_z)^2)]
So if we square both sides of the equation and bring the RHS over to the
left, and replacing the "delta" symbol with "d" for convenience, we get;
((dE/dp)dt)^2 - dx^2 - dy^2 -dz^2 > or = 0
or bring the LHS over to the right we get;
0 > or = -((dE/dp)dt)^2 + dx^2 + dy^2 + dz^2
So it leads to either signature of the Minkowski metric. But it is always a
timelike world line! Meaning that the uncertainty principle implies
velocities must be less than or equal to c.
Now classically in GR a particle at rest has a world line;
(cdt)^2 = ds^2
but because of zero point energy we can no longer speak of an arbitrary
reference frame where the particle is at rest. There is __ALWAYS__ an
uncertainty in its position and momentum. Even if you could choose a frame
where it is at rest now, a moment later it could be moving and you _can't
know for certain_ that it hasn't. So in any quantum theory of gravity, a
world line for a particle at rest;
(cdt)^2 = ds^2 Classically
should be replaced by the world line;
((dE/dp)dt)^2 - dx^2 - dy^2 - dz^2 = ds^2 which includes the
uncertainty intervals.
Since the Compton wavelength of a free particle is equivalent to
representing the particle like a photon with velocity c, (mc=h/lamda_C) and
the uncertainty implies an average velocity within the intervals
dE/dp, it implies that the world line of a free particle at rest, its
zitterbewegung motion is null, or its expectation value is null. Which can
be interpreted as "A wave packet at rest has minimum uncertainties, and is
composed of plane waves that travel null geodesics". Since the
zitterbewegung is a direct consequence of the Heisenberg equations of motion
for a free particle that is a solution to the Dirac equation, and the
dispersion relationship for these oscillations is identical to that of light
(the solutions are plane waves), it appears to me to not only "intuitive"
but the correct way to construct a stationary wave packet from the
superposition of light-like plane waves. This makes the above world line
equation "exactly the same expression as the dispersion relationship for
these waves"
frequency * wavelength = c, or
c(dt) = dr
Over the past few days I have read much in Wald about the shortcomings of
Quantum gravity, in particular the "path integral" approach. The failure is
because the "true dynamical variables" of General Relativity have not been
identified. (They are trying to quantize space-time, which IMO does not
exist) So if we take the displacement of the world line through space-time
to represent "dispersion", which is a displacement of the wavelength
(proper-length) and frequency (proper-time) of actual wave packets, and
equate that with rotations in a "mathematical construction called
space-time", wouldn't that define the dynamical variables that Wald says are
unidentified?
Comments?
Todd Desiato
Note dE/dp = (pc^2)/E which is 0<(dE/dp)<c, or equal to.
Kevin A. Scaldeferri
Todd Desiato <tde...@san.rr.com> wrote:
>Consider the zero point motion of a free particle. There is an uncertainty
>in its Energy, position, momentum and frequency. The expressions for
>Heisenbergs Uncertainty principle show the minimum uncertainty obtainable by
>an observation of a free particle/wave. They are;
>
>(delta_E)(delta_t) > or = h/2
>
>which reads the "probability that a particle has energy within the interval
>delta_E" times the "probability that a particle is measured within the
>interval delta_t" is greater than or equal to a constant which is the
>"minimum" action of the particle, i.e. 1/2 cycle of its oscillation.
>
>(delta_p)(delta_r) > or = h/2
>
>which reads the "probability that a particle has momentum within the
>interval delta_p" times the "probability that a particle is measured within
>the interval delta_r" (radial to the most probable location) is greater than
>or equal to the minimum action.
This is not a correct reading of either of these inequalities. Also,
in nonrelativistic quantum mechanics, the energy-time relation is not
a Heisenberg uncertainty relation since there is no time operator to
take the commutator with the energy operator.
>For a free particle its Energy is just mc^2, but its momentum is not zero
>as it would classically be considered. Because of zero point motion,
>"zitterbewegung" a particle is never at rest, it always has an intrinsic
>motion.
A free particle does not exhibit zero point energy and there are zero
momentum solutions for the free particle.
>So
>basically delta - _E,p,t,r_ are never zero if a particle actually exists.
This statement is not true
[rest snipped]
--
======================================================================
Kevin Scaldeferri Calif. Institute of Technology
The INTJ's Prayer:
Lord keep me open to others' ideas, WRONG though they may be.
Todd Desiato
Kevin A. Scaldeferri wrote in message >...
>Todd Desiato > wrote:
>>Consider the zero point motion of a free particle. There is an uncertainty
>>in its Energy, position, momentum and frequency. The expressions for
>>Heisenbergs Uncertainty principle show the minimum uncertainty obtainable
by
>>an observation of a free particle/wave. They are;
>>
>>(delta_E)(delta_t) > or = h/2
>>
>>which reads the "probability that a particle has energy within the
interval
>>delta_E" times the "probability that a particle is measured within the
>>interval delta_t" is greater than or equal to a constant which is the
>>"minimum" action of the particle, i.e. 1/2 cycle of its oscillation.
>>
>>(delta_p)(delta_r) > or = h/2
>>
>>which reads the "probability that a particle has momentum within the
>>interval delta_p" times the "probability that a particle is measured
within
>>the interval delta_r" (radial to the most probable location) is greater
than
>>or equal to the minimum action.
