Hi.  I wrote the following paper.  I understand that I claim the most
fundamental law of science to be incorrect.  Please give me some
*constructive* feedback.
----------------

NEW SCIENCE -  redefining science from the roots

CONTENTS:

    (1) Inventions
 Two inventions that propel themselves "internally".  A third
invention which can be set up to be a perpetual motion, free-energy
device.

    (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) Dark Energy
 A hypothesis that the effects of "dark energy" (the acceleration of
the expansion of the universe) is instead caused by gravitational
forces.

    (5) Work
 Redefining work intuitively, with the knowledge of that the Law of
Conservation of Energy is wrong.

    (6) Electricity
 An attempt at explaining electricity with the knowledge of "(5) Work".


*best read in a size 10, fixed-size font.


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (1) INVENTIONS -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Inventions:
 1) The "Seesaw" Newton Motor 
 2) The "Simple" Newton Engine
 3) The "Gravitational-density" Dynamo

The first two inventions work on Newton's law that "every action has
an equal and opposite reaction."  The idea is to harness the "action"
and eliminate the "reaction", or convert the "reaction" into something
useable.  The two 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 that comes inside the box.  It only reacts to the
environment in the box which it creates, which it uses to propel
itself.

The third invention is a dynamo which creates electricity.  It can be
set up to be a perpetual motion, free-energy device.


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


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

Top view:


   M1a---M2a   <--front electromagnets

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


   M1b---M2b   <--back electromagnets


Ideally, "M1a", "M1b", "M2a", "M2b", "m1", "m2" are all
electromagnets.  (Some of the electromagnets can be changed into
permanent magnets where it is deemed 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.  (It is possible
to construct this without the back electromagnets).

The way this invention works is somewhat hard to understand.  Here is
a simplified version:

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. 
Since all the electromagnets that are connected to the base experience
a force in the forward direction, the entire device will be propelled
forward as the seesaw keeps swinging about.

The above explanation of the workings of the Seesaw Newton Motor is
incomplete.  One should understand that the very first swing of the
seesaw does not create enough energy to propel the device.  On the
other hand, after that first swing, the momentum of the seesaw greatly
increases the amount of repulsive force experienced between the
electromagnets on the swing and the ones in front.  To use an analogy,
consider yourself motionless and on rollerblades.  The very first
swing is like throwing a basketball; you won't move much in the
opposite direction.  All the swings after the first one is like
pushing against a wall; you will move in the opposite direction.

If, as the seesaw swings, "m1" hits "M1b" or "m2" hits "M2b", then the
collision will slow the forward motion of the entire device.  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 (by friction).

It 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 force in the forward direction.  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 stop moving and the "backward momentum" of the
electromagnet will be conveyed to the base via the pivot.  One could
avoid this by changing the seesaw by bending it so as to make a 90
degree 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 undoubtedly look more like a
quarter-slice of pizza.  In such a configuration, most of the
"backward momentum" will swing around becoming "forward momentum".  A
better, but more complicated, way to avoid the "backward momentum" of
the electromagnets being conveyed to the base is by ensuring that the
electromagnets are activated such that the seesaw never has a chance
to collide with any of the forward electromagnets.  In such a case,
input sensors would need to be used so that the electromagnets could
be perfectly timed to avoid collisions.


-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-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 in the backward direction, down the
cylinder.  Since every action has an equal and opposite reaction, the
cylinder will then experience a force in the forward direction.  This
force is ideally created by using electromagnets.  Let us say that
there is an electromagnet on the piston which repels the magnet. 
(Also, one could make this similar to a Linear Induction Motor, with
the piston as the projectile.)

  Side-view (cross-section):
      
  |     ___cylinder
  |    ||
  |    \/         
  |/-------------
  ||           #*| <--magnet ("*")      forward -->
  |\-------------
  |            /\
  |            ||__piston ("#")
  |
  |<--start
/-----------/-----------/-----------/-----------/


 STEP 2:
\-----------\-----------\-----------\-----------\
The electromagnet on the piston is activated.  Thus, the piston, which
is repelled by the magnet, moves down the cylinder, as the magnet and
the cylinder accelerate forward.

                                  
  |       ___ The magnet and the cylinder 
  |      ||         move "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 that the cylinder has
gained.  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).  If friction
is used to stop the piston, the friction must cause the piston to lose
velocity in decrements; should the brake make the piston stop
abruptly, then the "negative" momentum of the piston will be
transferred to the cylinder.

  |
  |
  |
  |    /-------------
  |    | #          *|
  |    \-------------
  |     /\
  |     ||__The piston must be stopped before
  |          it hits the "back" of the cylinder
  |<--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
/-----------/-----------/-----------/-----------/


--------------------------------------------------
Magnetic Propulsion for the "Simple" Newton Engine:

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 could be applied
to the Simple Newton Engine.  The wire coil would be the "piston" and
the magnets would be part of the "cylinder".


--------------------------------------------------
It should be noted that the Simple Newton Engine creates a small
amount of force for a relatively minute amount of time.  In my mind,
it would only be effective if many are used simultaneously.  For
example, I imagine that it wouldn't be too hard for the Simple Newton
Engine 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 getting a good force-to-mass ratio (acceleration); if
you can get acceleration greater than 10 m/s² then you can pretty much
launch any vehicle, no matter how massive, into space.  If the vehicle
is really massive, then all you need to do is add more individual
engines to the unit, and eventually it should lift off the ground.  If
such high accelerations cannot be made, then I'm sure this invention
can compete with ion propulsion.


