Henri Wilson wrote:
On Tue, 04 Apr 2006 13:12:38 +0200, "Paul B. Andersen"
wrote:


Henri Wilson wrote:
A moving charge constitutes a current that creates a 'back emf' in a closed
circuit (or between two plates). The field created by an accelerating charge
must therefore have the effect of reducing the applied field. Its effect will
be local to the charge. It must require considerable energy to maintain a field
gradient in space.

Yes, it is indeed quite simple.
If you have a static electric field with no energy supply,
this field will obviously be diminished somewhat when a charge
move along the field because the energy gained by the particle
must be taken from PE of the field. The gained energy of the charge
is qV, so the field must be diminished accordingly.


that's correct. The 'reverse field' causes it to diminish.


Look at this drawing:
V
E
|+ -- -|
|+ -- -| drift tube
-----|+ -- -|----------
-
q+
-----|+ -- -|-----------
|+ -- -|
|+ -- -|

It is basically a charged condenser with a tube through it.
(No tube between the condenser plates, though).
The tube on each side is at the same potential as the plates.
The potential difference between the plates is V.
In the tube a positive charge q is moving. There is
no field inside the tube, but there will be a surface
charge inside the tube with the opposite polarity,
following the moving charge q. (It is a local field
around the charge in the tube)

Count the charges, Henri.
On the left side there are 7 positive and one negative charge,
that is net 6 positive charges.
On the right side there are 6 negative charges.
The charge on the condenser is 6q.

The potential energy stored in the condenser is:
PE0 = 0.5*C*V^2, C = Q/V = 6q/V
PE0 = 3qV



(That field will become distorted into the 'reverse field bubble')


Now the situation after the charge has passed:
V

|+ -- -|
|+ -- -| drift tube
-----| |----------
-
q+
-----|+ -- -|-----------
|+ -- -|
|+ -- -|

The field have diminished somewhat.
The point is that the moving positive charge is
a part of the whole condenser system, and when
this charge has moved from the positive side of
the system to the negative side, the net result must
be a small discharge of the condenser.
(In my drawings the condenser charge has diminished

from 6q to 5q.)

I would dispute that.

Then you are wrong.
Count the charges.
(I feel a bit embarrassed of having to explain this in such a detail,
one would expect that everybody could count to 6 without help.)

On the left side there were originally 7 positive and one negative charge.
One positive charge has moved to the other side.
That leaves 6 positive and one negative charge, and the one negative
charge is no more tied up to the q+, and will neutralise one of
the positive charges on the plate, leaving 5 positive charges.

On the right side there were originally 6 negative charges.
One positive charge, the q+, is added. So we end up with
6 negative and one positive, or net 5 negative charges.
One of the 6 negative charges on the plates will be tied up to
the q+, leaving 5 charges on the plates.

Note that if the original potential difference was V,
the potential must now have dropped to (5/6)V.

So the potential energy in the condenser has dropped to:
PE1 = 0.5*C*((5/6)V)^2 = (25/12)qV

I would say that V momentarily drops as q passes between the plates but returns
to its original value after it has gone.

It doesn't matter what you think happens while the charge
passes between the plates, we only have to see the difference
before and after.
We have PE0 = 3qV, PE1 = (25/12)qV
d_E = PE0 - PE1 = (11/12)qV

Note that the average potential before - after is:
Vavg = (V + (5/6)V)/2 = (11/12)V

So the condenser has lost the energy d_E = q*Vavg
Since the total energy of the system must be conserved,
the moving charge must have gained the potential energy q*Vavg

And the original speed of the charge doesn't matter at all.

I cannot see any sign of the condensor circuit passing current. The charge q
came from a separate circuit, CMIIW.

Don't be silly. Of course the moving charge IS
the current transporting one positive charge
from one side of the system to the other.

But you are quite funny. :-)
Above, you said:
| A moving charge constitutes a current that creates a 'back emf'
| in a closed circuit (or between two plates).
Now , you say:
| The charge q came from a separate circuit, CMIIW.
and your point with that remark seems to be that
it shouldn't affect the field between the plates.

But it does. It diminishes the field between the plates.
Your former statement is partly right. (Even if it is
imprecisely put, and your reasoning probably is wrong.)

Where did your 6th '+' go, eh, Paul?
You started with seven and ended with six...strange that.

Quite. +7 + (-1) = +6. Strange that.


I realise you are trying to simulate a cyclotron but remember the field between
the D's is AC.

But as the field is set up before the charge enters it,
it can be considered stationary for the short time
it takes the charge to pass it.

Why the hell do I have to state this over and over?


The energy of the whole system is unchanged.

