Complete Collection of Ideas

/----------------------------------------------------------------------------*
| --------------| A COLLECTION OF IDEAS | by Raheman Velji |
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* * * [must use a fixed-width font to view diagrams properly] * * *

CONTENTS:

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(1) Inventions
A) The Seesaw Newton Motor
B) The Simple Newton Engine

Two inventions which will have a lasting effect on transportation,
especially in space exploration.

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(2) Law of Conservation of Energy
A) Gravitational-density Dynamo
B) Potential Energy

Two examples which clearly demonstrate that the Law of Conservation of
Energy is wrong.

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(3) Work and Energy
A) Creating and Destroying Energy
B) Defining Work and Energy

Energy can be created and destroyed, so the Law of Conservation of
Energy is wrong. Also, work and energy will be defined in different -
more intuitive - ways.

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(4) Special Relativity
A) Einstein's 2 Postulates
B) Understanding the Michelson-Morley Experiment
C) Time Dialation, Length Contraction and the Rest

This section will demonstrate that special relativity's postulates are
contradictory. Also, it will be explained why the Michelson-Morley
experiment returns a "null" result. And, it will be shown that
Einstein's thought-experiments for Time Dialation and Length
Contraction are wrong.

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

Inventions:
1) The Seesaw Newton Motor
2) The Simple Newton Engine

Devices that use "self-sufficient propulsion" 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 useable energy. Thus, within the device, the
"reaction" is lost allowing the "action" to propel the device. All
devices that use "self-suffiecient propulsion" 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. Thus, they get the name "self-sufficient propulsion"
because they are self-sufficient. In other words, 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. (I propose that any device that uses self-suffiecient
propulsion should have the name "Newton" added to its full-name so that
we remember how it relates to Newton's law. I will use that convention
here; whether this convention should be followed is debatable.)

The idea of "self-suffiecient propulsion" will have a lasting effect on
transportation (especially in space exploration).

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-A) 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 explain. 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, the electromagnets
connected to the base, "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.
Notice that the seesaw does *not* rotate, it simply moves back and
forth, like a seesaw.

It should be noted that as the seesaw swings about, a bit of the
"backward" energy of the electromagnets on the seesaw will be conveyed
to the base via the pivot, thus slowing down the entire device. That
loss of speed, though, is negligible.

The above explanation of the workings of the Seesaw Newton motor is
incomplete. One must understand the following:

Every action has an equal and opposite reaction. The main idea of the
Seesaw Newton motor is to harness the "action" by converting the
"reaction" into useable energy. When the front electromagnet, back
electromagnet and the electromagnet on the seesaw are activated, the
front and back electromagnets experience a "positive" force by being
forced forward. The electromagnet on the seesaw, however, experiences
a "negative" force as it moves in the backward direction. One must get
rid of the "negative" energy of the electromagnet on the seesaw. If
the "negative" energy is not rid of, then it will somehow be
transferred to the entire device, thus not allowing the device to gain
velocity and move forward. The Seesaw Newton motor does not only get
rid of the "negative" energy, it in fact uses it to propel the device
further. Consider the following scenario: a Seesaw Newton motor at
rest, and set-up similar to the diagram above. Now, let us initiate a
current through "M1a", "M1b", and "m1". The electromagnets on the base
("M1a" and "M1b") will experience a "positive" force by being forced
forward. The electromagnet on the seesaw ("m1"), however, will
experience a "negative" force by being forced backward. However, at
the other end of the seesaw, the electromagnet ("m2") seems to be
approaching the front electromagnet ("M2a") and receding from the back
electromagnet ("M2b"). Thus, at the other end of the seesaw, when
those electromagnets are activated, the repulsive force between the
electromagnet on the seesaw and the front electromagnet will be
greater, thus propelling the device further forward. Also, at the
other end of the seesaw, when those electromagnets are activated, the
attractive force between the electromagnet on the seesaw and the back
electromagnet will be greater, again propelling the device further
forward. The fact that both magnets ("M2a" and "M2b") experience a
greater forward force is due to the initial "negative" energy of the
electromagnet on the seesaw ("m1"). Thus, both the "action" and the
"reaction" are harnessed to propel the entire device forward. Thus, in
a sense this invention is more effective than a space shuttle because
it harnesses both the "action" and "reaction", unlike a shuttle which
only uses the "action".

If both "action" and "reaction" are to be harnessed, one must ensure
that the electromagnets on the seesaw should not hit either the front
electromagnets or the back electromagnets. That is because any
collision will slow the forward motion of the entire device. One must
avoid collisions by ensuring that the electromagnets are activated such
that the seesaw never has a chance to collide. Thus, input sensors
would need to be used to calculate the speed of the seesaw so that the
electromagnets can be perfectly timed to avoid collisions. By avoiding
collisions, both "action" and "reaction" are harnessed.

Notice that for this invention to actually move the electromagnets must
be very strong and the entire device must be light. Otherwise, the
device will stay in the same spot and just wiggle about instead of
moving. In any case, this invention can definetely compete with
devices that use ion propulsion.

Also, the entire Seesaw Newton motor can (with a battery) be put into a
box and the box would move without interacting with the environment
outside the box. Thus, we say it moves using "self-sufficient
propulsion".

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-B) The Simple Newton Engine-=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The Simple Newton engine works using "self-sufficient propulsion".

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

\-----------\-----------\-----------\-----------\
Every action has an equal and opposite reaction. The main idea of the
Simple Newton engine 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 friction, which is discussed in "Step 3".

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
("X") that is connected to the front of the cylinder. (Also, one could
make this similar to a Linear Induction Motor, with the piston as the
projectile.)

Side-view (cross-section):

| ___cylinder
| ||
| \/
|/-------------
|| #X| --magnet ("X") forward --
|\-------------
| /\
| ||__piston ("#")
|
|--start
/-----------/-----------/-----------/-----------/

STEP 1:
\-----------\-----------\-----------\-----------\
Now, activate the electromagnet on the piston. So 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...
| \/ --
| /-------------
| | # X|
| \-------------
| /\ --
| ||__ ...as the piston moves backward
| through the cylinder
|--start
/-----------/-----------/-----------/-----------/

STEP 2:
\-----------\-----------\-----------\-----------\
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. Consider the following analogy: 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. Thus, in the Simple Newton engine the "reaction" is
lost due to friction (as heat and possibly sound) while the "action" is
harnessed to propel the cylinder forward.

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

STEP 3:
\-----------\-----------\-----------\-----------\
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. Or perhaps the piston can
slowly move back on its wheels towards the front of the cylinder. Or
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 the cylinder.

|
|
|
| /-------------
| |# X|
| \-------------
|
|
|
|--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.

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

Also, like the Seesaw Newton motor, the entire Simple Newton engine can
(with a battery) be put into a box and the box would move without
interacting with the environment outside the box. Thus, we say it uses
"self-sufficient propulsion".

It should be noted that the Simple Newton engine creates a small amount
of force for a relatively minute amount of time. Nonetheless, I'm sure
this invention can compete with ion propulsion.

