Initially Einstein chose c'=Lf':
http://www.physlink.com/Education/AskExperts/ae13.cfm :
"So, it is absolutely true that the speed of light is _not_ constant in
a gravitational field [which, by the equivalence principle, applies as
well to accelerating (non-inertial) frames of reference]. If this were
not so, there would be no bending of light by the gravitational field
of stars. One can do a simple Huyghens reconstruction of a wave front,
taking into account the different speed of advance of the wavefront at
different distances from the star (variation of speed of light), to
derive the deflection of the light by the star.
Indeed, this is exactly how Einstein did the calculation in:
"On the Influence of Gravitation on the Propagation of Light," Annalen
der Physik, 35, 1911.
which predated the full formal development of general relativity by
about four years. This paper is widely available in English. You can
find a copy beginning on page 99 of the Dover book "The Principle of
Relativity." You will find in section 3 of that paper, Einstein's
derivation of the (variable) speed of light in a gravitational
potential, eqn (3). The result is,
c' = c0 ( 1 + V / c2 )
where V is the gravitational potential relative to the point where the
speed of light c0 is measured."
However later Einstein had to camouflage the fact that the frequency
shift is due to variable speed of light and introduced gravitational
time dilation - a concept extremely dangerous for human rationality.
Two identical clocks in identical conditions (identical gravitational
fields) allegedly have different rates. Rationality is immediately
destroyed and the victim starts worshipping both the miracle and its
creator.
Pentcho Valev
"Pentcho Valev" <pva...@yahoo.com> wrote in message
news:1161408957.7...@e3g2000cwe.googlegroups.com...
> Photons move in a gravitational field
No. Photons = quantum. No spacetime, no curavture, no gravity.
> and either undergo acceleration
What part of "c" don't you understand?
David A. Smith
It is a well known fact that this line of thought
give the wrong result. Considering light to be "falling"
like Einstein did in this paper, does _not_ give the same
prediction for the bending of light as GR does, it
predicts only half the angle.
Later observations have proven GR to give the correct
prediction.
This shows that the bending of light cannot be explained
by light falling in a Newtonian gravitational field.
Einstein's paper:
"On the Influence of Gravitation on the Propagation of Light,"
Annalen der Physik, 35, 1911.
was simply wrong.
He got it right in 1916.
> However later Einstein had to camouflage the fact that the frequency
> shift is due to variable speed of light and introduced gravitational
> time dilation - a concept extremely dangerous for human rationality.
In the 1911 paper, Einstein wrote in paragraph 4:
"A ray of light going past the Sun would accordingly undergo
deflection to the amount of 4*10^-4 = 0.83 second of arc"
While he in:
"Die Grundlage der allgemeinen Relativtatstheorie",
Annalen der Physik, 49, 1916
wrote in paragraph 22:
"According to this, a ray of light going past the Sun
undergoes a deflection of 1.7 second of arc."
The former prediction was wrong, the latter was correct.
He didn't "camouflage the fact that the frequency
shift is due to variable speed of light".
He had realized that this line of thought was simply wrong.
Which is experimentally verified.
You seem to think that you can prove GR wrong by
pointing out that Einstein contradicted himself.
Of course much of what Einstein wrote was wrong.
He was wrong in 1911, and he was wrong about much of
what he wrote after 1916. Einstein used most of his
life on a theory which led nowhere.
The special and general theories of relativity
are still valid theories because they are never
falsified and are experimentally very well confirmed.
That Einstein was wrong about a lot of things is
utterly irrelevant to the valididy of SR and GR.
Paul
It is a well known fact that you are a shithead who thinks E-fields are in
phase
with B-fields, getting your well-known facts wrong.
In fact the more well known a fact is, the less chance it has of being a
fact.
So Einstein's 1911 formula
c' = c0 ( 1 + V / c2 )
is wrong? Give the correct one (then we may test the compatibility of
both formulas with the redshift f'=f(1+V/c^2)).
