 |
|
| Science Forum Index » Physics - Relativity Forum » MMX Null = Anisotropy = Ether... |
|
Page 2 of 3 Goto page Previous 1, 2, 3 Next |
|
| Author |
Message |
| Tom Roberts... |
 Posted: Mon Nov 02, 2009 1:56 am |
|
|
|
Guest
|
Peter Riedt wrote:
[quote]Miller did in 1926 the
interferometer
experiment at Mt Wilson. He predicted a fringe shift of 1.12 but got
0.088.
This is enough to prove c-v+c+v.
[/quote]
No, it isn't enough to "prove" anything, because the errorbar on that
result of Miller's is larger than 0.088. That means that his measurement
is consistent with no fringe shift at all, and also with a fringe shift
considerably larger than 0.088 -- that does not "prove" anything,
because his result is insufficient to distinguish between SR and the
aether model he used in his analysis.
A modern analysis of Miller's raw data gives a value of 0.000 +- 0.015
fringe, giving an upper bound (90% confidence) of 6 km/sec. This
analysis models his background MUCH more accurately than Miller's did,
and thus results in a considerably smaller errorbar.
See: http://www.arxiv.org/abs/physics/0608238
Tom Roberts |
|
|
| Back to top |
|
|
|
| Jerry... |
| Posted: Mon Nov 02, 2009 3:25 am |
|
|
|
Guest
|
On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
[quote]On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is 0.000000036670
seconds and downstream 0.000000036663 seconds for a total time of
0.000000073333 seconds which gives a speed of 299999997m/sec for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
[/quote]
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry |
|
|
| Back to top |
|
|
|
| doug... |
| Posted: Mon Nov 02, 2009 9:18 am |
|
|
|
Guest
|
Peter Riedt wrote:
[quote]On Nov 1, 11:30 am, doug <x... at (no spam) xx.com> wrote:
But, in any case, you notice that the average speed, even when you
just do the numbers to obscure the result show that the average
velocity is not c and thus there should have been an effect as
has been known for well over a century. Thus you are just showing
that you are a crank and do not even understand the results when
you take your numbers.
Doug, the difference from c is 3m/sec. Miller did in 1926 the
interferometer
experiment at Mt Wilson. He predicted a fringe shift of 1.12 but got
0.088.
This is enough to prove c-v+c+v.
Your two speed car formula is 2m/((1m/30miles/hr)+(1m/90miles/hr))
=45miles/hr.
My formula for MMX is 22m/((11m/(c-v))+(11m(c+v)))=299999997m/sec.
[/quote]
See, now you have changed your claim. In the part you snipped,
you claimed that it was (c+v+c-v)/2=c. Now you are saying something
totally different.
Even though you now are admitting you initial claim was wrong, you
continue on to demonstrate your ignorance of both error bars and
history. Nothing can be gotten from Miller's data now and that is
not important since other experiments have been done with accuracy
perhaps a billion times better and no signals are seen. You are
aware that science has continued to improve in the last century?
[quote]
Peter Riedt
[/quote] |
|
|
| Back to top |
|
|
|
| Peter Riedt... |
| Posted: Mon Nov 02, 2009 6:00 pm |
|
|
|
Guest
|
On Nov 2, 10:18 pm, doug <x... at (no spam) xx.com> wrote:
[quote]Peter Riedt wrote:
On Nov 1, 11:30 am, doug <x... at (no spam) xx.com> wrote:
But, in any case, you notice that the average speed, even when you
just do the numbers to obscure the result show that the average
velocity is not c and thus there should have been an effect as
has been known for well over a century. Thus you are just showing
that you are a crank and do not even understand the results when
you take your numbers.
Doug, the difference from c is 3m/sec. Miller did in 1926 the
interferometer
experiment at Mt Wilson. He predicted a fringe shift of 1.12 but got
0.088.
This is enough to prove c-v+c+v.
Your two speed car formula is 2m/((1m/30miles/hr)+(1m/90miles/hr))
=45miles/hr.
My formula for MMX is 22m/((11m/(c-v))+(11m(c+v)))=299999997m/sec.
See, now you have changed your claim. In the part you snipped,
you claimed that it was (c+v+c-v)/2=c. Now you are saying something
totally different.