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

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-A) Gravitational-Density Dynamo-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

---------------------------------------
The following is what I call a "Gravitational-density dynamo":

_____
| \_____
semi- __\ |_____ \_____
permeable / | | \_____ \_______
material | | \_____ |
(dialysis | | \___ |
tubing) | | | |
| |
| | |
| | |
| | |
| | ------*------ --\
| | | |
| | | turbine
| | |
tube B -- | |
(contains | | | |
perfluoro- | | | |
octane) | | | |
| | | | -- tube A
| | | | (contains
| |_________________| | water)
| | |
|____________|____________|

/|\
\_ semi-permeable
material
(dialysis
tubing)


Tube A contains 250ml of water. Tube B contains 750ml of
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 water is insoluble in perfluorooctane,
and since water is less dense than perfluorooctane, the water will rise
to the top of tube B. The water that has risen will permeate through
the dialysis tubing at the top of tube B. Once enough water has
accumulated at the top of tube B, it 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 (and
generator, which is not shown), so long as the perfluorooctane does not
seep into tube A through the semi-permeable material. Eventually, the
perfluorooctane will seep through the dialysis tubing, and so this
invention is not a perpetual motion machine.

But how can this dynamo generate electricity without any input
energy? First, let's observe that the water at the top of tube B has a
gravitational potential energy. When it falls, the gravitational
potential energy is realized and is converted into electricity by the
turbine (and generator, which is not shown). But how did the water
initially get its gravitational potential energy? It got its
gravitational potential energy by being displaced upward in a fluid
(perfluorooctane) that is more dense than it. Thus, we must conclude
that insoluble objects immersed in fluids that are more dense gain
gravitational potential energy by being displaced upwards. However,
where is that energy coming from? By the Law of Conservation of Energy
something must lose energy so that another can gain energy. Since we
cannot find anything losing energy, we must conclude that the Law of
Conservation of Energy is wrong, and that gravity creates forces which
then create/destroy energy; in this case it created energy in the final
form of electricity.

As mentioned before, enough perfluorooctane will eventually seep
through the dialysis tubing causing the level of the liquid in tube B
to lower such that the water cannot escape through the top of the tube.
And so, the turbine will stop spinning. At such a point we can easily
"unmix" both liquids by pouring all the liquid into a tall cylinder.
If we leave the two liquids in the tall cylinder for awhile then the
water will accumalate at the top and the perflourooctane will gather at
the bottom. We know that originally there was 250ml of water. So, we
need only take the top 250ml of liquid (water) from the cylinder and
put it into tube A; the rest of the 750ml of liquid (perfluorooctane)
can be dispensed back into tube B.

Thus, this dynamo can continually produce electricity; when the
turbine stops turning because the two liquids mix, then we need only
unmix the two liquids and restart the dynamo.

Notice again that this dynamo creates electricity without using
any input energy! Some may argue that we used energy to unmix the two
liquids. That is true, *but* even though we used energy to unmix the
two liquids we did not *give* the two liquids energy. That is, two
liquids in separate beakers have the same amount of energy as the same
two liquids in the same beaker.

Notice that the upward force on the water is balanced; that is,
the perfluorooctane experiences an equal force downwards. This,
however, does not change the fact that this dynamo creates energy (in
the form of electricity) without input energy.

We can conclude by noting that energy is being created/destroyed
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 do work on the system. If "positive work" is done, then the
system will gain energy. If "negative work" is done, then the system
will lose energy. Should these forces be sustained for a longer
duration of real time, then the forces might be found to have not done
any work on the system (that is, it added the same amount of energy
that was removed). Whether "positive work" or "negative work" is done
is relative.

I must make it clear that I do not consider using the
Gravitational-density dynamo to create electricity in any situation
because it is slow and ineffecient. I am discussing the
Gravitational-density dynamo here simply to demonstrate that the Law of
Conservation of Energy is wrong.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-B) Potential Energy-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Suppose we have two magnets with like-charges "q" and "q0". The
space between the two charges is "r". Let the potential energy between
the charges be "U". Consulting a physics textbook we find that

1 q*q0
U = ------ ------
4*pi*E r

where "pi" equals 3.14
"E" is the permittivity of free space

As the two magnets are moved closer to each other, potential
energy will be gained and kinetic energy will be lost. As the two
magnets move away from each other, potential energy will be lost and
kinetic energy will be gained.

Say, initially, that both magnets are far apart. Now, let us do
work by moving the charges closer together. When we are done and the
magnets are close to each other, the potential energy will have
increased. The increase will be equivalent to the work we did pushing
them together.

Now, let's say that we took two hammers and pounded both magnets
until they lost their magnetism. Then, the potential energy between
the two magnets will dissappear. Thus, the system has lost energy
without any part of the system gaining energy. Thus, we have
demonstrated that the Law of Conservation of Energy is wrong.

Let me recap: First, we did work to move two repelling magnets
together. Thus, we lost kinetic energy while the magnets gained
potential energy. We then destroyed the magnetism of the magnets, thus
losing the potential energy. Thus, all-in-all, we lost energy.

This idea, which works on magnetism, can also be applied to
gravity.

Consider two stationary gaseous planets, both made entirely of
deutrium. Let's do work on the planets, increasing the gravitational
potential energy between the planets, by moving them apart. The
increase in gravitational potential energy will be equivalent to the
amount work we did separating the planets.

Now, let's say that the deutrium of both planets began to fuse by
the following equation:

deutrium atom + deutrium atom = helium atom + neutron + 3.27 MeV
(from http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html)

(It is true that I didn't include the initial energy to start the
fusion. However, the above equation is properly balanced, so we do not
have to consider the initial energy required.)

Now, it is obvious that mass is being converted into energy.
Since the masses of both planets are decreasing, the gravitational
potential energy between both planets will also decrease. Thus, the
work we did moving the planets apart (which is now graviational
potential energy) will diminish. We have again demonstrated that the
Law of Conservation of Energy is wrong.

Let me recap: First, we did work by moving the two planets apart.
Thus, we lost kinetic energy while the planets gained gravitational
potential energy. We then converted some of the mass of the planets
into energy. Thus, we lost mass and in the process we lost
gravitational potential energy. Thus, all-in-all, we lost energy.

(One might oversimplify the above to say, "What goes up does not
*necessarily* come down.")

Or, since mass and energy are interchangeable, what if the mass of
both planets suddenly converted into energy. I don't know exactly how
this could happen, but nonetheless, it is within the realm of
possibilities. Thus, the mass of both planets would dissappear and so,
the gravitational potential energy would also dissapear.

-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (3) WORK AND ENERGY-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

Mechanical energy is the energy which is possessed by an object due to
its motion or its stored energy of position. When I use the term
"mechanical energy" in this section, I am reffering solely to "the
energy which is possessed by an object due to its motion" *not* "its
stored energy of position". (I won't use the term "kinetic energy"
because that term is related to the equation "½mv²", and I do not
want to imply that I am using that equation.)

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-A) Creating and Destroying Energy-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Let's say we have two electromagnets (coils of wire) with air cores.

Now, let's set them next to each other. And then, let's send an
electrical current through them so that they repel each other. Because
they repel each other they will begin to move away from each other.
The two electromagnets were stationary and now they are moving - now
they have mechanical energy. Thus we have created energy.

Now, let's have the two electromagnets move towards each other. Again,
let's send an electrical current through them so that they repel each
other. They will stop moving. The two electromagnets had mechanical
energy and then they stopped. Thus we have destroyed energy.

-- Some may argue that for both scenarios above the total energy of
the system is zero because the momentum of both electromagnets when
taken together is zero. However, the mechanical energy of both
electromagnets can be turned into another form of energy; for example,
we can let both electromagnets rub against a surface like ashphalt.
The heat and sound which is produced is due to friction and it is
energy. Thus, we must conclude that the electromagnets initially also
had energy. Thus, the total energy of the system is not zero! We
cannot simply add the momentum of the objects in the system and derive
a conclusion from that. The energy of a system depends on the addition
of the *individual* energies of the objects in the system, not just the
addition of the energies of the objects in the system.

