Dark Energy hypothesis, etc.

Contents:

(1) Dark Energy
(2) Conservation of Energy
(3) Bird & Earth
(4) Inventions
(5) Work
(6) Electricity

Today's science is on the wrong track. Mark my words, sometime in the
next 30 years science will crash. And when it does, people will laugh
so hard at how we see the universe today. (Potential energy,
hahahahah.) And then, people will be *forced* to become open to new
ideas, such as what follows.

(1) Dark Energy
A hypothesis that the effects of "dark energy" are instead a
manifestation of gravitational forces.

(2) Conservation of Energy
Two reasons why the Law of Conservation of Energy is wrong.

(3) Bird & Earth
An example which demonstrates that the Law of Conservation of Energy
is wrong.

(4) Inventions
Three inventions that propel themselves "internally".

(5) Work
Examination of present day's definition of work.

(6) Electricity
An attempt at explaining electricity with the knowledge of (2)
Conservation of Energy and (5) Work.


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (1) DARK ENERGY |-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Two masses (e.g. stars) with sufficient velocities can pass by each
other without colliding and both gain speed. (As I say in this paper,
gravity can create energy.) I believe that that might be the cause
for the seeming acceleration of the expansion of the universe, not
"dark energy".

To illustrate, consider the following diagram; two masses (indicated
by asterisks "*") A and B each with respective velocities in the
direction of vA and vB.

(Diagram must be read using a "fixed-size font".)


* -- mass A

||
\/ vA

vB /\
||

mass B -- *


If the speed of mass A and mass B aren't great enough then,
eventually, the two will collide. On the other hand, if the speed is
truly great, then the two will speed by each other without much
gravitational interaction. However, if the speed is great, but not
too great, then the two masses will come together, but instead of
colliding they will "swing around each other" and exit at a greater
speed than they entered. The whole incident, entering, "swinging
around each other", and exiting, would certainly look like a dance;
thus I call such a phenomenon a "gravitational dance". Since the two
masses gained speed, it is obvious that I'm implying that the law of
conservation of energy is wrong. Yes, it's wrong.

As I said, the speed of the masses must be quite great in order for a
gravitational dance to succeed. Our universe is full of masses all
hurtling by each other at great speeds. There is sufficient room
between stars and planets so that they do not collide often; yet, that
sufficient room (and great speed of masses) is ideal for gravitational
dances. Thus, I hypothesize that there isn't anything like a "dark
energy". The fact that the expansion of the universe is accelerating
may simply be a manifestation of gravitational interactions between
the masses of the universe.


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (2) CONSERVATION OF ENERGY -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

The law of conservation of energy is wrong!

There are two reasons for this:

1) Attributing potential energy to objects is usually wrong
2) Gravity, magnetism, etc., create instaneous forces which then
create/destroy energy

1)
"Potential energy" should only be called that so long as the
potential cannot disappear without being realized. Consider a balloon
of hydrogen a meter above the ground. The hydrogen has a mass of M.
Now, if we cause all the hydrogen to undergo fusion, then we'd be left
with a balloon full of helium and a whole lot of energy. The mass of
the helium would be approximately 0.992*M. There's a drop in mass.
But gravitational potential energy is proportional to mass. So, where
did that minute, but measurable, amount of potential energy go?!? It
got turned into various forms of energy, e.g. heat, light, sound. Do
these forms of energy have a gravitational potential energy? I don't
think so; sound definitely doesn't and I don't think heat does either.
So where did that gravitational potential energy go?!? I don't know.
There's definitely less. So, either we say that potential energy was
destroyed without being realized (quite ridiculous), or we say that
the hydrogen balloon never truly had a "potential".

2)
In reality, energy is being created all around us
instantaneously. (I have never seen it be destroyed instantaneously.)
When energy is created instantaneously, its immediate affect on the
system is nothing (e.g. for forces, the vectors cancel each other
out). After the immediate effect, and after a minute amount of real
time, this instantaneous energy will be found to have either done
"positive work" on the system or "negative work"; that is, energy will
be added to the system, or removed. Should this instantaneous energy
be sustained for a longer duration of real time, then the energy might
be found to have not added or removed any energy from the system (that
is, it added the same amount of energy that was removed). (The
following bit on "Bird & Earth" is an example which relates to this
paragraph.)


