let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D). let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD be two
points of a vertical line that represents L0, so that M is north of FD. let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according to S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long, straight
track. the train moves at constant speed v relative to S. let S' carry an
identical timing device, consisting of FD and M, so that M - FD = L0. let M
and FD be two points of a vertical line (a line that is perpendicular to the
length of the track) that represents L0, so that M is north of FD. in other
words, L0 is represented by a line (inside the train) that extends across
the train . let the bulb emit a ray of light that travels north to M from F.
and when the reflected light returns to D, the clock ticks and another flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the clock is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by S' on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal base of
ABC, so that A and C are the base angles of ABC. let C be east of A . let B
be the angle opposite AC. and let B be north of AC. let P be the midpoint of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the train). a
light signal (at that instant) alerts S that the ray of light is reflected
at B.

the ray of light is detected at C (at the detector D inside the train). and
a light signal (at that instant) alerts S that the ray of light is detected
at D.

according to S, delta t is the time interval between the light signal at A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC = AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta t =
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or equal
to 1, thus delta t is greater or equal to delta t0. that is, S measures a
greater interval between ticks. the effect is called time dilation. the time
interval delta t0, measured by S', who is at rest relative to the clock
inside the train is called the proper time.

francisco wrote:
let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D). let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD be two
points of a vertical line that represents L0, so that M is north of FD. let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according to S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long, straight
track. the train moves at constant speed v relative to S. let S' carry an
identical timing device, consisting of FD and M, so that M - FD = L0. let M
and FD be two points of a vertical line (a line that is perpendicular to the
length of the track) that represents L0, so that M is north of FD. in other
words, L0 is represented by a line (inside the train) that extends across
the train . let the bulb emit a ray of light that travels north to M from F.
and when the reflected light returns to D, the clock ticks and another flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the clock is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by S' on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal base of
ABC, so that A and C are the base angles of ABC. let C be east of A . let B
be the angle opposite AC. and let B be north of AC. let P be the midpoint of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the train). a
light signal (at that instant) alerts S that the ray of light is reflected
at B.

the ray of light is detected at C (at the detector D inside the train). and
a light signal (at that instant) alerts S that the ray of light is detected
at D.

according to S, delta t is the time interval between the light signal at A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC = AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta t =
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or equal
to 1, thus delta t is greater or equal to delta t0. that is, S measures a
greater interval between ticks. the effect is called time dilation. the time
interval delta t0, measured by S', who is at rest relative to the clock
inside the train is called the proper time.

The result depends on the angle of the light pulse to the direction
of the two observers relative motion. The choice of the light clock and
its orientation arbitrarily selects a particular direction which gives
the correct result. We do not prove that the square of the longest side
of any triangle is equal to the sum of the squares of the other two
sides by drawing a right angled triangle, proving that it works in that
case and then generalising. That is mathematically unsound and it is
mathematically unsound to pick a particular direction for the light
pulse to travel in when the result depends on the direction. In short,
it is a mathematical fiddle.
http://users.powernet.co.uk/bearsoft/LtClk.html

Sue...