-- Some may argue that the electrons in the current of electricity
are affected such that they balance the scales so that energy is not
created or destroyed. But what if we inserted iron into the cores of
the electromagnets. Then the repulsive force between the
electromagnets will be greater. Thus, more energy will be created and
destroyed but we used the same amount of current. Thus, we can only
conclude that the electrons in the current of electricity do not
balance the scales, because iron cores can amplify the strength of the
electromagnets without any change in the applied current.

-- Some may argue that mechanical energy is being transformed into
potential energy. But we know from the previous section that potential
energy can disappear without being realized, and so, we again come to
the conclusion that energy can be created and destroyed.

Because mechanical energy can be created and destroyed we can conclude
that the Law of Conservation of Energy is wrong.

--------------
Now, mechanical energy depends on mass and velocity. But velocity is
relative; so, we must conclude that the mechanical energy in a system
is relative to the frame of reference you are in.

For example, consider a skydiver plumetting to the Earth such that he
has reached his terminal velocity. Someone on the Earth will observe
the skydiver falling and say that the mechanical energy of his own
system depends on the mass of the skydiver and the speed at which he is
falling at. On the other hand, the skydiver will observe the Earth to
be moving towards him. He will say that the mechanical energy of his
system depends on the mass of the Earth and the speed at which the
Earth is approaching him. Both are obviously different; the energy
viewed from the skydiver's position is much greater. But both
observations are correct when we include the frame of reference from
which we are observing things.

Also, since velocity and mechanical energy are relative we can no
longer distinguish between two objects that collide. For example,
consider a tennis ball colliding with a basketball. If you are in the
same frame of reference as the tennis ball than you will say that the
basketball hit the tennis ball. If you are in the same frame of
reference as the basketball then you will say that the tennis ball hit
the basketball. We cannot distinguish between two objects in a
collision if you do not distinguish between frames.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-B) Defining Work and Energy-=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

First, let's realize that force has two equations, or rather, that
it can be observed in two different ways. First, there is "ineffective
force":

f_i = pA

where "f_i" is ineffective force ("i" stands for "ineffective")
"p" is pressure
"A" is area

And then there is "effective force":

f_e = ma

where "f_e" is effective force ("e" stands for "effective")
"m" is mass
"a" is acceleration

Effective force is ineffective force which is allowed to cause a change
in mechanical energy.

--------------
Consider the following scenario: 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. Now
work is already defined. So it looks like we have two different ways
of defining work. Let us distinguish between the two buy giving them
names. Let the traditional meaning for work - force multiplied by
displacement - be called "productive work" whereas the "new" definition
for work - force multiplied by time - will be called "general work".
The following is the equation for "general work":

W_g = f*t

where "W_g" is general work ("g" stands for "general")
"f" is a force (it can be effective or ineffective)
"t" is a period of time

I propose that the unit for "general work" (which is force
multiplied by time) should be "P", for Prescott, Joule's middle name.
Thus, one prescott equals one newton second.

Now, work defined as it is today (productive work) is wrong
intuitively but it is *very* useful in making calculations. It
calculates work, where work is defined as causing an object to be
displaced in a certain direction. I should make it clear that any form
of work can be considered useful or useless depending on the situation
and its application. Here is the equation for productive work:

W_p = f*s

where "W_p" is general work ("p" stands for "productive")
"f" is a force (it can be effective or ineffective)
"s" is displacement

(I realize that force multiplied by time is called an impulse.
However, the term "general work" is more fitting because it relates to
"productive work".)

--------------
Of course, just as force has "effective force", work has
"effective work". The term "effective" means that the work is allowed
to cause a change in mechanical energy. If the work does not cause a
change in mechanical energy then the work is called "ineffective work".
Also, "effective general work" causes a change in "general energy" of
the system while "effective productive work" causes a change in
"productive energy" of the system.

We will be using Newtonian mechanics in all of our examples below.

To find out effective general work, take the term "f" and make it
effective, that is, change it into "f_e".

W_g = f*t
W_ge = f_e*t
= ma*t

* where "W_ge" is an amount of general energy
("g" stands for "general"
"e" stands for "energy")

And since "a*t" equals "v" (where "v" is velocity) we can simplify the
equation for general energy to the following:

W_ge = mv

Momuntum is equal to "mv". Thus, in Newtononian mechanics, effective
general work causes a change in general energy equivalent to a change
in momentum.

Now let's consider productive work.

w_p = f*s
W_pe = f_e*s
= ma*s

* where "W_pe" is an amount of productive energy
("p" stands for "productive"
"e" stands for "energy")

Notice that "a*s" equals "½v²". Thus, the equation for effective
productive reduces to:

W_pe = ½mv²

Kinetic energy is equal to "½mv²". Thus, in Newtononian mechanics,
effective productive work causes a change in productive energy
equivalent to a change in kinetic energy.

--------------
Force has never (in my opinion) been defined properly. I define
it as the "magnitude of work". Now, we have defined work in two ways:
general work and productive work. General work depends on force and
time while productive work depends on force and displacement. Now,
when we are considering "time" and "displacement" from the point of
view of work, we will call them the "duration" of work. Thus, work has
a "magnitude" and a "duration".

When we are looking at general energy and general work of a system
then we will use the function "Mg". When we are looking at productive
energy and productive work of a system then we will use the function
"Mp".

Now let's define "Mg" and "Mp" as we would is Newtonian mechanics.

The parameters of "Mg" are "m","v","F","t" such that:

"Mg(m,v,F,t) = mv + Ft"

The parameters of "Mp" are "m","v","F","s" such that:

"Mp(m,v,F,s) = ½mv² + Fs"

Notice that by creating the above functions we can easily alter the
function or add more parameters if other theories demand it to be
changed.

Which equation should we use during a situation? Well, it's useful to
use the equation in which energy is conserved. Notice that if energy
is conserved in one system, it is likely not conserved in the other.
This leads us to a kind of conundrum; energy is only conserved
depending on how you look at the situation.

---------------------------------------
ASIDE:

Power equals productive work divided by time. Productive work equals
"F*s" where "F" is force and "s" is displacement. Time is "t". So
power equals "F*s/t".

Let's apply a force on an object. Let "v_i" be the initial velocity
and "v_f" be the final velocity of the object. Now, "s/t" is equal to
a change in velocity of the object, which is equal to "v_f-v_i". So
the average power of productive work done on the object is
"F*(v_f-v_i)".

Now, let "v_f" be "v_i+dv", where "dv" is a small (infinitesmal)
increase in velocity. So power now becomes "F*(v_i+dv-v_i)" which
simplifies to "F*dv". Now, "dv=a*dt" so power equals "F*a*dt". This
means that over a very small (infinitesmal) amount of time (dt)
instantaneous power increases at the rate "F*a".

-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-\-
-|-|-| (4) SPECIAL RELATIVITY -|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-|-
-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-A) Einstein's 2 Postulates=-=-=-=-=-=-=-=-=-=-=-=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Here are Einstein's first two postulates of special relativity:

#1) The laws of physics are the same in every inertial frame of
reference. That is, one cannot distinguish one inertial frame from the
others or make one frame somehow more "correct" than another.