>
>This is not a correct reading of either of these inequalities.
I know everybody words this differently. What's your version?
>Also,
>in nonrelativistic quantum mechanics, the energy-time relation is not
>a Heisenberg uncertainty relation since there is no time operator to
>take the commutator with the energy operator.
>
I am talking relativistic QM, using the Dirac equation. In relativity Energy
and momentum, as well as space and time have an intimate relationship
involving c. The Energy-time relationship I gave holds true for any
relativistic wave-packet, independent of having a "time operator", with
which to observe it. Look it up in any QM text. The time interval here is
related to the wavelength interval by the dispersion relationship of the
wave.
>>For a free particle its Energy is just mc^2, but its momentum is not zero
>>as it would classically be considered. Because of zero point motion,
>>"zitterbewegung" a particle is never at rest, it always has an intrinsic
>>motion.
>
>A free particle does not exhibit zero point energy and there are zero
>momentum solutions for the free particle.
>
Classically you are correct. Relativistic QM you are wrong. All fermions
have ZPE, or not depending on how you order the operators, "normal,
symmetric, anti-symmetric". However, the Einstein spontaneous emissoin
coefficients will only be 1/2 of the correct value if you do not take into
accout the ZPE of both the vacuum and the particle. ZPE is a necessary
attribute of both the particle source and the field.
Reference "Dirac, P.A.M., "The Quantum Theory of the Electron", Proc. Roy.
Soc. Lond. A126, 360, 1930. Or my reference "Milonni, P.W. "The Quantum
Vacuum", Accademic Press Inc., Ch. 9, 323, 1994. "..rapid oscillations
(zitterbewegung) superimposed on this motion..." of the free electron.
"..zitterbewegung is associated with the fluctuations of the particle
position on the order of a Compton wavelength." "...it is found that the
zitterbewegung results precisely from the interference of the state
amplitudes of positive- and negative-energy plane-wave solutions of the
Dirac equation."
>>So
>>basically delta - _E,p,t,r_ are never zero if a particle actually exists.
>
>This statement is not true
>
The statement is easily shown to be true by examining the Heisenberg
equations of motion for the free particle solutions of the Dirac equation.
Keep an open mind.
Todd Desiato
Ken McLean
"Todd Desiato" <tde...@san.rr.com>
Regarding your post to sci.physics.particle re Uncertainty and Relativity
>Now delta_r was chosen here for a reason. In Cartesian coordinates we get;
>
>(delta_r) = sqrt[(delta_x)^2 + (delta_y)^2 + (delta_z)^2)]
This is wrong. Bad calculus. Chain rule: df(g(x))/dx= df(g(x))/d(g(x))*dg(x)/dx
r=SQRT(X**2+Y**2+Z**2)
implies
dr/dx= X/R
and therefore
dr = (xdx+ydy+zxz)/r
If you think a bit, it makes perfect sense that a radial field has a gradient
that is a radial vector (i.e. Grad.R)=(x,y,r)/r very similarly.
Trust me. You can check it numerically on a calculator if you want.
Also,
(delta_p)(delta_r) > or = h/2
refers to LINEAR momentum and Linear displacement. This relation is usually
demonstrated with plane waves and packets thereof, for delta_r to be a
radial measure one would have to deal with spherical symmetric situations,
which generally lends itself to the treatment of bound states where one can
talk about energy levels and angular momentum etc.
>
>Now here's the good part. Take the ratio of the two uncertainties, we get;>\
>
>(delta_E)
>----------- (delta_t) > or = (delta_r)
>(delta_p)
>
Yeah well, the math and interpretation mistakes render the rest of the argument
moot. Sorry.
Ken McLean
Todd Desiato
Ken McLean wrote in message <6oo1jr$cit$1...@lns572.lns.cornell.edu>...
>
>"Todd Desiato" >
>>Now delta_r was chosen here for a reason. In Cartesian coordinates we get;
>>
>>(delta_r) = sqrt[(delta_x)^2 + (delta_y)^2 + (delta_z)^2)]
>
>This is wrong. Bad calculus. Chain rule: df(g(x))/dx=
df(g(x))/d(g(x))*dg(x)/dx
>
>r=SQRT(X**2+Y**2+Z**2)
>
No it's not wrong. I specified that delta_r is measured wrt the most
probable location, i.e. the particle is centered (on average) at the origin
of the coordinate system I chose. Then delta_x is just (x-0)=x, so
(delta_x)^2 = x^2 etc., just as you wrote it. If I were talking about a
change wrt some other coordinate system, then the chain rule would apply. In
this case I was only refering to the deviation in the position of the
particle about its average value at x=y=z=t=0.
>dr = (xdx+ydy+zdz)/r
>
>If you think a bit, it makes perfect sense that a radial field has a
gradient
>that is a radial vector (i.e. Grad.R)=(x,y,z)/r very similarly.
(typo's corrected above) What you say is fine if I were not speaking about
the uncertainty in its position about the origin of coordinates in its rest
frame. I didn't want to complicate the post with a coordinate independent
approach.