Pentcho Valev
Yes.
> Give the correct one (then we may test the compatibility of
> both formulas with the redshift f'=f(1+V/c^2)).
There is no such equation in GR.
You don't calculate what GR predicts for the red shift that way.
Use the Schwarzschild metric.
Paul
You are absolutely right, I know that the E-fields and B-field
in an EM-wave are in phase.
Please prove your claim that they are not by pointing out
the error in the following:
| Androcles wrote:
| quote:
| > curl E = -dB/dt -- Maxwell, 90 degree phase shift.
| end quote
|
| Ah. You _did_ look it up. :-)
| So you have finally found the correct equation.
| (Which happens to be the equation I have shown you
| umpteen times. Some coincidence, isn't it? :-))
| So we can forget the stupid E = -dB/dt, and see
| what the correct equation has to tell us.
|
| Assume we have a plane EM-wave propagating along
| the x-axis with the B-field:
| B_x = 0
| B_y = 0
| B_z = Bo*sin(wt - kx) where k = w/c
|
| So we get:
| (curl E)_x = -@B_x/@t = 0
| (curl E)_y = -@B_y/@t = 0
| (curl E)_z = -@B_z/@t = -w*Bo*cos(wt - kx)
|
| So you are absolutely right - it is 90 degrees phase shift
| between (curl E) and B.
|
| But what about E?
|
| (curl E)_z = @E_y/@x = -w*Bo*cos(wt - kx)
|
| integrating:
| E_y = (w/k)*Bo*sin(wt - kx), w/k = c
|
| So the E-field of the wave is:
| E_x = 0
| E_y = c*Bo*sin(wt - kx)
| E_z = 0
|
| which shows that according to Maxwell's curl E = -@B/@t,
| the E-field and B-field in an EM-wave are in phase.
Please try again, you are so funny when you do.
Like this:
Androcles wrote:
| A comedy of errors, Tusselad.
| Hilarious.
| "Partial derivatives are defined as derivatives of a function of multiple
| variables when all but the variable of interest are held fixed during the differentiation. "
|
| ref:
| http://mathworld.wolfram.com/PartialDerivative.html
|
| The variable of interest is t.
|
| "E_y = c*Bo*sin(wt - kx)." -- Tusselad.
|
| The variables of no interest are Bo, c and x.
| Making those constant, say one, one and zero,
| "E_y = sin(wt)."
Hilarious, no? :-)
Drink your Glenlivet an give it another try, Androcles.
I am sure it will be fun.
Paul
As I've stated, I don't understand assertions.
[snip "asked and answered" text]
|| So you are absolutely right - it is 90 degrees phase shift
|| between (curl E) and B. - Tusselad.
Right, right and right again.
|
| Hilarious, no? :-)
|
| Drink your Glenlivet an give it another try, Androcles.
| I am sure it will be fun.
Since you've demonstrated your ability to look up earlier text, look it up.
No error found in anything *I* wrote, pity you don't understand complex
algebra.
It's quite simple,
(a+ib)^2 = a^2 + 2a(ib) + (ib)^2
but i^2 = -1, so
(a+ib)^2 = a^2 + 2a(ib) -b^2
= a^2-b^2 + i2ab
= (a+b)(a-b) + i2ab
What are the three cube roots of 1, Tusselad, and how are they related to
3-phase?
http://www.ewh.ieee.org/soc/es/Nov1997/09/B_WOU1.JPG
http://www.du.edu/~jcalvert/tech/threeph.htm
Have fun, do your homework, I'll relax with a Glenlivet.
Hilarious, yes?
Androcles
A hypnotist would have confused the issue in a much better way (there
are only hypnotists and zombies in Einstein's criminal cult).
Pentcho Valev
Indeed.
B_z = Bo*sin(wt - kx) where k = w/c
(curl E)_z = -@B_z/@t = -w*Bo*cos(wt - kx)
Definitely 90 degrees phase shift between (curl E)_z and B_z.