The PE lost by the field must be equal to the gain
of the KE of the moving charged particle.

QED.


Well something doesn't add up.

Then you cannot add.


No charge from the condensor circuit moves from one plate to the other.
I think you are saying one is 'pushed back' into the battery by the field of
the moving q+. It is replenished to 6q with current from the battery when q
passes through.
In other words, there is s small blip of current going in and out of the
battery as q passes. That's where the loss of energy occurs. ..half a cycle of
AC current x the circuit impedance.

You are babbling.
I stated in the very beginning:
| If you have a static electric field with no energy supply,
| this field will obviously be diminished somewhat when a charge
| move along the field because the energy gained by the particle
| must be taken from PE of the field. The gained energy of the charge
| is qV, so the field must be diminished accordingly.

Which is exactly what I have demonstrated.
There is no power supply, and the charges on
the condenser plates can go nowhere.
It is but one charge moving from one side
of the system to the other.

One would expect that the few charges in my simplistic
scenario was easy to add up, but I will take your word for
your inability to do so.


AhA! I think I have discovered another reason for the difficulty in
accelerating charges.

The energy gained is not qV. It reduces with particle speed.

Time to burn all those books and start again, I think.

Quite.
That is good description of the situation.
If you are right, then Coulomb, Faraday, Maxwell et al
must all have been horribly wrong.

So you better burn their books, because there
is no way YOU can be wrong.

There is no need to see what happens to the local
field around the charge while the particle is
in transit when we know the net result.
And the net result is independent of the speed of the charge.
That means that the electric field is diminished by the same
amount regardless of what the speed of the particle is!
It is no additional cancelling of the electric field due
to the speed of the charge.

The "considerable energy to sustain the field" is qV
per passing charge. This "considerable energy" do
not change with the speed of the particle.

Of course the situation in an RF-cavity is much
more complicated that this, an RF-cavity is not
without power supply. There will be a current flowing
into it, and the field will not be diminished in
the same way as above.

The bottom line is still:
Every time the particle passes through the RF-cavity,
it gains the same amount of energy, regardless of
the speed of the particle.


No it doesn't Paul. Tear up all the books that says it does.
The energy gained falls of with speed.

Quite.

You can twist an turn as much as you want,
experimental evidence show that the accelerating
field is never cancelled by anything related
to the speed of the particle.

The whole "reverse field bubble" idea is ridiculous.
It simply does not add up.


Can you not see that the distortion of the charge's own natural field
constitutes that bubble?

Quite.
There is obviously a reverse field bubble on your mind.
It does indeed seem to neutralize its power.

Paul

"PD" wrote in message
ups.com...
|
| Hexenmeister wrote:
| "PD" wrote in message
| ups.com...
| |
| | Hexenmeister wrote:
| | "PD" wrote in message
| | ups.com...
| | |
| | | Henri Wilson wrote:
| | | On 4 Apr 2006 01:50:07 -0700, "PD"
| wrote:
| | |
| | |
| | | Henri Wilson wrote:
| | | On 3 Apr 2006 06:37:52 -0700, "PD"

| wrote:
| | |
| | |
| | |
| | | One puts the beam in spatially separated bunches and times
their
| | travel
| | | between two points that are reasonably distant apart.
| | | This is what is done routinely at both linear and circular
| | beamlines.
| | |
| | | Are you *completely* unaware of all the rather commonplace
| | | verifications of SR?
| | |
| | | That's hardly a 'free beam'. It is confined by a strong
magnetic
| | field.
| | |
| | | A linear beamline is hardly a strong magnetic field
| | |
| | | As usual, you're talking incomprehensible crap PD.
| | |
| | | Really?
| | | You asked how one measures the speed of a free electron beam.
| | | I told you it was routinely done in a linear beamline, and told
you
| | | how.
| | | You said this is hardly a free electron beam, due to the presence
of a
| | | strong magnetic field.
| | | I said a linear beamline does not have a strong magnetic field.
| | |
| | | What's incomprensible about that, Henri?
| | | When you get to facts you don't like, you announce that they are
| | | incomprehensible?
| | |
| | | Is this similar to your approach to the scientific method, wherein
if
| | | experiment disagrees with your theory, then the experiment is
| obviously
| | | suspect?
| |
| | Yeah... Of course. Free cosmic muons travel 62 miles in 2.2 usec
| | but cannot exceed the speed of light, which plays the part of an
| infinitely
| | great velocity in our theory, so they must be time dilated and the
| distance
| | length contracted.
| |
| | What's stupid about that, Phuckwit Duck?
| | When you get to facts you don't like, you announce that 62
| miles/2.2usec
| | is stupid?
| |
| | Except it doesn't go for 2.2 usec in a muon beam. It goes longer than
| | that, according to *timers*.
|
| *Timers*, Duck?
| I'm still waiting to see an inertial observer and a clock that moves at
| 0.9c. So far you've handwaved about muon beams.
| "One puts the beam in spatially separated bunches and times their
| travel between two points that are reasonably distant apart."
| 62 miles is "reasonably distant apart".
|
| So is 1725 m, ducky.
|
| Now you say a *timer*
| is used.
|
| Yup. Mind you, not a sundial.
|
| Ok, what duration of time does it take a bunch of muons to
| travel 62 miles by your *timer*?
|
| I don't use a timer for the ones in the atmosphere, ducky.