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-|-|-| (3) BIRD & EARTH -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Consider an Earth that is stationary and is not affected by any
external forces. Alone on the Earth is a hummingbird sitting in its
nest in the world's last tree. The rest of the Earth is totally
lifeless and motionless. Suddenly, the hummingbird, which has a mass
of 5 grams, begins to hover 5 kilometers off the ground. The downward
gravitational force on the hummingbird is given by the equation

F = G*m_b*m_e / (r+5)²

where G is the gravitatiional constant
(6.673 * 10^(-11) Nm²/kg²)
m_b is the mass of the bird (0.005 kg)
m_e is the mass of the Earth (5.97 * 10^24 kg)
r is the radius of the Earth (6.38 * 10^6 m)

Now, this hummingbird is resilient and has enough energy to hover
above the ground for 10^19 years. It is obvious that the hummingbird
is converting chemical energy into kinetic energy. As it flaps its
wings, two things happen; one, the hummingbird is pushed upward, and
two, air is pushed downward. Since the hummingbird is a fair enough
distance from the Earth (5km to be exact), the downward force on the
air molecules never actually reach the ground because it gets
distributed amongst other air particles. And so, as this force is
distributed amongst billions of molecules, none of them ever gain a
sufficient velocity to reach the ground, and so the force isn't
conveyed to the ground.

So, we took care of all the forces, right? Wrong! We only considered
the gravitational force of the Earth on the bird. But what about the
gravitational force of the bird on the Earth? That force creates an
acceleration of

a = G*m_b / (r+5)²
= 8.196889698 * 10^(-27) meters/second²

After 10^19 years, when the hummingbird returns to its nest, the Earth
will be traveling at a velocity of

t = 10^19 years
= 3.15576 * 10^26 seconds

v = a * t
= 2.586741663 meters/second

The Earth was stationary and now it's moving at more than two
meters per second! Can you account for that? Where did the energy to
move the Earth come from? Some of you may argue that the bird's
chemical energy was converted to the Earth's kinetic energy. That's
quite ridiculous because, as we saw earlier, the chemical energy of
the bird was transferred to kinetic energy of its wings and then of
air particles; in simpler terms, the bird's energy simply pushed air,
nothing more.

I hope you can clearly see and appreciate that gravity (and other
forces like magnetism) create kinetic energy instantaneously out of
nothing. But notice that at any "instance", the instantaneous energy
"cancles out". You see, as the bird was hovering, we could say that
the bird was perpetually falling to the Earth. Likewise, the Earth
was perpetually falling toward the hummingbird. The forces on each
(bird and Earth) when taken together, cancel out. However, when that
instantaneous force is sustained for a real duration of time, it
effects its environment by adding or removing energy from the system.
In this case, energy was added to the system; that's why the Earth is
moving.

What does all of this mean? It means that the law of
conservation of energy is wrong! It means that perpetual motion and
free-energy devices do not contradict reality!


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-|-|-| (4) INVENTIONS -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Inventions:
1) The "Seesaw" Newton Motor
2) The "Simple" Newton Engine
3) The "Horseshoe" Newton Engine

These three inventions work on Newton's law that "every action has an
equal and opposite reaction." The idea is to harness the "action" and
elimenate the "reaction", or convert the "reaction" into something
useable. All inventions work without affecting the environment. That
is, they don't need a road to push off of like cars, they don't have
to push air like planes or spew out gases like space shuttles. They
propel themselves *internally*. That is, you can put a box around the
entire device and the box would move, and nothing would enter or exit
the box, and the device itself wouldn't react with the environment
inside the box.

(must be read using a "fixed-size font" to view diagrams)


-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-1) The "Seesaw" Newton Motor=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Top view:


M1a---M2a

m1
\
\ /\
\ ||
o --seesaw ||
\ forward
\
\
m2


M1b---M2b


Ideally, "M1a", "M1b", "M2a", "M2b", "m1", "m2" are all
electromagnets. (Some of the electromagnets can be changed into
permanent magnets where it is fit.) "M1a", "M1b", "M2a", and "M2b" are
fastened to the base, while "m1" and "m2" are connected to a "seesaw"
whose pivot ("o") is connected to the base.

When "M1a" and "m1" are nearly touching an electric current is sent
through "M1a", "M1b", and "m1". "M1a" should repel "m1" while "M1b"
should attract "m1". Thus, both "M1a" and "M1b" will experience a
force in the forward direction, while the seesaw swings around
bringing "m2" close to "M2a". As "M2a" and "m2" are close now, an
electric current will pass through "M2a", "M2b", and "m2". "M2a"
should repel "m2" while "M2b" should attract "m2". Again, "M2a" and
"M2b" will experience a force in the forward direction while the
seesaw swings back to its starting position to repeat the cycle.
Thus, the base will experience forward propulsion as the seesaw
continually swings about.

