THE GENERAL RELATIVITY.
Stamenin
AN ARGUMENT FOR THE GENERAL POSTULATE OF RELATIVITY.
At the page 68 of his book Relativity Einstein writes about the General
postulate of relativity. He imagines a large portion of empty space, so
far removed from the stars, that we have before us approximately the
conditions required by the fundamental law of Galilei. It is then
possible to choose a Galileian reference body for this part of space
relative to which points at rest remain at rest and points at motion
continue permanently in uniform rectilinear motion.
From this supposition is visible that Einstein uses the name
Galileian system only to not make use of the Newton's absolute
coordinate system, because between them in this portion of space they
are both of them valid and both are impossible to be determined.
And by there definitions they represent the same thing, a coordinate
system that is staying in a large portion of the cosmos where do not
exist gravitational forces.
In continuation he says: In this large space he puts a chest and a man
in it. To the lid of the chest is fixed a hook with a rope and by an
imaginary being is pulled "upwards". But how does the man in the
chest regard the process, asks Einstein.
The acceleration at the chest will be transmitted to him by the
reaction of the floor of the chest. This situation is complete similar
with the situation in a room at the earth and that there is a
gravitational field.
To the end of the page 69 Einstein says: "Ought we to smile at the
man and say that he errs in his conclusion? I do not believe we ought
to if we wish to remain consistent, we must rather admit that his mode
of grasping the situation violates neither reason nor known mechanical
laws. Even though it is being accelerated with respect to the
"Galileian space" first considered, we can nevertheless regard the
chest as being at rest. We have thus good grounds for extending the
principle of the relativity to include bodies of reference which are
accelerated with respect to each other, and as a result we have gained
a powerful argument for a generalized postulate of relativity".
So we can see that Einstein by using the behavior of the inertial and
the gravitational forces to acct upon every atom of the material body
reached the conclusion that an accelerated coordinate system is an
absolute inertial system.
Evidently this conclusion is in contradiction:
1) With the starting supposition that the chest is in an accelerated
motion.
2) With the principle of the relativity that says that is not valid for
the accelerated systems of coordinates because it is valid only for
systems of coordinates that are in right line motion and with a
constant speed.
3) By this he makes confusion between the gravitational and inertial
forces considering them as being of the same nature. But we know that
that they are different, the gravitational force is an external force
of the body and inertial force is an internal force in the body. The
only thing that is common to these forces is the behavior to act upon
every atom of the material body.
4) And by this he concludes that has gained a powerful argument for the
generalization of this principle of the relativity!!!!
This kind of logic could be named, the negation of the affirmation.
And after that Einstein continues: " We must note carefully that the
possibility of this mode of interpretation rests on the fundamental
property of the gravitational field of giving all bodies the same
acceleration, or, what comes to the same thing, on the law of the
equality of inertial and gravitational mass"!!!
This final argumentation shows how difficult was for Einstein trying
in such a way to trick the reader by using such an illogical
argumentation.
What I can conclude after all that is said, is the fact that the
General theory of the relativity has a tricky base and not powerful
arguments and fundamental laws of the nature.
2/10/2006.
Bill Hobba
"Stamenin" <tas...@hotmail.com> wrote in message
news:1159834957.3...@b28g2000cwb.googlegroups.com...
That is not the conclusion he reached - he reached the correct conclusion -
it is non inertial - not absolute.
Bill
Paul B. Andersen
Einstein's book "Relativity" is a popular book, not a scientific
paper.
It can at the very best give you a vague idea of what GR is about.
Believing that you after having read a popular book are competent
to critisize the theory described in that book is a rather naive idea.
And you failed to grasp the basic idea which Einstein tried to mediate.
You say: " the gravitational force is an external force of the body"
Is it?
If it is an external force, why can't you feel it?
The only external force you now can feel is the force
from your chair, pushing you upwards.
If you jump out the window, you feel nothing.
That is because no external forces are acting on you while
you are in free fall.
Paul
Sorcerer
"Paul B. Andersen" <paul.b....@hia.no> wrote in message
news:1160036583.6...@m73g2000cwd.googlegroups.com...
Yes, science fiction.
| It can at the very best give you a vague idea of what GR is about.
