uncertainty principle

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Matthew Konicki

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Oct 20, 1999, 3:00:00 AM10/20/99
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I would like to know if someone could explain two things to me. First, why
doesn't the uncertainty principle imply that we cannot know the position of
light with any certainty, since we know it's velocity with precise
certainty? Second, if E = mc^2, then why does light have energy, despite
being massless? Any insight would be appreciated.

Nathan Urban

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Oct 20, 1999, 3:00:00 AM10/20/99
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In article <7ul6f0$p2e$1...@nntp6.atl.mindspring.net>, "Matthew Konicki" <kon...@mindspring.com> wrote:

> I would like to know if someone could explain two things to me. First, why
> doesn't the uncertainty principle imply that we cannot know the position of
> light with any certainty, since we know it's velocity with precise
> certainty?

The uncertainty relation is between position and _momentum_, and though
photons have a single _speed_, they can have any momentum.

> Second, if E = mc^2, then why does light have energy, despite
> being massless?

Because E = mc^2 is a special case of a more general formula, and that
special case only applies in the rest frame of a massive particle.
The general formula valid in all frames for all particles is
E^2 = (mc^2)^2 + (pc)^2. For the case of a photon, it reduces to E = pc.
Photons have energy and momentum, but no mass.

Bill Rowe

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Oct 20, 1999, 3:00:00 AM10/20/99
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In article <7ul6f0$p2e$1...@nntp6.atl.mindspring.net>, "Matthew Konicki"
<kon...@mindspring.com> wrote:

>I would like to know if someone could explain two things to me. First, why
>doesn't the uncertainty principle imply that we cannot know the position of
>light with any certainty, since we know it's velocity with precise

>certainty? Second, if E = mc^2, then why does light have energy, despite
>being massless? Any insight would be appreciated.

Think about a wave for a moment. How do you define the position of a wave
at an instant in time? Would you use the crest? the trough? some other
feature of the wave? The point is position of a wave is not well defined.

Postion of a photon is defined by its wavefunction. The wavefunction
defines the probability of finding something at a particular point. It has
similar properties to physical waves. Hence defining postion is equally
difficult.

There is a second issue as well in determing position. In order to locate
a particle you have to do something to detect it. That means interacting
with it in some way. That interaction affects the motion of the particle.

E = mc^2 says nothing about light. It doesn't apply to light. Instead, it
applies to particles with mass at rest.

The complete equation expresses a relationship between momentum, energy
and mass. E = mc^2 is a special case of the complete equation.

The complete equation is

E^2 = (pc)^2 + (mc^2)^2

which applies to all particles.

For light, m = 0 and this equation reduces to E = pc.

--
-
PGPKey fingerprint: 6DA1 E71F EDFC 7601 0201 9243 E02A C9FD EF09 EAE5


Charles Francis

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Oct 21, 1999, 3:00:00 AM10/21/99
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In article <7ul6f0$p2e$1...@nntp6.atl.mindspring.net>, Matthew Konicki
<kon...@mindspring.com> writes

>I would like to know if someone could explain two things to me. First, why
>doesn't the uncertainty principle imply that we cannot know the position of
>light with any certainty, since we know it's velocity with precise
>certainty?

The short answer is that it does. We cannot detect a photon in an
eigenstate of position, only where it has been absorbed.


>Second, if E = mc^2, then why does light have energy, despite
>being massless? Any insight would be appreciated.
>

This is an approximation to the mass shell condition

m^2 = E^2 - P^2

where P is momentum.

For light we have E^2 = P^2

A rapid demonstration of the mass shell condition is given in

Conceptual Foundations of Special and General Relativity
http://xxx.lanl.gov/abs/physics/9909051

--
Charles Francis
cha...@clef.demon.co.uk

z@z

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Oct 22, 1999, 3:00:00 AM10/22/99
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Nathan Urban wrote:
| "Matthew Konicki" wrote:

| > I would like to know if someone could explain two things to me. First,
| > why doesn't the uncertainty principle imply that we cannot know the
| > position of light with any certainty, since we know it's velocity with
| > precise certainty?
|

| The uncertainty relation is between position and _momentum_, and though
| photons have a single _speed_, they can have any momentum.

At least the original photons (postulated by Einstein) have both
position and momentum. This is easy to understand. Assume a small
source of photons in a dark room and a small detector. If a photon
is detected, we know not only its position but also its momentum.
The magnitude of the momentum is

p = v * m = c * [h*f/c^2] = h*f/c

and its direction is exactly given by the vector from the source
to the detector. To deny such simple reasonings is pure sophistry.