>Also,
>
> (delta_p)(delta_r) > or = h/2
>
>refers to LINEAR momentum and Linear displacement. This relation is usually
>demonstrated with plane waves and packets thereof,
Which is what I am refering to, as a free particle is a wave-packet of
superimposed plane-waves. Even when it is "at rest", the resulting
interferrence pattern of the probability amplitudes is not infitesimally
narrow. It has a width in both space and time, as well as energy and
momentum.
>for delta_r to be a
>radial measure one would have to deal with spherical symmetric situations,
The action of a free particle, in a vacuum, at the origin of the coordinate
system is a sphericaly symmetric situation. The uncertainty in its position
applies in all directions. The ZPE spectrum of both the particle and the
vacuum are symmetrical and Lorentz invariant, therefore the width of each
distribution is spherically symmetric. The exchange of momentum between the
particle and the vacuum is in equalibrium and is spherically symmetrical,
otherwise its average position would not remain at the origin.
>which generally lends itself to the treatment of bound states where one can
>talk about energy levels and angular momentum etc.
>>
Forget about classical QM. Think relativistic QED, and the relativistic
Dirac equation who's solutions are plane waves and free particles are a
superposition of those solutions.
>>Now here's the good part. Take the ratio of the two uncertainties, we
get;>\
>>
>>(delta_E)
>>----------- (delta_t) > or = (delta_r)
>>(delta_p)
>>
>
>
>Yeah well, the math and interpretation mistakes render the rest of the
argument
>moot. Sorry.
There is nothing wrong with the math or the interpretation under
relativistic QM, other than it requires a little effort to see something old
in a new light and associate it with phenomenon that were unrelated
classically.
Todd Desiato
Kevin A. Scaldeferri
Todd Desiato <tde...@san.rr.com> wrote:
>
>Kevin A. Scaldeferri wrote in message >...
>>Todd Desiato > wrote:
>>>Consider the zero point motion of a free particle. There is an uncertainty
>>>in its Energy, position, momentum and frequency. The expressions for
>>>Heisenbergs Uncertainty principle show the minimum uncertainty obtainable
>by
>>>an observation of a free particle/wave. They are;
>>>
>>>(delta_E)(delta_t) > or = h/2
>>>
>>>which reads the "probability that a particle has energy within the
>interval
>>>delta_E" times the "probability that a particle is measured within the
>>>interval delta_t" is greater than or equal to a constant which is the
>>>"minimum" action of the particle, i.e. 1/2 cycle of its oscillation.
>>>
>>>(delta_p)(delta_r) > or = h/2
>>>
>>>which reads the "probability that a particle has momentum within the
>>>interval delta_p" times the "probability that a particle is measured
>within
>>>the interval delta_r" (radial to the most probable location) is greater
>than
>>>or equal to the minimum action.
>>
>>This is not a correct reading of either of these inequalities.
>
>I know everybody words this differently. What's your version?
The product of the statistical uncertainties in the position and
momentum of a particle is greater than or equal to hbar/2. Your
version claims that it the the product of the probability densities
which is being considered.
>
>>Also,
>>in nonrelativistic quantum mechanics, the energy-time relation is not
>>a Heisenberg uncertainty relation since there is no time operator to
>>take the commutator with the energy operator.
>>
>
>I am talking relativistic QM, using the Dirac equation. In relativity Energy
>and momentum, as well as space and time have an intimate relationship
>involving c. The Energy-time relationship I gave holds true for any
>relativistic wave-packet, independent of having a "time operator", with
>which to observe it. Look it up in any QM text. The time interval here is
>related to the wavelength interval by the dispersion relationship of the
>wave.
>
>>>For a free particle its Energy is just mc^2, but its momentum is not zero
>>>as it would classically be considered. Because of zero point motion,
>>>"zitterbewegung" a particle is never at rest, it always has an intrinsic
>>>motion.
>>
>>A free particle does not exhibit zero point energy and there are zero
>>momentum solutions for the free particle.
>>
>
>
>Classically you are correct. Relativistic QM you are wrong.
No, it is still true in relativistic quantum mechanics. The following
describes issues in quantum field theory.
> All fermions
>have ZPE, or not depending on how you order the operators, "normal,
>symmetric, anti-symmetric". However, the Einstein spontaneous emissoin
>coefficients will only be 1/2 of the correct value if you do not take into
>accout the ZPE of both the vacuum and the particle. ZPE is a necessary
>attribute of both the particle source and the field.
If you want to talk about spontaneous emission, you have coupled the
particle to the photon field. It is no longer a free particle. It is
an interacting particle.
>>>So
>>>basically delta - _E,p,t,r_ are never zero if a particle actually exists.
>>
>>This statement is not true
>>
>
>The statement is easily shown to be true by examining the Heisenberg
>equations of motion for the free particle solutions of the Dirac equation.
I find it very easy to write down solutions of the Dirac equation with
well defined energies and momenta.
Todd Desiato
Kevin A. Scaldeferri wrote in message <6ou8a9$5...@gap.cco.caltech.edu>...
I can see from my statement above that using the word "probability" was
incorrect and could be interpreted to mean "probability density". I thought
the statement "within the interval" would suffice. These ARE the probability
"widths" along each axis, E,t,p and r. The uncertainty in position has a
range (width) of possible values along the position axis r. I called that
width delta_r, etc.