Doug, it is not different. You changed it by adding distance. It still[/quote]
works
for c, v and d or just c and v. The relationships of distance, speed
and time
are immutable and have nothing to do with length contraction and time
dilation.
Peter Riedt |
|
|
| Back to top |
|
|
|
| Peter Riedt... |
| Posted: Mon Nov 02, 2009 6:01 pm |
|
|
|
Guest
|
On Nov 2, 9:25 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
[quote]On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is 0.000000036670
seconds and downstream 0.000000036663 seconds for a total time of
0.000000073333 seconds which gives a speed of 299999997m/sec for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry, using your values for c and v, the perpendicular[/quote]
light path distance according to Michelson&Morley
(American Journal of Science 203/1887) is 22.000000108725m
giving a perpendicular time of 0.000000073384101306sec. The
difference between the parallel and perpendicular times is
0.0000000000000003627sec which is half the difference you
calculated.
Peter Riedt |
|
|
| Back to top |
|
|
|
| Koobee Wublee... |
| Posted: Mon Nov 02, 2009 10:55 pm |
|
|
|
Guest
|
On Oct 29, 5:15 pm, doug <x... at (no spam) xx.com> wrote:
[quote]Peter Riedt wrote:
MMX Null = Anisotropy = Ether
In 1887 Michelson and Morley predicted an interferometer experiment
(MMX) would show interference of light due to the presence of the
ether. When their experiment produced a null result, an explanation
was required. Lorentz suggested an answer after others pointed the
way. The light path across the parallel and perpendicular arms of the
interferometer equipment had been assumed to be of unequal length due
to the movement of the apparatus through space. The difference in
light path length was equalised in Lorentz’s theory by a contraction
of the parallel arm proportional to the speed of the earth (v) on
which the lab was located. However, Lorentz was wrong. The null result
of the experiment (MMX) is caused by the anisotropy of light which in
turn is caused by the ether. The speed of light over the perpendicular
arm is c, unimpeded and unassisted by the ether. On the parallel arm,
the speed upstream is c-v, retarded by the ether. Downstream, it is c
+v, assisted by the ether. The null result is determined by the
formula
perpendicular c = parallel (c+v+c-v)/2 = null result.
And here you make an elementary math error...
[/quote]
It is amazing when these college dropout were first fed with bull$hit,
more bull$hit would come out of them. Mr. Riedt is quite correct.
Just consult with Professor Roberts. The null results actually
supported the emission theory of light as Mr. Riedt has brought up.
Unfortunately, the emission theory of light fails at
electromagnetism. <shrug> |
|
|
| Back to top |
|
|
|
| doug... |
| Posted: Mon Nov 02, 2009 11:31 pm |
|
|
|
Guest
|
Peter Riedt wrote:
[quote]On Nov 2, 10:18 pm, doug <x... at (no spam) xx.com> wrote:
Peter Riedt wrote:
On Nov 1, 11:30 am, doug <x... at (no spam) xx.com> wrote:
But, in any case, you notice that the average speed, even when you
just do the numbers to obscure the result show that the average
velocity is not c and thus there should have been an effect as
has been known for well over a century. Thus you are just showing
that you are a crank and do not even understand the results when
you take your numbers.
Doug, the difference from c is 3m/sec. Miller did in 1926 the
interferometer
experiment at Mt Wilson. He predicted a fringe shift of 1.12 but got
0.088.
This is enough to prove c-v+c+v.
Your two speed car formula is 2m/((1m/30miles/hr)+(1m/90miles/hr))
=45miles/hr.
My formula for MMX is 22m/((11m/(c-v))+(11m(c+v)))=299999997m/sec.
See, now you have changed your claim. In the part you snipped,
you claimed that it was (c+v+c-v)/2=c. Now you are saying something
totally different.
Doug, it is not different. You changed it by adding distance. It still
works
for c, v and d or just c and v. The relationships of distance, speed
and time
are immutable and have nothing to do with length contraction and time
dilation.
[/quote]
Apparently you have no clue what you are claiming. You say that average
speed is the average of the speeds (which is wrong in any case) and
then you work out a calculation showing that your claim is wrong. Do
you realize that you showed your claim to be wrong?
