jem wrote:

Another way to get the result is to note that the speed of the light
pulse (measured in the accelerated coordinate system) is e^t (c=1, g~1),
so after N secs the distance from the origin is approx. e^N - 1 light-secs.

Interestingly, e^N - 1 N for N 0 so the distance to the light pulse
after accelerating in its direction is greater than it would be with no
acceleration. It takes a hard acceleration in the opposite direction of
the light pulse in order to remain "close" to it.

Re. inuition - suppose the initial acceleration were instantaneous to v
= tanh N. After N secs. the light pulse would then be at a distance of
N light secs, but since the specified acceleration rate is less than
that, the distance to the pulse after N secs must be greater than N
light secs (i.e. approaches infinity for increasing N).

Correction: in units where c=1, g~1/year (not 1/sec), so in the first
paragraph above replace 'secs' and light-secs' with 'years' and
'light-years'.

In message ,
writes
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

"It is important to emphasise that special relativity purports to
describe certain physical phenomena only relative to (or from the point
of view of) inertial reference systems, and the speed of a clock
relative to one of these systems determines its timekeeping behaviour
(G. Builder, 1958)."

My understanding is that present dogma says that you cannot apply SR
unless you define an inertial frame to which all measurements can be
referred. Then it is only measurements relative to that inertial frame
which are relevant. If at to=0 there was an observer stationary w.r.t
the craft (before it accelerated) you can define how far away the pulse
is from that observer. It is not until the acceleration stops and the
moving observer becomes an inertial observer that you can treat it as a
mathematical problem covered by relativity. The history involving
acceleration is junked. You have two inertial FoR and you know how far
the pulse of light is in one and can work out where it is in the other.

But then I could be wrong. Bilge says I have no understanding of
relativity.

Nailing relativity is like trying to nail fog.
--
John Kennaugh
to email convert the number from hex to decimal

"John Kennaugh" wrote in message .uk...
In message ,
writes
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

"It is important to emphasise that special relativity purports to
describe certain physical phenomena only relative to (or from the point
of view of) inertial reference systems, and the speed of a clock
relative to one of these systems determines its timekeeping behaviour
(G. Builder, 1958)."

My understanding is that present dogma says that you cannot apply SR
unless you define an inertial frame to which all measurements can be
referred. Then it is only measurements relative to that inertial frame
which are relevant. If at to=0 there was an observer stationary w.r.t
the craft (before it accelerated) you can define how far away the pulse
is from that observer. It is not until the acceleration stops and the
moving observer becomes an inertial observer that you can treat it as a
mathematical problem covered by relativity. The history involving
acceleration is junked. You have two inertial FoR and you know how far
the pulse of light is in one and can work out where it is in the other.

But then I could be wrong. Bilge says I have no understanding of
relativity.

He is right.


Nailing relativity is like trying to nail fog.

Nailing idiots is like nailing nails.

Dirk Vdm

"John Kennaugh" wrote in message
.uk...
In message ,
writes
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

"It is important to emphasise that special relativity purports to
describe certain physical phenomena only relative to (or from the point
of view of) inertial reference systems, and the speed of a clock
relative to one of these systems determines its timekeeping behaviour
(G. Builder, 1958)."

My understanding is that present dogma says that you cannot apply SR
unless you define an inertial frame to which all measurements can be
referred. Then it is only measurements relative to that inertial frame
which are relevant. If at to=0 there was an observer stationary w.r.t
the craft (before it accelerated) you can define how far away the pulse
is from that observer. It is not until the acceleration stops and the
moving observer becomes an inertial observer that you can treat it as a
mathematical problem covered by relativity. The history involving
acceleration is junked. You have two inertial FoR and you know how far
the pulse of light is in one and can work out where it is in the other.

But then I could be wrong. Bilge says I have no understanding of
relativity.

Nailing relativity is like trying to nail fog.
--
John Kennaugh
to email convert the number from hex to decimal

Hi John:
Before I give you the "formula" I'll write a few words to maybe straighten
out some of your SR misconceptions.
Yes, the Lorentz transformation equations apply only to transformations
between inertial coordinate frames. Therefore the frame-fixed observers are
inertial. But these observers can observe (measure and record) the motions
of ALL objects, inertial and accelerated, that are moving with respect to
their inertial frame. At each event of an accelerating object there is an
instantaneous comoving inertial frame.

In the problem you pose, when the acceleration, g, ends at time t (by the
spacecraft's clock), that event's comoving frame allows you to make the
transformation from the initial frame (actually you need to do some algebra
to find an intersection between two lines first, then apply the contraction
function, and not just plug values into the standard Lorentz equations).