"Sue..." wrote in message
ups.com...
|
| francisco wrote:
| let observer S be at rest on the ground. let S carry a timing
device,
| consisting of FD (a flashing lightbulb F attached to a detector D).
let a
| mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD
be two
| points of a vertical line that represents L0, so that M is north of
FD. let
| the bulb emit a ray of light that travels north to M from F. when
the
| reflected light returns to D, the clock ticks and another flash is
| triggered. let delta t0 be the time interval between ticks according
to S,
| so that delta t0 = 2L0/c.
|
| let observer S' be in a train moving horizontally east on a long,
straight
| track. the train moves at constant speed v relative to S. let S'
carry an
| identical timing device, consisting of FD and M, so that M - FD =
L0. let M
| and FD be two points of a vertical line (a line that is
perpendicular to the
| length of the track) that represents L0, so that M is north of FD.
in other
| words, L0 is represented by a line (inside the train) that extends
across
| the train . let the bulb emit a ray of light that travels north to M
from F.
| and when the reflected light returns to D, the clock ticks and
another flash
| is triggered. again, let delta t0 be the time interval between ticks
| according to S', so that delta t0 = 2L0/c.
|
| the time interval delta t0 is observed by either S or S' when the
clock is
| at rest with respect to that observer.
|
| let's consider now the situation when S looks at the clock carried
by S' on
| the moving train. .
|
| imagine that ABC is an isosceles triangle. let AC be the horizontal
base of
| ABC, so that A and C are the base angles of ABC. let C be east of A
.. let B
| be the angle opposite AC. and let B be north of AC. let P be the
midpoint of
| AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.
|
| let the train move east, parallel to AC
|
| according to S, the following takes place:
|
| the bulb in the train emits a ray of light at A. a light signal (at
that
| instant) alerts S that the bulb has emitted a ray of light at A.
|
| the ray of light is reflected at B (from the mirror M inside the
train). a
| light signal (at that instant) alerts S that the ray of light is
reflected
| at B.
|
| the ray of light is detected at C (at the detector D inside the
train). and
| a light signal (at that instant) alerts S that the ray of light is
detected
| at D.
|
| according to S, delta t is the time interval between the light
signal at A
| and the light signal at C, and
|
| AC = v*(delta t).
|
| also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and
PC = AC/2
| = v*(delta t)/2
|
| so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.
|
| S observes that the light beam travels a distance of 2L, so that
delta t =
| 2L/c
|
| so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c
|
| substituting for L0 from delta t0 = 2L0/c and solving delta t =
| 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the
following:
|
| delta t = (delta t0)/SQRT(1 - v^2/c^2)
|
| the factor in the denominator of this equation is always less than
or equal
| to 1, thus delta t is greater or equal to delta t0. that is, S
measures a
| greater interval between ticks. the effect is called time dilation.
the time
| interval delta t0, measured by S', who is at rest relative to the
clock
| inside the train is called the proper time.
|
| The result depends on the angle of the light pulse to the direction
| of the two observers relative motion. The choice of the light clock
and
| its orientation arbitrarily selects a particular direction which gives
| the correct result. We do not prove that the square of the longest
side
| of any triangle is equal to the sum of the squares of the other two
| sides by drawing a right angled triangle, proving that it works in
that
| case and then generalising. That is mathematically unsound and it is
| mathematically unsound to pick a particular direction for the light
| pulse to travel in when the result depends on the direction. In short,
| it is a mathematical fiddle.
| http://users.powernet.co.uk/bearsoft/LtClk.html
|
| Sue...
LOL! Right on! Stumble over the flashlight, kicking it though
arctan(v/c) and change time!
Androcles.

concerning the following site:

http://users.powernet.co.uk/bearsoft/LtClk.html

Typical Anti-Relativty garbage, not worth your time.
__________________
Janus

The whole problem with the world is that fools and fanatics are always so
certain of themselves, and wiser people so full of doubts.--Bertrand Russell

"Sue..." wrote in message
ups.com...

francisco wrote:
let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D). let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD be
two
points of a vertical line that represents L0, so that M is north of FD.
let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according to
S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long,
straight
track. the train moves at constant speed v relative to S. let S' carry an
identical timing device, consisting of FD and M, so that M - FD = L0. let
M
and FD be two points of a vertical line (a line that is perpendicular to
the
length of the track) that represents L0, so that M is north of FD. in
other
words, L0 is represented by a line (inside the train) that extends across
the train . let the bulb emit a ray of light that travels north to M from
F.
and when the reflected light returns to D, the clock ticks and another
flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the clock
is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by S'
on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal base
of
ABC, so that A and C are the base angles of ABC. let C be east of A . let
B
be the angle opposite AC. and let B be north of AC. let P be the midpoint
of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the train).
a
light signal (at that instant) alerts S that the ray of light is
reflected
at B.

the ray of light is detected at C (at the detector D inside the train).
and
a light signal (at that instant) alerts S that the ray of light is
detected
at D.

according to S, delta t is the time interval between the light signal at
A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC =
AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta t
=
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or
equal
to 1, thus delta t is greater or equal to delta t0. that is, S measures a
greater interval between ticks. the effect is called time dilation. the
time
interval delta t0, measured by S', who is at rest relative to the clock
inside the train is called the proper time.

