Einstein's addition of velocities theorem gives an erroneous result.

It can easily be tested by using the conservation of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

To be consistent, we must be certain that both velocities are truly
..75c -- and we
must also be certain that the final momentum represents the correct
amount and
that it also represents a conserved amount.

To accomplish this, we set up a two case scenario.
We declare m = 1 and c= 1. This makes computation simpler with no
jeopardy to veracity.


CASE 1


( )-------.75c---------------------.75c--------------( )
p=1.13 p= 1.13


This certifies that we have two velocities of .75c.

Now the complication here is that according to custom, the final
momentum is zero.

So to acquire a meaningful momentum resultant, we simply change the
perspective and declare the right hand system as inertial and all the
velocity and momentum to reside in the left hand system.

This also guarantees that the resultant momentum is a conserved
quantity.

According to Einstein the answer is Case 2.


CASE 2

----------------------.96c---------------------------( )( )

p = 3.43


Next, we note that we have changed nothing between Case 1 and Case 2
except the viewpoint. Therefore, in order maintain conservation, the
final momentum must be the sum of the momenta in Case 1, i.e., 2.26.

But it isn't. Therefore, the answer is wrong.

According to my theory, written up in my book, the correct answer is
..915c.

*********************************************



Vertner H. Vergon

Dear vertvergon:

wrote in message
oups.com...

Einstein's addition of velocities theorem gives an
erroneous result.

It can easily be tested by using the conservation
of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

Conservation of momentum applies to interactions observed from a
single frame. It does not require anything of what other
observers see, other than they also observe momentum conserved
over any interaction. But should you be worried about it, apply
gamma, and all is well.

The list of things you don't understand about even Newtonian
physics just keeps getting longer, doesn't it?

David A. Smith

wrote in message oups.com...

Einstein's addition of velocities theorem gives an erroneous result.

Yes, we know, your understanding of this theorem gives
an erroneous result:
http://users.telenet.be/vdmoortel/di...ablyWrong.html
Thanks for reminding us.

Dirk Vdm

N:dlzc D:aol T:com (dlzc) wrote:
Dear vertvergon:

wrote in message
oups.com...

Einstein's addition of velocities theorem gives an
erroneous result.

It can easily be tested by using the conservation
of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

Conservation of momentum applies to interactions observed from a
single frame. It does not require anything of what other
observers see, other than they also observe momentum conserved
over any interaction. But should you be worried about it, apply
gamma, and all is well.

The list of things you don't understand about even Newtonian
physics just keeps getting longer, doesn't it?

David A. Smith

vergon
Speak for yourself, Dave.

That garbage you spew above does not addres the issue. and is a
nonsequitur.

The conservation of momentum exists in a closed system. My procedure
(which obviously is too clever for you to understand) involves momentum
in a closed system.

And for your information gamma WAS used. If you knew what you were
talking about you would have perceived that.

wrote:
Einstein's addition of velocities theorem gives an erroneous result.

It can easily be tested by using the conservation of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

To be consistent, we must be certain that both velocities are truly
.75c -- and we
must also be certain that the final momentum represents the correct
amount and
that it also represents a conserved amount.

To accomplish this, we set up a two case scenario.
We declare m = 1 and c= 1. This makes computation simpler with no
jeopardy to veracity.


CASE 1


( )-------.75c---------------------.75c--------------( )
p=1.13 p= 1.13


This certifies that we have two velocities of .75c.

Now the complication here is that according to custom, the final
momentum is zero.

So to acquire a meaningful momentum resultant, we simply change the
perspective and declare the right hand system as inertial and all the
velocity and momentum to reside in the left hand system.

This also guarantees that the resultant momentum is a conserved
quantity.

According to Einstein the answer is Case 2.


CASE 2

----------------------.96c---------------------------( )( )

p = 3.43


Next, we note that we have changed nothing between Case 1 and Case 2
except the viewpoint. Therefore, in order maintain conservation, the
final momentum must be the sum of the momenta in Case 1, i.e., 2.26.

But it isn't. Therefore, the answer is wrong.

According to my theory, written up in my book, the correct answer is
.915c.

*********************************************



What about the transformation of the mass?

Dirk Van de moortel wrote:
wrote in message oups.com...

Einstein's addition of velocities theorem gives an erroneous result.

Yes, we know, your understanding of this theorem gives
an erroneous result:
http://users.telenet.be/vdmoortel/di...ablyWrong.html
Thanks for reminding us.

Dirk Vdm

VERGON
Admittedly the quote you gave as a link was poorly written. It was not
complete.

Let me see if I can correct that:

It seems everybody agrees observed longitudinal length in an observed
frame decreases with an increase in velocity.

Postulate 1: Longitudinal length and "distance" are the same thing as
longitudinal length is also a distance.

Postulate 2: Velocity is distance per time.

Conclusion: Since longitudinal length reduces with an increase in
velocity,
then *observed* velocity reduces with an increase in velocity of the
moving frame.

Restated: The velocity of the moving frame (with respect to the
observer's frame) is greater than the observed velocity.

