Re: I'm Not Sure if this Worries Me or Not.

Henri Wilson wrote:
Paul B. Andersen wrote:
Henri Wilson wrote:
Paul B. Andersen wrote:
Henri Wilson wrote:
It is true.
A charge cannot be made to move at >c between the plates of a condensor (which
is effectively what an accelerator is).
It requires infinfite energy to raise its speed to c wrt those plates.

[snip obfuscating quibble]

We KNOW that the field transfers the same amount of energy
to the particle every time the particle passes through it,
so the field cannot be cancelled by anything.

It requires a lot of energy to cancel the applied field, even locally.

You are evading the point.
Whatever your "field bubble" does, it cannot cancel the accelerating field.


Why not. It is a 'reverse' field. It MUST partly neutralize it.

Refuted below.

The associated energy is contained in 'Wilson's reverse
field bubble' that is carried along with the charge.

Exactly.
You claim that the kinetic energy of a charged particle moving
in space free of field is contained in a "reverse field bubble".
We know the kinetic energy of the particle + your bubble
is m*gamma*v. So any moving charged particle must carry
a "field bubble" along with it, and this field bubble
has the remarkable property that its kinetic energy is
independent of the particle's charge, it depends only on
its mass and speed.

I know that electrons behave as though their masses increase according to
something like gamma....but has this been proven correct for, say, protons? I
doubt it.

All charged particles behave in such a way that
p = m*gamma*v and E = m*gamma*c^2.
Both protons and ions are routinely accelerated in accelerators.


As I have pointed out elsewharer

And of course the reason why you cannot make the particle + bubble
move faster than c is that its kinetic energy approaches
infinity when its speed approaches c.

If you want to include the bubble energy in with the mechanical KE, that is
true...but I would consider the bubble energy potential rather than kinetic.

Here is why the "bubble energy" must be kinetic energy:
It is YOU that include the "bubble energy" in the KE of of the particle.
The particle is KNOWN to have the KE = m*(gamma-1)*c^2.
YOU claim that the part of this energy that exceeds mv^2/2 is
contained in the "field bubble".
KE is by definition the energy that depend on the speed.
So the "bubble energy" MUST be KE.

So why are you saying that the reason is that the accelerating
field is cancelled by your field bubble?
If it were, how could it then transfer energy to the particle
+ your bubble?

You must be a moron not to see the giant stupidity in this.

The bubble's existence is related to the travel time of electric fields.

There is no "travel time of electric fields" in an accelerator.
The electric field is established _before_ the charged particle
enters it, and it can be considered a static field during the very
short time the particle is in it.
You KNOW this because you have been told many times,
so why do you keep repeating this stupidity?


The faster the electron, the bigger the bubble and the smaller the effect of
the applied field. One could speculate that the bubble field was always less
than the applied field gradient.

You are repeating your giant self contradiction.
Read again what I have told you over and over,
and think just a little bit, for once.
1. We KNOW that the particle gains the same amount of
KE every time it passes through the accelerating field (RF-cavity),
regardless of what the speed is. The gained energy is simply q*V,
where q is the charge and V is the potential drop in the accelerating
field (integral E ds). We know this because when the accelerator is
in steady state (at the top of its performance), the KE gained in
the RF-cavities is lost as synchrotron radiation in the bends.
This energy is easily measured, and MUST be carried from the RF-cavity
to the bends as KE in the particles. That's why high performance
accelerators have to be big, the bends must be gentle to loose less
energy per revolution.
Electrons are accelerated to 100 GeV, that is gamma = 200000,
v = 0.999999999987c, or v = (c - 3.8 mm/sec).
Even at this speed, the electric field transfers as much energy
to the particle as at any other speed.
2. This proves that nothing cancels the electric field at any speed,
the effect of the electric field is the same at any speed.
3. Since we know that the KE = m*gamma*c^2 (gamma-1 ~= gamma)
even when the particle is NOT in the electric field, and you
claim that the vast majority of this KE is contained in
the "field bubble", it follows that this "field bubble"
has nothing with the accelerating field to do, but is only
determined by the mass and the speed of the particle.


The bubble has to move with the charge. This is where the field reaction time
enters the picture.

Oh? How? What is it?

4. The fact that the KE of your "field bubble" is independent
of the charge of the particle shows that it is not of
electromagnetic nature.


I said the bubble has PE not KE.

And I showed above why your bubble energy is KE.


Henri, you haven't even tried to refute any of my arguments.
Why is that?
Didn't you read them?
Didn't you understand them?
Or do you agree with them?


(How could it be? If an electron and a proton are moving
with the same speed, we have - according to you - two
unity charges moving with the same speed, but the "field bubble"
of one of them contains much more energy than the other.)


I have several views on this. One goes along the lines of Len Gassenbeek's spinning particle idea.
(the 'bubble' involves the spinning of the particle)

So the KE is now stored as spin? No bubble? :-)
When you realize that one "explanation" is wrong, invent a new,
entirely different, and even more idiotic explanation, eh? :-)

Another says the 'gamma' increase has only been accurately verified with
electrons.

Which illustrates your desperate seach for explanations. :-)

So what is the nature of you bubble? What kind of field is it?
Could I suggest WUBEBW field?
(Wilsonian Unknown By Everybody But Wilson - field.)
Another world shattering discovery by Henri Wilson.


It stands to reason that it must be true. It is simple stuff really.

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.

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)


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.)

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.

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.

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.


Paul
.

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