Bulshit, it tells you exactly.
| Believing that you after having read a popular book are competent
| to critisize the theory described in that book is a rather naive idea.
|
Learn to spell "critic" then go on from there, critic.
| And you failed to grasp the basic idea which Einstein tried to mediate.
| You say: " the gravitational force is an external force of the body"
| Is it?
Yes.
| If it is an external force, why can't you feel it?
If there are radio waves, why can't you see them?
| The only external force you now can feel is the force
| from your chair, pushing you upwards.
| If you jump out the window, you feel nothing.
| That is because no external forces are acting on you while
| you are in free fall.
Hahahaha!
The moron Andersen still doesn't understand Newton's third law.
You'll be saying next that he doesn't accelerate!
Hilarious!
Paul B. Andersen
One question, Androcles.
If you put yourself behind the wheel in a F1 car,
you can accelerate at 1g in the horizontal direction.
The will feel the 1000N force pushing your 100 kg body
in the back extremely well.
If you jump out of the window, you feel nothing
pushing you while you are falling.
If the acceleration is the same in the two cases,
why is then what you feel so very different?
Paul
Sue...
I feel it.
Mach's principle
Course taught
http://farside.ph.utexas.edu/teaching.html
"Space time"
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
http://farside.ph.utexas.edu/teaching/jk1/lectures/node13.html
"Retarded potential"
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html
"Transformations"
http://arxiv.org/abs/physics/0204034
========
> http://www.physik.unizh.ch/groups/grouposterwalder/new/timeresolved/spgun.html
> http://www.slac.stanford.edu/pubs/slacpubs/5250/slac-pub-5420.pdf
Sue...
RCMP
http://www.okmilmuseum.ca/images/RCMP%20uniform.jpg
Sorcerer
"Paul B. Andersen" <paul.b....@hiadeletethis.no> wrote in message
news:eg2o5n$ph1$1...@dolly.uninett.no...
Hmm.. this is quite an interesting subject.
Let us hypothesize a rigid rod aligned with the x-axis, ends
at 0 and 1, and strike the end of the rod with a hammer to
accelerate the rod in the positive x-direction at t = 0.
The impulse moves along the rod at the speed of sound
(in rod material) and the end which is at 1 begins to move
at t = 1.
Conclusion:
The hypothesis is false, the rod is not rigid.
To be rigid, it is so dense as to be a cylindrical black hole,
which will not move when struck by a hammer.
Ever had that "butterfly in the tummy" feeling that children
love when riding roller coasters? It's because your internal
organs bend your diaphragm relative to your skeleton.
In gravitational free fall ALL of your body accelerates
together simultaneously, but in a push your fat arse is
compressed between your skin and pelvis, which has inertia.
It is this compression that results in bruises when I beat
your hollow head in with a baseball bat. In other people
the same effect causes brain damage, but they have a
brain to be damaged.
Gravity acts THROUGH matter, there are no gravity shields.
Mathematicians use the term "field" for this phenomenon,
but that's only because they have not coined a word for it
and have borrowed from agriculture. An area field of grass
for cattle to graze is the same everywhere (ideally), a volumetric
field is also the same everywhere. If I were a Shakespeare
I'd call it a grield, if you were an Ibsen you'd be just as
confused as you are now. The language needs to develop
to convey the ideas.
Humpty Roberts let out a great sigh.
" <sigh>", he said.
"The nuances of English. I was discussing the usage of words and
not the concepts they represent."
I'm discussing the concepts that have no words. Alas, I am
no Shakespeare or even Ibsen.
The written page has encouraged 2 dimensional thinking.
Rubik had a very useful educational tool in his cube, a
mathematical group with the operation of rotation on a set.
Every child should own one. We have words such as volume
and area, but we apply "field" to both.
In the case of the "rigid" rod, both ends and all matter between
moves simultaneously in a grield. Should the grield not be uniform
(inverse square law) the rod will turn as one end is accelerated
more than the other, and so the Moon maintains one face toward
the Earth, or did until Commander David Scott hit it with a hammer
and a feather simultanously, the hammer and feather being
rigidly connected as far as I could tell.