NB. The following is too embarrassing for you not to snip:
(curl E)_z = @E_y/@x
so:
@E_y/@x = -w*Bo*cos(wt - kx)
Can you find what E_y is from this, Androcles?
No?
Too hard for you?
Didn't you say you had a degree in mathematics?
I will have to tell you then:
E_y = (w/k)*Bo*sin(wt - kx)
which shows that according to Maxwell's curl E = -@B/@t,
the E-field and B-field in an EM-wave are in phase.
Frustrating, being unable to prove me and
the rest of the world wrong, isn't it?
Now you can snip all this, and repeat something utterly irrelevant,
like:
> Since you've demonstrated your ability to look up earlier text, look it up.
> No error found in anything *I* wrote, pity you don't understand complex
> algebra.
>
> It's quite simple,
> (a+ib)^2 = a^2 + 2a(ib) + (ib)^2
> but i^2 = -1, so
> (a+ib)^2 = a^2 + 2a(ib) -b^2
> = a^2-b^2 + i2ab
> = (a+b)(a-b) + i2ab
>
> What are the three cube roots of 1, Tusselad, and how are they related to
> 3-phase?
That will indeed make you look so intelligent, and it will
be a very convincing argument for your mantra:
"The E-field and B-field in an EM-wave look like this:"
http://tinyurl.com/we5fu
This Woodbury fellow who is responsible for this blunder
seems to be a biochemist, which kind of excuses him.
I bet he wouldn't argue much if an electrical engineer
had made him aware of his blunder.
> http://www.ewh.ieee.org/soc/es/Nov1997/09/B_WOU1.JPG
> http://www.du.edu/~jcalvert/tech/threeph.htm
> Have fun, do your homework, I'll relax with a Glenlivet.
> Hilarious, yes?
Na. Rather boring.
Too much, too often.
Paul
Perhaps Pentcho Valev also ignores the difference between
special realtivity and general relativity. That is a new one.
Updated list of differences, as ignored by Pentcho Valev:
- special realtivity and general relativity,
- teachers and hypnotists,
- laymen and zombies,
- a person being right and a theory being right,
- students and imbeciles,
- bad science and bad engineering,
- bad engineering and bad cost management,
- honing the foundations of a theory and fighting it,
- physics and linguistics,
- an article written in 1905 and a theory created in 1915,
- understanding a book and turning its pages,
- speed and relative (aka closing) speed,
- doing algebra and randomly writing down symbols,
- real life and a Usenet hobby group,
- receiving a detailed reply and being ignored,
- everyday concepts and scientific concepts in physics,
- the three things that smell like fish,
- inertial and non-inertial,
- speed and velocity,
- an article and a book,
- relativity and disguised ether addiction,
- algebra and analytic geometry,
- kneeling down and bending over,
- local and global,
- a sycophant in English and in French,
- a relation and an equation,
- massive and massless particles,
- a Mexican poncho and a Sears poncho,
- implication and equivalence,
- group velocity and phase velocity,
- science and religion
Dirk Vdm
> Pentcho Valev
>
General Relativity was falsified in 1978 when observational evidence
revealed that galaxies do not rotate in accordance with Keplerian
dynamics (i.e. they had flat rotation curves).
References:
[1 ] ^ A. Bosma, "The distribution and kinematics of neutral hydrogen
in spiral galaxies of various morphological types", PhD Thesis,
Rejksuniversiteit Groningen, 1978, available online at the Nasa
Extragalactic Database
Ok then, mission accomplished, case closed.
Now, what is it you are struggling with?
We'll go one little step at a time.
| B_z = Bo*sin(wt - kx)
Definitions:
Bo = mag field strength, call it 1 for simplicity.
w is radians/sec
t is time
x is distance
k = ?
