Bill Owen -- Fourth Attempt

"Bill Owen" wrote in message ...

[snip]

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.
"Because the expression on the right-hand side uses tau, not t. " -- Bill
Owen, above.

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 yours or
Savard's ASSUMPTION that upsilon = v when length xxi and time t tau,
that way lies insanity.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the contemptible
arrogant *******.

-- 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.

Please be patient ... I didn't get a chance to check the newsgroup over
the weekend. More below.

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

[snip]

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.
"Because the expression on the right-hand side uses tau, not t. " -- Bill
Owen, above.

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 yours or
Savard's ASSUMPTION that upsilon = v when length xxi and time t tau,
that way lies insanity.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the contemptible
arrogant *******.

-- 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.

I'll have to think about this a bit more, but I *suspect* that you may
be confusing v -- the relative velocity of the two coordinate systems --
with the velocity of an object as measured in one or the other of the
coordinate systems. The latter is indeed dx/dt or dxi/dtau, but that's
not necessarily the same as v.

BTW, my earlier post was pure algebra, showing how one can take two
equations tau(x,t) and xi(x,t) and invert them to get t(xi,tau) and
x(xi,tau). That's straightforward, and the math holds regardless of the
meaning behind the symbols.

-- Bill

P.S. I see that I accidentally omitted a slash in step 2 of my earlier
post. Obviously the last term should have been (v^2/c^2) x.

"Bill Owen" wrote in message ...

Please be patient ... I didn't get a chance to check the newsgroup over
the weekend. More below.

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

[snip]

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.
"Because the expression on the right-hand side uses tau, not t. " -- Bill
Owen, above.

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 yours or
Savard's ASSUMPTION that upsilon = v when length xxi and time t tau,
that way lies insanity.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the
contemptible
arrogant *******.

-- 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.

I'll have to think about this a bit more, but I *suspect* that you may
be confusing v -- the relative velocity of the two coordinate systems --
with the velocity of an object as measured in one or the other of the
coordinate systems. The latter is indeed dx/dt or dxi/dtau, but that's
not necessarily the same as v.

BTW, my earlier post was pure algebra, showing how one can take two
equations tau(x,t) and xi(x,t) and invert them to get t(xi,tau) and
x(xi,tau). That's straightforward, and the math holds regardless of the
meaning behind the symbols.
================================================== =====

You'll have to think about this a bit more, but I *suspect* that you may be
confusing v, the velocity of a point travelling a distance from zero to x in
duration t with upsilon, the velocity of a point travelling a distance from
(omega - xi) to omega in duration tau, which is definitely not the same as
|v| in relativistic drivel but is the same in the real world.

zero(0)--------------------x velocity v = x/t
xi----------------omega(0) velocity upsilon = xi/tau


I refer you to
http://www.fourmilab.ch/etexts/einstein/specrel/www/
There is no tau(x,t) or xi(x,t) in Einstein's inequalities, perhaps you mean
tau(x',t) and xi(x',t) where x' = x-vt.
In particular, I refer to the inequality
http://www.fourmilab.ch/etexts/einst...ures/img22.gif
You'll have to think about this a bit more, without the "but".

-- 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

P.S. I see that I accidentally omitted a slash in step 2 of my earlier
post. Obviously the last term should have been (v^2/c^2) x.
============================================
Fair enough.

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

Please be patient ... I didn't get a chance to check the newsgroup over
the weekend. More below.

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

[snip]

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.
"Because the expression on the right-hand side uses tau, not t. " -- Bill
Owen, above.

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 yours or
Savard's ASSUMPTION that upsilon = v when length xxi and time t tau,
that way lies insanity.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the
contemptible
arrogant *******.

-- 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.

I'll have to think about this a bit more, but I *suspect* that you may
be confusing v -- the relative velocity of the two coordinate systems --
with the velocity of an object as measured in one or the other of the
coordinate systems. The latter is indeed dx/dt or dxi/dtau, but that's
not necessarily the same as v.

BTW, my earlier post was pure algebra, showing how one can take two
equations tau(x,t) and xi(x,t) and invert them to get t(xi,tau) and
x(xi,tau). That's straightforward, and the math holds regardless of the
meaning behind the symbols.
================================================== =====

You'll have to think about this a bit more, but I *suspect* that you may be
confusing v, the velocity of a point travelling a distance from zero to
x in
duration t with upsilon, the velocity of a point travelling a distance from
(omega - xi) to omega in duration tau, which is definitely not the same as
|v| in relativistic drivel but is the same in the real world.

zero(0)--------------------x velocity v = x/t
xi----------------omega(0) velocity upsilon = xi/tau


I refer you to
http://www.fourmilab.ch/etexts/einstein/specrel/www/
There is no tau(x,t) or xi(x,t) in Einstein's inequalities, perhaps you
mean
tau(x',t) and xi(x',t) where x' = x-vt.
In particular, I refer to the inequality
http://www.fourmilab.ch/etexts/einst...ures/img22.gif
You'll have to think about this a bit more, without the "but".