Androcles
Stamenin
From the Einstein book I took this topic. Your problem is in the fact
that you do not take in account that only these two forces the
gravitational and the inertial force act upon every atom of the
marerial body in motion or at a state of staying. this is the cause wy
you don't feel the gravitational force. For this I recomend to read the
point (3) of this mine topic. For more information about this fenomenon
I send and the followig article:
THE INERTIAL AND GRAVITATIONAL MASS
Is possible to exist two masses, the inertial and the gravitational
mass? Einstein
considers that they exist and are the basic argument for his general
theory of the relativity. Is that true?
Einstein in his book RELATIVITY in page 65 and 66 makes a parallel
analysis between the gravitational field, the electric field and the
magnetic field. Everything that is said there represents known facts
about the fields. But at the end of the 66th page he has written:
"According to Newton's law of motion, we have:
(Force)=(inertial mass) x (acceleration), or [F=(mi). a.]
Where the "inertial mass" is a characteristic constant of the
accelerated body. If now gravitation is the cause of the acceleration,
we than have:
(Force)=(gravitational mass) x (intensity of the gravitational field),
or [F=(mg). g] where the "gravitational mass" is likewise a
characteristic constant for the body.
From these two relations follows:
a=(mg/mi).g
Where g is the intensity of the gravitational field.
If now, we find from experience, the acceleration is to be independent
of the nature and the conditions of the body and always the same for a
given gravitational field, then the ratio of the gravitational to the
inertial mass must likewise be the same for all bodies. By a suitable
choice of units we can thus make this ratio equal to unity. We then
have the following law: The gravitational mass of a body is equal to
its inertial mass. It is true that this important law had hitherto been
recorded in mechanics, but it had not been interpreted. A satisfactory
interpretation can be obtained only if we recognize the following fact:
The same quality of a body manifests itself according to circumstances
as "inertia" or as "weight". In the following section we shall
show to what extent this is actually the case, and how this question is
connected with the general postulate of relativity".
This is what Einstein says about the two masses. Everything seems to
be right, but it isn't. Let us analyze the mathematical relations
shown above by Einstein.
The second law of the mechanics is:
F=m.a...........(1). And the law of the universal attraction
is:
G=kMm/r^2.....(2)
We can say that G is the gravitational force of attraction between
the two material bodies with mass M and m. But we can't say that F of
the second law of the mechanics is an inertial force or an external
force without describing and the phenomenon of the motion of the
material body, because the accelerated motion doesn't appear without
the action of an external force. According to the third law of the
mechanics the two forces are opposed and equal between them. So if we
write the second law as being:
Fi=m.a ........(1a) ,then it should be considered as a relation for
the calculation of the inertial force when we know the value of the
acceleration. And is very important to be mentioned that this force is
an internal force which appears in the material body. But the
acceleration doesn't appear if there do not exist the external force
Fe that determines the material body to move in an accelerated motion.
In the case of the gravitational forces, if we let a material body
with mass m to fall down free it will be attracted by the earth's
gravitational force G, which is an external force for the material
body. From the relation (1) will have:
G=m.a........(3) and, results that the acceleration is:
a=G/m=(k.M.m/r^2)/m=k.M/r^2=g.
Results that g is, simply an acceleration and not the intensity of
the gravitational field.
This is absolute correct because we know the gravitational force by the
Newton's law of the universal attraction. So the mass of this example
evidently is an inertial mass and the work done by other scientists to
demonstrate that the inertial mass and gravitation mass are equal
between them was done in vain. As a result we can write:
G=m.g. ......(4).
By this we can conclude that the term (m.g) represents in fact the
inertial force that appears in the material body that is falling down
and opposes to the gravitational force expressed with the Newton's
relation (2).
The relations (1) and (2) are in fact of such a huge importance that
we can conclude that with the aid of these relations we can calculate
the inertial and the gravitational forces that act upon all the stars
and planets. At the same time these two notions the inertial and
gravitational forces because they act upon every atom of the material
body enable the existence of the cosmos in the shape in which they are
known now. If one of them disappears everything will be destructed.
Out of that we can name the second law of the mechanics as being the
general law of the inertia, because by making the (a=0) we obtain the
first law of the mechanics as a particular case.
The relation, G=m.g is an application of the second law of the
mechanics.
And the most important conclusion is that do not exist gravitational
mass and inertial mass but exists only one mass of the material
bodies. The supposition that Einstein uses as a fundamental law of the
nature for his general relativity is a false supposition and because of
that we can conclude that his general theory of the relativity is false
theory.