#2) The speed of light in a vacuum is the same in all inertial frames
of reference and is independent of the motion of the source.

In the future, when I use the term "system" I mean it as a short way of
saying "inertial frame of reference".

Now, for postulate #2 to be true all observers *inside* a system should
agree with all observers *outside* the system that the speed of light
coming from a source *inside* the system is the constant "c".

For convience sake let us make the following defenitions:

* An "Outsider System" is true when all observers *outside* a system
measure the speed of light which emanates from a source *inside* the
system to be the constant "c".

* An "Insider System" is true when all observers *inside* a system
measure the speed of light which emanates from a source *inside* the
system to be the constant "c".

So, for postulate #2 to be true both the Outsider System and the
Insider System should be true; that is, all observers inside and
outside the system should agree that the speed of light (which emanates
from a source inside the system) is the constant "c".

Generally, we will call the person inside the system (where the light
source is) the "insider" while the person outside the system will be
the "outsider".

Now, when we measure a length using a ruler or when we measure time
using a clock then we will say that those quanties are "measured". On
the other hand, you could figure out the distance or time of an event
by using the equation "d=v*t" - where "d" is distance, "v" is velocity,
and "t" is time. If we use that equation to determine a length or a
duration of time then we will say that those quantities are "derived".

Also, we will be using three different devices, what I call "SD
devices" and "SMD devices", and "light-clocks". All three aparatus may
have a light-source, a mirror and a light-detector. To simplify
verbiage, the "light-source" will be called the "source" and the
"light-detector" will be called the "detector".

A "SD device" is an apparatus consisting of a clock, a source and a
detector. The apparatus is set up such that the clock starts when the
source emits a flash of light. The light then gets registered by the
detector which causes the clock to stop. The device is called an "SD"
device because light goes from the light-(S)ource to the
light-(D)etector. For this entire section the distance between the
source and the detector in a SD device will be "L".

A "SMD device" is very similar to a "SD device" except that it has a
mirror. The apparatus is set up such that the clock starts when the
source emits a flash of light. The light is then reflected off the
mirror. The light returns to the source where it is registered by the
detector which causes the clock to stop. The device is called an "SMD"
device because light goes from the light-(S)ource to the (M)irror and
back to the light-(D)etector. For this entire section the distance
between the source/detector and the mirror in a SMD device will be "L".

It should be noted that SMD devices differ from "light-clocks".
Einstein used light-clocks in his famous thought-experiments. A
light-clock is an apparatus set up like a SMD device but without the
clock. The crucial difference between the two is that a SMD device
*measures* an amount of time while a light-clock *derives* an amount of
time. How does a light-clock derive time? Well, when you look at a
light-clock in action you will see the light traverse a certain
distance "d". A user using a light-clock assumes that the speed of
light is the constant "c". Thus, the light clock - using distance and
the speed of light - derives the time "t" elasped by using the equation
"t=d/c". For this entire section the distance between the
source/detector and the mirror in a light-clock (when observed at rest)
will be "L".

We will be demonstrating later on in this section that Einstein's
thought-experments for Time Dialation and Length Contraction are false.
So, we will assume below that time doesn't dialate and length doesn't
contract. However, you can allow time to dialate and length contract
and still be led to the same conclusions below so long as you let the
velocity be much less than "c" so that "1/(1-(v/c)²)^½" (a factor in
Einstein's equations) is essentialy equal to "1".


---------------------------------------
We will call the person outside the system the "outsider" while the
person inside the system will be the "insider". The outsider will be
watching the insider who is in a space ship.

We will be analyzing two situations.

For now, let us assume that the Outsider System is true.

---------------------------------------
Situation #1: (assuming Outsider System is correct)

On the space ship is a SD device secured such that the source is at the
back of the space ship and the detector is at the front.

Now, to start off the space ship is at rest with the outsider.

It is the insider's job to start the SD device when we decide to do the
experiment. Let the insider start the experiment.

The outsider will say he saw light traverse a distance "L" in a time
"t". Also, since we are using an Outsider System he will say that the
speed of light is "c". Thus, the outsider will say that:

"L = c*t"

Now, the insider will also say that he saw a flash of light travel a
distance "L" in a time of "t". Thus, the insider will also say that
the light was travelling at a speed of "L/t". From the above equation
we can say that "L/t" equals "c" and so the insider will agree with the
outsider that the speed of light was "c".

Now, let's accelerate this space ship forward so that it ends up with a
speed of "v" relative to the outsider. Let's have the insider do the
experment once more.

--- WHAT THE OUTSIDER SEES:
|
| c*t
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| L v*t
\_________________________________

Notice that this round it will take more time than "t" for the light to
be detected. This is because the ship is moving forward, and so, the
front of the ship will have moved forward by a factor of "vt" before
the flash of light could reach the detector. So this time the outsider
will say that he saw a flash of light travelling at a speed of "c"
traverse a distance "L+vt" in a time of "t". Thus, we arrive at the
following equation:

"L + v*t = c*t"

Again, the insider will say that he saw a flash of light travel a
distance "L" in a time of "t". Thus, the insider will say that the
light was travelling at a speed of "L/t". From the above equation we
can say that "L/t" equals "c-v" and so the insider will not agree with
the outsider that the speed of light was "c"; he will say that the
speed of the flash of light was "c-v". Thus, when we use an Outsider
System somebody inside the system where the source of the light is will
not agree with someone outside the system that the speed of light is
the constant "c".

But notice that the above equation can be solved for "v":

"v = c - L/t"

So far we have said that "v" is the relative velocity of the outsider
and the space ship. But we have a little problem. The insider will
measure the time elasped during the experiment to be some "fixed
value". This fixed value has nothing to do with the relative velocity
of the outsider and the space ship! Even though the above equation is
what the outsider observes, the insider can conduct the experiment on
his own and thus get a value for "v" without any aid or reference to
the outsider! Now, what exactly is this velocity relative to? It must
be a velocity that is measured relative to some "absolute frame of
reference"! Put another way: If the time (in the above equation) is
some fixed value then we find that the velocity must be some fixed
value! Now what exactly does a "fixed value" for the velocity mean?
Again, a "fixed value" for velocity must mean that the velocity is
relative to some "absolute frame of reference". And since we said
above that "v" is the relative velocity of the outsider and the space
ship then we must notice that we have inadvertently put the outsider at
rest with the "absolute frame of reference". (From now on I will refer
to the "absolute frame of reference" as the "absolute system". Also, a
velocity measured relative to the absolute system will be called an
"absolute velocity".)

And so we have just proved that the first postulate is wrong! Look!
We have distinguished one inertial frame from the others! We've
derived an equation that determines the velocity of the ship relative
to an absolute system.

Now let's consider an outsider that is travelling at a speed of "u"
relative to the absolute system. If the outsider's velocity is in the
same direction as the space ship then the outsider will see the light
traverse a distance "L+(v-u)t" in a time "t". Using the above equation
we can say that the outsider will say the light travelled at a speed of
"c-u". Now, if the outsider's velocity is in the opposite direction of
the space ship then he will see the light traverse a distance
"L+(v+u)t" in a time "t". Thus, he will then say he saw light travel
at the speed of "c+u". Thus, only the outsider at rest with the
absolute system will measure the speed of light to be the constant "c".

---------------------------------------
Now, we need to add to the definition above of the "Outsider System"
and the "Insider System" because they are incomplete. We avoided
mentioning this before to avoid confusion.