[quote]
Peter Riedt[/quote] |
|
|
| Back to top |
|
|
|
| Jerry... |
| Posted: Tue Nov 03, 2009 1:15 am |
|
|
|
Guest
|
On Nov 2, 10:01 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
[quote]On Nov 2, 9:25 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is 0.000000036670
seconds and downstream 0.000000036663 seconds for a total time of
0.000000073333 seconds which gives a speed of 299999997m/sec for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry, using your values for c and v, the perpendicular
light path distance according to Michelson&Morley
(American Journal of Science 203/1887) is 22.000000108725m
giving a perpendicular time of 0.000000073384101306sec. The
difference between the parallel and perpendicular times is
0.0000000000000003627sec which is half the difference you
calculated.
[/quote]
Correct. I was running out the door to catch a plane, and
calculated the perpendicular time incorrectly. YOU, however, have
NO SUCH EXCUSE.
Multiply 3.627e-16 seconds by the speed of light, and you get
1.087e-7 meters which you may compare with 5.4e-7 meters for
the wavelength of yellow light.
In other words, Michelson and Morley made no mistake in their
calculations, and the predicted fringe shift should have been
easily discernable in their apparatus.
STOP BEING SO STUPID. LEARN SOME ALGEBRA.
Jerry |
|
|
| Back to top |
|
|
|
| doug... |
| Posted: Tue Nov 03, 2009 9:51 am |
|
|
|
Guest
|
Koobee Wublee wrote:
[quote]On Oct 29, 5:15 pm, doug <x... at (no spam) xx.com> wrote:
Peter Riedt wrote:
MMX Null = Anisotropy = Ether
In 1887 Michelson and Morley predicted an interferometer experiment
(MMX) would show interference of light due to the presence of the
ether. When their experiment produced a null result, an explanation
was required. Lorentz suggested an answer after others pointed the
way. The light path across the parallel and perpendicular arms of the
interferometer equipment had been assumed to be of unequal length due
to the movement of the apparatus through space. The difference in
light path length was equalised in Lorentz’s theory by a contraction
of the parallel arm proportional to the speed of the earth (v) on
which the lab was located. However, Lorentz was wrong. The null result
of the experiment (MMX) is caused by the anisotropy of light which in
turn is caused by the ether. The speed of light over the perpendicular
arm is c, unimpeded and unassisted by the ether. On the parallel arm,
the speed upstream is c-v, retarded by the ether. Downstream, it is c
+v, assisted by the ether. The null result is determined by the
formula
perpendicular c = parallel (c+v+c-v)/2 = null result.
And here you make an elementary math error...
It is amazing when these college dropout were first fed with bull$hit,
more bull$hit would come out of them. Mr. Riedt is quite correct.
Just consult with Professor Roberts. The null results actually
supported the emission theory of light as Mr. Riedt has brought up.
Unfortunately, the emission theory of light fails at
electromagnetism. <shrug
[/quote]
As usual koobee demonstrates his ignorance and hatred. This time
he shows that he does not know elementary algebra. |
|
|
| Back to top |
|
|
|
| Peter Riedt... |
| Posted: Tue Nov 03, 2009 2:37 pm |
|
|
|
Guest
|
On Nov 3, 7:15 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
[quote]On Nov 2, 10:01 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 2, 9:25 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is 0.000000036670
seconds and downstream 0.000000036663 seconds for a total time of
0.000000073333 seconds which gives a speed of 299999997m/sec for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry, using your values for c and v, the perpendicular
light path distance according to Michelson&Morley
(American Journal of Science 203/1887) is 22.000000108725m
giving a perpendicular time of 0.000000073384101306sec. The
difference between the parallel and perpendicular times is
0.0000000000000003627sec which is half the difference you
calculated.
Correct. I was running out the door to catch a plane, and
calculated the perpendicular time incorrectly. YOU, however, have
NO SUCH EXCUSE.
Multiply 3.627e-16 seconds by the speed of light, and you get
1.087e-7 meters which you may compare with 5.4e-7 meters for
the wavelength of yellow light.
In other words, Michelson and Morley made no mistake in their
calculations, and the predicted fringe shift should have been
easily discernable in their apparatus.
STOP BEING SO STUPID. LEARN SOME ALGEBRA.
Jerry- Hide quoted text -
- Show quoted text -
[/quote]
Jerry, the calculations of M&M are correct but their logic is
incorrect.