In some respect, QM was an attempt not to admit that Einstein was
right after experiments (Compton 1923, Bothe and Geiger 1925)
had shown that Bohr was wrong (e.g. Bohr, Kramers and Slater) and
Einstein right. The fathers of QM tried to save as much as possible
from their previously advocated but now experimentally refuted
positions, taking refuge with obscure mathematics.

| > Second, if E = mc^2, then why does light have energy, despite
| > being massless?
|

| Because E = mc^2 is a special case of a more general formula, and that
| special case only applies in the rest frame of a massive particle.
| The general formula valid in all frames for all particles is
| E^2 = (mc^2)^2 + (pc)^2. For the case of a photon, it reduces to E = pc.
| Photons have energy and momentum, but no mass.

Modern physics has given up or never taken seriously the mass
energy equivalence. The distinction between particles with mass
and particles without mass leads to a lot of contradictions.

The formula

E^2 = [mc^2]^2 + [pc]^2

does not make a lot of sence because the momentum p depends on
the mass-engergy equivalence itself.

p(v) = m(v) * v = m(0)/sqrt(1-v^2/c^2) * v

So if rest mass m(0) = m we get

E^2 = [mc^2]^2 + [pc]^2 = [mc^2]^2 + [mvc]^2 / [1 - v^2/c^2]

= ... = [mc^2]^2 / [1 - v^2/c^2 ]

The version

E = mc^2 / sqrt(1-v^2/c^2)

is simpler and does not implicitly assume something which is not
explicitely stated.

In the case of photons the momentum relation p = m*v is fundamental
because it is a necessary condition for consistent momentum
conservation in all cases.


Wolfgang Gottried G.

Relationality instead of Relativity:
http://members.lol.li/twostone/E/physics1.html

francis Rey

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Oct 30, 1999, 3:00:00 AM10/30/99
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Objet: Re: uncertainty principle
Date: 20 Oct 1999 16:09:04 -0400
De: nur...@crib.corepower.com (Nathan Urban)
Répondre-A: nur...@vt.edu
Société: InSystems Technologies, Inc.
Forums: sci.physics.relativity

In article <7ul6f0$p2e$1...@nntp6.atl.mindspring.net>, "Matthew Konicki"

<kon...@mindspring.com> wrote:

> I would like to know if someone could explain two things to me. First,
why
> doesn't the uncertainty principle imply that we cannot know the position
of
> light with any certainty, since we know it's velocity with precise
> certainty?

The uncertainty relation is between position and _momentum_, and though
photons have a single _speed_, they can have any momentum.

> Second, if E = mc^2, then why does light have energy, despite
> being massless?

Because E = mc^2 is a special case of a more general formula, and that
special case only applies in the rest frame of a massive particle.
The general formula valid in all frames for all particles is
E^2 = (mc^2)^2 + (pc)^2. For the case of a photon, it reduces to E = pc.
Photons have energy and momentum, but no mass.

Nathan Urban
-----------------

Francis Rey

Momentum ? what is a momentum ? Just, by definition:

p = m v

Then for the photon, if we believe Nathan:

E= p c = m v c !!

Obviously not ! The photon velocity is c , then p = m c
Then:
E = m c c = m c^2

With p = m c the general energy formula than becomes:

E^2 = (mc^2)^2 + (pc)^2. = (mc^2)^2 + (m c c)^2.

E^2 = 2 (mc^2)^2 E = (mc^2)sqrt 2 ????

But Nathan also says the photon has got no mass !
No mass ?
then m = 0
What about p = 0 c ?
No mass can also give p = c
What about the general formula for energy ?

--
francis Rey

The fact much to learn don't teach intelligence
Heraclite of Ephesus.

franc...@wanadoo.fr
http://perso.wanadoo.fr/francis.rey/

francis Rey

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Oct 31, 1999, 2:00:00 AM10/31/99
to

Francis Rey

ZAZ would have to learn a bit of Special Relativity before to give lesson to
poor
Matthew Konicki, I bet who knows perfectly the points he want to be explained.
ZAZ must be a teacher who, as it is the use, teaches things he does not know.

Objet: Re: uncertainty principle
Date: Fri, 22 Oct 1999 18:55:23 +0200
De: "z@z" <z...@z.lol.li>
Société: Swisscom IP+ (post doesn't reflect views of Swisscom)
Forums: sci.physics.relativity

Nathan Urban wrote:
| "Matthew Konicki" wrote:

| > I would like to know if someone could explain two things to me. First,
| > why doesn't the uncertainty principle imply that we cannot know the
| > position of light with any certainty, since we know it's velocity with
| > precise certainty?
|
| The uncertainty relation is between position and _momentum_, and though
| photons have a single _speed_, they can have any momentum.