If we are using an "Outsider System" then the direction of the light
follows the direction the source is pointing in as seen by an outsider
(someone outside the system). Now, all the outsiders are in different
systems so that they will all (usually) disagree as to what the actual
direction of the light is. Because only one outsider can be "right" as
to what the actual direction of the light is, then we are led to the
conclusion that only one frame of reference is "right". This leads us
directly back to the idea and necessity to create an absolute system.
This means that only one outsider in a unique system will see light
follow from the source in a "straight" line. Everyone else will see
light "bend", that is, the light will not follow from the source in a
straight line.

On the other hand, if we are using an "Insider System" then the
direction of the light follows the direction the source is pointing in
as seen by an insider. Now, since all insiders are in the same system
then they will all agree as to what the actual direction the light is
moving in. So, we have no need in this case to create an absolute
system. This means that only the insiders will see light follow from
the source in a "straight" line. Everyone else will see light "bend",
that is, the light will not follow from the source in a straight line.

It seems to be (as we have seen and shall see in what follows) that the
observer that sees light travel from the source in a straight line is
also the observer who witnesses light travel at the constant speed "c".

The reason why we could leave these points out of the definitions
before is because in Situation #1 all outsiders and all insiders will
agree as to what the direction the light is heading in.

---------------------------------------
Situation #2: (assuming Outsider System is correct)

On the space ship is another SD device such that the source is secured
on the floor of the train and the detector is fastened above so that it
will (hopefully) register the light from the source.

(The beginning of Situation #2 is similar to Situation #1.)

To start off the space ship is at rest with the outsider.

It is the insider's job to start the SD device when we decide to do the
experiment. Let the insider start the experiment.

The outsider will say he saw light traverse a distance "L" in a time
"t". Also, since we are using an Outsider System he will say that the
speed of light is "c". Thus, the outsider will say that:

"L = c*t"

Now, the insider will also say that he saw a flash of light travel a
distance "L" in a time of "t". Thus, the insider will also say that
the light was travelling at a speed of "L/t". From the above equation
we can say that "L/t" equals "c" and so the insider will agree with the
outsider that the speed of light was "c".

Now, let's accelerate this space ship forward so that it ends up with a
speed of "v" relative to the outsider. Let's have the insider do the
experment once more.

Now, the outsider will see exactly what he saw before. That is, he
will see the light emanate from the source an move upward.

However, while the flash of light is heading upwards towards the
detector, the space ship has moved forward by a factor of "vt". Thus,
if the space ship is fast enough then it may have moved forward enough
such that the flash of light might not even hit the detector! The
light may not hit the detector because the light is travelling upwards
as seen from outside the system, not inside.

--- WHAT THE INSIDER SEES:
|
| vt
| ______
| · |
| · |
| · | ct forward --
| c*(1+(v/c)²)^½ * t · |
| · |
| ·|
\_________________________________

The insider will say he saw light travel a distance
"((vt)²+(ct)²)^½" in a time "t". Thus, he will say he saw light
travel at the speed of "c*(1+(v/c)²)^½". So, the insider will
measure the speed of light to be greater than or equal to the constant
"c", but never less.

Now we can measure the length "vt". Let that length be "Z". Then we
can create an equation that solves for v:

"v = Z/t"

So, since we are using an Outsider System we can solve for "v" which is
the velocity relative to the absolute system. Also, above we have
inadvertently put the outsider at rest with the absolute system.

Also, notice that during Situation #2 the insider will see light bend!
The light is travelling upwards as seen from outside the system so
inside the system the light will appear to bend, that is, it will not
follow from the source in a straight line!

---------------------------------------
Before we move on, it should be noted that in Situation #1 and
Situation #2 we only examined the velocity in one dimension. So, if we
are to try to actually implement the thought-experiments in real life
then one would have to consider the other dimensions of the velocities
of the spaceship and the flash of light.

---------------------------------------
CONCLUSIONS: From the above, if we are to say that the Outsider System
is true then we are led to three inevitable conclusions:

-- (1) Postulate #1 is wrong! There must be some absolute system
for "v" to be relative to, and so, we have distinguished one system
from the others.

-- (2) Postulate #2 has errors! We've used the Outsider System and
we've found that the original definition of the Outsider System is
wrong! The speed of light is only the constant "c" when it is measured
from the absolute system.

-- (3) When observed from inside the system where the light source
is, the flash of light will seem to bend, that is, it will not follow
from the source in a straight line.

We have seen above that the Outsider System is ridden with pitfalls.
Now, many experiments have been done where the light source and the
experimenter are inside the system. In such experiments the speed of
light has never deviated from "c" and light has never appeared to bend.
So with these problems it is likely that we started with the wrong
assumption.

So instead let us assume that the Insider System is right.

---------------------------------------
Situation #1: (assuming Insider System is correct)

Now, to start off the space ship is at rest with the outsider.

It is the insider's job to start the SD device when we decide to do the
experiment. Let the insider start the experiment.

The insider will see the light traverse a distance "L" in a time "t".
Also, since we are using an Insider System the insider will say the
light travelled at the speed of "c". So,

"L = c*t"

Now let's consider an outsider that is travelling at a speed of "v"
relative to the space ship. Let the insider do the experiment once
more.

--- WHAT THE OUTSIDER SEES:
|
| (c-v)*t
| |----------|
|
| |··········| forward --
|
| |-------|
| v*t
| |------------------|
| L
\_________________________________

If the outsider's velocity is in the forward direction of the space
ship, the outsider will see the light traverse a distance "L-vt" in a
time "t". Using the above equation we can say that the outsider will
see the light travel at a speed of "c-v".

--- WHAT THE OUTSIDER SEES:
|
| (c+v)*t
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| L v*t
\_________________________________


Now, if the outsider's velocity is in the opposite direction of the
forward direction of the space ship, then he will see the light
traverse a distance "L+vt" in a time "t". Thus, he will then say he
saw light travel at the speed of "c+v". This means that someone
outside the system will not agree with the insider that the speed of
light is the constant "c"!

---------------------------------------
Situation #2: (assuming Insider System is correct)

Now, to start off the space ship is at rest with the outsider.

It is the insider's job to start the SD device when we decide to do the
experiment. Let the insider start the experiment.

The insider will see the light traverse a distance "L" in a time "t".
Also, since we are using an Insider System the insider will say the
light travelled at the speed of "c". So,

"L = c*t"

Now let's consider an outsider that is travelling at a speed of "v"
relative to the space ship. Let the insider do the experiment once
more.

--- WHAT THE OUTSIDER SEES:
|
| v*t
| ______
| · |
| · |
| · | L forward --
| (c²+v²)^½ * t · |
| · |
| ·|
\_________________________________

If the outsider's velocity is in the forward direction of the space
ship, the outsider will see the light traverse a distance
"(L²+(v*t)²)^½" in a time "t".

--- WHAT THE OUTSIDER SEES:
|
| ·|
| · |
| (c²+v²)^½ * t · | L forward --
| · |
| · |
| ·_____|
| v*t
\_________________________________

If the outsider's velocity is in the opposite direction of the forward
direction of the space ship, the outsider will see the light traverse a
distance "(L²+(v*t)²)^½" in a time "t".

Both observations above are symetrical. Thus, we find that the
outsider will measure the speed of light to be "c*{1+(v/c)²}^½"

---------------------------------------
CONCLUSIONS: From the above, if we are to say that the Insider System
is true then we are led to two conclusions:

-- (1) Postulate #1 is right! We have no need to create an absolute
system and so postulate #1 is liberated.