Let's avoid units of metres or feet, throw in c, value 1.
c = x/t = 1 and we'll make k =c = 1 and w = 2pi for simplicity.
So at x = 0, B_z = sin(wt)
and at x = 2pi, B_z = sin(wt +2pi)
Well waddaya know, B_z at x = 0 is in phase with B_z at x = 2pi
for all t.
Ignore rant below.
What *IS* it you do not understand?
Androcles.
? :-0 ???
Of course the engineer with a degree in mathematics knows that
? = w/c
This was so obvious that you didn't find it necessary to mention.
Right?
> Let's avoid units of metres or feet, throw in c, value 1.
> c = x/t = 1
t = x/c
c = 1
t = x
B_z = Bo*sin(wt - wx/c)= Bo*sin(wx - wx) = 0
> and we'll make k =c = 1 and w = 2pi for simplicity.
> So at x = 0, B_z = sin(wt)
> and at x = 2pi, B_z = sin(wt +2pi)
> Well waddaya know, B_z at x = 0 is in phase with B_z at x = 2pi
> for all t.
Even better.
You have shown that sin(wt-kx) = 0 for all x and t.
A major mathematical achievement!
Was it possibly this which gave you your degree in mathematics?
Now, when the first small step is done, we can take the next.
> What *IS* it you do not understand?
What I *DON'T* understand *IS* how to find E_y.
I *KNOW* that when:
B_z = Bo*sin(wt - kx)
then:
(curl E)_z = -@B_z/@t = -w*Bo*cos(wt - kx)
and *I* know that:
(curl E)_z = @E_y/@x
so:
@E_y/@x = -w*Bo*cos(wt - kx)
But *HOW* do I find E_y from this?
Can *YOU* please help me?
My attempt so far is:
E_y = (w/k)*Bo*sin(wt - kx)
But since this proves you wrong, it can't be right, right?
It's wrong, right?
Or is it right?
This struggling confuses me.
But you can help me, right?
Or is it wrong?
Paul
| > || So you are absolutely right - it is 90 degrees phase shift
| > || between (curl E) and B. - Tusselad.
| >
| > Right, right and right again.
|
| Indeed.
Ok then, mission accomplished, case closed.
If you want me to help you understand physics and mathematics
then answer the questions.
|| B_z = Bo*sin(wt - kx)
|
| Definitions:
| Bo = mag field strength, call it 1 for simplicity.
| w is radians/sec
| t is time
| x is distance
| k = ?
Androcles
That's what I did.
Your question was:
| What *IS* it you do not understand?
My answer was:
| What I *DON'T* understand *IS* how to find E_y.
|
| I *KNOW* that when:
| B_z = Bo*sin(wt - kx)
| then:
| (curl E)_z = -@B_z/@t = -w*Bo*cos(wt - kx)
| and *I* know that:
| (curl E)_z = @E_y/@x
| so:
| @E_y/@x = -w*Bo*cos(wt - kx)
|
| But *HOW* do I find E_y from this?
| Can *YOU* please help me?
|
| My attempt so far is:
| E_y = (w/k)*Bo*sin(wt - kx)
|
| But since this proves you wrong, it can't be right, right?
| It's wrong, right?
| Or is it right?
|
| This struggling confuses me.
| But you can help me, right?
| Or is it wrong?
It's wrong! Right?
Paul
Falsified is the wrong word I claim. For one, we don't know what the
mass distribution in a galaxy is. If we assume all the mass is in the
stars, you're right - GR was proven not to apply over galactic
distances. Still, this isn't a falsification but a realm of
application.
Even so, the mass isn't all in the stars. What is the density and
distribution of neutral hydrogen in these galaxies? What is the
percentage of ionization and the magnetic field?
Is there any dark matter?
Cheers - shevek
In answer to
Definitions:
Bo = mag field strength, call it 1 for simplicity.
w is radians/sec
t is time
x is distance
k = ?
you wrote:
quote:
? :-0 ???
unquote.