"Tom Roberts" wrote in message
om...
| GSS wrote:
| Tom Roberts wrote:
| I repeat: that is not really "speed".
| Let us elaborate this point.
|
| Imagine a train leaving one city at 12:00 and arriving in a city 60
| miles to its west at 12:01. Do you really think that train traveled
| 3,600 miles per hour? Of course not! This example used two _different_
| coordinate systems for "time", the two timezones of those two cities. To
| obtain the speed you _must_ use a single coordinate system; then you'll
| realize it traveled just under 60 miles per hour.
|
|
| If a time interval *dt* is measured by using UTC (or TAI) time
| standards in reference frames K1, K2, K3 etc. in relative motion within
| our solar system, will you regard this time interval as real or not
| real?
|
| "real" has nothing to do with it.
|
|
| If a distance interval *ds* is measured by using a standard meter rod
| as per SI standards in reference frames K1, K2, K3 etc. in relative
| motion within our solar system, will you regard this distance interval
| as real or not real?
|
| "Real" has nothing to do with it.
|
| To obtain a speed, you must divide the distance traveled by the travel
| time, and _all_ quantities _must_ be measured in a single coordinate
| system. In Newtonian mechanics and SR, the coordinate system must be
| inertial, using standard clocks and rulers. In GR (or other coordinate
| systems) this merely yields coordinate speed.
|
| _Nothing_ else is speed. Because that is what we mean by the word. shrug
|
|
| Tom Roberts


| I use a
| timer for the ones in the beamline.

No you don't, you lying ****. You've never been within a mile of a
muon beamline.


So you don't know cosmic muons travel 62 miles in 2.2 usec.
Funny, every other dolt relativist calls this "proof" of SR, squawking
"nothing can exceed the speed of light."



| Or did you get lost in the
| discussion? Or are you still maintaining that "tame" muons in beamlines
| live longer than "feral" muons in cosmic rays?

I'm maintaining that feral muons travel 62 miles in 2.2 usec.
What are you worried about, anyway?
That's still a lot less than infinity.

"an observer approaching a source of light with the velocity c,
this source of light must appear of infinite intensity."

"Thence we conclude that a balance-clock at the equator must go
more slowly, by a very small amount, than a precisely similar clock
situated at one of the poles under otherwise identical conditions."

"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity."

"Longitudinal mass = m / sqrt(1-v²/c²)^3
Transverse mass = m / (1-v²/c²)"


These are all clues to division by zero and only the clueless would
believe them.

The actual zero may be found in
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c
to be a universal constant--the velocity of light in empty space."

Notice that the light goes from A to A in time (t'A-tA), so c = 0/0.

|
| I'll tell you one thing, Phuckwit Duck. You are a lying ****.
|
| Yeah,

Ok, so you agree.

| you tell me all sorts of nonsense.

I'm only reporting what shrugging Roberts says shrug:
"To obtain a speed, you must divide the distance traveled by the travel
time".
"Real" has nothing to do with it.
and what Einstein says:
"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity."
I don't know where he gets "our" from, though. He must mean you,
I don't share the idiot's theory.

I had high hopes -- really -- that perhaps one misguided soul would
read something sensible and say, "Oh... Really?...Oh. I see I was
confused. OK, I get it now. Now what about...?" My head knew better,
my heart does not.

[sitting in the duck blind, waiting with a shotgun for a phuckwit duck to
appear]

Of course, I have my achievements.
From: JanPB - view profile
Date: Fri, Mar 10 2006 7:10 am
Email: "JanPB"
Groups: sci.physics, sci.physics.relativity


kk's postings about physical content of Einstein's postulates made me
look again at his 1905 paper and I noticed what I think is another
experimental assumption.