-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-3) The "Gravitational-density" Dynamo-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The general idea of the Gravitational-density Dynamo is not a
perpetual motion, free-energy device.  Here is the general idea of the
Gravitational-density Dynamo:

First, an object (the object could be a liquid) on the ground is
inserted into a tall tube which contains a fluid which is more dense
than the object.  Also, the object should be insoluble in the fluid. 
Due to the density difference, the submerged object is displaced
upwards.  When the object reaches the top, it is removed from the tube
and is dropped.  Due to the force of gravity, it will fall.  The
energy of the falling object will somehow be harnessed to create
electricity.  Once the object has reached the ground, it must be
reinserted back into the tube to repeat the process.  (Notice that the
tube should be as tall as possible so as to maximize the amount of
energy of the object when it falls.)


--------------------------------------------------
Here is a specific and practical version of the Gravitational-density
Dynamo.  This version is a perpetual motion, free-energy device.


             _________________________
            |                         | 
            |    _________________    |
            |   |                 |   |  
            |   |
            |   |                       |      
            |   |                       |      
            |   |                       |                   
            |   |                 ------*------  <--\
            |   |                       |           |
            |   |                       |       turbine
            |   |                       |                       
 tube B --> |   |                  
  (contains |   |                 |   |
 perfluoro- |   |                 |   |
 octane)    |   |                 |   |
            |   |                 |   | <-- tube A
            |   |                 |   |      (contains 
            |   |_________________|   |       water) 
            |            |            |
            |____________|____________|
                          
                        /|\ 
                          \_ semi-permeable
                             material               
                              (dialysis
                               tubing)       


Tube A contains water.  Tube B contains perfluorooctane.  Tube A and
Tube B are connected to each other by dialysis tubing, which is a
semi-permeable material.  Water can permeate through the dialysis
tubing, but perfluorooctane can't.  Due to osmotic pressure, the water
in tube A will pass through the dialysis tubing entering tube B. 
Since perfluorooctane is insoluble in water, and since water is less
dense than perfluorooctane, the water will rise to the top of tube B. 
Once enough water has accumulated at the top of tube B, the water will
fall, turning the turbine, and returning back into tube A.  Notice
that this dynamo didn't require any input energy, and it will continue
to work, creating electricity by turning the turbine.  It is a
perpetual motion, free-energy device.  It is a totally renewable
energy resource; it renews itself.

This invention truly demonstrates that gravity creates forces which
can then create/destroy energy, and that the Law of Conservation of
Energy is wrong.


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

The Law of Conservation of Energy is wrong!  

There are two reasons for this:

  1) Attributing potential energy to objects can be wrong (esp. when
the potential is due to a particular geometrical arrangement.)
  
  2) Gravity, magnetism, etc., create 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 deuterium a meter above the ground.  The deuterium has a mass of 1
kilogram and the balloon has a gravitational potential of 10 joules. 
Now, if we make all the deuterium undergo fusion, then we'd be left
with a balloon full of helium and neutrons, and a whole lot of energy.
 The mass of the helium and the neutrons would be approximately
0.99999826 kilograms and the gravitational potential of the balloon
(with the helium and neutrons) would be 9.9999826 joules.  There's a
drop in mass.  And since gravitational potential energy is
proportional to mass, there is also a drop in gravitational potential
energy.  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.  Some may argue
that since heat and sound relate to the speed of the particles, the
relativistic mass increases thus recovering the "lost" gravitational
potential energy.  This argument is erroneous since gravity's force
depends only on rest mass, not relativistic mass.  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 deuterium balloon never truly had a "potential", at least not in
any real sense.  I'm going with the latter explanation.

     Thus, we can say that lifting an object in gravity does not endow
the object with "potential energy".  However, it does endow the object
with a potential *for* energy.  That distinction, "for", clarifies the
entire issue.  It means that should certain events occur, that
potential for energy can be realized.  However, it also means that
there is no reality in that potential; should other events occur, the
potential can disappear completely, without being realized.  The
potential is in our minds, on paper, in our calculations, but in
reality it is nothing.  A similar argument can be made for magnetism. 
In general, when a potential energy is a consequence of a particular
geometrical arrangement, then the potential energy is most likely a
potential *for* energy.

2)
     Energy is being created all around us.  Gravity and magnetism are
prime examples.  Both create forces.  The immediate effect of the
forces on the system is nothing (the vectors of the forces cancel each
other out).  However, after the immediate effect, and after a minute
amount of real time, the forces 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 these forces be sustained
for a longer duration of real time, then the forces 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 portions on
"(3) Bird & Earth" and "(4) Dark Energy" provide examples which relate
to this paragraph.)


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (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 gravitational 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 robust and has enough energy to hover
above the ground for 10^19 years.  Also, the hummingbird will at all
times maintain a distance of 5 kilometers between it and the ground. 
As the bird flies, it is obvious that the bird is converting chemical
energy into kinetic energy.  As it flaps its wings, two things happen;
one, air is pushed downward, and two, it is pushed upward.  Since the
hummingbird is a fair 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 a
minute 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 into 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 pushed air, nothing
more.

