Quadibloc -- Fourth Attempt

"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
===============================================
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
===============================================
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.


Hint: where did the t go and where did the tau come from in the inverse
transform?

Bud

"William Hamblen" wrote in message
m...

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway
wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
===============================================
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.


Hint: where did the t go and where did the tau come from in the inverse
transform?

Bud
==========================================
Hint:
I want an answer from Quadiblockhead, not a ****ing hint from a ****head.
Hint:
Answer the question or shut the **** up.

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

Lord Androcles, Zeroth Earl of Medway wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
[snip]
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

Because the expression on the right-hand side uses tau, not t. It's
certainly true that x = xi/beta + vt.

Here's the trivial derivation.

1) Solve tau = beta(t-vx/c^2) for t:
t = (tau/beta) + vx/c^2.

2) Multiply both sides by v:
vt = (v tau)/beta + (v^2c^2) x.

3) Solve xi = beta(x-vt) for (vt):
vt = x - xi/beta.

4) Equate the right-hand sides of (2) and (3):
(v tau)/beta + (v^2/c^2) x = x - xi/beta.

5) Solve for x:
(v tau)/beta + xi/beta = x - (v^2/c^2) x
= (1 - v^2/c^2) x
= x / beta^2

6) Multiply both sides by beta^2:
beta(v tau + xi) = x
which is half of the inverse transformation. Solve for t in terms of xi
and tau similarly, by solving the equations for xi(x,t) and tau(x,t) for
x, equating the two expressions for x, and solving for t. The result is
t = beta (tau + v xi/c^2).
And if you substitute this expression for t into x = xi/beta + vt, you
get x as a function of xi and tau:
x = (xi/beta) + v beta (tau + v xi/c^2)
= (xi/beta) + v beta tau + beta v^2 xi/c^2
= beta (v tau) + xi (1/beta + beta v^2/c^2)
= beta (v tau) + beta xi (1/beta^2 + v^2/c^2)
= beta (v tau) + beta xi ((1 - v^2/c^2) + v^2/c^2)
= beta (v tau) + beta xi (1)
x = beta (xi + v tau).
The math really does work, and the inverse expressions are as claimed.


Perhaps an analogy to a rotation of the coordinate axes in plain old
Cartesian analytic geometry may help. If you rotate the x- and y-axes
about the z axis by some angle theta, the new {x',y',z'} are related to
the old {x,y,z} by
x' = x cos theta + y sin theta
y' = -x sin theta + y cos theta
z' = z
and the old are related to the new by
x = x' cos theta - y' sin theta
y = x' sin theta + y' cos theta
z = z'.
Note how cos theta multiplies x on the RHS of the equation for x', AND
it multiplies x' on the RHS of the equation for x. Something similar is
happening here.

-- Bill Owen

P.S. John, you're welcome.

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway wrote:
"William Hamblen" wrote in message
m...

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway
wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
============================================== =
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.


Hint: where did the t go and where did the tau come from in the inverse
transform?

Bud
==========================================
Hint:
I want an answer from Quadiblockhead, not a ****ing hint from a ****head.
Hint:
Answer the question or shut the **** up.

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

QED

Good job, Androcles.

Bud

On Jan 10, 6:32*pm, Bill Owen wrote:

P.S. *John, you're welcome.

Since I reply to Oriel, I won't criticize you for responding to such a
one. I do thank you for trying... but he is impervious to rational
argument, which, of course, shows, even more definitively than the
mere fact of his being in the minority, being opposed to the
recognized consensus of the scientific community does, that he's the
crank, though he uses rhetorical devices to try to shift the onus of
crank-hood elsewhere.

John Savard

On Jan 10, 7:13*pm, William Hamblen
wrote:

Good job, Androcles.

While I have broken down and responded to his posts on a few
occasions, you can see why, unlike Oriel, I regard him as beneath
contempt.

Back when I took science, one of the first things they taught us was
that you can't just plug physics equations together like building
blocks. Even if you obey the rules of algebra perfectly, that doesn't
help if the "v" in one equation doesn't stand for the same thing as
the "v" in the other equation you're trying to substitute in.

Thus, E = mc^2 does not mean you can derive the power consumption of
New York City in a year from the mass of sturgeon roe produced in the
Caspian Sea by multiplying it by the speed of light squared. Despite
one being an energy and the other a mass.

John Savard

"Lord Androcles, Zeroth Earl of Medway" wrote in message
...