In the rest of that first post I just manipulated the equations by taking
the ratio of the two and squaring both sides. Then I had probability
densities. Where is the error in that? I find the idea of deriving the
metric for flat space-time from the Zitterbewegung motion of a free particle
a fascinating subject. It implies in curved space-time the uncertainty
widths are distorted from spherical symmetry.
Todd D.
Kevin A. Scaldeferri
before discovering the most egregious error in this post, which was
pointed out to me in a private email message. (Thanks, Ken)
In article <UObr1.246$E5.23...@proxye1.san.rr.com>,
Todd Desiato <tde...@san.rr.com> wrote:
>
>(delta_E)(delta_t) > or = h/2
>
>(delta_p)(delta_r) > or = h/2
>
>
>Now here's the good part. Take the ratio of the two uncertainties, we get;
>
>(delta_E)
>----------- (delta_t) > or = (delta_r)
>(delta_p)
>
You can't do this! A>C and B>C implies nothing about A/B.
Todd Desiato
Kevin A. Scaldeferri wrote in message <6p0dvi$j...@gap.cco.caltech.edu>...
>Todd Desiato wrote:
>>
>>(delta_E)(delta_t) > or = h/2
>>
>>(delta_p)(delta_r) > or = h/2
>>
>>
>>Now here's the good part. Take the ratio of the two uncertainties, we
get;
>>
>>(delta_E)
>>----------- (delta_t) > or = (delta_r)
>>(delta_p)
>>
>
>You can't do this! A>C and B>C implies nothing about A/B.
>
It's fine as long as you have A>B, and since;
E^2 = m^2 + p^2
I'm pretty confident that the magnitude of E > p, or equal to it. I was
considering only the magnitudes of the amplitudes. Furthermore if delta_r is
larger than delta_t it would imply velocity greater than 1 (faster than
light). So this implies
delta_t > delta_r
(or equal to it), assuring that A>B, and;
(delta_E)(delta_t) > or = (delta_p)(delta_r)
Sorry, no error other than not stating this explicitly in the first place.
This is the flat time-like space-time metric formulated from the uncertainty
principle. In a curved space-time the probability distributions are
distorted, and the uncertainty relations can be formulated from the metric
for each axis.
Todd Desiato
Todd Desiato
Kevin A. Scaldeferri wrote in message <6ou8a9$5...@gap.cco.caltech.edu>...
>No, it is still true in relativistic quantum mechanics. The following
>describes issues in quantum field theory.
>
Right
>> All fermions
>>have ZPE, or not depending on how you order the operators, "normal,
>>symmetric, anti-symmetric". However, the Einstein spontaneous emissoin
>>coefficients will only be 1/2 of the correct value if you do not take into
>>accout the ZPE of both the vacuum and the particle. ZPE is a necessary
>>attribute of both the particle source and the field.
>
>If you want to talk about spontaneous emission, you have coupled the
>particle to the photon field. It is no longer a free particle. It is
>an interacting particle.
>
No, that was just an example that ZPE is a real quantity, it has real
effects that can't be transformed away. The ZP field is coupled to the Dirac
field. That's as free as a particle can get. Check out the Dirac velocity
operator. There are two eigenvalues, +/- c. So when is a particle "at rest"?
Never. The ziterbewegung is an intrinsic motion. It is a variation in the
interferrence pattern of the superposition of positive and negative
amplitude plane waves with velocity +/- c, AND it is a pointlike particle
moving about within its probability widths. It has a group velocity that
averages zero in its rest frame.
>I find it very easy to write down solutions of the Dirac equation with
>well defined energies and momenta.
The "well defined" comes after you measure it, not before. You choose which
parameter to measure, and even then its only true for that event. When you
measure it again and again you get a statistical average, which is the
average motion of a particle "at rest" in flat space-time. Its really quite
simple yet fascinating.
Todd Desiato
Kevin A. Scaldeferri
In article <pgSs1.1139$E5.65...@proxye1.san.rr.com>,
Todd Desiato <tde...@san.rr.com> wrote:
>
>Kevin A. Scaldeferri wrote in message <6p0dvi$j...@gap.cco.caltech.edu>...
>>Todd Desiato wrote:
>>>
>>>(delta_E)(delta_t) > or = h/2
>>>
>>>(delta_p)(delta_r) > or = h/2
>>>
>>>
>>>Now here's the good part. Take the ratio of the two uncertainties, we
>get;
>>>
>>>(delta_E)
>>>----------- (delta_t) > or = (delta_r)
>>>(delta_p)
>>>
>>
>>You can't do this! A>C and B>C implies nothing about A/B.
>>
>
>
>It's fine as long as you have A>B, and since;
>
>E^2 = m^2 + p^2
>
But E > p does not imply Delta_E > Delta_p.
>I'm pretty confident that the magnitude of E > p, or equal to it. I was
>considering only the magnitudes of the amplitudes. Furthermore if delta_r is
>larger than delta_t it would imply velocity greater than 1 (faster than
>light). So this implies
>
>delta_t > delta_r
No, you are confusing statistical uncertainty with the change in value
of a variable
Todd Desiato
Kevin A. Scaldeferri wrote in message <6p17ng$7...@gap.cco.caltech.edu>...