-- (2) Postulate #2 has errors! We've used the Insider System and
we've found that the speed of light is the constant "c" only when
measured from inside the system where the light source is.

---------------------------------------
So, when we use the Outsider System then light travels at the constant
"c" from the absolute system but not "c" from all other systems. When
we use the Insider System then light travels at the constant "c" from
inside the system but not "c" from all other systems. Thus, we can
conclude that the Outsider System is incompatible with the Insider
System. Postulate #2 is wrong no matter which way you look at it!
Either the Outsider System is right or the Insider System is right, not
both! The Outsider System means that the speed of light does not
depend on the motion of the source of light while the Insider System
means that the speed of light does depend on the motion of the source
of light; contradiction ensues.

---------------------------------------
ASIDE:

Sound propagates through air using an Outsider System.

Consider two people, a pilot and a co-pilot, both sitting in the
cockpit of a plane. The co-pilot is behind the pilot. The plane is
travelling faster than the speed of sound relative to the ground and
atmosphere.

Now, if the cockpit is closed then when the co-pilot says something the
sound of his voice will travel forward to the pilot. The speed of the
sound of his voice will be travelling at the speed of sound relative to
the air in the cockpit.

However, if the cockpit is open and the co-pilot says something the
sound of his voice will *not* travel forward to the pilot. The speed
of the sound of his voice will be travelling at the speed of sound
relative to the air of the atmosphere. But since the plane is
travelling faster than the speed of sound relative to the atmosphere,
the co-pilot's voice will not be heard by the pilot.

*(I am interested in knowing how open the cockpit can be such that the
pilot still hears the co-pilot's voice.)*

Notice that if the cockpit is open then we can determine the velocity
of the plane relative to the atmosphere as we did above with light.
The velocity is zero when the plane is stationary with the atmosphere,
the atmosphere being the medium through which sound propagates through.


Thus, if we are to assume that the Outsider System for light is true,
then we can say that when the space ship's absolute velocity is zero
then it is stationary with the "ether", the medium through which light
(supposedly) propagates through.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-B) Understanding the Michelson-Morley Experiment=-=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Now, the Michelson-Morley experiment attempts to find the Earth's speed
relative to the ether. The experiment has its errors. I will now
point out those errors and explain why we get a "null result" when we
perform the experiment. My experiment is different from
Michelson-Morley's setup but it essentially demonstrates the same
thing.

I will be assuming that light propagates through space using the
Insider System.

We will have two SMD devices and we will do this experiment on the
equator of the Earth. One SMD device is set up perpendicular to the
equator while the other SMD device is set up parallel to the equator.
The SMD device perpendicular to the equator has the source/detector
south and the mirror north, while the SMD device parallel to the
equator has the source/detector west and the mirror east.

And there are two people, an insider and an outsider. The insider is
on the ground next to the two SMD devices. The outsider is in a space
ship above the Earth such that he observes the SMD devices to be
directly below him every 24 hours. (This last bit of information is
unneccesary but I include it so that you can visualize the situation
better; that is, the time "24 hours" can really be anything..)

Now, when we are doing experiments on Earth we are not in a inertial
frame of reference because we are accelerating due to gravity. But,
for an experiment that lasts a brief period of time - like this one -
we can assume that the Earth is not changing inertial frames of
reference. So we can assume that the insider and the SMD devices
remain in the same inertial frame of reference.

Now, let's do the experiment; let's activate the SMD devices.

Since we are using an Insider System we can say that the insider will
see the light in both SMD devices to be travelling at a constant speed
"c". For both the SMD devices the insider will say he saw the light
traverse a distance "L" twice. Since the speed is constant and the
distance traversed is constant we can say that for both SMD devices it
takes the same amount of time for the light to go from the source to
the mirror as it does for the light to go from the mirror to the
detector. We will call that time "t". So, "t" is the time it takes
for a one way trip from the source/detector to the mirror. We get the
following equation:

"t = L/c"

The outsider, on the other hand sees the experiment differently. The
Michelson-Morley experiment has its errors here. The equations I get
for the observations of the outsider differs from what Michelson-Morley
use in their experiment.

Before we go on, it should be noted that the outsider sees the Earth to
be rotating at a speed of "v". (If the time for one complete orbit is
"24 hours" as said above, then "v" equals approximatly "3*10^4" meters
per second. Again, this information is unnecessary and I include it
only so that you may visualize the situation better.)

Now, for the apparatus perpendicular to the equator:

--- WHAT THE OUTSIDER SEES:
|
| ·|
| · |
| c*{1+(v/c)²}^½ * t · | L forward --
| · |
| · |
| ·_____|
| v*t
\_________________________________

When the light is travelling towards the mirror the outsider sees the
light travel at a speed of "c*{1+(v/c)²}^½" traversing a distance
"(L²+(v*t)²)^½" in a time "t".

--- WHAT THE OUTSIDER SEES:
|
| |·
| | ·
| | · c*{1+(v/c)²}^½ * t forward --
| L | ·
| | ·
| |_____·
| v*t
\_________________________________

When the light is returning back to the detector he sees light travel
at a speed of "c*{1+(v/c)²}^½" traversing a distance
"(L²+(v*t)²)^½" in a time "t".

Now, the above two situations are symetrical. Thus we only derive one
equation which works for both situations:

(1) "c*{1+(v/c)²}^½ * t = (L² + (v*t)²)^½"

For the apparatus parallel to the equator:

--- WHAT THE OUTSIDER SEES:
|
| (c+v)*t
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| L v*t
\_________________________________

When the light is travelling towards the mirror the outsider sees light
travel at a speed of "c+v" traversing a distance "L+v*t" in a time "t".

Thus we get the following equation:

(2) "(c+v) * t = L + v*t"

--- WHAT THE OUTSIDER SEES:
|
| (c-v)*t
| |----------|
|
| |··········| forward --
|
| |-------|
| v*t
| |------------------|
| L
\_________________________________

When the light is returning back to the detector he sees the light
travel at a speed of "c-u" traversing a distance "L-v*t" in a time "t".

Thus we get the following equation:

(3) "(c-v) * t = L - v*t"

Now, I will leave it to you to simplify the above three equations.
They all simplify to the following equation:

"L = c*t"

Thus, the equations of the outsider agree with the insider! The
insider and the outsider will agree with each other that "L=c*t" is
true.

Since they agree we can see why we get a null result when we do the
Michelson-Morley experiment: We chose the outsider arbitrarily. The
outsider could have been in any inertial frame of reference. Thus,
from the above we can say that *any* outsider in *any* inertial frame
will agree with the insider that "L=c*t". That is why we get a "null"
result from the Michelson-Morley experiment because everybody who looks
at the experiment simply observes that the following equation is true:
"L=c*t".

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
=-=-=-C) Time Dialation, Length Contraction and the Rest=
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

The next two sections will discuss the two thought-experiments used to
derive the equations for special relativity's Time Dialation and Length
Contraction. Both experiments assume that postulate #2 is correct,
that is, it assumes that both the Outsider System and the Insider
System are true and compatible. We know better though, that the two
systems are incompatible.

So:

(1) First, it will be shown that when the Outsider System is used the
experiments are fail.

(2) Secondly, it will be shown that even if postulate #2 is right
(that is, the Outsider System and the Insider System are compatible)
then the thought-experiments fail.