-- 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

P.S. I see that I accidentally omitted a slash in step 2 of my earlier
post. Obviously the last term should have been (v^2/c^2) x.
============================================
Fair enough.

Wholeheartedly agreed that I'll have to do some more thinking. It's
been over 20 years since I've done any coursework in relativity (either
special or general) and as an astrometrist I don't often need to use it.
For my JPL colleagues who do radio navigation it's an entirely
different story.

So again I will ask for your patience while I try to find the time to
get the dust off the textbooks, not to mention the neurons.

-- Bill

"Bill Owen" wrote in message ...

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

Please be patient ... I didn't get a chance to check the newsgroup over
the weekend. More below.

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

[snip]

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.
"Because the expression on the right-hand side uses tau, not t. " -- Bill
Owen, above.

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 yours or
Savard's ASSUMPTION that upsilon = v when length xxi and time t tau,
that way lies insanity.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the
contemptible
arrogant *******.

-- 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.

I'll have to think about this a bit more, but I *suspect* that you may
be confusing v -- the relative velocity of the two coordinate systems --
with the velocity of an object as measured in one or the other of the
coordinate systems. The latter is indeed dx/dt or dxi/dtau, but that's
not necessarily the same as v.

BTW, my earlier post was pure algebra, showing how one can take two
equations tau(x,t) and xi(x,t) and invert them to get t(xi,tau) and
x(xi,tau). That's straightforward, and the math holds regardless of the
meaning behind the symbols.
================================================== =====

You'll have to think about this a bit more, but I *suspect* that you may
be
confusing v, the velocity of a point travelling a distance from zero to x
in
duration t with upsilon, the velocity of a point travelling a distance
from
(omega - xi) to omega in duration tau, which is definitely not the same
as
|v| in relativistic drivel but is the same in the real world.

zero(0)--------------------x velocity v = x/t
xi----------------omega(0) velocity upsilon = xi/tau


I refer you to
http://www.fourmilab.ch/etexts/einstein/specrel/www/
There is no tau(x,t) or xi(x,t) in Einstein's inequalities, perhaps you
mean
tau(x',t) and xi(x',t) where x' = x-vt.
In particular, I refer to the inequality
http://www.fourmilab.ch/etexts/einst...ures/img22.gif
You'll have to think about this a bit more, without the "but".

-- 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

P.S. I see that I accidentally omitted a slash in step 2 of my earlier
post. Obviously the last term should have been (v^2/c^2) x.
============================================
Fair enough.

Wholeheartedly agreed that I'll have to do some more thinking. It's
been over 20 years since I've done any coursework in relativity (either
special or general) and as an astrometrist I don't often need to use it.
For my JPL colleagues who do radio navigation it's an entirely
different story.

So again I will ask for your patience while I try to find the time to
get the dust off the textbooks, not to mention the neurons.

-- Bill
================================================== ===
It's been over 40 years for me, Bill, and over twenty five years (1987)
since I discovered why stars vary in intensity, so I have a great deal of
patience for those with intelligence and an open mind but little patience
for bigots like Savard who think they know it all but can't do the math.
My mathematical model reproduces this curve as well as that of eclipsing
variables and cepheids, all of which have a common cause.
http://www.britastro.org/vss/gifc/00918-ck.gif

You could call it a follow up to Swan Leavitt's work.
The point is, if the star is a constant emitter and moves in an elliptical
orbit as per Kepler, and if space has no properties, then the star varies in
intensity with the period of its orbit.

-- 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 Jan 15, 3:01*am, "Lord Androcles, Zeroth Earl of Medway"
wrote:
I have a great deal of
patience for those with intelligence and an open mind but little patience
for bigots like Savard who think they know it all but can't do the math.

It's clear that _you_ don't have the right attitude, and thus it's not
really worth the effort of doing the math for your sake.

And, as proof of that, I posted a reply to a post of yours where
somebody *did* the math, and you just ignored it.

John Savard

"Quadibloc" wrote in message
...

On Jan 15, 3:01 am, "Lord Androcles, Zeroth Earl of Medway"
wrote:
I have a great deal of
patience for those with intelligence and an open mind but little patience
for bigots like Savard who think they know it all but can't do the math.

It's clear that _you_ don't have the right attitude

===========================================
Answer the question or **** off, Oriel Savard.

So why is Savard multiplying xi by beta instead of dividing?
Answer: To deliberately obfuscate and pretend he's slick, the
contemptible arrogant snipping *******.

-- 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.