28/09/2006.
> Paul
Paul B. Andersen
You have got it backwards.
In relativity, there is but one type of mass.
In Newtonian mechanics, there are two - inertial and gravitational.
Exactly.
Note that the two types of mass isn't Einstein's invention.
He points out that there has always been two types of mass in
Newtonian mechanics, but it has been taken for granted that
they always are equal.
(You could call equality of the masses an empirical law,
but it doesn't follow from the basic laws of Newtonian mechanics.)
Einstein states:
"The same quality of a body manifests itself according to circumstances
as "inertia" or as "weight"."
That is, the inertial mass and gravitatonal mass is one and the same
quality of the body. There is but one mass.
This is the equivalence principle on which GR rests.
> This is what Einstein says about the two masses. Everything seems to
> be right, but it isn't. Let us analyze the mathematical relations
> shown above by Einstein.
> The second law of the mechanics is:
> F=m.a...........(1). And the law of the universal attraction
> is:
> G=kMm/r^2.....(2)
Note that these are two different laws.
There is nothing in Newtonian mechanics that say the constant
of proportionality m in the former law and ditto in the latter law
have to be the same constant.
Experiments indicate they are the same, but in Newtonian mechanics,
this is accidental, there is no law which say they have to be the same.
If I understand you right, your point is that according to Newtonian
mechanics, gravitational mass and inertial mass must be the same,
there is but one kind of mass, and the equivalence principle is true.
I don't think you have proved that, your proof is basically:
"if we assume the two m's in the two laws are the same m,
then there is but one m."
And your conclusion is weird indeed:
> And the most important conclusion is that do not exist gravitational
> mass and inertial mass but exists only one mass of the material
> bodies. The supposition that Einstein uses as a fundamental law of the
> nature for his general relativity is a false supposition and because of
> that we can conclude that his general theory of the relativity is false
> theory.
> 28/09/2006.
You assure that the equivalence principle is true,
and conclude that general relativity, which rest on
the validity of the equivalence principle, is a false.
An extraordinary achievement! :-)
Paul
Paul B. Andersen
>
> Hmm.. this is quite an interesting subject.
If you think so, why don't you give a serious response?
[snip babble]
Paul
Sorcerer
| [snip babble]
|
Ok, mission accomplished as requested.
Now to the serious stuff.
Stamenin
I agree with you that exists only one mass but do not agree that this
is a cause for believing that because of that could be concluded the
existence of the principle of the equivalence.
In reality exists the behavior of the mass to act as a gravitational
force and as an inertial force. The situations in which it acts so are
three:
1) Is the case where do not exist external forces. In this case the
material body tends to keep his motion according to the first law of
the mechanics.
2) Is the case when upon the material body acts a mechanical force. In
this case the mass acts as an inertial mass tending to oppose to the
external mechanical force. Such a case is the example with the chest in
the large portion of the cosmos where do not exist gravitational
forces. The fact that in the chest the man feels the inertial force as
a gravity is not a cause to conclude that there exists an gravitational
field an to say that the man could conclude that the chest is in a
motion with constant speed because Einstein knew that the Galilean
coordinate system from which he was observing the chest was in a
inertial motion. This is an evident contradiction.
3) And the final case is when upon the material body acts an external
gravitational force. This is the most important case when the mass acts
in two ways, as an inertial mass according to the second law of the
mechanics, and as a gravitational mass according to the law of the
universal attraction. In this case are hidden the most of the not
understood phenomena, for example, why a man in falling at the earth do
not feel the gravitation. And this case explains why are valid the laws
of the mechanics at the earth in spite of the fact that we have here
gravitational and inertial forces. The earth is rotating around the sun
and upon the earth acct gravitational and inertial forces but we do not
feel them because this forces act upon every atom of the material
bodies and upon us.
About these phenomena I give the following explanations:
The nature of the inertial force is not the same with the nature of
the gravitational force. One is exterior the other is interior in the
material body. And this is an additional argument against the
acceptance of the replacing the force with a gravitational field. What
permitted to Einstein, to do this change is the fact, that this two
forces act upon every atom of the body's mass (m). But there is not
any reason to not make a difference between a force and a gravitational
field.