The result depends on the angle of the light pulse to the direction
of the two observers relative motion. The choice of the light clock and
its orientation arbitrarily selects a particular direction which gives
the correct result. We do not prove that the square of the longest side
of any triangle is equal to the sum of the squares of the other two
sides by drawing a right angled triangle, proving that it works in that
case and then generalising. That is mathematically unsound and it is
mathematically unsound to pick a particular direction for the light
pulse to travel in when the result depends on the direction. In short,
it is a mathematical fiddle.
http://users.powernet.co.uk/bearsoft/LtClk.html

Sue...

francisco wrote:
concerning the following site:

http://users.powernet.co.uk/bearsoft/LtClk.html

Typical Anti-Relativty garbage, not worth your time.

Then you should have no problem addressing the issues it
raises with logical argument instead of disparaging remarks.

Your opinion of the website does explain how a relative
moving observer can alter the path between the mirrors
to permit the appliction of special propagation calculations.

Rain doesn't follow a longer path, nor take a longer period
of time to reach the ground, simply because you view it
from a moving vehicle.

Sue...

__________________
Janus

The whole problem with the world is that fools and fanatics are always so
certain of themselves, and wiser people so full of doubts.--Bertrand Russell

"Sue..." wrote in message
ups.com...

francisco wrote:
let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D). let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD be
two
points of a vertical line that represents L0, so that M is north of FD.
let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according to
S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long,
straight
track. the train moves at constant speed v relative to S. let S' carry an
identical timing device, consisting of FD and M, so that M - FD = L0. let
M
and FD be two points of a vertical line (a line that is perpendicular to
the
length of the track) that represents L0, so that M is north of FD. in
other
words, L0 is represented by a line (inside the train) that extends across
the train . let the bulb emit a ray of light that travels north to M from
F.
and when the reflected light returns to D, the clock ticks and another
flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the clock
is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by S'
on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal base
of
ABC, so that A and C are the base angles of ABC. let C be east of A . let
B
be the angle opposite AC. and let B be north of AC. let P be the midpoint
of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the train).
a
light signal (at that instant) alerts S that the ray of light is
reflected
at B.

the ray of light is detected at C (at the detector D inside the train).
and
a light signal (at that instant) alerts S that the ray of light is
detected
at D.

according to S, delta t is the time interval between the light signal at
A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC =
AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta t
=
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or
equal
to 1, thus delta t is greater or equal to delta t0. that is, S measures a
greater interval between ticks. the effect is called time dilation. the
time
interval delta t0, measured by S', who is at rest relative to the clock
inside the train is called the proper time.

The result depends on the angle of the light pulse to the direction
of the two observers relative motion. The choice of the light clock and
its orientation arbitrarily selects a particular direction which gives
the correct result. We do not prove that the square of the longest side
of any triangle is equal to the sum of the squares of the other two
sides by drawing a right angled triangle, proving that it works in that
case and then generalising. That is mathematically unsound and it is
mathematically unsound to pick a particular direction for the light
pulse to travel in when the result depends on the direction. In short,
it is a mathematical fiddle.
http://users.powernet.co.uk/bearsoft/LtClk.html

Sue...

a scientific theory usually begins with general statements called
postulates, which attempts to provide a basis for the theory. from these
postulates we can obtain a set of mathematical laws in the form of equations
that relate physical variables. finally, we test the predictions of the
equations in the laboratory. the theory stands until contradicted by
experiment, after which the postulates may be modified or replaced, and the
cycle is repeated.

the critical test of any theory is of course how well it agrees with
experiment. einstein's special theory of relativity has been subjected to
exhaustive tests over the past 98 years and has passed every one. where
classical physics and relativity predict different results, experiment has
been found to agree with relativity theory.

do you know of an experiment that does not agree with relativity theory,
that proves that relativity theory is wrong?

the author of the following site:
http://users.powernet.co.uk/bearsoft/LtClk.html,
has not been able to prove with experimet that relativity theory is wrong.
he has not been able to prove with experimet that his opinion is correct.
thus, it is a waist of time to speculate with him.



"Sue..." wrote in message
ups.com...

francisco wrote:
concerning the following site:

http://users.powernet.co.uk/bearsoft/LtClk.html

Typical Anti-Relativty garbage, not worth your time.