Recall Einstein's definition of synchronisation of two clocks A and B:
send a light pulse from A to B, reflect the light at B back to A,
register the emmission and the arrival times: t_A, t_B, and t'_A. The
clocks A and B are synchronised if:


t_B - t_A = t'_A - t_B (*)


(Equivalently, one can send two light rays, from A to B and from B to A
and then the requirement is that:


t_B - t_A = t'_A - t'_B, (**)


where t'_B is the second ray's emission time at B.)


He then says: "We assume that this definition of synchronism is free
from contradictions" and notes that the following two conditions are
satisfied (thus guaranteeing the "freedom from contradictions"):


"1. If the clock at B synchronizes with the clock at A, the clock at
A synchronizes with the clock at B.
2. If the clock at A synchronizes with the clock at B and also with
the clock at C, the clocks at B and C also synchronize with each
other."


This portion of the 1905 paper is frequently glossed over (I've never
seen it discussed on this NG) but the second condition is in fact
non-trivial and it demands that certain experiment yield certain
result. Otherwise Einstein's synchronisation of clocks is vacuous.


Try to prove condition 2 yourself: sync clock at A with clock at B
(condition (*) or (**) satisfied), and also clock at A with clock at C
(ditto), and based on this try to prove that B is in sync with C
(again, this means that (*) or (**) must hold for B and C). While doing
this remember that A, B, C are in 3D space, i.e., not necessarily
colinear.


You'll see you cannot prove it unless you know that the following is
also true: the time it takes for a light ray to complete a triangular
roundtrip A-B-C-A equals the time it takes for a light ray to complete
the reverse triangular trip: A-C-B-A (imagine mirrors positioned at B
and C to make the light ray go around). This must be verified
experimentally, otherwise there is a chance the definition of Einstein
sync is vacuous. -- Bilewacky


Androcles.




| PD
|
| Androcles.
|
|
|
|
|
| | This, of course, is counter to Androcles's mythology, so it must have
| | only gone 2.2 usec anyway, or so he would have us believe.
| |
| | Is this similar to your approach to the scientific method, wherein
if
| | experiment disagrees with your theory, then the speed of cosmic
muons
| | is obviously suspect, you ****ing moron?
| |
| | Androcles.
| |
| |
| |
| |
| | | PD
| | |
| | |
| | |
| | | It doesn't tell us what we want to know anyway. We want to
see if
| | there is an
| | | 'apparent mass increase'.
| | |
| | | Why do you want to know that? You wanted to know how to measure
the
| | | *speed* of an electron beam. Why do you need to know an
"apparent
| mass
| | | increase" to do that?
| | |
| | | Attempt to change the subject noted.
| | |
| | | As usual, you're talking incomprehensible crap PD.
| | |
| | | Do you have anything intelligent to contribute?
| | |
| | | PD
| | |
| | |
| | |
| | | PD
| | |
| | | I suppose one could put it through a magnetic field but
that
| would
| | still give
| | | an ambiguous answer. If it was going c due to the
collapse
| of
| | the bubble it
| | | wouldn't bend as much....but the same would apply if one
| accepted
| | SR's mass
| | | increase.
| | |
| | | So I would say the odds on a major scientific discovery
are at
| | worst evens at
| | | best about 1000 to one on.
| | |
| | |
| | |
| | | Paul
| | |
| | |
| | | HW.
| | | www.users.bigpond.com/hewn/index.htm
| | |
| | |
| | | HW.
| | | www.users.bigpond.com/hewn/index.htm
| | |
| | |
| | | HW.
| | | www.users.bigpond.com/hewn/index.htm
| | |
| |
|

Hexenmeister wrote:
"PD" wrote in message
ups.com...
|
| Hexenmeister wrote:
| "PD" wrote in message


| I use a
| timer for the ones in the beamline.

No you don't, you lying ****. You've never been within a mile of a
muon beamline.

That would be an error on your part.



So you don't know cosmic muons travel 62 miles in 2.2 usec.
Funny, every other dolt relativist calls this "proof" of SR,

I don't think so. I don't know of a physicist that claims a cosmic muon
travels 62 miles in 2.2 usec.

squawking
"nothing can exceed the speed of light."



| Or did you get lost in the
| discussion? Or are you still maintaining that "tame" muons in beamlines
| live longer than "feral" muons in cosmic rays?

I'm maintaining that feral muons travel 62 miles in 2.2 usec.

That's inconsistent with measurement in scintillator telescopes.
Haven't we been through this lovely discussion before? Do you want to
dredge up your idiocies from that conversation?

What are you worried about, anyway?
That's still a lot less than infinity.

"an observer approaching a source of light with the velocity c,
this source of light must appear of infinite intensity."