     I hope you can now clearly see and appreciate that gravity (and
other forces like magnetism) create kinetic energy instantaneously out
of nothing.  (Instantaneous energy is synonymous with force.)  But
notice that at any "instance", the instantaneous energy "cancels 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 the forces are
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!


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (4) 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 A and B
(indicated by asterisks "*") 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 speeds
are 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 at which 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 "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.


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

	At this point I assume you have read enough of this paper to convince
yourself that the Law of Conservation of Energy is wrong.  Once one
has realized that energy is not conserved, the big question that
arises is how did something so obvious allude us, and for so long. 
The answer to that has many reasons.  One reason is that we did not
define work intuitively.  I will now attempt to rectify that.

     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.  Thus, intuitively, work should equal force multiplied by time.

     Notice, that this means that work done on an object does not
necessarily 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 force 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 for
"general" work.

     Let us now consider "effective" work, a term I've coined to mean
general work that increases/decreases velocity unhindered by other
forces.  First, let's look at it from a Newtonian point of view; let's
start with "F=ma".  Now, if we allow that acceleration to
increase/decrease velocity (unhindered by other forces), then force
multiplied by time (which is general work) becomes equivalent to
"W=mv" (which is momentum).  Thus, effective work is equal to
momentum.  Now, due to Relativity, momentum and effective work are
found to be corrected to equal "ymv" where "y=1/(1-v²/c²)^½".

I have found the following equation in "Introduction to the Relativity
Princeple" by Gabriel Barton (pg. 189):

         ya = 1/m ( F - 1/c² V(V?F) )

where y is equal to "1/(1-v²/c²)^½"
      a is acceleration
      m is mass
      F is the vector for force
      V is the vector for velocity
      c is the speed of light

let µ be the angle between force and velocity
    v be velocity
    W_g be an amount of general work
    W_e be an amount of effective work

The above equation can be written as
  
         ya = |F|/m ( 1 - v²/c² * cos(µ) )
     
Since |F| = dW_g/dt   (***NOTE Below)

         ya = dW_g/dt/m ( 1 - v²/c² * cos(µ) )

     yma*dt = dW_g ( 1 - v²/c² * cos(µ) )

And since yma*dt = ym*dv = dW_e

       dW_e = dW_g ( 1 - v²/c² * cos(µ) )

From the above equation, we can infer two things: the effectiveness of
general work decreases..

1) ..as velocity increases 
2) ..as the angle between force and velocity decreases

In other words, general work is the most effective when velocity is
low when compared with the speed of light, and when the angle between
force and velocity nears or equals 90 degrees.

     As I said before, effective work is equal to momentum.  And so I
assert that effective work and momentum should be synonymous.  And I
propose that the real unit for work (that is, general work, which 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 "productive work".

     Now, work defined as it is today (productive work) 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 let me clarify the vocabulary I have used here:

"General work" (or just "work"): "=Ft".  (calculated in prescotts)

"Effective work": "=ymv" where "y=1/(1-v²/c²)^½".  Effective work is
general work which is allowed to create/destroy motion, unhindered by
other forces.  It is synonymous with momentum.  Notice that the
effectiveness of general work decreases as velocity increases and as
the angle between force and velocity decreases.  (calculated in
prescotts)

"Productive work": "=Fs" where "s" is displacement.  (calculated in joules)

Useful/Useless work:  Any work can be considered useful or useless
depending on the situation.

You may be asking, "Why allow general work to be called "work" for
short, instead of allowing that short form to indicate "effective
work" or "productive work"?"  I kept it that way because "general
work" is a more general term compared with "effective work" and
"productive work".  It's kind of like how a comedian put it, "Why is
it 'corn on the cob' and 'corn'?  Instead, it should be 'corn' and
'corn off the cob'."

Of course, one could use the short-form "work" to indicate either
"general work", "effective work" or "productive work" by its use in
context.
---------

---------
***NOTE:

Notice that I let "F" equal "dW_g/dt" and not "W_g/t". This is because
in the following steps "a*dt" is found to equal "dv".  On the other
hand, "a*t" does *not* equal "v", because, over a period of real time,
the equation for acceleration depends on velocity and the equation for
velocity depends on acceleration.  The equation for velocity under an
acceleration over a period of real time is more complicated than the
equation for velocity under an acceleration over an infinitesmal
period of time.
---------


-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (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 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 wire  (C/s)
  dq is an infinitesimal amount of charge  (C)
  dt is an infinitesimal amount of time  (s)
  dN is an infinitesimal 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 the voltage  (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)
  -------------------------
                          F
                     V = ---  
                         enA 

                         P
                       = --
                         en
                        
So 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

Notice that in this case "W_i" does not equal "mv/(1-v²/c²)^½".  This
is because over the period of time "t", which is greater than the
average change in time between electron collisions, the acceleration
of the electron is hindered when the electron loses its energy during
a collision.


***************************
  -------------------------
  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  (ohm*m)
  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)
  -------------------------

(Remember, we are using prescotts.)  Each electron gains
"m_e*2v/(1-v²/c²)^½" of energy before it makes a collision and losses
it's energy.  Since "v", the average drift velocity, is well below
"c", the speed of light, I will omit the denominator.  The collision
will take place in "t_c" seconds.  "U" is the amount of work to move a
Coulomb "l" meters along the wire.  And, 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.