But the property of the gravitational and inertial forces to act upon
every atom of the mass of the material body is very important for
something else. This property is that, that allows us the explanation,
why the principle of the relativity is valid and for coordinate system
rigidly attached to the center of the earth or to the earth's
surface. This property can say that, stays on the bases in which is the
world constructed, and without it Galilei and Newton wouldn't be able
to discover the mechanical laws and the cosmos couldn't exist in the
form of the suns and planets. It is very important to mention that the
laws of the mechanics and the principle of the relativity are related
to the motion of the material bodies with a known mass (m). Because of
that Einstein wasn't right to apply the principle of the relativity
for a light ray, for which we don't know its mass (m), we don't
know its nature if it is ondulatory or corpuscular. We can see with the
aid of the light but we can't see a light ray.
But let us continue with the relativity and give a better explanation
why the principle of the relativity is valid on the earth. Until now we
have been speaking about the principle of the relativity as being valid
relatively to the absolute inertial system or a relative system which
is in motion with a constant speed and in a right line relatively to
him. But in the same time we said that such coordinate systems are
impossible to be determined. It seams that we enter in a miraculous
circle. From this miraculous circle we can get out by taking in
consideration the property of the gravitational and inertial forces to
act upon every atom of the material body.
In order to explain the action of the inertial and gravitational
forces upon the material bodes let us take an example with a cosmic
cabin satellite of the earth. In the inside of the cabin there is a
state of weightlessness in spite of the fact that there act two forces,
one exterior as a result of the gravitation of the earth, called
centripetal, and one centrifugal, as a result of the inertia of the
cabin mass. In this small space, we have completely a similar
situation, as it should be if the cabin is placed in so-called large
portion in cosmos where do not exist gravitational forces. But there
the cabin will be in a motion in a right line and with constant speed,
while here in the earth it will be in a circular motion with constant
linear speed. If the cabin satellite of the earth is moving through an
elliptic orbit the motion will be with variable speed and direction. If
it falls directly toward the earth its motion will be in a right line
and with accelerated speed. And just if we suppose that the cabin is
coming with a high speed from the cosmos the cabin will pass through a
hyperbolic trajectory changing the speed upward when is coming and
diminishing it at the part of the trajectory when it is departing from
the earth. All these motions are referred to a Galileian system of
coordinates with his origin in the center of the earth and with the
coordinates oriented toward the fix stars. In all these motions, in the
cabin exists the situation of weightlessness.
If we will throw a pencil in the cabin it would move in a right line
if the cabin will not have a motion of rotation.relatively to the fix
stars. In such a way, we can say that in the cabin, relatively to a
coordinate system rigidly attached to the cabin with his coordinates
oriented toward the fix stars, the Newton's laws are valid in the
same form, as they are in the cabin, when it is far away in cosmos in a
large portion of space were doesn't exist gravitational forces.
If we now make a comparison of this example with the earth's
surface, we can realize that everything would be similar if here
wouldn't be the earth's gravitation. Because of this the second
Newton's law has a same form only in horizontal plane and not in
vertical direction. That means that the principle of the relativity is
valid only for motions in a horizontal direction. The rotation of the
earth around his axis doesn't have any influence because, this motion
is done with a constant speed and the centrifugal force which appears
as a result of the rotation, is annulled by the gravitation of the
earth. The centripetal force which appears because of the gravitational
attraction of the sun is annulled with the centrifugal force which
appears as a consequence of the earth's rotation around the sun and
because this two forces act upon every atom of everything that exits at
the earth. That is why we do not feel the influence of the sun
attraction, and the attraction of the galaxy and nobody of us is aware
that is separately a satellite of the sun. And this is the cause why
we can say that the Galilean coordinate system as had been defined by
me is a good approximation of the absolute coordinate system, defined
by Newton. But we have to note that, the inertia of the material bodies
creates inertial forces only when they change their speed relatively to
the absolute coordinate system. In this way we can say now that by this
is solved the miraculous circle that was created by the discovery of
the first law of the mechanics by Galileo Galilei where the notions of
a material body is staying or is moving uniform and in a right line
weren't defined properly.