Then you should have no problem addressing the issues it
raises with logical argument instead of disparaging remarks.

Your opinion of the website does explain how a relative
moving observer can alter the path between the mirrors
to permit the appliction of special propagation calculations.

Rain doesn't follow a longer path, nor take a longer period
of time to reach the ground, simply because you view it
from a moving vehicle.

Sue...

__________________
Janus

The whole problem with the world is that fools and fanatics are always so
certain of themselves, and wiser people so full of doubts.--Bertrand
Russell

"Sue..." wrote in message
ups.com...

francisco wrote:
let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D).
let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD
be
two
points of a vertical line that represents L0, so that M is north of
FD.
let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according
to
S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long,
straight
track. the train moves at constant speed v relative to S. let S' carry
an
identical timing device, consisting of FD and M, so that M - FD = L0.
let
M
and FD be two points of a vertical line (a line that is perpendicular
to
the
length of the track) that represents L0, so that M is north of FD. in
other
words, L0 is represented by a line (inside the train) that extends
across
the train . let the bulb emit a ray of light that travels north to M
from
F.
and when the reflected light returns to D, the clock ticks and another
flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the
clock
is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by
S'
on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal
base
of
ABC, so that A and C are the base angles of ABC. let C be east of A .
let
B
be the angle opposite AC. and let B be north of AC. let P be the
midpoint
of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at
that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the
train).
a
light signal (at that instant) alerts S that the ray of light is
reflected
at B.

the ray of light is detected at C (at the detector D inside the
train).
and
a light signal (at that instant) alerts S that the ray of light is
detected
at D.

according to S, delta t is the time interval between the light signal
at
A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC
=
AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta
t
=
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or
equal
to 1, thus delta t is greater or equal to delta t0. that is, S
measures a
greater interval between ticks. the effect is called time dilation.
the
time
interval delta t0, measured by S', who is at rest relative to the
clock
inside the train is called the proper time.

The result depends on the angle of the light pulse to the direction
of the two observers relative motion. The choice of the light clock and
its orientation arbitrarily selects a particular direction which gives
the correct result. We do not prove that the square of the longest side
of any triangle is equal to the sum of the squares of the other two
sides by drawing a right angled triangle, proving that it works in that
case and then generalising. That is mathematically unsound and it is
mathematically unsound to pick a particular direction for the light
pulse to travel in when the result depends on the direction. In short,
it is a mathematical fiddle.
http://users.powernet.co.uk/bearsoft/LtClk.html

Sue...

francisco wrote:
a scientific theory usually begins with general statements called
postulates, which attempts to provide a basis for the theory. from these
postulates we can obtain a set of mathematical laws in the form of equations
that relate physical variables. finally, we test the predictions of the
equations in the laboratory. the theory stands until contradicted by
experiment, after which the postulates may be modified or replaced, and the
cycle is repeated.

So in this case you can offer some proof that moving observers alter
a path which they do not interact with?


the critical test of any theory is of course how well it agrees with
experiment. einstein's special theory of relativity has been subjected to
exhaustive tests over the past 98 years and has passed every one.
classical physics and relativity predict different results, experiment has
been found to agree with relativity theory.

do you know of an experiment that does not agree with relativity theory,
that proves that relativity theory is wrong?

Sure,
H&K and GPS disproves *time dilation*.
Pound Rebeka and GPS disproves *gravitational redshift*.


the author of the following site:
http://users.powernet.co.uk/bearsoft/LtClk.html,
has not been able to prove with experimet that relativity theory is wrong.
he has not been able to prove with experimet that his opinion is correct.
thus, it is a waist of time to speculate with him.

I don't know that he is even an experimentalist. His
description of the light clock absurbity is better that
most so I excerped for that purpose. Experiments are
not necessary to prove or disprove mathematical absurbdities.
They stand failed as conceived and experimentally impossible.


Elements of a scientific method
[...]
3. Prediction (logical deduction from the hypothesis)
http://en.wikipedia.org/wiki/Scientific_method

Sue...





"Sue..." wrote in message
ups.com...

francisco wrote:
concerning the following site:

http://users.powernet.co.uk/bearsoft/LtClk.html

Typical Anti-Relativty garbage, not worth your time.