"Thence we conclude that a balance-clock at the equator must go
more slowly, by a very small amount, than a precisely similar clock
situated at one of the poles under otherwise identical conditions."

"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity."

"Longitudinal mass = m / sqrt(1-v²/c²)^3
Transverse mass = m / (1-v²/c²)"


These are all clues to division by zero and only the clueless would
believe them.

The actual zero may be found in
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c
to be a universal constant--the velocity of light in empty space."

Notice that the light goes from A to A in time (t'A-tA), so c = 0/0.

|
| I'll tell you one thing, Phuckwit Duck. You are a lying ****.
|
| Yeah,

Ok, so you agree.

| you tell me all sorts of nonsense.

[rants about other posts in other threads removed]

| PD
|
| Androcles.
|
|
|
|

"PD" wrote in message
oups.com...

Hexenmeister wrote:
"PD" wrote in message
ups.com...
|
| Hexenmeister wrote:
| "PD" wrote in message


| I use a
| timer for the ones in the beamline.

No you don't, you lying ****. You've never been within a mile of a
muon beamline.

That would be an error on your part.



So you don't know cosmic muons travel 62 miles in 2.2 usec.
Funny, every other dolt relativist calls this "proof" of SR,

I don't think

We know that.


so. I don't know of a physicist that claims a cosmic muon
travels 62 miles in 2.2 usec.

squawking
"nothing can exceed the speed of light."



| Or did you get lost in the
| discussion? Or are you still maintaining that "tame" muons in beamlines
| live longer than "feral" muons in cosmic rays?

I'm maintaining that feral muons travel 62 miles in 2.2 usec.

That's inconsistent with measurement in scintillator telescopes.
Haven't we been through this lovely discussion before? Do you want to
dredge up your idiocies from that conversation?

I want you to **** off. You are a dumb ****.


What are you worried about, anyway?
That's still a lot less than infinity.

"an observer approaching a source of light with the velocity c,
this source of light must appear of infinite intensity."

"Thence we conclude that a balance-clock at the equator must go
more slowly, by a very small amount, than a precisely similar clock
situated at one of the poles under otherwise identical conditions."

"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity."

"Longitudinal mass = m / sqrt(1-v2/c2)^3
Transverse mass = m / (1-v2/c2)"


These are all clues to division by zero and only the clueless would
believe them.

The actual zero may be found in
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c
to be a universal constant--the velocity of light in empty space."

Notice that the light goes from A to A in time (t'A-tA), so c = 0/0.

|
| I'll tell you one thing, Phuckwit Duck. You are a lying ****.
|
| Yeah,

Ok, so you agree.

| you tell me all sorts of nonsense.

[rants about other posts in other threads removed]

No counter argument to light's speed being infinite.
You lose, Phuckwit Duck. Game, set and match. **** off.
Androcles.



| PD
|
| Androcles.
|
|
|
|

Hexenmeister wrote:
"PD" wrote in message
oups.com...

Hexenmeister wrote:
"PD" wrote in message
ups.com...
|
| Hexenmeister wrote:
| "PD" wrote in message


| I use a
| timer for the ones in the beamline.

No you don't, you lying ****. You've never been within a mile of a
muon beamline.

That would be an error on your part.



So you don't know cosmic muons travel 62 miles in 2.2 usec.
Funny, every other dolt relativist calls this "proof" of SR,

I don't think

We know that.


so. I don't know of a physicist that claims a cosmic muon
travels 62 miles in 2.2 usec.

squawking
"nothing can exceed the speed of light."



| Or did you get lost in the
| discussion? Or are you still maintaining that "tame" muons in beamlines
| live longer than "feral" muons in cosmic rays?

I'm maintaining that feral muons travel 62 miles in 2.2 usec.

That's inconsistent with measurement in scintillator telescopes.
Haven't we been through this lovely discussion before? Do you want to
dredge up your idiocies from that conversation?

I want you to **** off. You are a dumb ****.


What are you worried about, anyway?
That's still a lot less than infinity.

"an observer approaching a source of light with the velocity c,
this source of light must appear of infinite intensity."

"Thence we conclude that a balance-clock at the equator must go
more slowly, by a very small amount, than a precisely similar clock
situated at one of the poles under otherwise identical conditions."

"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity."

"Longitudinal mass = m / sqrt(1-v2/c2)^3
Transverse mass = m / (1-v2/c2)"


These are all clues to division by zero and only the clueless would
believe them.

The actual zero may be found in
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c
to be a universal constant--the velocity of light in empty space."

Notice that the light goes from A to A in time (t'A-tA), so c = 0/0.

|
| I'll tell you one thing, Phuckwit Duck. You are a lying ****.
|
| Yeah,

Ok, so you agree.

| you tell me all sorts of nonsense.