-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

by Raheman Velji
July 19, 2004

http://www.angelfire.com/rebellion2/rahemanvelji/

---

Request for Question Clarification by hedgie-ga on 21 Jul 2004 13:00 PDT

Hi Bloch-ee

You are not asking  a question.
You say:
         Please give me some *constructive* feedback.

But it is not clear what that means.

Commenters did take issue with some of your statements. 
That's a feeback,
but not  positive.

Do you want someone to say :" very interesting " - for $30?

Do you want the paper published, admired, get a Nobel prize?

Please do explain your motives and goals.

hedgie

Hi,

I'm not a physicist, just a simple country mathematician, but I
believe your statement:

This argument is erroneous since gravity's force
depends only on rest mass, not relativistic mass.

contains an incorrect premise.

Light, which has no rest mass, nevertheless bends through a measurable
angle as it passes the Sun.  Such a measurement performed by Sir
Arthur Eddington in 1919 was an early experimental verification of for
Einstein's General Theory of Relativity:

[Putting relativity to the test]
http://archive.ncsa.uiuc.edu/Cyberia/NumRel/EinsteinTest.html

Of course by its nature "relativistic mass" depends on the frame of
observation, but surely despite the flexibility this gives to
different observers, it cannot be argued that gravitational force
depends only on "rest mass" because light doesn't have rest mass.

I guess the real test of your theory is whether you and others will be
able to produce experimental verification as well.

regards, mathtalk-ga

This comment is in reply to the above comment.

Light trajectory only *seems* to bend.  Our space, when there is no
gravity, is Euclidean; that is, the fastest way to go from point A to
point B is a straight-line.  However, in the presence of gravity,
space-time curves.  Thus, in the presense of gravity, the fastest
point from A to B *seems* to have a curved trajectory.  However, if
you define a straight-line as the fastest way between two points, then
you realize that light travels in a straight-line even in the presense
of gravity; it doesn't appear straight to us because we cannot see the
curvature of space-time.

You're right; light does not have a rest mass.  Thus, gravity does not
affect light directly.  Gravity distorts space-time.  And a change in
space-time can make the trajectory of light seem to curve.  Indirect
relationship..

Gravity's force depends *only* on rest mass, not relativistic mass.

If I damage a car with a bat, there is an absolute amount of damage I
made.  We can observe that the damage may cost this much money or that
much money.  Nonetheless, the damage I made on the car is absolute. 
Perhaps it is one flat tire and a broken window.  Likewise, the force
of gravity between two objects is absolute.  From different vantage
points we may observe it to be this or that, but nonetheless, the
force between two objects is absolute.  Perhaps it is 5 Newtons.  But,
since relativistic mass depends on the frame of observation, it is
obvious that gravity's force cannot depend on relativistic mass.  In
other words, if gravity's force depended on relativistic mass, the
force of gravity between two objects would *not* be absolute.

According to Einstein's theory of General Relativity, the "force" of
gravity is not distinguishable from the corresponding "distortion" of
space-time.

So your point that the bending of light in a gravitational field is
due to space-time "curvature" is well taken, but I cannot allow you on
this account to reify some alternative "absolute" force of gravity
that depends only on rest mass.

regards, mathtalk-ga

Hey, this response hould help you understand some fundamental laws of
physics and clarify some of the examples you have given. I started off
college as a physics major, but it was too boring, so I majored in
neuroscience instead, but I still remember enough physics to respond
to your ideas.

1) The seesaw motor

This is pretty much how a motor works. I actually built the exact
device you constructed when I was a kid. It was a wheel with magnets
around it. I put four electromagnets around it, and you could hit
buttons that turned them on and off, and if you did it right it made
the wheel spin. In your case however, it doesn't spin in one
direction, but goes back and forth, but same idea-- it just depends on
your timing of the electromagets being on or off.

You believe that this construction will move forward. The
electromagnets M1a/b and M2a/b WOULD move forward IF they were
separate from the pivot connection of m1/m2. This is because of
Newton's law that everything has an equal and opposite reaction-- the
M's exert a force on the m's, pushing them back. The m's exert an
equal and opposite reaction, pushing the M's forward. However, this
would of course only work until the M1s passed the pivot point, i.e.,
for the length of the pivot arm. You suggest that this is all in an
enclosed box, i.e., a connected system. The forward movement, then,
that you are thinking of is in the form of rotation-- the rotation of
the pivot arm will exert and equal and opposite rotation of the box.
The rotational momentums of the two will be equal and opposite. But
the box will never move forward, it will just wiggle back and forth!