Then you should have no problem addressing the issues it
raises with logical argument instead of disparaging remarks.

Your opinion of the website does explain how a relative
moving observer can alter the path between the mirrors
to permit the appliction of special propagation calculations.

Rain doesn't follow a longer path, nor take a longer period
of time to reach the ground, simply because you view it
from a moving vehicle.

Sue...

__________________
Janus

The whole problem with the world is that fools and fanatics are always so
certain of themselves, and wiser people so full of doubts.--Bertrand
Russell

"Sue..." wrote in message
ups.com...

francisco wrote:
let observer S be at rest on the ground. let S carry a timing device,
consisting of FD (a flashing lightbulb F attached to a detector D).
let a
mirror M be a distance L0 from FD, so that M - FD = L0. let M and FD
be
two
points of a vertical line that represents L0, so that M is north of
FD.
let
the bulb emit a ray of light that travels north to M from F. when the
reflected light returns to D, the clock ticks and another flash is
triggered. let delta t0 be the time interval between ticks according
to
S,
so that delta t0 = 2L0/c.

let observer S' be in a train moving horizontally east on a long,
straight
track. the train moves at constant speed v relative to S. let S' carry
an
identical timing device, consisting of FD and M, so that M - FD = L0.
let
M
and FD be two points of a vertical line (a line that is perpendicular
to
the
length of the track) that represents L0, so that M is north of FD. in
other
words, L0 is represented by a line (inside the train) that extends
across
the train . let the bulb emit a ray of light that travels north to M
from
F.
and when the reflected light returns to D, the clock ticks and another
flash
is triggered. again, let delta t0 be the time interval between ticks
according to S', so that delta t0 = 2L0/c.

the time interval delta t0 is observed by either S or S' when the
clock
is
at rest with respect to that observer.

let's consider now the situation when S looks at the clock carried by
S'
on
the moving train. .

imagine that ABC is an isosceles triangle. let AC be the horizontal
base
of
ABC, so that A and C are the base angles of ABC. let C be east of A .
let
B
be the angle opposite AC. and let B be north of AC. let P be the
midpoint
of
AC, so that BP is the perpendicular bisector of AC, and PC = AC/2.

let the train move east, parallel to AC

according to S, the following takes place:

the bulb in the train emits a ray of light at A. a light signal (at
that
instant) alerts S that the bulb has emitted a ray of light at A.

the ray of light is reflected at B (from the mirror M inside the
train).
a
light signal (at that instant) alerts S that the ray of light is
reflected
at B.

the ray of light is detected at C (at the detector D inside the
train).
and
a light signal (at that instant) alerts S that the ray of light is
detected
at D.

according to S, delta t is the time interval between the light signal
at
A
and the light signal at C, and

AC = v*(delta t).

also, according to S, L = SQRT[(BP)^2 + (PC)^2], where BP = L0 and PC
=
AC/2
= v*(delta t)/2

so, L = SQRT{(L0)^2 + [v* (delta t)/2]^2}.

S observes that the light beam travels a distance of 2L, so that delta
t
=
2L/c

so, delta t = 2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c

substituting for L0 from delta t0 = 2L0/c and solving delta t =
2*SQRT{(L0)^2 + [v* (delta t)/2]^2}/c for delta t gives the following:

delta t = (delta t0)/SQRT(1 - v^2/c^2)

the factor in the denominator of this equation is always less than or
equal
to 1, thus delta t is greater or equal to delta t0. that is, S
measures a
greater interval between ticks. the effect is called time dilation.
the
time
interval delta t0, measured by S', who is at rest relative to the
clock
inside the train is called the proper time.

The result depends on the angle of the light pulse to the direction
of the two observers relative motion. The choice of the light clock and
its orientation arbitrarily selects a particular direction which gives
the correct result. We do not prove that the square of the longest side
of any triangle is equal to the sum of the squares of the other two
sides by drawing a right angled triangle, proving that it works in that
case and then generalising. That is mathematically unsound and it is
mathematically unsound to pick a particular direction for the light
pulse to travel in when the result depends on the direction. In short,
it is a mathematical fiddle.
http://users.powernet.co.uk/bearsoft/LtClk.html

Sue...