At least the original photons (postulated by Einstein) have both


position and momentum. This is easy to understand. Assume a small
source of photons in a dark room and a small detector. If a photon
is detected, we know not only its position but also its momentum.
The magnitude of the momentum is

p = v * m = c * [h*f/c^2] = h*f/c

and its direction is exactly given by the vector from the source
to the detector. To deny such simple reasonings is pure sophistry.

In some respect, QM was an attempt not to admit that Einstein was
right after experiments (Compton 1923, Bothe and Geiger 1925)
had shown that Bohr was wrong (e.g. Bohr, Kramers and Slater) and
Einstein right. The fathers of QM tried to save as much as possible
from their previously advocated but now experimentally refuted
positions, taking refuge with obscure mathematics.

| > Second, if E = mc^2, then why does light have energy, despite


| > being massless?
|
| Because E = mc^2 is a special case of a more general formula, and that
| special case only applies in the rest frame of a massive particle.
| The general formula valid in all frames for all particles is
| E^2 = (mc^2)^2 + (pc)^2. For the case of a photon, it reduces to E = pc.
| Photons have energy and momentum, but no mass.

--------------
Francis Rey
I already replied to that statement of Nathan Urban.
What means "no mass" ? It is necessary to express that by an equation:
a) m = 0 ?
Then p = 0 c = 0
b) no letter to eexpress a mas in momentum ?

then p = c
then
E^2 = (mc^2)^2 + (mc c)^2 = 2 (mc^2)^2 ?????
------------------
Wolfgang:

Modern physics has given up or never taken seriously the mass
energy equivalence. The distinction between particles with mass
and particles without mass leads to a lot of contradictions.

The formula

E^2 = [mc^2]^2 + [pc]^2

does not make a lot of sence because the momentum p depends on
the mass-engergy equivalence itself.

p(v) = m(v) * v = m(0)/sqrt(1-v^2/c^2) * v

---------------
Francis Rey

Very curious abundancy of mass ! Three mass in the problem ?
m, m(v), and m(o)
In fact it never existed any m(v). In SR one writes merely:

m = m(o) / sqrt(1-v^2/c^2)

And it is from this formula for the "relativistic mass" that the Energy formula
is
built: here the three needed elements

E = mc^2 , m = m(o) /sqrt(1-v^2/c^2) , p = mv

Just a little approximation and you can obtain the general energy formula.
That is in the good books on relativity.

Now, it exists another fine approximation, used first by Einstein himself given

here above by Wolfgang :
--------
Wolfgang text:

**So if rest mass m(0) = m we get**

E^2 = [mc^2]^2 + [pc]^2 = [mc^2]^2 + [mvc]^2 / [1 - v^2/c^2]

= ... = [mc^2]^2 / [1 - v^2/c^2 ]

---------
Francis Rey
I like very much that kind of approximation, because we have absolute right to
use it anywhere in the derivation. Then return to the relativistic mass:
m = m(o) /sqrt(1-v^2/c^2) = m /sqrt(1-v^2/c^2)

Shouldn't that give something like:

1 = 1 /sqrt(1-v^2/c^2)

1 - v^2/c^2 = 1 ????

v = 0

Is that a strange condition ? No ! All is perfectly right.
The Lorentz formulae are done for a light ray running in a moving frame with a

velocity v .
They cannot be used for any massive particle. The result v = 0, means that it
does not exist such moving frame.
In matter of fact the formula for realtivistic mass is quite wrong, as well as
the
formula of the general energy.
--------
Wolfgang:

The version

E = mc^2 / sqrt(1-v^2/c^2)

is simpler and does not implicitly assume something which is not
explicitely stated.

---------------
Francis Rey

poor wolfgang ! **In SR** this version is the main original formula to obtain
the
E^2 = (mc^2)^2 + (pc)^2
---------------
Wolfgang:


In the case of photons the momentum relation p = m*v is fundamental
because it is a necessary condition for consistent momentum
conservation in all cases.

Wolfgang Gottried G.
---------------
Francis Rey
p = m v is fundamental for the photon, when he has just said that the photon
is
massless and has a velocity c .