At that time, t, the spacecraft speed is V = c*tanh(gt/c). In the initial
frame the acceleration ends at
T = (c/g)*sinh(gt/c) and X = (c^2/g)*[cosh(gt/c)-1]. At that instant the
distance (as measured by the comoving frame's observer, and therefore, the
spacecraft's observer) from the spacecraft to the light-pulse is delta-x =
(Vc/g)*sqrt[(1+V/c)/(1-V/c)] (provided I haven't made some algebraic error).

Keep at it
Eli Botkin

"Eli Botkin" wrote in message
news:1095816613.Od8QGXcw1pPHBDAQwR4O9g@teranews...
|
| "John Kennaugh" wrote in message
| .uk...
| In message ,
| writes
| Would someone please be kind enough to post the formula for the
| following problem? A space craft is on the x-axis. At time t0=0 as
| measured by the craft's clock, the craft fires a laser in the
| x-direction, and simultaneously starts accelerating along the x-axis.
| The accelerometer on the craft reads g. The acceleration stops at
| time t1= N seconds as measured by an ideal clock on board the craft.
| How far away is the pulse of light from this craft (as measured by
| observers on the craft) when the acceleration stops?
|
| "It is important to emphasise that special relativity purports to
| describe certain physical phenomena only relative to (or from the point
| of view of) inertial reference systems, and the speed of a clock
| relative to one of these systems determines its timekeeping behaviour
| (G. Builder, 1958)."
|
| My understanding is that present dogma says that you cannot apply SR
| unless you define an inertial frame to which all measurements can be
| referred. Then it is only measurements relative to that inertial frame
| which are relevant. If at to=0 there was an observer stationary w.r.t
| the craft (before it accelerated) you can define how far away the pulse
| is from that observer. It is not until the acceleration stops and the
| moving observer becomes an inertial observer that you can treat it as a
| mathematical problem covered by relativity. The history involving
| acceleration is junked. You have two inertial FoR and you know how far
| the pulse of light is in one and can work out where it is in the other.
|
| But then I could be wrong. Bilge says I have no understanding of
| relativity.
|
| Nailing relativity is like trying to nail fog.
| --
| John Kennaugh
| to email convert the number from hex to decimal
|
| Hi John:
| Before I give you the "formula" I'll write a few words to maybe straighten
| out some of your SR misconceptions.
| Yes, the Lorentz transformation equations apply only to transformations
| between inertial coordinate frames. Therefore the frame-fixed observers
are
| inertial. But these observers can observe (measure and record) the
motions
| of ALL objects, inertial and accelerated, that are moving with respect to
| their inertial frame. At each event of an accelerating object there is an
| instantaneous comoving inertial frame.
|
| In the problem you pose, when the acceleration, g, ends at time t (by the
| spacecraft's clock), that event's comoving frame allows you to make the
| transformation from the initial frame (actually you need to do some
algebra

Oh... 'Do' some algebra. Wow!
Is that rigorous algebra, or just some sloppy old algebra?

For quotations following, reference:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
("On the Electrodynamics of Moving Bodies" by Albert Einstein)

1) "light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body",
a totally unproven assumption without any evidence to support it.

2) "In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c to be a universal constant- the velocity of light in empty
space.",
an admitted assumption that is quite worthless when there is any
relative motion between A and B, yet essential to the derivation of the
remainder of Einstein's nonsense.

3) The equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) ,
the ½ of which is derived from 2) above and is tantamount to saying
(1/3 + 2/3)/2 = 1/3.

4) The missing 0' from that equation, since x' = x-vt, hence 0' = 0-vt,
and the equation should be
½[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
at the very least.

5) The further assumption "IF we place x' = x-vt ... " without considering
IF we place x' = x+vt, from which we derive (using Einstein's method)
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

6) The statements
"But the ray moves relatively to the initial point of k,
when measured in the stationary system, with the velocity c-v..."
and
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we obtain
V = (c+w)/(1+w/c) = c."
which are contradictory, the first being Galilean, the second being
contrary to the vector addition of velocities, an axiom of a vector space.

7) The lack of a check to verify the theory is self-consistent by feeding
the new PoR given in 6) into the equation given in 3) and finding a total
failure.
Check:
(t1-t)/(t2-t)*[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/V+x'/V)] = tau(x',0,0,t+x'/V)

Androcles.




| to find an intersection between two lines first, then apply the
contraction
| function, and not just plug values into the standard Lorentz equations).
|
| At that time, t, the spacecraft speed is V = c*tanh(gt/c). In the
initial
| frame the acceleration ends at
| T = (c/g)*sinh(gt/c) and X = (c^2/g)*[cosh(gt/c)-1]. At that instant
the
| distance (as measured by the comoving frame's observer, and therefore, the
| spacecraft's observer) from the spacecraft to the light-pulse is delta-x =
| (Vc/g)*sqrt[(1+V/c)/(1-V/c)] (provided I haven't made some algebraic
error).
|
| Keep at it
| Eli Botkin
|
|

On Wed, 22 Sep 2004 04:44:42 GMT, "Androcles"
wrote:

[snip]

Oh... 'Do' some algebra. Wow!
Is that rigorous algebra, or just some sloppy old algebra?

http://users.pandora.be/vdmoortel/di...ythagoras.html

Which kind is that? We need something as a reference.