(3) Thirdly, it will be shown that when we are using the Insider
System the thought-experiments fail.

---------------------------------------
Time Dialation..
---------------------------------------

There are two people, an insider and an outsider. The outsider is
standing on the Earth while the insider is sitting on a train. The
train is travelling forward at a velocity "v" relative to the Earth.

There is a SMD device on the train such that the source/detector is
secured on the floor of the train while the mirror is fastened above
the source such that it will (hopefully) reflect the light from the
source directly back down to the detector. This Time Dialation
thought-experiment is similar to the thought-experiment described in
Situation #2.

(1) Now, if we are to use an Outsider System the outsider will see
the flash of light emanate from the source and move upward. While the
flash of light is heading upwards towards the mirror, the train has
moved forward a bit. Thus, if the train is fast enough then it may
have moved forward enough such that the flash of light might not even
hit the mirror at all! The light may not hit the mirror because the
light is travelling upwards as seen from outside the system, not
inside. The experiment as stated by special relativity requires that
the light gets reflected back to the ground, and so, this experment
does not produce proper results when we use the Outsider System.

For the moment let us assume that the experiment was simply to have the
source send a flash of light upward (i.e. it didn't need to get
reflected downward by the mirror). In this case, the outsider will say
he saw light travel a distance "L" in a time "t". And since we are
using an Outsider System he will say that he saw light travel at the
speed of "c".

--- WHAT THE INSIDER SEES:
|
| "vt"
| ______
| · |
| · |
| · | "ct" forward --
| "c*(1+(v/c)²)^½ * t" · |
| · |
| ·|
\_________________________________

On the other hand, the insider will say he saw light travel a distance
"((vt)²+(ct)²)^½" in a time "t". Thus, he will say he saw light
travel at the speed of "c*(1+(v/c)²)^½". So, the insider will
measure the speed of light to be greater than or equal to the constant
"c", but never less. (Again, notice that since we are using an
Outsider System we can solve for "v" which is the velocity relative to
the absolute system. Also, above we have inadvertently put the
outsider at rest with the absolute system.)

(2) Now, the way this Time Dialation experiment goes is that the
Outsider System and the Insider System are assumed to be compatible;
that is, both the insider and the outsider measure the speed of light
to be "c". Also, it assumes that the light from the source does infact
hit the mirror and get reflected downward back to the detector. We
know that these assumptions are wrong and contradictory, but let us go
forward anyway. Also, instead of using a SMD device, the Time
Dialation experiment uses a light-clock.

Let the insider derive a time of "tI" elaspe while the outsider derives
a time of "tO" to have elasped.

The insider sees the light travel a distance "2L".

--- WHAT THE OUTSIDER SEES:
| ___
| | ·|·
| | · | ·
| L | · | · forward --
| | · | ·
| | · | ·
| _|_ ·_____|_____·
|
| |-----------|
| "vtO"
\_________________________________

Meanwhile the outsider sees the light travel a distance
"2*[{(vtO/2)²+L²}^½]".

And we assumed that both the insider and the outsider see light travel
at the constant "c". Now, we will use the equation "t=d/c", where "t"
is an amount of derived time, "d" is distance, and "c" is the constant
at which light travels at. So the insider derives a time

"tI = 2L/c"

while the outsider derives a time

"tO = 2*[{(vtO/2)²+L²}^½]/c"

Since the outsider sees the light travel a greater distance than the
insider, Einstein (and his friends) then use the equation "t=d/c" to
claim that the outsider will measure a greater amount of time to elapse
than the insider.

Using the two equations, the Time Dialation experiment goes on to
derive the following equation:

"tO = y*tI"

* where "y" equals "1/(1-(v/c)²)^½"

Now, this equation is supposed to demonstrate that "time dialates".
Notice that the outsider sees the flash of light travel a greater
distance than the insider. This is directly responsible for the fact
that we then get an equation which demonstrates that time dialates.
This is wrong! All this says is that the light *seemed* to travel a
greater distance as seen by the outsider. The time dialation equation
means that the *derived* quantity for time has dialated; this does not
imply that the *measured* quantity of time has dialated. Einstein (and
his friends) often make the mistake of saying measured time dialates
because derived time dialates; this is wrong.

Let me put it another way: What if I were with the insider on the
train and I was looking at the light-clock's reflection in a concave
mirror. Because the mirror is concave I would not see the light-clock
properly; the light-clock would seem to be larger. Thus, it would seem
to me that the light in the light-clock travels a greater distance.
Can I then conclude that time has dialated because it seems that the
light has travelled a greater distance for me? Of course not!

In any case, this is how most physics textbooks leave the subject.

However, what if we moved the light-clock down to Earth beside the
outsider? Then, if you repeat the Time Dialation experiment one will
derive the following equation:

"tI = y*tO"

Now both equations demonstrate that time dialates! If we are to say
that derived time dialates then there is no problem. But if we mean
that measured time dialates then we have the following problem: Which
equation is true and which is false? Both the insider and the outsider
have equal rights to have there measured time dialate with respect to
the other. In essence, both equations together mean that "My time is
faster than your time which is faster than my time which is faster than
your time which is, etc..." Now, physics books and experiments often
allow one of the equations to be true while the other equation is
dismissed (e.g. the famous "Twin Paradox" experiment, special
relativity's thought-experiment for Length Contraction); such action is
unjustified.

Another problem which arises when you use only one equation is that we
do not know what the velocity "v" (in the equation) is relative to. To
get to that problem we could have simply analyzed the tenets of special
relativity. Einstien's first postulate of special relativity is that
"the laws of physics are the same in every inertial frame of
reference." That is, "one cannot distinguish one inertial frame from
the others or make one frame somehow more "correct" than another."
However, when only one of the equations above is used then we require a
*unique* velocity relative to a *unique* frame of reference; that
contradicts the first postulate.

(3) Now, if we are to use an Insider System then the insider will
observe that a flash of light emanates from the source and moves upward
such that it hits the mirror affixed above, and gets reflected back
down to the detector. We can be assured that the light hits the mirror
because the light is travelling upwards as seen from inside the system.

In this case, the insider will say he saw light travel a distance "2L"
in a time "t". And since we are using an Insider System he will say
that he saw light travel at the speed of "c". So: "2L=c*t".

--- WHAT THE OUTSIDER SEES:
| ___
| | ·|·
| | · | ·
| L | · | · forward --
| | · | ·
| | · | ·
| _|_ ·_____|_____·
|
| |-----------|
| "vt"
\_________________________________

On the other hand, the outsider will say he saw light travel a distance
"2*((vt/2)²+L²)^½" in a time "t". Thus, he will say he saw light
travel at the speed of "c*(1+(v/c)²)^½. So, the outsider will
measure the speed of light to be greater than or equal to the constant
"c", but never less.

Notice that if we take the three observations made by the outsider -
(1) the distance the light traversed, (2) the time it took, and (3) the
speed of the light - and make an equation out of them then the equation
will simplify to become "2L=c*t".

In any case, when we are using an Insider System for light we no longer
find that time dialates.

Now, finally, does time really dialate? Some experiments demonstrate
that it does. If time does dialate, I don't think it is because of
Einstein's thought experiment for Time Dialation. In any case, more
experiments need to be done to get the big picture in focus.
Thought-experiments like the ones generated by Einstein are riddled
with pitfalls, and should not *define* physics, but rather be used to
*explain* physics.