[rants about other posts in other threads removed]

At some point Androcles, you always end up saying nothing of value,
instead only saying "**** off." About the same time, your newsreader
seems to screw up the attribution marks. I wonder what the correlation
is.


No counter argument to light's speed being infinite.

That's an *argument*??

You lose, Phuckwit Duck.

Lose what?

Game, set and match. **** off.

Killfile me, and killfile any message that responds to me.

Androcles.



| PD
|
| Androcles.
|
|
|
|

On Tue, 04 Apr 2006 13:31:41 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Fri, 31 Mar 2006 23:29:21 +0200, "Paul B. Andersen"
wrote:


Yes, I get it.
You are saying that a charged particle will keep accelerating
after it has left the RF-cavity. Since the energy of a charged
particle+bubble going close to c can be thousands of times
the Newtonian kinetic energy, it will accelerate to tens or hundreds
times the speed of light as the bubble dissipates.

Hilarious or a major scientific discovery? :-)


How does one measure the speed of a free electron beam?

The frequency at which the particles pass is pretty obvious
since the frequency of the RF-cavities must be exactly the same
(or a multiple ).
The circumference of the circuit is known.

What about a straight beam?

I suppose one could put it through a magnetic field but that would still give
an ambiguous answer. If it was going c due to the collapse of the bubble it
wouldn't bend as much....but the same would apply if one accepted SR's mass
increase.

So I would say the odds on a major scientific discovery are at worst evens at
best about 1000 to one on.

And your latest major scientific discovery is that
the particles in an accelerator go faster than c,
but nobody notices it? :-)

Where did I say that?
Maybe Androcles is right!!!!

Paul


HW.
www.users.bigpond.com/hewn/index.htm

Henri Wilson wrote:
On Tue, 04 Apr 2006 13:31:41 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Fri, 31 Mar 2006 23:29:21 +0200, "Paul B. Andersen"
wrote:


Yes, I get it.
You are saying that a charged particle will keep accelerating
after it has left the RF-cavity. Since the energy of a charged
particle+bubble going close to c can be thousands of times
the Newtonian kinetic energy, it will accelerate to tens or hundreds
times the speed of light as the bubble dissipates.

Hilarious or a major scientific discovery? :-)


How does one measure the speed of a free electron beam?

The frequency at which the particles pass is pretty obvious
since the frequency of the RF-cavities must be exactly the same
(or a multiple ).
The circumference of the circuit is known.

What about a straight beam?

I suppose one could put it through a magnetic field but that would still give
an ambiguous answer. If it was going c due to the collapse of the bubble it
wouldn't bend as much....but the same would apply if one accepted SR's mass
increase.

So I would say the odds on a major scientific discovery are at worst evens at
best about 1000 to one on.

And your latest major scientific discovery is that
the particles in an accelerator go faster than c,
but nobody notices it? :-)

Where did I say that?
Maybe Androcles is right!!!!

The same Androcles who cannot understand why an example of a Poisson
equation is useful?


Paul


HW.
www.users.bigpond.com/hewn/index.htm

On Wed, 05 Apr 2006 23:22:25 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 04 Apr 2006 13:12:38 +0200, "Paul B. Andersen"
wrote:


Henri Wilson wrote:
A moving charge constitutes a current that creates a 'back emf' in a closed
circuit (or between two plates). The field created by an accelerating charge
must therefore have the effect of reducing the applied field. Its effect will
be local to the charge. It must require considerable energy to maintain a field
gradient in space.

Yes, it is indeed quite simple.
If you have a static electric field with no energy supply,
this field will obviously be diminished somewhat when a charge
move along the field because the energy gained by the particle
must be taken from PE of the field. The gained energy of the charge
is qV, so the field must be diminished accordingly.


that's correct. The 'reverse field' causes it to diminish.


Look at this drawing:
V
E
|+ -- -|
|+ -- -| drift tube
-----|+ -- -|----------
-
q+
-----|+ -- -|-----------
|+ -- -|
|+ -- -|

It is basically a charged condenser with a tube through it.
(No tube between the condenser plates, though).
The tube on each side is at the same potential as the plates.
The potential difference between the plates is V.
In the tube a positive charge q is moving. There is
no field inside the tube, but there will be a surface
charge inside the tube with the opposite polarity,
following the moving charge q. (It is a local field
around the charge in the tube)

Count the charges, Henri.
On the left side there are 7 positive and one negative charge,
that is net 6 positive charges.
On the right side there are 6 negative charges.
The charge on the condenser is 6q.