2)The simple engine

You say, "If friction is used to stop the piston, the friction must
cause the piston to lose velocity in decrements; should the brake make
the piston stop abruptly, then the negative momentum of the piston
will be transferred to the cyclinder." It does not matter how abruptly
or unabruptly you slow the piston down-- it will transfer all its
momentum back into the cylinder. You suggest pulling the piston back
with a motor, but pulling it back with a motor will take some sort of
outside energy (unless you use the energy of the piston engine, but
you can't get more energy out than you put in!), but that is beside
the point. The point here is that... let me explain with an example.
Essentially you are saying that you shoot a gun in space. Your
backwards momentum equals the forward momentum of the bullet
(m1v1=m2v2). If you had that bullet attached to a cable and you pulled
it back to you, your momentum will now be zero, regardless of how fast
or slow you pulled it back to you. If you instead used a space shuttle
or something to pull the bullet back to you, then yes, you are still
moving, and you could shoot the bullet again and go back faster, but
then you also have to include the momentum of the space shuttle in the
calculations (because the shuttle pulling back the bullet is going to
change the momentum of each), and then you cannot say that this is an
independent system, or a perpetual motion machine, or anything like
that. You are forgetting to include the calculations for all
components of a system. In this case, momentum and energy are
conserved, and the piston and cylinder by themselves will go back and
forth, not forward. If you include some other system like a motor that
pulls the piston back, then the momentum of that motor must be
considered (and the energy needed to run that motor!).

You state that the electronic analog of this could "compete with ion
propulsion." Not so-- ion propulsion can exert a continuous force, but
this device could only exert a force for as long as the engine is
tall. Think about it.

3)The gravitational density dynamo

Now you are getting somewhere. This device is as close as you get to
perpetual motion. However, it too will not work indefinitely. The
reason is that as water flows through the system, a few molecules of
perfluorooctane will be carried in the water as a result of brownian
motion and entropy. Assuming this doesn't happen is like assuming
there is no friction. So eventually the perfluorooctane:water gradient
will be evened out, and no more water will flow. Also, I might add,
that you say that it "didn't require any input energy," but it did
require energy at some point in the history of the earth to make a
water:octane gradient. For example, a muscle cell uses energy to make
a chemical gradient, and then uses the energy of that chemical
gradient to contract. It cannot be said that it doesn't require energy
to set up a gradient. But in your case, that is more beside the point.
I also should add that many theoretical devices have been designed
that would be perpetual motion machines were it not for friction or
other fundamental behaviors of the universe.

You also state that "gravity creates forces which can then
create/destroy energy." Well of course that's true-- anytime you apply
a force over a distance it will equal work, which is energy. This is
no way shows that the law of conservation of energy is wrong.

Now, on to your more theoretical discussion. 

CONSERVATION OF ENERGY

1) First, of all, you should realize that potential energy is the
total energy that can be extracted from a system. You give the example
of a deuterium balloon with m=1kg and p.e.=10J. The potential energy
is the gravitational potential energy, while the mass in that equation
is also a form of potential energy, as given by einstein, e=mc^2. So
you can see that the gravitational p.e. was converted to pure energy.
There is no flaw there, or in your words, the potential was
"realized," i.e., potential energy was converted to actual energy.

2)You state that "the immediate effect of the forces on the system is
nothing (the vectors of the forces cancel each other out)." I hope you
are talking about equal and opposite forces, because if not, then you
need to take calculus.

3) Bird and Earth example: You say that as the bird flaps its wings,
the force downward is distributed to the air molecules, but never
reaches the earth. Think of another fluid body (air acts as a fluid
body), e.g., a box of water sitting on a scale. If you put a floating
ball into that water, the weight is distributed through the water
molecules, and the scale will show an increased weight, exactly equal
to that of the ball. What you are saying has no basis whatsoever. So
you can see that the earth would not be moving 2.58 m/s, but rather, 0
m/s, because the F of the bird downward is such that it is equal and
opposite to the earth's force on it plus its force on the earth.

4) Dark Energy: Two stars approaching each other will attract
gravitationally and speed up, yes, but you forgot about the gravity
when the stars are moving away from each other! This will cause the
stars to slow back down to their original speed.

5) Work: First, you need to realize that physics' definition of work
is not the same as our daily use of the term. It does not mean effort
was expended. By your definition, work is identical to momentum, which
makes no sense at all. This redefines newtons law of "an object in
motion tends to stay in motion" to "an object in motion must exert
work to stay in motion," or "an object in motion tends to decelerate."
(This, of course, is disregarding friction.) Also, by your definition,
the table with the book sitting on it would do infinite work, just by
sitting there forever, resisting the force of gravity! If you did
infinite work, that releases an infinite amount of energy.
Nonsensical.

Then, you just take einstein's gamma factor and plug it into your
equations with no mathematical basis, proofs, or derivations
whatsoever. Also, you say the equation W=.5*mv^2 is non-intuitive--
well if you know calculus, it could not be more intuitive.

I was going to comment on the rest of your paper, but since I have
already disproved everything so far, maybe you should revise some
things first. Well, ok, to disprove I guess I would have to go through
the math of it all, but since you didn't do that either, I'm not going
to take the time. Now, I don't think you're dumb by any means-- at
least you are being creative and thinking about things. I'd recommend
taking a 1st year physics class for physics majors. The reason is that
when you take a GE physics class you learn just enough to be
dangerous, ha ha, that is, you learn the concepts (which you are
trying to refute), but you don't learn the math and theory behind it.
Anyway, I know I could be more complete, but it's not like I'm getting
paid for this, and I think what I've given you will help a lot.
Cheers!

Nice job, purkinje!  I think you did a rather nice job of answering this question.

Another thing to note is that it is rather simple to get around the
first law of thermodynamics, which is that energy/mass is neither
created or destroyed in a system, and most of your ideas can allow the
energy to remain constant.