>No, you are confusing statistical uncertainty with the change in value
>of a variable
Why do you think there is statistical uncertainty? Because the value of the
variable is changing. You get one of many possible values each time you
observe it. It is the wave nature of the particle. I'm not confused in any
way.
Do you reject wave-particle duality as being a fact of nature? Are you
implying that the value of the variable is independent of the statistical
uncertainty in the observation? If you are, that's absurd.
One observation gives you a well defined momentum for only that one event.
You must make many observations to determine the distribution of values that
variable can take on. There is a distribution of possible values because the
value of the variable is constantly changing. If you have some prejudice
against treating a particle as a wave, then say so. But there is nothing in
QFT or Relativity that is contradictory to what I posted.
Todd Desiato
Kevin A. Scaldeferri
Todd Desiato <tde...@san.rr.com> wrote:
>
>Kevin A. Scaldeferri wrote in message <6ou8a9$5...@gap.cco.caltech.edu>...
>
>>
>>> All fermions
>>>have ZPE, or not depending on how you order the operators, "normal,
>>>symmetric, anti-symmetric". However, the Einstein spontaneous emissoin
>>>coefficients will only be 1/2 of the correct value if you do not take into
>>>accout the ZPE of both the vacuum and the particle. ZPE is a necessary
>>>attribute of both the particle source and the field.
>>
>>If you want to talk about spontaneous emission, you have coupled the
>>particle to the photon field. It is no longer a free particle. It is
>>an interacting particle.
>>
>
>No, that was just an example that ZPE is a real quantity, it has real
>effects that can't be transformed away.
In a non-gravitational, free field theory, the zero point energy has
no effect. Our world does not correspond to such a theory, but that
doesn't change the truth of that statement.
> The ZP field is coupled to the Dirac field.
The zero point field?
>Check out the Dirac velocity
>operator. There are two eigenvalues, +/- c.
Only for massless particles.
> So when is a particle "at rest"? Never.
Wrong.
>>I find it very easy to write down solutions of the Dirac equation with
>>well defined energies and momenta.
>
>The "well defined" comes after you measure it, not before. You choose which
>parameter to measure, and even then its only true for that event. When you
>measure it again and again you get a statistical average, which is the
>average motion of a particle "at rest" in flat space-time. Its really quite
>simple yet fascinating.
Yes, it is quite simple.
( sqrt(p.sigma) xi ) exp(-i p.x )
( sqrt(p.sigmabar) xi )
where xi is a two-component spinor, is a solution to the Dirac
equation with a well defined energy and momentum. This is trivial to
show. Take the Fourier transform of the above. The resulting
momentum space wavefunction is non-zero for only a single value of the
4-momentum.
Kevin A. Scaldeferri
Todd Desiato <tde...@san.rr.com> wrote:
>
>Kevin A. Scaldeferri wrote in message <6p17ng$7...@gap.cco.caltech.edu>...
>>No, you are confusing statistical uncertainty with the change in value
>>of a variable
>
>
>Why do you think there is statistical uncertainty? Because the value of the
>variable is changing. You get one of many possible values each time you
>observe it. It is the wave nature of the particle. I'm not confused in any
>way.
You are very confused. There is statistical uncertainty because a
quantum mechanical particle is described by a wavefunction not a
location in phase space. The uncertainty is a function of the
wavefunction _at a given time_. It has nothing to do with the change
in the wavefunction with time.
BTW, by "value" I meant "expectation value". I assumed you did too
since that was the only way your statement made any sense.
Anyways, this is my last post in this thread. I've got better things
to do than continue in a pointless attempt to explain your errors to
you. I've done my duty to point them out to anyone reading who might
not have noticed them immediately.
Todd Desiato
Kevin A. Scaldeferri wrote in message <6p3a2r$s...@gap.cco.caltech.edu>...
>Todd Desiato > wrote:
>>No, that was just an example that ZPE is a real quantity, it has real
>>effects that can't be transformed away.
>
>In a non-gravitational, free field theory, the zero point energy has
>no effect. Our world does not correspond to such a theory, but that
>doesn't change the truth of that statement.
>
Since the first post of this thread I've been talking about the space-time
metric. It should be obvious I am referring to a Quantum field theory that
implies gravitation. Noticing that the probability distributions are
intimately related to the space-time metric is a fundamental step in the
formulation of such a theory.
>> The ZP field is coupled to the Dirac field.
>
>The zero point field?
>
Sorry, I meant zero point electromagnetic field, and the zero point fermion
field, and the zero point gluon field and the zero point quark field, and
...
>>Check out the Dirac velocity
>>operator. There are two eigenvalues, +/- c.
>
>Only for massless particles.
>
The velocity operator I am talking about operates on the probability waves,
not the wave packet. The waves that comprise a wave packet move at c, even
when the group velocity is (on average) zero. It is valid for the Dirac
equation. The text says nothing about it only applying to only massless
particles and infact mentions it while discussing a free electron. So..a
matter of interpretation perhaps?
>> So when is a particle "at rest"? Never.
>
>Wrong.
>
Can you prove that? For every observation of the same free particle you get
the same value, with no variations, no errors, no uncertainty?
>>>I find it very easy to write down solutions of the Dirac equation with
>>>well defined energies and momenta.