By what have said we until now, results that all these curbed
motions are in fact inertial motions. The material bodies being, the
cabin or the planets, all they are in a state of equilibrium without
any resultant force to be realized by man. These bodies are moving in
an empty space without braking forces and because of that their motion
is permanent. From this results, that the cosmos is in permanent motion
and that there aren't material bodies which are in a state of
staying. By this ones again we can say that not having such bodies, we
haven't any possibility to find out an absolute inertial system. But
the inertia of the material bodies creates inertial forces only when
the material bodies change their speed and their direction relatively
to the absolute inertial system.
Koobee Wublee
> Androcles wrote:
| Stamenin wrote:
> Einstein's book "Relativity" is a popular book, not a scientific
> paper.
Yes, it is so true. Einstein's book is full of errors. Do you like
how Einstein pulled out two of the equations that build up the Lorentz
transform from another two equations equating zero with zero? It is
not even good magic trick.
> It can at the very best give you a vague idea of what GR is about.
Do you, Professor Andersen? Let's find out.
> If you put yourself behind the wheel in a F1 car,
> you can accelerate at 1g in the horizontal direction.
> The will feel the 1000N force pushing your 100 kg body
> in the back extremely well.
>
> If you jump out of the window, you feel nothing
> pushing you while you are falling.
>
> If the acceleration is the same in the two cases,
> why is then what you feel so very different?
>
> > Ever had that "butterfly in the tummy" feeling that children
> > love when riding roller coasters? It's because your internal
> > organs bend your diaphragm relative to your skeleton.
> > In gravitational free fall ALL of your body accelerates
> > together simultaneously, but in a push your fat arse is
> > compressed between your skin and pelvis, which has inertia.
> > [...]
> > Gravity acts THROUGH matter, there are no gravity shields.
>
> ...why don't you give a serious response?
Androcles gave you an excellent answer and a well thought-out response.
Feeling is a biological substance. From that, the feeling of free
fall shall be explained through the biological feeling. Chair pushing
you up applied to your arse first. Thus, you feel the gravity with
your arse first. Again, this is a subject best left for biologists to
answer. In the meantime, let's get back to our discussions in
physics.
> [...]
>
> That is, the inertial mass and gravitatonal mass is one and the same
> quality of the body. There is but one mass.
> This is the equivalence principle on which GR rests.
I am sure you don't understand GR.
> | The second law of the mechanics is:
> | F=m.a...........(1). And the law of the universal attraction
> | is:
> | G=kMm/r^2.....(2)
>
> Note that these are two different laws.
> There is nothing in Newtonian mechanics that say the constant
> of proportionality m in the former law and ditto in the latter law
> have to be the same constant.
Yeah, tell me about it. As one of the infinite solutions to the
Einstein field equations was solved, we have the
semi-proto-Schwarzschild metric described in the following spacetime
equation.
ds^2 = c^2 (1 + K / r) dt^2 - dr^2 / (1 + K / r)...
Only through the Newtonian gravitational law (2) equating with the
Euler-Lagrange equation associated with r which is basically (1) that
you have the following
K = - 2 G M / c^2
> Experiments indicate they are the same, but in Newtonian mechanics,
> this is accidental, there is no law which say they have to be the same.
True, but the standard interpretation to GR does not address this issue
either. Only through proper interpretations to Noether's theorem,
that the energy conservation is indeed a universal and fundamental
phenomenon in which un-mistakenly identifies gravity as a force,
Galileo's principle of Equivalence can be proven.
> [...]
>
> You [Stamenin] assure that the equivalence principle is true,
> and conclude that general relativity, which rest on
> the validity of the equivalence principle, is a false.
Einstein Equivalence Principle is a dead end in the development of GR.
"Hey, look, mom. No hands. Me and Marcel [Grossmann] just jumped
out of the windows, and we expect to understand gravity." Yeah,
right! Einstein and Grossmann's work based on this principle went
nowhere. On the contrary, Newton came up with the law of gravity by
observing a falling apple. Einstein in a free fall could not do any
better than Newton observing objects in a free fall. As Dr. Roberts
has ingeniously pointed out, it would eventually meet the ground in a
tragic end.
You may ask then where the Einstein field equations come from. Well,
the Einstein field equations can only be derived through the Lagrangian
Hilbert patched together like that monster by Doctor Frankenstein.