The potential energy stored in the condenser is:
PE0 = 0.5*C*V^2, C = Q/V = 6q/V
PE0 = 3qV



(That field will become distorted into the 'reverse field bubble')


Now the situation after the charge has passed:
V

|+ -- -|
|+ -- -| drift tube
-----| |----------
-
q+
-----|+ -- -|-----------
|+ -- -|
|+ -- -|

The field have diminished somewhat.
The point is that the moving positive charge is
a part of the whole condenser system, and when
this charge has moved from the positive side of
the system to the negative side, the net result must
be a small discharge of the condenser.
(In my drawings the condenser charge has diminished

from 6q to 5q.)

I would dispute that.

Then you are wrong.
Count the charges.
(I feel a bit embarrassed of having to explain this in such a detail,
one would expect that everybody could count to 6 without help.)

On the left side there were originally 7 positive and one negative charge.
One positive charge has moved to the other side.
That leaves 6 positive and one negative charge, and the one negative
charge is no more tied up to the q+, and will neutralise one of
the positive charges on the plate, leaving 5 positive charges.

On the right side there were originally 6 negative charges.
One positive charge, the q+, is added. So we end up with
6 negative and one positive, or net 5 negative charges.
One of the 6 negative charges on the plates will be tied up to
the q+, leaving 5 charges on the plates.

Note that if the original potential difference was V,
the potential must now have dropped to (5/6)V.

So the potential energy in the condenser has dropped to:
PE1 = 0.5*C*((5/6)V)^2 = (25/12)qV

I would say that V momentarily drops as q passes between the plates but returns
to its original value after it has gone.

It doesn't matter what you think happens while the charge
passes between the plates, we only have to see the difference
before and after.
We have PE0 = 3qV, PE1 = (25/12)qV
d_E = PE0 - PE1 = (11/12)qV

Note that the average potential before - after is:
Vavg = (V + (5/6)V)/2 = (11/12)V

So the condenser has lost the energy d_E = q*Vavg
Since the total energy of the system must be conserved,
the moving charge must have gained the potential energy q*Vavg

And the original speed of the charge doesn't matter at all.

I cannot see any sign of the condensor circuit passing current. The charge q
came from a separate circuit, CMIIW.

Don't be silly. Of course the moving charge IS
the current transporting one positive charge
from one side of the system to the other.

You are joking surely.


But you are quite funny. :-)
Above, you said:
| A moving charge constitutes a current that creates a 'back emf'
| in a closed circuit (or between two plates).
Now , you say:
| The charge q came from a separate circuit, CMIIW.
and your point with that remark seems to be that
it shouldn't affect the field between the plates.

But it does. It diminishes the field between the plates.

Only momentarily.

Your former statement is partly right. (Even if it is
imprecisely put, and your reasoning probably is wrong.)

Where did your 6th '+' go, eh, Paul?
You started with seven and ended with six...strange that.

Quite. +7 + (-1) = +6. Strange that.


I realise you are trying to simulate a cyclotron but remember the field between
the D's is AC.

But as the field is set up before the charge enters it,
it can be considered stationary for the short time
it takes the charge to pass it.

Why the hell do I have to state this over and over?


The energy of the whole system is unchanged.

The PE lost by the field must be equal to the gain
of the KE of the moving charged particle.

QED.


Well something doesn't add up.

Then you cannot add.


No charge from the condensor circuit moves from one plate to the other.
I think you are saying one is 'pushed back' into the battery by the field of
the moving q+. It is replenished to 6q with current from the battery when q
passes through.
In other words, there is s small blip of current going in and out of the
battery as q passes. That's where the loss of energy occurs. ..half a cycle of
AC current x the circuit impedance.

You are babbling.
I stated in the very beginning:
| If you have a static electric field with no energy supply,
| this field will obviously be diminished somewhat when a charge
| move along the field because the energy gained by the particle
| must be taken from PE of the field. The gained energy of the charge
| is qV, so the field must be diminished accordingly.


This is becoming rather fun...

How can it be diminished if it is static and isolated.

Your configuration has one plate with a dearth of negative charges and the
other with a surplus. There is no battery connecting the two plates.
When an extraneous charge passes though the gap, none of the original charges
move from one plate to the other...or dissappear from the plates.


Which is exactly what I have demonstrated.
There is no power supply, and the charges on
the condenser plates can go nowhere.
It is but one charge moving from one side
of the system to the other.

But not one of the charges on the plates.

One would expect that the few charges in my simplistic
scenario was easy to add up, but I will take your word for
your inability to do so.

Will you please explain how the passage of the q+ charge somehow caused an
electron to jump from right plate to the left?