However, and this is a big however, you totally violate the second law
of thermodynamics, which basically states that entropy always
increases.  Most of your ideas assume entropy remains constant, which
is unfortunately not how this world works.

Furthermore, you need to take into account things like friction,
electrical resistance, etc.

Imagine if you will, the following device:  2 motors with the shafts
welded together.  The + terminal on each motor is connected to the -
terminal on the other.  Now, you give that motor a good ripstart, much
like you would on a lawnmower.  Everybody knows that a motor acts as a
generator.  So, (theoretically), one of your motors is generating
electricity and the other is consuming electricity and generating
work.  Wow, there you have it -- a perpetual motion machine.  Imagine
bigger motors, attaching gears to it, and placing these in cars.  No
more pollution, you can leave them on 24-7, and you have the perfect
device.

Oh, except for that whole thing about friction and electrical
resistance means that every cycle, you get probably only 70%
efficiency, meaning after about 10 cycles, you'll have less than 3% of
your original evergy in your system.

Also remember that your electromagnets need to somehow be powered...if
they require power from an external source, then they are not truly
perpetual motion machines.

Sorry to burst your bubble - you have some great thoughts, and even
though some are physical impossibilities, don't stop -- keep thinking
up ideas like this and keep working on it, because I'm sure when
Edison introduced the idea for a "lamp that burns without oil (light
bulb)", people told him that it was also a physical impossibility, or
when Otto introduced the idea for the "carriage that runs without
horses (automobile)", people probably ridiculed him and said it was
the stupidedst idea they've ever heard.

Who knows?  Lots has left to be discovered in science, so don't be discouraged.

When I said, "electromagnets need to be powered", I meant those for
your seesaw Newton motor.

In response to:

Subject: Re: Physics paper requires constructive feedback
From: purkinje-ga on 21 Jul 2004 12:40 PDT

---
You don't understand the "Seesaw" Newton Motor.

You say,

"You believe that this construction will move forward. The
electromagnets M1a/b and M2a/b WOULD move forward IF they were
separate from the pivot connection of m1/m2. This is because of
Newton's law that everything has an equal and opposite reaction-- the
M's exert a force on the m's, pushing them back. The m's exert an
equal and opposite reaction, pushing the M's forward.  However, this
would of course only work until the M1s passed the pivot point, i.e.,
for the length of the pivot arm. You suggest that this is all in an
enclosed box, i.e., a connected system. The forward movement, then,
that you are thinking of is in the form of rotation-- the rotation of
the pivot arm will exert and equal and opposite rotation of the box. 
The rotational momentums of the two will be equal and opposite. But
the box will never move forward, it will just wiggle back and forth!"

The seesaw *doesn't* rotate.  It simply swings back and forth like a
seesaw.  Do you understand it now?
---

---
You say: "It does not matter how abruptly or unabruptly you slow the
piston down-- it will transfer all its momentum back into the
cylinder."

No it won't.  If I'm on a bike and I stop abrubtly by pushing down
hard on my brakes, I (my body) will go hurtling forth until I hit a
wall.  In the presence of gravity, I might hit the ground before I hit
a wall, but the point remains the same.  However, if I push on my
brakes and slowing come to a stop, I can avoid being thrown forward. 
And moreover, by coming to a stop slowing, the momentum of me and the
bike is dissipated as heat, and perhaps sound.

That is the *main* idea of the "Simple" Newton Engine.  Every action
has an equal and opposite reaction.  The idea is to harness the action
by getting rid of the reaction.  How do we get rid of the momentum of
the reaction?  One way is by using brakes as said above.  The idea is
to get rid of the momentum of the reaction by dissipating it as heat
and sound, and thus letting the action propel the device forward.

Do you understand it now?
---

---
You're right.  I can't make the gravitational-density dynamo into a
perpetual motion machine since, as you say, "the perfluorooctane:water
gradient will be evened out".
---

---
You say, 

"1) First, of all, you should realize that potential energy is the
total energy that can be extracted from a system. You give the example
of a deuterium balloon with m=1kg and p.e.=10J. The potential energy
is the gravitational potential energy, while the mass in that equation
is also a form of potential energy, as given by einstein, e=mc^2. So
you can see that the gravitational p.e. was converted to pure energy. 
There is no flaw there, or in your words, the potential was
"realized," i.e., potential energy was converted to actual energy."

You need to re-read the definition of 'gravitational potential
energy'.  Here it is:

"Gravitational potential energy is energy an object possesses because
of its position in a gravitational field."
(I got that from http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html)

For the law of conservation of energy to be correct, the amount of
potential energy, kinetic energy, and the energy from mass from
*before* and *after* must be equal.

( letting g, the gravity, equal 9.8 J/(kg*m) )

Before:
  Energy from mass of balloon full of deutrium: 1 kg which equals...
    E = mc^2 = (1 kg) * c^2 = c^2 Joules
  Gravitational potential energy of balloon full of deutrium: 1 kg, 1
meter above ground...
    E = mgh = (1 kg) * (9.8 J/(kg*m)) * (1 m) = 9.8 Joules

Between before and after, the balloon full of deutrium undergoes
fusion, leaving behind helium, neutrons, and a whole lot of energy.  I
realize that I haven't taken into account the initial amount of energy
to initiate the fusion.  Nonetheless, the equation I use, which is
stated below, is balanced:

deutrium atom + deutrium atom => helium atom + neutron + 3.27 MeV
(found the above equation at
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html)

After:
  Energy from mass of balloon full of helium and neutrons:  
  0.99999826 kg which equals...
    E = mc^2 = 0.99999826 * c^2 Joules
  Energy "liberated" from nuclear fusion (heat, light, sound, etc.)..
    E = mc^2 = 0.00000174 * c^2 Joules
  Gravitational potential energy of balloon full of helium and neutrons...
    E = mgh = (0.99999826 kg) * (9.8 J/(kg*m)) * (1 m) = 9.799982948 Joules

Adding up the before and after, we find that..