There is no other way, professor. Don't feel bad. Almost (near
100%) of the experts in this field do not understand GR either. Dr.
Hartle, with great endorsement from professor Draper on a crusade to
teach the calculus of Variations applied to the geodesics first before
Riemannian differential geometry, may just spell doom for GR.
Currently, with Riemannian differential geometry, I can safely say not
too many really understand this subject. However, the peer pressure is
forcing them to see the nonexistence of the Emperor's new clothes
'til this day. And this is the modern science. Very sad, indeed.
Sorcerer
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:1160283124.7...@b28g2000cwb.googlegroups.com...
BTW, Andersen is an ASSistant professor in a county of 150,000
souls and will never understand mathematics.
I call him Tusselad, Norwegian for "troll", which is all he is.
Androcles
Mike
You feel your reaction to the force exterted on you by the chair that
is essentially taking you forward. This is an inertial force. No chair,
no motion. make sure it is bolted well.
>
> If you jump out of the window, you feel nothing
> pushing you while you are falling.
Assuming negligible air friction, you feel nothing because there is no
chair as in the car case. There is no connection between the two
examples because you are trying to make Force a directly observable
quantity, which IS NOT and Einstein failed to understand this and
ridiculed himself although by accident the EP is valid locally.
> If the acceleration is the same in the two cases,
> why is then what you feel so very different?
You know, there are deaf, blind, autistic and all sorts of people. Are
you going to base a science on what you feel?
Then, you should understand that Force is not any directly measurable
quantity but it is only measured through its effect, acceleration in
the case of inertial forces and displacement in the case of potential
forces. When the second derivative of position wrt time is not zero we
know there is a force acting on a body. Nobody cares what you feel.
Thought experiments are the curse of modern science. You are wasting
your time dealing with issues you do not understand.
Mike
>
> Paul
Paul B. Andersen
So we better use instruments.
What does an accelerometer measure if you put it in the F1 car?
In your chair?
While free falling?
>
> Then, you should understand that Force is not any directly measurable
> quantity but it is only measured through its effect, acceleration in
> the case of inertial forces and displacement in the case of potential
> forces. When the second derivative of position wrt time is not zero we
> know there is a force acting on a body. Nobody cares what you feel.
>
> Thought experiments are the curse of modern science. You are wasting
> your time dealing with issues you do not understand.
Do you think an accelerometer in a real experiment
would show a different acceleration than I stated
in my thought experiments above?
Paul
Paul B. Andersen
The man in the chest knows but one thing.
The floor is pushing him with a measurable external force.
He can then conclude that he must be accelerating.
What's contradicting about that?
> 3) And the final case is when upon the material body acts an external
> gravitational force. This is the most important case when the mass acts
> in two ways, as an inertial mass according to the second law of the
> mechanics, and as a gravitational mass according to the law of the
> universal attraction.
But the man in the chest standing on the ground knows but one thing.
The floor is pushing him with a measurable external force.
He can then conclude that he must be accelerating.
There are no differences which can be measured locally.
> In this case are hidden the most of the not
> understood phenomena, for example, why a man in falling at the earth do
> not feel the gravitation.
Not understood phenomena?
According to GR, it is simple.
A free falling body feels no force pushing him because
no force is pushing him and he isn't accelerating.
He is in inertial motion.
> And this case explains why are valid the laws
> of the mechanics at the earth in spite of the fact that we have here
> gravitational and inertial forces. The earth is rotating around the sun
> and upon the earth acct gravitational and inertial forces but we do not
> feel them because this forces act upon every atom of the material
> bodies and upon us.
Yes, of course that's the way Newtonian gravitation explains it.
No need to explain what has been known for centuries.
> About these phenomena I give the following explanations:
>
>
> The nature of the inertial force is not the same with the nature of
> the gravitational force. One is exterior the other is interior in the
> material body.
Stop right there.
There is no "inertial force", because it isn't really a force.
You are probably referring to the reaction force to the external
force that is accelerating the body. But this reaction force
"act upon every atom of the material bodies and upon us", just
like the Newtonian gravitational force does.
So if the latter is "internal", so is the former.