>>
>>The "well defined" comes after you measure it, not before. You choose
which
>>parameter to measure, and even then its only true for that event. When you
>>measure it again and again you get a statistical average, which is the
>>average motion of a particle "at rest" in flat space-time. Its really
quite
>>simple yet fascinating.
>
>Yes, it is quite simple.
>
> ( sqrt(p.sigma) xi ) exp(-i p.x )
> ( sqrt(p.sigmabar) xi )
>
>where xi is a two-component spinor, is a solution to the Dirac
>equation with a well defined energy and momentum. This is trivial to
>show. Take the Fourier transform of the above. The resulting
>momentum space wavefunction is non-zero for only a single value of the
>4-momentum.
>
Observing a particle with a relative momentum is not "at rest" now is it?
From the other post Todd wrote;
>>Why do you think there is statistical uncertainty? Because the value of
the
>>variable is changing. You get one of many possible values each time you
>>observe it. It is the wave nature of the particle. I'm not confused in any
>>way.
>
>You are very confused. There is statistical uncertainty because a
>quantum mechanical particle is described by a wavefunction not a
>location in phase space.
Gee really?
>The uncertainty is a function of the
>wavefunction _at a given time_. It has nothing to do with the change
>in the wavefunction with time.
>
The uncertainty is a determined by the wavefunction for every observation _
at all times_. If you make many observations, each at a separate event in
space-time, and end up with a distribution of values, then what's the
difference? Your statement is contradictory.
Todd Desiato
Anil Trivedi
it may be helpful, at least to the rest of us, if you took a
step back and summarized in a paragraph or two just what you
are arguing about? :)
-Anil Trivedi
SDRodrian
>....
>From: "Dave Morgan" <mor...@physics.wm.edu>
>Date: Sun, Jul 26, 1998 18:55 EDT
>Message-id: <6pgqk7$1...@sjx-ixn4.ix.netcom.com>
>
>----------
>In article <199807270035...@ladder01.news.aol.com>,
>sdro...@aol.com (SDRodrian) wrote:
>
>>Motion is defined as change (you cannot have motion without
>>change nor change without motion--this is self-evident).
>>
>
>Right, and I'm saying CHANGE WITH RESPECT TO WHAT?!
>
You're thinking of CHANGE as an absolute! Quit it!
Listen carefully: one change (one item of existence
'moving' into some another shape) and with respect to
all/one some other (item of existence 'moving' into another
shape of their/its own NOT related to the localized
original change you're talking about).
"All is change" is one of our most ancient axioms: But
you cannot think of the universe as an absolute singularity
(the universe is merely the sum of everything in it): It is
NOT one thing but all things.... (Hint.)
>In calculus terms, velocity is dx/d??
>
>You use the words motion and change, neither of which has any meaning
>without a solid concept of time.
>
>Dave
It is NOT time that produces change. Change is the nature
of existence (cause/effect). The concept of Time is strictly
speaking ONLY the method the human mind uses to
describe the changes of which our universe is composed:
Every single item of existence, from the least quark to
galaxies and super-massive black holes, is in the process
of changing (thusly it is impossible to describe reality
as anything other than change):
Every atom in the universe is in the process of changing (call it
decaying if you like, or call it whatever else you like)... that's
how they gained existence, that's how they exist, that's how
they will cease to exist--But language betrays us: An atom
does not cease to exist, it ceases to exist as an atom and
becomes something else. This ought not to be all that hard
to grasp.
Well, in the same way, Existence itself cannot have had
a beginning (or it would not exist). But if we describe
the form/shape of the universe as existence, then AND
ONLY THEN can we describe existence as having
begun with the Big Bang. However, this is only because
we are really speaking of the universe when we speak
of existence.
Because Time is only an idea, and not a force acting
upon the items of reality, we can pretty much dispose
of it (or, disregard it) when we describe existence:
Thusly, if we had (theoretically) frozen the motions of
all of the subatomic particles which constituted the city
of Springfield when Abe Lincoln was alive and living there
and unfroze them today, then today we could step
'back in time' and visit Lincoln in Springfield.
And yet our journey would not really be backwards
through time, would it? It would merely be from here
to there (probably by jeep).
Now consider an alternate theoretical scenario: Suppose
that instead of freezing the subatomic particles that
constituted Springfield in the 1840s we instead re-created
them exactly... quark by quaek by quark...
Obviously Abe would still be there as well and we could still
chew the fat with him, we would know it's 1998 and Lincoln
would have to be convinced it was not the 1840s.... so what
does all that do to the notion of Time? What difference then
would there be between Springfield 1840, the Springfield we
just unfroze, and the Springfield we recreated? (e. g. What
difference would there be between the 1840s Lincoln, the
unfrozen Lincoln, and the re-created Lincoln?
Perhaps now you can see that the only thing that exists
is the Present. And that the differences between the
so-called Past and the Present are differences of shape,
not Time. The zillion things which constitute existence
change--every last single one of them according to the
peculiar circumstances in which they find themselves
(where they are, but NOT as they are in Time=time is
only our description of how much change has taken place).
Abe Lincoln did not (live in the past) come into existence
and then vanish from existence... he came to exist as
Abe Lincoln by a number of changes and then changed
into something else--but he certainly didn't cease to exist.