"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
===============================================
Not only is Savard hopeless at simple algebra, he quotes the drool of some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

"William Hamblen" wrote in message
m...

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway
wrote:
"William Hamblen" wrote in message
m...

On 2013-01-10, Lord Androcles, Zeroth Earl of Medway
wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
============================================== =
Not only is Savard hopeless at simple algebra, he quotes the drool of
some
unnamed moron who is equally hopeless.
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.


Hint: where did the t go and where did the tau come from in the inverse
transform?

Bud
==========================================
Hint:
I want an answer from Quadiblockhead, not a ****ing hint from a ****head.
Hint:
Answer the question or shut the **** up.

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

QED

Good job, Androcles.

Bud
========================================
Thanks. As you can see, the contemptible Savard behaves just as Kelleher
does, giving no explanation for his irrational beliefs.
Surely it is not to difficult to explain why he multiplies instead of
dividing?
Does he really believe that if half of 5-1 is 2, then half of 2+1 is 5?

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

"Bill Owen" wrote in message ...

Lord Androcles, Zeroth Earl of Medway wrote:
"Quadibloc" wrote in message
...
(begin quote)
At the end of Section 3 we find the transformation derived:

tau=beta(t-vx/c^2),
xi=beta(x-vt),
eta=y,
zeta=z,
where beta=1/sqrt(1-v^2/c^2).

With trivial algebraic manipulation we can derive the inverse
transformation:

t=beta(tau+v(xi)/c^2),
x=beta(xi+v(tau)),
y=eta,
z=zeta.
(end quote)
[snip]
Perhaps he can show, step-by-step, his trivial derivation, like this:
xi = beta(x-vt)
Divide both sides of the equation by beta
xi/beta = beta(x-vt)/beta
Since beta/beta = 1,
xi/beta = 1*(x-vt)
Add vt to both sides of the equation
xi/beta +vt = (x-vt)+vt
Since vt - vt = 0,
x = xi/beta +vt

Why is Savard multiplying xi by beta instead of dividing?

Because the expression on the right-hand side uses tau, not t. It's
certainly true that x = xi/beta + vt.

Here's the trivial derivation.

1) Solve tau = beta(t-vx/c^2) for t:
t = (tau/beta) + vx/c^2.

2) Multiply both sides by v:
vt = (v tau)/beta + (v^2c^2) x.

3) Solve xi = beta(x-vt) for (vt):
vt = x - xi/beta.

4) Equate the right-hand sides of (2) and (3):
(v tau)/beta + (v^2/c^2) x = x - xi/beta.

5) Solve for x:
(v tau)/beta + xi/beta = x - (v^2/c^2) x
= (1 - v^2/c^2) x
= x / beta^2

6) Multiply both sides by beta^2:
beta(v tau + xi) = x
which is half of the inverse transformation. Solve for t in terms of xi
and tau similarly, by solving the equations for xi(x,t) and tau(x,t) for
x, equating the two expressions for x, and solving for t. The result is
t = beta (tau + v xi/c^2).
================================================== ==============
v = x/t, so
t = beta (tau + x/t xi/c^2).

Incomplete, you still have 1/t on the RHS, Compo.


v = dx/dt (which can be integrated to x plus a constant) and since v is a
constant too we have v = x/t anywhere on the x-axis.
This the velocity of the train/ship/plane/car in the Earth's Euclidean
coordinate system.

Unfortunately for you the velocity of Earth anywhere on the xi-axis of the
vehicle's coordinate system is upsilon = dxi/dtau = xi/tau and upsilon does
not equal v.

To demonstrate, let v = 0.866c so that beta = 1/sqrt(1-0.866^2) = 2
Choosing x = 0.866, we have t = 1 for simplicity.
Hence tau = t * sqrt(1-0.866^2) = 0.5

xi = beta(0.866 - 0.866*1) = 0, the Earth is at the origin of the vehicle's
coordinate system at tau = 0.5, so it must have been at beta(0-vt) at tau =
0 and travelled a distance dxi = 0-upsilon.tau in time dtau = 0.5-0 with
velocity upsilon = xi/tau = 1.732/0.5 = 3.464c

Perhaps you can explain how the Earth exceeds the speed of light in the
frame of the vehicle, but I'm certainly not going to accept your's or
Savard's ASSUMPTION that upsilon = v.


-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

P.S. Kelleher, you're welcome.