If, as the seesaw swings, "m1" hits "M1b" or "m2" hits "M2b", then the
collision will slow the forward motion. Thus, such a collision is
undesirable. One could avoid this collision by keeping the back
electromagnets far enough from the seesaw (as I have in the diagram),
or a brake could be installed in the pivot to stop the complete swing
of the seesaw.

In may seem that if the seesaw swings so hard that "m1" hits "M1a" or
"m2" hits "M2a" then the force of the collision will cause the base to
experience a forward force. This is wrong. Only the "forward
momentum" of the seesaw will "push" the base forward. However, when
the seesaw hits the front electromagnets, the entire seesaw will
"buckle" and the "backward momentum" of the electromagnet will be
conveyed to the base through the pivot. One could avoid this by
changing the seesaw by bending it so as to make a corner where it
attaches with the pivot. Then, connect both ends of the seesaw
together, ideally, the connection should be a curve. After doing
that, the seesaw will undoubtly look more like a slice of pizza. In
such a configuration, most of the "backward momentum" will swing
around becoming "forward momentum". Another way to avoid the
"backward momentum" of the electromagnets being conveyed to the base
is by ensuring that the swinging of the seesaw never stops. This can
be done by properly timing the electromagnets. If the seesaw is
continually in motion, then it will never have the chance to "buckle"
and so the "backward momentum" of the electromagnets will never be
conveyed to the base, via the pivot.


-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-2) The "Simple" Newton Engine-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The Simple Newton Engine is a cylinder with a piston in it. The
piston may require wheels to move inside the cylinder.


STEP 1:
\-----------\-----------\-----------\-----------\
The idea is to force the piston down the shaft, ideally by using
electromagnets. Also, one could make this similar to a Linear
Induction Motor, with the projectile as the piston.

Side-view (cross-section):

| ___cylinder
| ||
| \/
|/-------------
|| #| forward --
|\-------------
| /\
| ||__ piston ("#")
|
|--start
/-----------/-----------/-----------/-----------/


STEP 2:
\-----------\-----------\-----------\-----------\
As the piston moves down the cylinder, the cylinder itself will
accelerate and gain speed, and thus move forward.

--
| ___ The cylinder moves "forward"...
| ||
| \/
| /-------------
| | # |
| \-------------
| /\
| ||__ ...as the piston moves "back" through the cylinder
| --
|--start
/-----------/-----------/-----------/-----------/


STEP 3:
\-----------\-----------\-----------\-----------\
In fractions of a second, the piston will have arrived at the "back"
of the cylinder. The piston must be stopped before it slams into the
back of the cylinder, because if it does, then the energy of the
piston will cancel out the "forward" velocity of the cylinder. So,
the energy of the piston must be removed (by friction, e.g. brakes on
the wheels) or harnessed (a method which converts the "negative"
energy of the piston into something useable).

|
|
|
| /-------------
| | # |
| \-------------
| /\
| ||__The piston must be stopped before it hits the "back"
|
|--start
/-----------/-----------/-----------/-----------/


STEP 4:
\-----------\-----------\-----------\-----------\
When the piston has reached the end, and has been brought to a stop,
it must then be moved to the front of the cylinder, perhaps by hooking
it to a chain which is being pulled by a motor. Perhaps, the piston
can be removed from the cylinder when it is being transferred to the
front, and thus leave the cylinder free so that another piston can
"shoot" through it.

|
|
|
| /-------------
| |# |
| \-------------
|
|
|
|--start
/-----------/-----------/-----------/-----------/


Return to STEP 1:
\-----------\-----------\-----------\-----------\
The piston has been returned to the front. Overall, the engine has
moved and gained velocity. Now it is ready to restart at STEP 1.

|
|
|
| /-------------
| | #|
| \-------------
|
|
|
|--start
/-----------/-----------/-----------/-----------/


-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-3) The "Horseshoe" Newton Engine=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The Horseshoe Newton engine is like the Simple Newton engine, except
that the chamber is a semi-circular loop. The piston may require
wheels to move inside the chamber.


STEP 1:
\-----------\-----------\-----------\-----------\
Again, the idea is to force the piston through the chamber, ideally by
using electromagnets. Also, one could make this similar to a Linear
Induction Motor, with the projectile as the piston. The piston should
only experience a force when it is going opposite the forward
direction; thus, the force on the chamber would be opposite that, that
is, in the forward direction.