Abe is still with us, it's just that he has changed a great deal
by now.
I hope this helps,
Dan
SDRodrian
<< Subj: Re: GOD, TIME, and The BIBLE ....
Date: 98-07-27 09:25:49 EDT
From: mor...@physics.wm.edu (Dave Morgan)
To: sdro...@aol.com (SDRodrian)
-----Original Message-----
From: SDRodrian <sdro...@aol.com>
>You're thinking of CHANGE as an absolute! Quit it!
>Listen carefully: one change (one item of existence
>'moving' into some another shape) and with respect to
>all/one some other (item of existence 'moving' into another
>shape of their/its own NOT related to the localized
>original change you're talking about).
>
> Whoa! Care to find me the subject and verb in that "sentence"?!
I'll leave that to my old English teacher (even at her age she NEVER fails).
> All I'm asking is this. you have used the word MOTION over and over. Please
> provide me with a DEFINITION of this word without either explicitly or
> implicityly referring to TIME. I say it can't be done.
> Dave
>>
No sooner asked. But please READ carefully and re-read whatever
you find lacking in meaning a few times before giving up on it...
1. Forget Time, all references to it--you never heard of the animal--Period.
2. Now, how would you, a brain that has never heard of Time, describe
the changes you notice in the 'real' world around you? As a brain, it
would certainly occur to you to ask yourself... WHY are they happening
(what makes them 'happen').....?
3. Now, reflect for a moment: Would it even occur to you that how fast
they are happening is of any pertinent consequence at all--except maybe
as a form of amusement or entertainment for you? (Remember that
Science has not yet been invented.)
You can see that 1 and 2 are things we share with the 'lower' animals,
and it is ONLYwhen we cross over into the human brain that 3 comes
into play at all.
Time is not concerned with the description of change, nor is it involved
with why changes occur: The mind describes change as away (motion)
from whatever it was before. This does not involve the concept of Time
but only the motions of subatomic particles... an ice-cube melts into
water NOT because time melts it (keep it cold enough and no amount of
time will ever have the slightest effect on it). I hope this makes it obvious
to you--I have tried my best to make nothing but self-evident statements.
The notion of time is exclusively confined to measuring how faster and/or
slower one change happens in relation to something else (it is ONLY a
means by which the human mind--and ONLY the human mind until we
come across intelligent extraterestrial life--quantifies such rates of change).
The notion of Time is useful whenever you wish to toss out mathematical
formulas to explain physical phenomena (such as, "I'll be with you in an
hour."), but an hour does not exist outside the human mind--it is merely
an agreement between you and me that it equals the motion of this
little stick from here to there.
S. D. Rodrian
Jim Carr
>
>The product of the statistical uncertainties in the position and
>momentum of a particle is greater than or equal to hbar/2. Your
>version claims that it the the product of the probability densities
>which is being considered.
Sometimes it can be helpful to say it is the product of the
variances, if the reader is familiar with that term for a
statistical uncertainty.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Jim Carr
}
} No, you are confusing statistical uncertainty with the change in value
} of a variable
"Todd Desiato" <tde...@san.rr.com> writes:
>
>Why do you think there is statistical uncertainty? Because the value of the
>variable is changing.
That is not why I think it is uncertain. I think it is uncertain
because it is not sharply defined. In this case it is intrinsically
uncertain. You toss a pair of dice many times and determine the
mean and variance of the probability distribution. That is what you
know about those dice. Nothing is "changing" to produce some blurry
distribution -- that is the distribution. It is an intrinsic property
of those dice, if you like. If they are not "fair" dice, that will
show up in those moments being different from some other set of dice.
>You get one of many possible values each time you
>observe it. It is the wave nature of the particle. I'm not confused in any
>way.
You are when you say something is changing.
Todd Desiato
Jim Carr wrote in message <6qamfg$4mg$1...@news.fsu.edu>...
>Kevin A. Scaldeferri wrote:
>}
>} No, you are confusing statistical uncertainty with the change in value
>} of a variable
>
>>
>>Why do you think there is statistical uncertainty? Because the value of
the
>>variable is changing.
>
> That is not why I think it is uncertain. I think it is uncertain
> because it is not sharply defined. In this case it is intrinsically
> uncertain. You toss a pair of dice many times and determine the
> mean and variance of the probability distribution. That is what you
> know about those dice. Nothing is "changing" to produce some blurry
> distribution -- that is the distribution. It is an intrinsic property
> of those dice, if you like.
Each time you roll the dice there is a probability of getting a different
pair of faces, facing up. The pair of faces facing up "changes" within the
allowable values each time you roll the dice. The dice are not changing, but
the outcome is. That is all I was saying.
>If they are not "fair" dice, that will
> show up in those moments being different from some other set of dice.
>
>>You get one of many possible values each time you
>>observe it.
See?
>It is the wave nature of the particle. I'm not confused in any
>>way.
>
> You are when you say something is changing.
>
The outcome of the observation is not the same every time. Otherwise you
would have a well defined value and not a distribution of values. That all.