[snip]

"John Kennaugh" wrote in message
.uk...
In message ,
writes
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

"It is important to emphasise that special relativity purports to
describe certain physical phenomena only relative to (or from the point
of view of) inertial reference systems, and the speed of a clock
relative to one of these systems determines its timekeeping behaviour
(G. Builder, 1958)."

My understanding is that present dogma says that you cannot apply SR
unless you define an inertial frame to which all measurements can be
referred. Then it is only measurements relative to that inertial frame
which are relevant. If at to=0 there was an observer stationary w.r.t
the craft (before it accelerated) you can define how far away the pulse
is from that observer. It is not until the acceleration stops and the
moving observer becomes an inertial observer that you can treat it as a
mathematical problem covered by relativity. The history involving
acceleration is junked. You have two inertial FoR and you know how far
the pulse of light is in one and can work out where it is in the other.

But then I could be wrong. Bilge says I have no understanding of
relativity.

Nailing relativity is like trying to nail fog.
--
John Kennaugh
to email convert the number from hex to decimal

Hi John:
Before I give you the "formula" I'll write a few words to maybe straighten
out some of your SR misconceptions.
Yes, the Lorentz transformation equations apply only to transformations
between inertial coordinate frames. Therefore the frame-fixed observers are
inertial. But these observers can observe (measure and record) the motions
of ALL objects, inertial and accelerated, that are moving with respect to
their inertial frame. At each event of an accelerating object there is an
instantaneous comoving inertial frame.

In the problem you pose, when the acceleration, g, ends at time t (by the
spacecraft's clock), that event's comoving frame allows you to make the
transformation from the initial frame (actually you need to do some algebra
to find an intersection between two lines first, then apply the contraction
function, and not just plug values into the standard Lorentz equations).

At that time, t, the spacecraft speed is V = c*tanh(gt/c). In the initial
frame the acceleration ends at
T = (c/g)*sinh(gt/c) and X = (c^2/g)*[cosh(gt/c)-1]. At that instant the
distance (as measured by the comoving frame's observer, and therefore, the
spacecraft's observer) from the spacecraft to the light-pulse is delta-x =
(Vc/g)*sqrt[(1+V/c)/(1-V/c)] (provided I haven't made some algebraic error).

Keep at it
Eli Botkin

John:
Indeed, when I reviewed what I wrote, I noted that I left out a term, the
position of the spacecraft at acceleration cut-off at time t. Too hasty.

The term is Xo = (c^2/g)*[cosh(gt/c)-1].

So the distance from the spacecraft to the pulse becomes
delta-x = (V*c/g)*sqrt[(1+V/c)/(1-V/c)] - Xo*sqrt[1-(V/c)^2].

Another form, maybe more instructive, is
delta-x = {V*c/[g*(1-V/c)] - Xo}*sqrt[1-(V/c)^2].


" Hi John:
Before I give you the "formula" I'll write a few words to maybe straighten
out some of your SR misconceptions.
Yes, the Lorentz transformation equations apply only to transformations
between inertial coordinate frames. Therefore the frame-fixed observers
are
inertial. But these observers can observe (measure and record) the
motions
of ALL objects, inertial and accelerated, that are moving with respect to
their inertial frame. At each event of an accelerating object there is an
instantaneous comoving inertial frame.

In the problem you pose, when the acceleration, g, ends at time t (by the
spacecraft's clock), that event's comoving frame allows you to make the
transformation from the initial frame (actually you need to do some
algebra
to find an intersection between two lines first, then apply the
contraction
function, and not just plug values into the standard Lorentz equations).

At that time, t, the spacecraft speed is V = c*tanh(gt/c). In the
initial
frame the acceleration ends at
T = (c/g)*sinh(gt/c) and X = (c^2/g)*[cosh(gt/c)-1]. At that instant
the
distance (as measured by the comoving frame's observer, and therefore, the
spacecraft's observer) from the spacecraft to the light-pulse is delta-x =
(Vc/g)*sqrt[(1+V/c)/(1-V/c)] (provided I haven't made some algebraic
error).

Keep at it
Eli Botkin