Peter Riedt |
|
|
| Back to top |
|
|
|
| Inertial... |
| Posted: Tue Nov 03, 2009 7:54 pm |
|
|
|
Guest
|
"Peter Riedt" <riedt1 at (no spam) yahoo.co.uk> wrote in message
news:54889b50-7b6f-45d1-8442-0c8101b230ee at (no spam) u25g2000prh.googlegroups.com...
[quote]On Nov 3, 7:15 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 10:01 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 2, 9:25 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is
0.000000036670
seconds and downstream 0.000000036663 seconds for a total time
of
0.000000073333 seconds which gives a speed of 299999997m/sec
for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry, using your values for c and v, the perpendicular
light path distance according to Michelson&Morley
(American Journal of Science 203/1887) is 22.000000108725m
giving a perpendicular time of 0.000000073384101306sec. The
difference between the parallel and perpendicular times is
0.0000000000000003627sec which is half the difference you
calculated.
Correct. I was running out the door to catch a plane, and
calculated the perpendicular time incorrectly. YOU, however, have
NO SUCH EXCUSE.
Multiply 3.627e-16 seconds by the speed of light, and you get
1.087e-7 meters which you may compare with 5.4e-7 meters for
the wavelength of yellow light.
In other words, Michelson and Morley made no mistake in their
calculations, and the predicted fringe shift should have been
easily discernable in their apparatus.
STOP BEING SO STUPID. LEARN SOME ALGEBRA.
Jerry- Hide quoted text -
- Show quoted text -
Jerry, the calculations of M&M are correct but their logic is
incorrect.
[/quote]
Have you come up yet with your proof the Lorentz transforms are inconsistent
and self-contradictory? You said you could show it, but never had. Were
you just lying all along? Or have you since realized your mistake? |
|
|
| Back to top |
|
|
|
| doug... |
| Posted: Tue Nov 03, 2009 8:08 pm |
|
|
|
Guest
|
Peter Riedt wrote:
[quote]On Nov 3, 7:15 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 10:01 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 2, 9:25 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Nov 2, 1:28 am, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
On Nov 1, 9:53 am, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
On Oct 29, 9:36 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
Doug, the time upstream for 11m of the parallel arm is 0.000000036670
seconds and downstream 0.000000036663 seconds for a total time of
0.000000073333 seconds which gives a speed of 299999997m/sec for
22m upstream and downstream.
By performing the computation to only five significant figures,
you miss understanding the effect.
Let the speed of light be 299,792,458 m/s and the average speed
of the Earth in orbit be 29,805 m/s
Time downstream = 0.0000000366884029556 s
Time upstream = 0.0000000366956987133 s
Total = 0.0000000733841016689 s for the parallel arm
Total = 0.0000000733841009436 s for the perpendicular
-----------------------------------------------------------------
Difference = 0.0000000000000007253 s
If had even a slight understanding of middle school algebra, you
wouldn't have made such a spectacle of yourself.
Jerry
Jerry, your calculations are good and almost correct. The time
for the perpendicular arm is not 22m/c but slighly more.
Even so, the difference of 0.0000000000000007253sec
is well covered by the observed fringe shift which of course
was less than the predicted fringe shift.
WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Jerry, using your values for c and v, the perpendicular
light path distance according to Michelson&Morley
(American Journal of Science 203/1887) is 22.000000108725m
giving a perpendicular time of 0.000000073384101306sec. The
difference between the parallel and perpendicular times is
0.0000000000000003627sec which is half the difference you
calculated.
Correct. I was running out the door to catch a plane, and
calculated the perpendicular time incorrectly. YOU, however, have
NO SUCH EXCUSE.
Multiply 3.627e-16 seconds by the speed of light, and you get
1.087e-7 meters which you may compare with 5.4e-7 meters for
the wavelength of yellow light.
In other words, Michelson and Morley made no mistake in their
calculations, and the predicted fringe shift should have been
easily discernable in their apparatus.
STOP BEING SO STUPID. LEARN SOME ALGEBRA.
Jerry- Hide quoted text -
- Show quoted text -
Jerry, the calculations of M&M are correct but their logic is
incorrect.
[/quote]
So you are saying that they did not see what they saw?