Todd
Jim Carr
|
| Kevin A. Scaldeferri wrote:
| } No, you are confusing statistical uncertainty with the change in value
| } of a variable
|
| "Todd Desiato" writes:
| >Why do you think there is statistical uncertainty? Because the value of
| >the variable is changing.
|
| That is not why I think it is uncertain. I think it is uncertain
| because it is not sharply defined. In this case it is intrinsically
| uncertain. You toss a pair of dice many times and determine the
| mean and variance of the probability distribution. That is what you
| know about those dice. Nothing is "changing" to produce some blurry
| distribution -- that is the distribution. It is an intrinsic property
| of those dice, if you like.
"Todd Desiato" <tde...@san.rr.com> writes:
>
>Each time you roll the dice there is a probability of getting a different
>pair of faces, facing up. The pair of faces facing up "changes" within the
>allowable values each time you roll the dice. The dice are not changing, but
>the outcome is. That is all I was saying.
Actually, the way I read your original remarks, you were saying that
something is moving around and the snapshots happen to catch it here
or there as it follows a path described by some (hidden) variable.
As this analogy tries to show, that variable takes on different
values but does not "change" in the way one uses the word classically.
| >It is the wave nature of the particle. I'm not confused in any
| >way.
|
| You are when you say something is changing.
>The outcome of the observation is not the same every time. Otherwise you
>would have a well defined value and not a distribution of values. That all.
OK, as long you don't mean that there is a hidden variable that is
changing smoothly, causing the seemingly random results we see. That
is not how QM works. My reaction is to the word "change" with usually
carries with it the baggage of a continuous variable with d/dt and
all that Newtonian time evolution business.
Todd Desiato
Jim Carr wrote in message <6rcqgu$3mr$1...@news.fsu.edu>...
>> "Todd Desiato" writes:
>>The outcome of the observation is not the same every time. Otherwise you
>>would have a well defined value and not a distribution of values. That
all.
>
> OK, as long you don't mean that there is a hidden variable that is
> changing smoothly, causing the seemingly random results we see. That
> is not how QM works. My reaction is to the word "change" with usually
> carries with it the baggage of a continuous variable with d/dt and
> all that Newtonian time evolution business.
OK. Interesting this thread came up again. I just read this week in Science
News magazine that using ultra low temperature Bose-Einstein condensates
induced by laser cooling, a group of scientist is working on a
gravity-gradient detector. It works by creating a single quantum state from
millions of cesium atoms, then observing interference in each clouds quantum
mechanical wave behavior over time, while separated by 1 meter. (Described
in Aug. 3 Physical Review Letters)
This is the same principle I was trying to show when I started this thread.
Todd Desiato
Jim Carr
|
| "Todd Desiato" writes:
| >The outcome of the observation is not the same every time. Otherwise you
| >would have a well defined value and not a distribution of values. That
| >all.
|
| OK, as long you don't mean that there is a hidden variable that is
| changing smoothly, causing the seemingly random results we see. That
| is not how QM works. My reaction is to the word "change" with usually
| carries with it the baggage of a continuous variable with d/dt and
| all that Newtonian time evolution business.
"Todd Desiato" <tde...@san.rr.com> writes:
>
>OK. Interesting this thread came up again. I just read this week in Science
>News magazine that using ultra low temperature Bose-Einstein condensates
>induced by laser cooling, a group of scientist is working on a
>gravity-gradient detector. It works by creating a single quantum state from
>millions of cesium atoms, then observing interference in each clouds quantum
>mechanical wave behavior over time, while separated by 1 meter. (Described
>in Aug. 3 Physical Review Letters)
Right. Standard quantum interference. You cannot predict which
atom will be in which place, but from the ensemble you get solid
information about the evolution of the QM state, specifically the
energy. This is a more precise elaboration of the "double slit"
experiments with neutrons that showed gravity affecting them.
>This is the same principle I was trying to show when I started this thread.
Not as I read you statement.
What evolves is the wavefunction.
Todd Desiato
Jim Carr wrote in message <6sd0hu$p72$1...@news.fsu.edu>...
>"Todd Desiato" <tde...@san.rr.com> writes:
>>
>>OK. Interesting this thread came up again. I just read this week in
Science
>>News magazine that using ultra low temperature Bose-Einstein condensates
>>induced by laser cooling, a group of scientist is working on a
>>gravity-gradient detector. It works by creating a single quantum state
from
>>millions of cesium atoms, then observing interference in each clouds
quantum
>>mechanical wave behavior over time, while separated by 1 meter. (Described
>>in Aug. 3 Physical Review Letters)
>
> Right. Standard quantum interference. You cannot predict which
> atom will be in which place, but from the ensemble you get solid
> information about the evolution of the QM state, specifically the
> energy. This is a more precise elaboration of the "double slit"
> experiments with neutrons that showed gravity affecting them.
>
>>This is the same principle I was trying to show when I started this
thread.
>
> Not as I read you statement.
>
> What evolves is the wavefunction.
>
And as it evolves the width of the probability density changes.
Todd Desiato
Todd Desiato
Metric Tensor of my Quantum wave dispersion interpretation of gravitation.
Comments?
This IS Quantum Gravity!
"Derivation of the Refractive Metric Tensor"
http://tjd-online.simplenet.com/Metric/Metric.html
"The Properties of Mass and Space-time Curvature as Wave Dispersion"
http://tjd-online.simplenet.com/Gravity/Dispersion.html
Thanks,
Todd Desiato