---------------------------------------
...Length Contraction..
---------------------------------------

There are two people, an insider and an outsider. The outsider is
standing on the Earth while the insider is sitting on a train. The
train is travelling forward at a velocity "v" relative to the Earth.

There is a SMD device on the train. The source/detector is secured at
the back of the train while the mirror is in the front of the train.
This Length Contraction thought-experiment is similar to the
thought-experiment described in Situation #1.

(1)

--- WHAT THE OUTSIDER SEES:
|
| c*t1
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| L v*t1
\_________________________________

Now, if we are to use an Outsider System the outsider will see the
flash of light emanate from the source and move forward. Let "t1" be
the duration of time it takes for the light to go from the source to
the mirror. While the flash of light is heading forward the train has
moved forward by a factor of "v*t1". Thus we derive the following
equation:

"L + v*t1 = c*t1"

--- WHAT THE OUTSIDER SEES:
|
| c*t2
| |----------|
|
| |··········| forward --
|
| |-------|
| v*t2
| |------------------|
| L
\_________________________________

Eventually, the light will hit the mirror and get reflected back toward
the detector. Let "t2" be the duration of time it takes for the light
to go from the mirror to the detector. Likewise, the train will have
moved forward by a factor of "v*t2" before the light from the mirror
returns back to the detector. So:

"L - v*t2 = c*t2"

Now we know above that since we are using an Outsider System the
outsider sees the flash of light travel at a constant velocity "c". On
the other hand, the insider sees the speed of the flash of light going
to the mirror and returning back to the detector to be different. When
the light is travelling towards the mirror the insider observes that it
traverses a distance "L" in a time "t1" and so he says the light
travelled (using the above equation) at a speed of "c-v". Meanwhile,
when the light is returning back to the detector the insider observes
that it traverses a distance "L" in a time "t2" and so he says the
light travelled (using the above equation) at a speed "c+v". (Again,
notice that since we are using an Outsider System we can solve for "v"
which is the velocity relative to the absolute system. Also, above we
have inadvertently put the outsider at rest with the absolute system.)

(2) Now, the way this Length Contraction experiment goes is that
postulate #2 is assumed to be true, which means that both the insider
and the outsider measure the speed of light to be "c". We know that
that cannot happen, but let us go forward anyway. Also, instead of
using a SMD device, the Length Contraction experiment uses a
light-clock.

Let the distance between the source and the mirror be "LI" as seen by
the insider and "LO" as seen by the outsider.

During the experiment, the insider will see light travel at a speed of
"c" traversing a distance "2LI". Let the insider derive the time of
the experiment to be "tI". Thus, "tI=2LI/c".

Now the outsider will see things differently. He will also see the
light travel at the speed "c". Now, let the derived time it takes for
the outsider to see the light go from the source to the mirror be "tO1"
while the derived time it takes for the outsider to see the light go
from the mirror to the detector be "tO2".

--- WHAT THE OUTSIDER SEES:
|
| c*t01
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| LO v*t01
\_________________________________

When the light is travelling towards the mirror the outsider will see
the light traverse a distance "d1" where "d1" equals "LO+vtO1" and "d1"
also equals "ctO1".

--- WHAT THE OUTSIDER SEES:
|
| c*t02
| |----------|
|
| |··········| forward --
|
| |-------|
| v*t02
| |------------------|
| LO
\_________________________________

When the light is travelling back to the detector the outsider will see
the light traverse a distance "d2" where "d2" equals "LO-vtO2" and "d2"
also equals "ctO2".

Let "tO" be the entire derived time it takes for the experiment to
complete as seen by the outsider; thus "tO=tO1+tO2" and so
"tO=2LO/[c{1-(v/c)²}].

Now, the Length Contraction experiement goes on to state that
"tO=y*tI", where "y" equals "1/(1-(v/c)²)^½". Using this equation by
itself is unjustified. We discussed this above in the section for Time
Dialation. We used the equation "tO=y*tI" and we've dismissed the
equation "tI=y*tO" without any justification!

In any case, you can use the above equations to get the following:

"LI = y*LO"

Einstein goes on to state that length has "contracted". Length has
*not* contracted! The fact that the "distance light travelled has
contracted" does not imply that "length has contracted"! That is, "LO"
and "LI" are derived quantities and they do not agree with measured
quantities of length. Likewise, "tO" and "tI" are derived quantities
and they do not agree with measured quantities of time.

In any case, we assumed above that postulate #2 holds, which we know it
doesn't.

(3) Now, if we are to use an Insider System then the insider will see
light travel at the constant speed "c" and traverse the distance "L"
when it is heading towards the mirror and a distance "L" when it is
returning back to the detector. Since the speed is constant and the
distance travelled is constant we can say it takes the same amount of
time for the light to reach the mirror as it does for the light to
return back to the detector; let this duration of time be "t". Thus we
get the following equation "L=c*t".

--- WHAT THE OUTSIDER SEES:
|
| (c+v)*t
| |--------------------------|
|
|
|··················|·······|
forward --
|
| |------------------|-------|
| L v*t
|
\_________________________________

Now, when the light is travelling towards the mirror the outsider
measures that the distance it traverses is "L+vt" in a period of time
"t". Dividing the two, we get the speed at which the outsider sees the
light travel at:

"(L+vt)/t = L/t + vt/t = c+v"

--- WHAT THE OUTSIDER SEES:
|
| (c-v)*t
| |----------|
|
| |··········| forward --
|
| |-------|
| v*t
| |------------------|
| L
\_________________________________

When the light is travelling back to the detector the outsider measures
that the distance it travels is "L-vt" in a period of time "t".
Dividing the two, we get the speed at which the outsider sees the light
travel at:

"(L-vt)/t = L/t - vt/t = c-v"

Thus, the outsider sees the light travel at a speed of "c+u" when it is
travelling towards the mirror and "c-u" when it is returning back to
the detector.

Notice that if we take the three observations made by the outsider -
(1) the distance the light traversed, (2) the time it took, and (3) the
speed of the light - and make an equation out of them then the equation
will simplify to become "L=c*t".

In any case, when we are using an Insider System we no longer find that
length contracts.

Now, finally, do lengths really contract? There may be some
experiments that demonstrate that it does (I personally haven't heard
of any). However, more experiments need to be done. I will repeat
myself: Thought-experiments like the ones generated by Einstein are
riddled with pitfalls, and should not *define* physics, but rather be
used to *explain* physics.

---------------------------------------
...and the Rest
---------------------------------------

---------------------------------------
It should be noted that when we have used the Insider System above then
all insiders and outsiders agreed that some equation looking like
"L=c*t" is true. I would like to know whether this is true for all
observers for all cases.

---------------------------------------
Now, the way velocities add under special relativity is wrong because
it assumes that time dialates and length contracts.

Also, the "Doppler effect" for light is wrong because it uses the fact
that time dialates.

---------------------------------------
ADDENDUM:

There must be many other people who have come to the same conclusions I
have here. The faults of special relativity are too obvious.

A great (but short) book which identifies the various failures of
special relativity:

"The Special Theory of Relativity" by Essen, L.

Also, Ardeshir Mehta has come up with many thought-experiments which
debunk special relativity:

http://homepage.mac.com/ardeshir/Relativity.html

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

by Raheman Velji


April 25, 2005

you can also view this paper (and updated versions) at...
....http://www.angelfire.com/un/rv

or a less updated copy can be found at...
....http://www.angelfire.com/rebellion2/rahemanvelji