Is a new type of fairy active here?

AhA! I think I have discovered another reason for the difficulty in
accelerating charges.

The energy gained is not qV. It reduces with particle speed.

Time to burn all those books and start again, I think.

Quite.
That is good description of the situation.
If you are right, then Coulomb, Faraday, Maxwell et al
must all have been horribly wrong.

No. Not completely.


So you better burn their books, because there
is no way YOU can be wrong.

There is no need to see what happens to the local
field around the charge while the particle is
in transit when we know the net result.
And the net result is independent of the speed of the charge.
That means that the electric field is diminished by the same
amount regardless of what the speed of the particle is!
It is no additional cancelling of the electric field due
to the speed of the charge.

The "considerable energy to sustain the field" is qV
per passing charge. This "considerable energy" do
not change with the speed of the particle.

Of course the situation in an RF-cavity is much
more complicated that this, an RF-cavity is not
without power supply. There will be a current flowing
into it, and the field will not be diminished in
the same way as above.

The bottom line is still:
Every time the particle passes through the RF-cavity,
it gains the same amount of energy, regardless of
the speed of the particle.


No it doesn't Paul. Tear up all the books that says it does.
The energy gained falls of with speed.

Quite.

But the 'bubble' gains more.

You can twist an turn as much as you want,
experimental evidence show that the accelerating
field is never cancelled by anything related
to the speed of the particle.

The whole "reverse field bubble" idea is ridiculous.
It simply does not add up.


Can you not see that the distortion of the charge's own natural field
constitutes that bubble?

Quite.
There is obviously a reverse field bubble on your mind.
It does indeed seem to neutralize its power.

Please explain the magical electron that jumps plates in your above theory.


Paul


HW.
www.users.bigpond.com/hewn/index.htm

On 5 Apr 2006 06:05:48 -0700, "PD" wrote:


Henri Wilson wrote:
On 4 Apr 2006 16:12:54 -0700, "PD" wrote:




Is this similar to your approach to the scientific method, wherein if
experiment disagrees with your theory, then the experiment is obviously
suspect?

You didn't describe your method.

Are you saying that the electrons are pulsed and you measure the time for
consecutive pulses to pass a detector?

No. That would be measuring the time between pulses. What is done is to
time the flight of a single pulse between two distant locations. One
can do this if the time of flight is short compared with the time
between bunches, or if there are distinguishing characteristics between
one pulse and the next.

That's a TW measurement.
I want a direct OW measurement.
It can be doenin a circular path such as a cyclotron.... but you said it can be
done in a free linear beam.


Did you not know that it is fairly straightforward to measure the time
of flight of a single pulse through two widely separated detectors? (It
is not my method, note. It is so mundane that it is commonly given as a
basic exercise to young HEP students to set up the equipment and the
electronics to measure that time of flight. I see that you have never
been give that opportunity, which is perhaps why you are so prone to
saying ridiculous things about how nature behaves in real life.)

Coincidence methods are TW. You have to synch the detectors with EM to get the
right delay times.



PD


HW.
www.users.bigpond.com/hewn/index.htm

On 6 Apr 2006 18:47:50 -0700, "Eric Gisse" wrote:


Henri Wilson wrote:
On Tue, 04 Apr 2006 13:31:41 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Fri, 31 Mar 2006 23:29:21 +0200, "Paul B. Andersen"
wrote:


Yes, I get it.
You are saying that a charged particle will keep accelerating
after it has left the RF-cavity. Since the energy of a charged
particle+bubble going close to c can be thousands of times
the Newtonian kinetic energy, it will accelerate to tens or hundreds
times the speed of light as the bubble dissipates.

Hilarious or a major scientific discovery? :-)


How does one measure the speed of a free electron beam?

The frequency at which the particles pass is pretty obvious
since the frequency of the RF-cavities must be exactly the same
(or a multiple ).
The circumference of the circuit is known.

What about a straight beam?

I suppose one could put it through a magnetic field but that would still give
an ambiguous answer. If it was going c due to the collapse of the bubble it
wouldn't bend as much....but the same would apply if one accepted SR's mass
increase.

So I would say the odds on a major scientific discovery are at worst evens at
best about 1000 to one on.

And your latest major scientific discovery is that
the particles in an accelerator go faster than c,
but nobody notices it? :-)

Where did I say that?
Maybe Androcles is right!!!!

The same Androcles who cannot understand why an example of a Poisson
equation is useful?

Learnt a new word have you geesey?



Paul


HW.
www.users.bigpond.com/hewn/index.htm


HW.
www.users.bigpond.com/hewn/index.htm