Before there is 
  (c^2 Joules + 9.8 Joules)

After there is
  (0.99999826*c^2 Joules + 0.00000174*c^2 Joules + 9.799982948 Joules) 
which equals
  (c^2 Joules + 9.799982948 Joules).

There is a loss of energy.  0.000017052 Joules disappeared.  Go figure..
---

---
You say:

"2)You state that "the immediate effect of the forces on the system is
nothing (the vectors of the forces cancel each other out)." I hope you
are talking about equal and opposite forces, because if not, then you
need to take calculus."

Yes, I am talking about equal and opposite forces.  What I'm saying at
that part is that gravity and magnetism create equal and opposite
forces.  But, at any instant, the force created by gravity and
magnetism cancel out.  But, after an amount of real time, the forces
will be found to have added or removed energy from the system.
---

---
You say:

"3) Bird and Earth example: You say that as the bird flaps its wings,
the force downward is distributed to the air molecules, but never
reaches the earth. Think of another fluid body (air acts as a fluid
body), e.g., a box of water sitting on a scale. If you put a floating
ball into that water, the weight is distributed through the water
molecules, and the scale will show an increased weight, exactly equal
to that of the ball. What you are saying has no basis whatsoever. So
you can see that the earth would not be moving 2.58 m/s, but rather, 0
m/s, because the F of the bird downward is such that it is equal and
opposite to the earth's force on it plus its force on the earth."

You may be right about this; I don't have any evidence against it. 
But, I really doubt that a bird five kilometers in the air has any
effect on the particles near the ground.  I mean, 5 kilometers! 
That's like 9 C.N. Towers!  But again, I have no evidence...
---

---
You say:

"4) Dark Energy: Two stars approaching each other will attract
gravitationally and speed up, yes, but you forgot about the gravity
when the stars are moving away from each other! This will cause the
stars to slow back down to their original speed."

Hehe..  And what if the stars don't get a chance to slow down?  What
if, two stars approached each other, increased their speed, and then,
soon after, they collided with other stars.  In that case, the
momentum gained by the stars would be converted into heat, light,
sound, etc., due to the collision.  Hehe..  Can't get rid of the heat,
light and sound, though..  Must conclude that energy was created..
---

---
You say:

"5) Work: First, you need to realize that physics' definition of work
is not the same as our daily use of the term. It does not mean effort
was expended. By your definition, work is identical to momentum, which
makes no sense at all."

It makes sense.  I'm not denying that how we've defined work is wrong.
 I figure it's incomplete.

"This redefines newtons law of "an object in motion tends to stay in
motion" to "an object in motion must exert work to stay in motion," or
"an object in motion tends to decelerate."

What I say does not mean that!  How did you get that?!

"(This, of course, is disregarding friction.) Also, by your
definition, the table with the book sitting on it would do infinite
work, just by sitting there forever, resisting the force of gravity!
If you did infinite work, that releases an infinite amount of energy.
Nonsensical."

No.  As I say, work does not necessarily mean it is "effective" work. 
You're right, a table with a book on it would do an infinite amount of
work.  But, the work isn't "effective" and so, in this case, an
infinite amount of energy ISN'T realeased.
---

---
You say:

"Then, you just take einstein's gamma factor and plug it into your
equations with no mathematical basis, proofs, or derivations
whatsoever. Also, you say the equation W=.5*mv^2 is non-intuitive--
well if you know calculus, it could not be more intuitive."

You're right about that.  I figure that if effective work is equal to
momentum in Newtonian terms, then effective work equals momentum in
Relavistic terms.  You're right; no mathematical basis, etc.
---

I feel obligated to clear up your remaining questions, mostly for your
own sake, so that if you get into a conversation with a physicist you
will not sound completely uneducated.

First, I did understand the seesaw motor. Read my response more
carefully-- the reason that the box will just "wiggle back and forth"
is because the rotating arm is wiggling back and forth. The system
will rotate back and forth with an equal and opposite momentum as that
of the rotating arm.

Second, again, it does not matter how abruptly or unabruptly you slow
the piston down-- it will transfer all its momentum back into the
cylinder. You give the example of braking a bik-- regardless of if you
slam on the brakes or apply them slowly, by the time you are stopped,
all of the momentum will have been transferred to the earth (not to
yourself, as you describe-- you are on the bike!). Since the earth is
so much more massive than you, you don?t even notice its change in
velocity.

I know what gravitational potential energy is. You are saying there is
some sort of "conservation of gravitational potential energy"-- i.e.,
all gravitational potential energy must be conserved. There is no such
law! It is only the conservation of total energy! Just because
gravitational potential energy is converted to kinetic energy when
something falls towards the earth doesn't mean there?s any violation
of law (as I'm sure you understand), and neither does mass being
converted into energy (as happens in your example) violate any law of
the conservation of energy.