Top view (cross-section):

__ __
piston -- |##| | |
("##") | | | |
| | | | --chamber
| | | |
| | | |
| \ / | /\
\ --____-- / ||
\_ _/ ||
--______-- forward
start -- -----------------
/-----------/-----------/-----------/-----------/


STEP 2:
\-----------\-----------\-----------\-----------\
As the piston moves through the chamber, the chamber itself will
accelerate and gain speed, and thus move forward.

__ __ /\
| | | | -- The chamber ||
| | | | moves forward.. ||
| | | |
..as the | | | |
piston -- |##| | |
moves | \ / |
through the \ --____-- /
chamber. \_ _/
--______--

start -- -----------------
/-----------/-----------/-----------/-----------/


STEP 3:
\-----------\-----------\-----------\-----------\
In fractions of a second, the piston will have arrived at the other
side of the chamber. Unlike the Simple Newton engine, the piston does
not have to be stopped from slamming into the chamber. Infact, when
the piston slams into the end of the chamber, the chamber will be
pushed forward.

__ __
| | |##| -- piston slamming
| | | | into end of chamber
| | | |
| | | |
| | | |
| \ / |
\ --____-- /
\_ _/
--______--


start -- -----------------
/-----------/-----------/-----------/-----------/


Return to STEP 1:
\-----------\-----------\-----------\-----------\
Also, note that the piston returns to a suitable position on its own,
unlike the piston in the Simple Newton engine which needs to be
"reloaded". Overall, the engine has moved and gained velocity. Now
it is ready to restart at STEP 1 (from the other side).

__ __
| | |##| -- piston slamming
| | | | into end of chamber
| | | |
| | | |
| | | |
| \ / |
\ --____-- /
\_ _/
--______--


start -- -----------------
/-----------/-----------/-----------/-----------/


It should be noted that all the inventions above create a small amount
of force for a relatively minute amount of time. In my mind, they'd
only be effective if many are used simultaneously. For example, I
imagine that it wouldn't be too hard for either Newton engines to have
a burst of 5N for a tenth of a second. Building a unit of ten
thousand of such Newton engines would create a combined force of
5000N, assuming that the engines can "reload" in 0.9 seconds. The
real problem is keeping a good force-to-mass ratio (acceleration); the
more force, the better, but the engine itself should ideally be light.
If you can get a force-to-mass ratio (acceleration) above 10, then
you'd have a brilliant method to launch vehicles into space, or to
lift (and propel) planes.


--------------------------------------------------
Magnetic Propulsion for the Newton Engines:

Cross-section:

mmmmmmmmmmmmmmmmmmmm
mmmmm ____ mmmmm -- "m" are magnets
mmmm /WWWWWW\ mmmm
mmm /W/ \W\ mmm
mm /W/ mm \W\ mm
m W mmmm W m -- "W" is a wire coil
m |W| mmmmmm |W| m
m |W| mmmmmm |W| m
m W mmmm W m X forward
mm \W\ mm /W/ mm (into paper)
mmm \W\____/W/ mmm
mmmm \WWWWWW/ mmmm
mmmmm mmmmm
mmmmmmmmmmmmmmmmmmmm

If the magnets "m" are arranged such that the field is perpendicular
to the wire, and if a current is set up in the wire coil, then the
wire coil will either move forward or backward. This setup can be
used in either of the Newton engines; the wire coil would be the
"piston" and the magnets would be part of the "cylinder" or "chamber".
The wire coil would need wheels on the side so that it could move
about inside the cylinder or chamber.


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-|-|-| (5) WORK -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Consider the following example: two classmates, Jack and Jill,
both able to hold a one kilogram brick. Naturally, holding that brick
on Earth is approximately equivalent to maintaining a force of 10
Newtons. Let's say that Jack held his brick for 20 seconds, and Jill
held her brick for 2 seconds. Now, without using any scientific
jargon, who did the most work? Jack obviously did more work than
Jill. (Even if you were to replace Jack and Jill with two tables, and
rested the bricks on the tables, work is still being done, as I
explain below.)