Nobody of sound mind can consider that incomprehensible Norwegian
Neanderthal Gibberish to be an answer to a question but since you insist,
B_z = Bo*sin(wt -(? :-0 ???)x)
Calculate E_y from there, I'm unable to assist you further.
Mission accomplished, case closed.
Androcles.
[completing real quote]
>
> Of course the engineer with a degree in mathematics knows that
> ? = w/c
> This was so obvious that you didn't find it necessary to mention.
> Right?
[completed real quote]
> unquote.
>
> Nobody of sound mind can consider that incomprehensible Norwegian
> Neanderthal Gibberish to be an answer to a question but since you insist,
> B_z = Bo*sin(wt -(? :-0 ???)x)
> Calculate E_y from there, I'm unable to assist you further.
> Mission accomplished, case closed.
Translation:
Androfart has no way out, so in a desperate attempt to
stop this conversation for once and for all, he resorts to
his dog shit trick.
End of Applied Village Idiot Psychology (AVIP) lesson #23.
Dirk Vdm
k = ?
Definition ? :-0> :-)
>
> you wrote:
> quote:
> ? :-0 ???
> unquote.
>
> Nobody of sound mind can consider that incomprehensible Norwegian
> Neanderthal Gibberish to be an answer to a question but since you insist,
> B_z = Bo*sin(wt -(? :-0 ???)x)
> Calculate E_y from there, I'm unable to assist you further.
> Mission accomplished, case closed.
But that wasn't the answer to your question:
| What *IS* it you do not understand?
This was:
| What I *DON'T* understand *IS* how to find E_y.
|
| I *KNOW* that when:
| B_z = Bo*sin(wt - kx)
| then:
| (curl E)_z = -@B_z/@t = -w*Bo*cos(wt - kx)
| and *I* know that:
| (curl E)_z = @E_y/@x
| so:
| @E_y/@x = -w*Bo*cos(wt - kx)
|
| But *HOW* do I find E_y from this?
| Can *YOU* please help me?
|
| My attempt so far is:
| E_y = (w/k)*Bo*sin(wt - kx)
|
| But since this proves you wrong, it can't be right, right?
| It's wrong, right?
| Or is it right?
I can see two possible reasons for your failure to answer:
1. You are able to do the math and thus you know that
I am right, but you find it too embarrassing to admit
that you were wrong.
2. You still believe that I am wrong, but you are unable to
do the math to prove it.
Which reason do you prefer, Androcles? :-0>
Paul
If the kinematics of the cosmos deviated only slightly from the GR
predictions, one has a case in asserting the a priori validity of that
theory and working with that theory to account for the observational
anomalies. That is not the case here. The observed large scale
structure and the motion of the cosmos is in stark contradiction with
that predicted by GR. In that case, the observational evidence suffices
as a reasonable falsification of GR. It's really as simple as that, and
it must be like that, as otherwise we could simply take the similar
approach with older, failed theories like and account for all anomalies
through inference. There's no limit to what we can do through unlimited
inference without falsification, but that's not science.
[...]
[Agreement by Tusselad]
| > | > || So you are absolutely right - it is 90 degrees phase shift
| > | > || between (curl E) and B. - Tusselad.
[Androcles]
| > | > Right, right and right again.
[Agreement by Tusselad]
| > | Indeed.
[Androcles]
| > Ok then, mission accomplished, case closed.
[Local village dog tord]
"you so obviously and miserably failed to get the point (which
for obvious reasons I will not even try to explain)".
Dork does so hates me winning since he tried to explain how he
got himself fired from his job, the poor jealous little fuck :-)
HAHAHAHAHAHA!
The behavior of a charged baloon in contact with the ceiling is also in
stark contradiction with that predicted by GR.
But this is not a falsification.. merely are realization that other
forces are at work.
Unless we have a better understanding of why galaxies rotate as they
do, we can't really say anything other than that we don't know.
> [...]