Yes, of course energy is going to be released (but not created!) if
two bodies (e.g., stars) approach each other and speed up. It is just
potential energy being converted to kinetic energy?- just like an
apple falling towards earth, its potential energy is converted to
kinetic energy, seen as velocity (and yes, the earth also accelerates
towards the apple, ever so slightly). If the star collides with
something after speeding up, that's just like the apple colliding with
the earth?s crust, or something like that. All well known principles.

You can't just answer my points with, "well, that?s not EFFECTIVE
work" or something like that. What is the mathematical difference
between effective work and 'non-effective work' in your theories? If
there's no mathematical difference, then there's no difference at all.
Ok, so if an object is just moving along through space with no forces
acting on it, it has a momentum. You say that that is the same as
work, i.e., the object is doing work. You are saying the object is
doing work to stay in motion, then it must decelerate, because it has
to give off energy to do work to stay in motion. Or maybe you are
saying that the object is giving off work as it moves through space,
so it should be accelerating. Either way, Newton showed how an object
in motion will stay in motion without exerting any effects on its
environment and with no external forces acting on it. I think what you
are trying to get at here is that the moving object has energy which
can be harnessed to do work, i.e., the kinetic energy can be made to
exert a certain force over a certain distance, thus doing work. Or a
book sitting on a table can be harnessed in a gravitational field to
do work. But the table just sitting there resisting a force is not
doing work. Work is a force applied for a distance through space. If
nothing is moved through space, then there was no work, although there
may have been energy needed to create the forces themselves. I admit
that the connection between our daily use of the term ?work? and the
physics use of the term is somewhat confusing. I?m sure there are many
good examples of why it is defined that way, but I?ll leave that up to
you to figure out. Just take a good, rigorous physics class, and it
will help a lot.

The problem with your views is that you are only considering two types
of energy:  kinetic and gravitational potential.  However there are
many many other forms of energy out there such as stored potential
(like a spring), chemical (like the energy release when you combust
gasoline), atomic (from atomic vibrations) and so on and so forth.

If you sum up ALL the energy in a closed system, it will always always
always remain constant.  You can convert one form to another as much
as you want, but you can never create energy; you can never destroy
energy.

Also, note that WORK by definition, can be positive or negative, and
is a VECTOR relationship, not scalar.  For instance, if you lift a one
newton block one meter against gravity, you have performed one joule
of work.  By bringing it back down, you effectively perform negative
one joule of work, for a total of zero work.

That's the problem with your seesaw - it just piddles back and forth;
does NO work...it's not that it is ineffective work, it's that it sums
up to NO work.

Hi blochee,

There are lots of areas where you could but your enthusiasm and
brainpower to good use.  Lots of areas of science are still
developing, and lots of work needs to be done.  This is not the case
with classical mechanics, or electromagnetism.  If you're interested
in doing scientific research, you could consider getting into one of
the younger fields, say environmental science.  If physics is what
you're into, you could do that too, but first you have to learn the
basics--one of which is that conservation of energy is NOT wrong, at
least not on a macroscopic scale.  If you would like to make a
contribution in the physics field, you are shooting yourself in the
foot by convincing yourself that centuries worth of carefully tested
theory is incorrect.  Do yourself a favor and go through some physics
books, read over purkinje's comments, and keep at it till you
understand.

Further to the other comments on this paper, there is a more
fundamental flaw in invention three: The Gravitational-density dynamo.

If we were to view the system as a simple tube of water and a tennis
ball. To start the process off we insert the tennis ball into the
bottom of the tube. This of course would have to be done in such a way
as to let none of the water out but let's assume for the moment that
we can do that. Once the ball is in the tube it will float to the top.

However...

When we put the ball into the tube we have to displace the same volume
of water. It can't come out of our trap door (we made it one-way only)
so the only way for it to go is upwards. In short, the entire column
of water is lifted up by the same volume as the tennis ball (other
wise there would never be room for it). It's this increase in the
water's potential energy which provides the force that lifts the ball
upwards - and it's precisely the same amount of energy as you gain
when the ball falls again.

In your example, this is all tucked away in that osmotic membrane.
Such membrane's use energy and in this case the amount of energy
necessary to push the water into the column of perfluorooctane (and
consequently displace the molecules of perfluorooctane upwards) is the
same as you would gain from the turbine in the first place.

Take into account the rest of the losses (friction etc.) and you very
slightly lose energy over the whole cycle.


Regarding the apparent loss of energy in the deuterium example, your
claim that the force of gravity depends only on non-rel mass cannot be
substantiated. The idea that the displacement of light due to curved
space-time is somehow fundamentally different from the application of
a gravitational force is absurd.

There are two (equally valid) ways of viewing it...
Either gravity is a force that affects all objects (light included) OR
gravity is a space-time distortion and all objects move across that
distortion via a geodesic (shortest) path. You can't say it's one view
for some objects and one view for the others.

Either way, the total potential for the system doesn't change. It's
just that the photons are now carrying that potential as they have a
mass (relativistic) and are in a gravitational field.

Some interesting ideas though. It takes a while to figure out where
the problems lie.

Hope that helps.

LJ.