Poor wolfgang ! Rationality in SR ???
-----------------

z@z

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Oct 31, 1999, 2:00:00 AM10/31/99
to
francis Rey wrote:
Wolfgang (z@z) wrote:

| > Modern physics has given up or never taken seriously the mass
| > energy equivalence. The distinction between particles with mass
| > and particles without mass leads to a lot of contradictions.
| >
| > The formula
| >
| > E^2 = [mc^2]^2 + [pc]^2
| >
| > does not make a lot of sence because the momentum p depends on
| > the mass-engergy equivalence itself.
| >
| > p(v) = m(v) * v = m(0)/sqrt(1-v^2/c^2) * v

| Very curious abundancy of mass ! Three mass in the problem ?


| m, m(v), and m(o)
| In fact it never existed any m(v). In SR one writes merely:
|
| m = m(o) / sqrt(1-v^2/c^2)

By using m(v) I want to stress that the mass is a function of
velocity. m(0) = m(velocity=0)

| > So if rest mass m(0) = m we get
| >

| > E^2 = [mc^2]^2 + [pc]^2 = [mc^2]^2 + [mvc]^2 / [1 - v^2/c^2]
| >
| > = ... = [mc^2]^2 / [1 - v^2/c^2 ]

Also the detailed version

[E(v)]^2 = [m(0) * c^2]^2 + [p(v) * c]^2
= [m(0) * c^2]^2 + [m(v)*v * c]^2
= [m(0) * c^2]^2 + [m(0)*gamma(v)*v * c]^2

shows that this formula is not more fundamental than the original

E(v) = m(0) * c^2 * gamma(v)

I think that lots of errors are caused in modern theoretical
physics because not enough attention is given to the distinction
between variables expressing constants and variables expressing
functions depending on other variables.

| > In the case of photons the momentum relation p = m*v is fundamental
| > because it is a necessary condition for consistent momentum
| > conservation in all cases.

| p = m v is fundamental for the photon, when he has just said that the


| photon is massless and has a velocity c .

Isn't it obvious that the consistent application of the mass-
energy-equivalence can only lead to the conclusion that massless
particles are impossible?

Cheers, Wolfgang

Relationality instead of relativity:
http://members.lol.li/twostone/E/physics1.html

Luc Bourhis

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Oct 31, 1999, 2:00:00 AM10/31/99
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Charles Francis <cha...@clef.demon.co.uk> wrote:

> This is an approximation to the mass shell condition
>
> m^2 = E^2 - P^2

^^^^^^^^^^^^^^^
Be careful charles, some newbies may be surprised by your formula where
c=1 !

--
Luc Bourhis
Center for Particle Physics / University of Durham
United Kingdom

Luc Bourhis

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Oct 31, 1999, 2:00:00 AM10/31/99
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francis Rey <franc...@wanadoo.fr> wrote:

> | > Second, if E = mc^2, then why does light have energy, despite
> | > being massless?
> |
> | Because E = mc^2 is a special case of a more general formula, and that
> | special case only applies in the rest frame of a massive particle.
> | The general formula valid in all frames for all particles is
> | E^2 = (mc^2)^2 + (pc)^2. For the case of a photon, it reduces to E = pc.
> | Photons have energy and momentum, but no mass.
> --------------
> Francis Rey
> I already replied to that statement of Nathan Urban.
> What means "no mass" ? It is necessary to express that by an equation:
> a) m = 0 ?
> Then p = 0 c = 0
> b) no letter to eexpress a mas in momentum ?
>
> then p = c
> then
> E^2 = (mc^2)^2 + (mc c)^2 = 2 (mc^2)^2 ?????

The simplest way to deal with these problems in SR is to *define* the
mass m of a particle with an energy E and a momentum by (I use c=1
everywhere and p is a 3-vector) :

m^2 = E^2 - p^2

and to require that m is the same in every frames.

For a massless particle m = 0 implies E = |p| (the norm of p) and there
is nothing more to say from an SR point of view. However for a photon
our knowledge of Quantum Mechanics gives a formula for E.

In order to continue one must remember a well-known mathematical result
which says that the set of transforms of (x0,x1,x2,x3) which let x0^2 -
(x1^2 + x2^2 + x3^2) invariant are the elements of the Lorentz group.
Therefore the transforms between observers for (E,p) are :

E' = gamma(E - v p)
p' = gamma(p - v E)

if I consider speeds v between frames which are // to the momentum.

If m is not 0 then |p| is not equal to E and I can find v with |v| < 1
so that p' = 0 (i.e. there is a frame where the momentum of the particle
is 0). We have then v = p/E and E' = gamma(E^2 - p^2)/E. This energy is
called rest energy and if I note it E0, I have E = m^2/E0 gamma(v). For
a small speed v one must recover the Newtonian behavior 1/2 m v^2 and
this implies that E0 = m. Therefore for a particle with a mass m :

E = m gamma and p = m gamma v

These relations are a consequece of SR and are independent of the nature
of the particle as long as it has a mass different from 0. The last step
consists in identifying v as the speed of the particle.

Game over.

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