Thus we have two velocities, the observed velocity -- and the true
velocity of the moving frame.

The two velocities are Lorentz variant. That is true velocity times the
Lorentz transform equals the observed velocity.

Where V is the velocity of the moving frame, v the observed velocity,
and R the Lorentz transform,
V x R = v

Thus there are two velocities, that extant -- and that observed.

It will be found that as V goes to infinity, v goes to c.
Therefore we conclude that superluminal velocities exist.
We also conclude that in p = mv/R, the R applies to v not to m.
So there is no such thing as relativistic mass. The mass is
invariable and it is v that goes to infinity as R goes to zero.

Now to add two velocities, we determine their actual velocities, add
them together, then multiply by R and obtain the resultant relative
velocity.

For v = .75c,

actual velocitity is V = v/R.

V + V = 2.267787

R = 1/sqrt(1 +[ 2.267787]^2) = 4.034733 x 10^-1

and

2.267787 x .4034733 = .9149915 c.

Igor wrote:
wrote:
Einstein's addition of velocities theorem gives an erroneous result.

It can easily be tested by using the conservation of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

To be consistent, we must be certain that both velocities are truly
.75c -- and we
must also be certain that the final momentum represents the correct
amount and
that it also represents a conserved amount.

To accomplish this, we set up a two case scenario.
We declare m = 1 and c= 1. This makes computation simpler with no
jeopardy to veracity.


CASE 1


( )-------.75c---------------------.75c--------------( )
p=1.13 p= 1.13


This certifies that we have two velocities of .75c.

Now the complication here is that according to custom, the final
momentum is zero.

So to acquire a meaningful momentum resultant, we simply change the
perspective and declare the right hand system as inertial and all the
velocity and momentum to reside in the left hand system.

This also guarantees that the resultant momentum is a conserved
quantity.

According to Einstein the answer is Case 2.


CASE 2

----------------------.96c---------------------------( )( )

p = 3.43


Next, we note that we have changed nothing between Case 1 and Case 2
except the viewpoint. Therefore, in order maintain conservation, the
final momentum must be the sum of the momenta in Case 1, i.e., 2.26.

But it isn't. Therefore, the answer is wrong.

According to my theory, written up in my book, the correct answer is
.915c.

*********************************************



What about the transformation of the mass?

VERGON

Mass is velocity invariant. It is the *observed* velocity that
undergoes change.

See my post to Dirk Van de moortel above.

Igor wrote:
wrote:
Einstein's addition of velocities theorem gives an erroneous result.

It can easily be tested by using the conservation of momentum.

To make it quick and easy, I shall use an example:
the addition of .75c to .75c. The theorem yields .96c.

To be consistent, we must be certain that both velocities are truly
.75c -- and we
must also be certain that the final momentum represents the correct
amount and
that it also represents a conserved amount.

To accomplish this, we set up a two case scenario.
We declare m = 1 and c= 1. This makes computation simpler with no
jeopardy to veracity.


CASE 1


( )-------.75c---------------------.75c--------------( )
p=1.13 p= 1.13


This certifies that we have two velocities of .75c.

Now the complication here is that according to custom, the final
momentum is zero.

So to acquire a meaningful momentum resultant, we simply change the
perspective and declare the right hand system as inertial and all the
velocity and momentum to reside in the left hand system.

This also guarantees that the resultant momentum is a conserved
quantity.

According to Einstein the answer is Case 2.


CASE 2

----------------------.96c---------------------------( )( )

p = 3.43


Next, we note that we have changed nothing between Case 1 and Case 2
except the viewpoint. Therefore, in order maintain conservation, the
final momentum must be the sum of the momenta in Case 1, i.e., 2.26.

But it isn't. Therefore, the answer is wrong.

According to my theory, written up in my book, the correct answer is
.915c.

*********************************************



What about the transformation of the mass?

wrote in message oups.com...

Dirk Van de moortel wrote:
wrote in message oups.com...

Einstein's addition of velocities theorem gives an erroneous result.

Yes, we know, your understanding of this theorem gives
an erroneous result:
http://users.telenet.be/vdmoortel/di...ablyWrong.html
Thanks for reminding us.

Dirk Vdm

VERGON

Yes, your name is Vergon.
We know.
Congratulations.

Dirk Vdm

I will not waste time finding the error in your calculations.
You see, this is not Einstein's error: we are dealing with Lorentz's
and Leibnitz's errors combined.
The "speed addition" is calculated starting from the Lorentz's
transforms via simple calculus:


x'=g(x-vt)
t'=g(t-vx/c^2) t=g(t'+vx/c^2)

u'=dx'/dt'=dx'/dt*dt/dt'=g(dx/dt-v)*g(1+v/c^2*dx'/dt')=

=g^2(u-v)(1+vu'/c^2)

Solving for u':

g^2(u-v)=u' [1-g^2(u-v)v/c^2]=u'(1-uv/c^2)g^2

Therefo

u'=(u-v)/(1-uv/c^2)

Take your complaints with Lorentz and Leibnitz, leave Einstein alone.