Or what are you complaining about?
[quote]
Peter Riedt[/quote] |
|
|
| Back to top |
|
|
|
| Jerry... |
| Posted: Wed Nov 04, 2009 3:08 am |
|
|
|
Guest
|
On Nov 3, 6:37 pm, Peter Riedt <rie... at (no spam) yahoo.co.uk> wrote:
[quote]On Nov 3, 7:15 pm, Jerry <Cephalobus_alie... at (no spam) comcast.net> wrote:
Correct. I was running out the door to catch a plane, and
calculated the perpendicular time incorrectly. YOU, however, have
NO SUCH EXCUSE.
Multiply 3.627e-16 seconds by the speed of light, and you get
1.087e-7 meters which you may compare with 5.4e-7 meters for
the wavelength of yellow light.
In other words, Michelson and Morley made no mistake in their
calculations, and the predicted fringe shift should have been
easily discernable in their apparatus.
STOP BEING SO STUPID. LEARN SOME ALGEBRA.
Jerry, the calculations of M&M are correct but their logic is
incorrect.
[/quote]
What is incorrect about their logic? Assuming a classical aether,
Michelson and Morley should have been able to detect a fringe
displacement. Since Michelson and Morley did not detect any
fringe displacement, the conclusion was that something was wrong
with the assumption of a classical aether.
Where is the error in logic?
Are you claiming that a classical aether should not have resulted
in a fringe displacement?
Jerry |
|
|
| Back to top |
|
|
|
| Koobee Wublee... |
| Posted: Wed Nov 04, 2009 9:51 pm |
|
|
|
Guest
|
On Nov 3, 6:51 am, doug <x... at (no spam) xx.com> wrote:
[quote]Koobee Wublee wrote:
It is amazing when these college dropout were first fed with bull$hit,
more bull$hit would come out of them. Mr. Riedt is quite correct.
Just consult with Professor Roberts. The null results actually
supported the emission theory of light as Mr. Riedt has brought up.
Unfortunately, the emission theory of light fails at
electromagnetism. <shrug
As usual koobee demonstrates his ignorance and hatred. This time
he shows that he does not know elementary algebra.
[/quote]
Professor Roberts had claimed in the past that the emission theory of
light also explains the null results of the MMX, and he was quite
correct. So, Koobee Wublee and Professor Roberts agree on this
account, and all of a sudden, the self-claimed phd, doug, ranted about
Koobee Wublee demonstrating His ignorance and hatred. That means
Professor Roberts was also demonstrating his ignorance and hatred by
claiming the emission theory of light also satisfies the null results
of the MMX. <shrug>
How can a self-proclaimed phd, doug, behave like a 5-year-old kid?
doug must be a pre-teen having the mental ability of 5-year-olds.
<shrug> |
|
|
| Back to top |
|
|
|
| doug... |
| Posted: Thu Nov 05, 2009 9:36 am |
|
|
|
Guest
|
Koobee Wublee wrote:
[quote]On Nov 3, 6:51 am, doug <x... at (no spam) xx.com> wrote:
Koobee Wublee wrote:
It is amazing when these college dropout were first fed with bull$hit,
more bull$hit would come out of them. Mr. Riedt is quite correct.
Just consult with Professor Roberts. The null results actually
supported the emission theory of light as Mr. Riedt has brought up.
Unfortunately, the emission theory of light fails at
electromagnetism. <shrug
As usual koobee demonstrates his ignorance and hatred. This time
he shows that he does not know elementary algebra.
Professor Roberts had claimed in the past that the emission theory of
light also explains the null results of the MMX, and he was quite
correct. So, Koobee Wublee and Professor Roberts agree on this
account, and all of a sudden, the self-claimed phd, doug, ranted about
Koobee Wublee demonstrating His ignorance and hatred. That means
Professor Roberts was also demonstrating his ignorance and hatred by
claiming the emission theory of light also satisfies the null results
of the MMX. <shrug
How can a self-proclaimed phd, doug, behave like a 5-year-old kid?
doug must be a pre-teen having the mental ability of 5-year-olds.
shrug
[/quote]
The comments were about Peter's algebra and had nothing to do
with the MMX. Koobee cannot read very well. |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Sat Dec 12, 2009 3:51 am
|
|