We saw from the example that, in plain English, Jack did more
work than Jill. Thus, work should be (intuitively speaking)
proportional to force and a duration of time. Using Occam's Razor,
the simplest equation we can then make using work, force and time is
"W=Ft". Notice, that this means that work done on an object does not
neccesarily have to create/destroy motion by increasing/decreasing
velocity. On the contrary, even if you placed a book on a table, work
is being done; the table is "maintaining" a force, and likewise, the
book is "maintaining" a force. The work of gravity between the two is
causing "stress" at the atomic level. Work, in general, does not
require a change in velocity. Thus, I call "W=Ft" the equation of
"general" work.

Let us now consider "effective" work, a term I've coined to mean
work that increases/decreases velocity unhindered by other forces. We
will see below that effective work is really just general work which
is allowed to create/destroy motion, unhindered by other forces.

Now, force equals mass multiplied by acceleration. Intuitively
speaking, it is obvious that force should be proportional to mass and
acceleration. However, why isn't there a "coefficient"? And why not
"mass squared" or "acceleration cubed"? The equation is how it is
because of two things; one, intuitively, it makes sense not to add
extra "factors" (Occam's Razor), and two, it simply gives the "right
answers".

Now, let's examine the equation for work as it stands today, that
is "W=½mv²". Intuitively speaking, "effective" work should be
proportional to mass and velocity. However, we added "factors" to the
equation. Without using scientific or mathematical jargon, I say that
we should be able to explain, in plain English, why we added factors
to the equation. And if we can't, then by Occam's Razor, we should
remove those factors. And, if we do remove all the extra factors, and
say that the equation for effective work is "W=mv", then we have again
arrived at the equivalent general equation for work, "W=Ft" (allowing
the force to cause motion, unhindered by other forces).

Now, consider the following example: Jack and Jill, each lifting
a one-kilogram brick from the ground to one meter above the Earth.
Jack lifts it in 20 seconds while Jill lifts it in 2 seconds. True,
the outcome is the same for either participant. However, in plain
English, Jack did more work; he did the same amount of "useful" work
(where usefulness is defined as causing the brick to be displaced one
meter above the ground), but he did a whole lot of "useless" work by
taking his time lifting the brick. Notice that Jack's power is 0.5
Watts, while Jill's power is 5 Watts. Since Jack and Jill did the
same amount of "useful" work, since Jack was at all time holding his
brick on the Earth, and since he has less power, we can only conclude
that somewhere along the line he did more "useless" work. What is
"useless" work in this case? Well, Jack did "useless" work either by
just holding his brick or by moving it such that it didn't get closer
to its destination (one meter above the Earth). Notice that the
former is general work while the latter is effective work. Thus,
general work and effective work can both be considered "useless" or
"useful" depending on the situation.

Now, work defined as it is today is wrong intuitively, but
nonetheless, it is a *VERY* *USEFUL* "measuring tool", and it *WORKS*
with the non-intuitive equation "W=½mv²". It calculates "useful"
work, where usefulness is defined as causing an object (I use that
term very loosely) to be displaced in a certain direction. (Power
calculates the rate at which this "useful" work is happening.) Now,
allow "useless" work to be defined as causing an object to be
displaced away from that direction, that is, away from its course.
Thus, if a force "F" is deviating from its course by an angle "A",
then we can say that

Ff = F*cos(A)
Fl = F*sin(A)

where "Ff" is "useful" force and "Fl" is "useless" force. We can also
say, given an amount of general work "Wg" we can find the amount of
work "Wj" accomplished (in joules) by using the following formula:

Wj = Wg²/(2m)

Thus, general work and work calculated in joules are related to each
other.

Notice that the rate at which general work or effective work is
happening is simply the force being applied. Whether that force is
causing displacement, etc., does not matter one bit.

I know that what I call work (effective work) is called momentum
and so I assert that work and momentum should be equivalent and
synonymous. And I propose that the real unit for work (that is, force
multiplied by time) should be "P", for Prescott, Joule's middle name.
Thus, one prescott equals one newton second. I relegate the old,
traditional meaning for work to the term "typical useful work" or just
"typical work".

If we allow work to equal mass multiplied by velocity then we can
say that force and work are both forms of energy, but they are
apparent in different "time frames". That is, work requires a
duration of real time for an effect to be experienced, while force
requires an infinitesimal amount of time to have an effect
experienced.


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-|-|-| (6) ELECTRICITY |-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Now, I am going to apply work using prescotts on an electrical
circuit.

***************************
Let's find the average drift velocity:
-------------------------
A is the average (weighted with respect to L)
cross-section of the wire (m²)
n is "free" electrons per unit volume (electrons/m³)
e is the magnitude of charge of an electron
(1.602 * 10^19 C/electron)
v is the average drift velocity of the electrons (m/s)
I is the current in the (C/s)
dq is an infinitesmal amount of charge (C)
dt is an infinitesmal amount of time (s)
dN is an infinitesmal number of electrons (electrons)
-------------------------
(1) dq = e*dN

dN = nAv*dt
(2) dt = dN/(nAv)

(1)/(2) dq/dt = e*dN/(dN/nAv)
I = enAv
v = I/(enA)


***************************
Let's find force:
-------------------------
W_j is the Work in Joules (N*m)
f is the force (N)
s is the distance (m)
V is volts (N*m/C)
-------------------------
W_j = F*s
dW_j = F*v*dt
dW_j/dt = F*v
V*I = F*v

V*I
F = -----
v

= VenA

-------------------------
P is pressure (Pa)
-------------------------
So,
F
V = ---
enA

P
= --
en

We can say that "Voltage is the electromagnetic-pressure (created by
an EMF source) per density of charge."

Notice that the pressure supplied by an EMF has nothing to do with the
length of the circuit. A battery hooked to a 1 meter circuit of 1cm²
wire uses the same pressure to start a current as a similar battery
hooked to a 10000 meter circuit of similar wire!

***************************
-------------------------
W_i is the Initial Work (in Prescotts) (N*s)
(the work done to start the electrical circuit)
t is a duration of time (s)
-------------------------
W_i = F*t
= VenA*t


***************************
-------------------------
U is Initial Work (in Prescotts) per Coulomb (N*s/C)
Q is an amount of charge (C)
p is the resistivity of the wire (ohms)
l is the length of the wire (m)
-------------------------
U = W_i/Q
= F/I
= (VenA)/(V/R)
= enAR
= enA*(p*l/A)
= enpl

Now, "U" is a constant for any given circuit. So, given any circuit,
a constant amount of work is done to move a Coulomb along the circuit.
Makes sense that it doesn't vary..


***************************
-------------------------
µ is Initial Work (in Prescotts) per Coulomb meter (N*s/(C*m))
-------------------------
µ = dU/dl
= enp

Thus, the rate at which work is done per unit distance depends only on
the material. Makes sense..


***************************
-------------------------
t_c is the average change in time between electron collisions (s)
m_e is the mass of an electron (9.109 * 10^(-31) kg/electron)
-------------------------

Each electron gains "m_e*2v" of energy (remember, we are using
prescotts) before it makes a collision and losses it's energy. The
collision will take place in "t_c" seconds. "U" is the amount of work
to move a Coulomb "l" meters along the wire. Thus, in "l" meters,
there will be "l/(v*t_c)" number of collisions. So,

l m_e*2v
----- * ------ = U
v*t_c e

2*l*m_e
--------- = enpl
t_c*e

2m_e
t_c = ----
e²np


which is correct.


***************************
*(I'm not 100% sure of the following part)*
-------------------------
W_t is the Total Work (in Prescotts) (N*s)
(the total amount of work done by all the electrons)
l_c is the average length between electron collisions (m)
a is acceleration (m/s²)
-------------------------

F
a = -----
m_e
------------------------------------------

v = a * t_c

F
v = ----- * t_c
m_e
------------------------------------------

l_c = 2*v*t_c

2F
= ----- * (t_c)²
m_e
------------------------------------------

W_t = F * (l/l_c) * t


l
= F * -------------- * t
2F
----- * (t_c)²
m_e


m_e
= ---------- * l * t
(2m_e)²
2(----)
(e²np)


m_e
= ----------- * l * t
8(m_e)²
---------
e^4n²p²


e^4n²p²
= --------- * l * t
8m_e
------------------------------------------

Even though the pressure by a source on two different circuits which
use the same wire is the same, it's obvious that more *work* is being
done in a longer circuit. The reason why the force/pressure is the
same while the work isn't is not hard to understand. An EMF source
creates "electromagnetic pressure" on the anode and/or cathode. Once
a circuit is started, this electromagnetic pressure is felt throughout
the circuit. You can imagine the electrons as being dominoes.
Whether you have 1 meter of dominoes falling or 10000 meters of
dominoes falling, the initial force to topple the first domino may be
the same, and yet, the amount of work done (the number of fallen
dominoes) can be very different. This obviously means that energy
*isn't* conserved. That's right.


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-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

by Raheman Velji

"Raheman" wrote in message
om...
Contents:


[snip crap]

You need to take a physics class. Why should
someone else correct your blatant errors when
you can do it yourself?