> And this is an additional argument against the
> acceptance of the replacing the force with a gravitational field. What
> permitted to Einstein, to do this change is the fact, that this two
> forces act upon every atom of the body's mass (m). But there is not
> any reason to not make a difference between a force and a gravitational
> field.
Einstein did not remove inertia, he removed the gravitational force.
According to GR there is no gravitational force.
A body accelerates if and only if an external force acts on it. Period.
All you have said to now is according to Newtonian mechanics.
As I said, there is no need to explain what has been known for centuries.
So what is your point?
In what way does that make GR inconsistent?
> By what have said we until now, results that all these curbed
> motions are in fact inertial motions. The material bodies being, the
> cabin or the planets, all they are in a state of equilibrium without
> any resultant force to be realized by man. These bodies are moving in
> an empty space without braking forces and because of that their motion
> is permanent. From this results, that the cosmos is in permanent motion
> and that there aren't material bodies which are in a state of
> staying. By this ones again we can say that not having such bodies, we
> haven't any possibility to find out an absolute inertial system. But
> the inertia of the material bodies creates inertial forces only when
> the material bodies change their speed and their direction relatively
> to the absolute inertial system.
Interesting statement.
Are you now arguing for GR? :-)
"all these curbed motions are in fact inertial motions" is
according to GR, you know.
I still don't get your point. Why do you claim that GR is inconsistent?
Both GR and Newtonian mechanics + gravitation are consistent theories,
you know. If they predicted the same for everything, they would
be two alternative interpretations of the same reality.
(Two different metaphysics.)
But they don't.
And in every case where there is a measurable difference between
the predictions of Newtonian gravitation and GR, GR has come out on top.
Newtonian gravitation is falsified.
GR is not - so far.
If it ever is, Newtonian gravitation is not an alternative.
Paul
Mike
You are wondering because you do not understand very basic physics and
at the same time you have been brain washed by icreadible
misconceptions and lies.
An accelerometer can measure only "by reacting" to a force. A explained
to you this already. Your body acts like an accelerometer when
contrained by a chair. Since action-reactions pairs act on different
bodies that act on each other, you need to contrain the accelerometr on
free fall to get the reading. That is not possible of course and you
get no reading. The same would happen if you let an accelerometr loose
in the F1 car.
Now, having reviewd basic physics, you can make an accelerometer using
a laser and measure position to a reference point. If you differentiate
twice carefully you get acceleration. Digital encoders attached on
motir shafts are used for this purpose all the time. You get a measure
of acceleration and then multiply by mass to get the force.
Remember, Force is not a directly measurable quantity but we measure
only its affects.
If you are clever enough and open minded you will understand the flaws
of the Einstein thought experiment and also way although he was totally
wrong and had misconceptions about basic physics it turns out the EP is
valid locally.
Mike
>
> Paul
Tom Roberts
> Remember, Force is not a directly measurable quantity but we measure
> only its affects.
Piezoelectric crystals and human nerve cells essentially measure force
directly.
Tom Roberts
LEJ Brouwer
What a complete pile of rubbish. You have to measure the resultant
voltage across the surface get an INDIRECT measure of the force, and
even that measurement ignores higher-order effects. As for human nerve
cells, there are probably few less direct ways to measure a force than
that. Who are you trying to kid now, Tom? And I'm still waiting for
your long list of 'glaringly obvious problems' with the composite
picture of the photon.
Mike
Really? Piezoelectric elements produce charges that are proportional to
the applied force. In piezoelectric accelerometers there is a seismic
mass attached to the element They operate according to Newton's 2nd
law.
Force is an "intellectual construction" useful in modeling problems of
mechanics. There is no way to measure force directly or devise
experiments that measure force directly. Those who based alternative
theories on thought experiments that questioned the reality of force
simply did not understand physics and what has been achieved before
them. There are other sound alternatives in place of force that result
in competitive metaphysics rather than insisting that mechanics is
plain geometry of spacetime. An example is power:
http://www.doaj.org/abstract?id=119444&toc=y/
Nevertheless, force mechanics are totally consistent and accurate even
in gravitational cases when second order effects are taken into
account.
Mike
>
>
> Tom Roberts
Paul B. Andersen
An accelerometer measures proper acceleration.
This is the _only_ acceleration that is measurable
without external references.
Doesn't that tell you something?
Put the accelerometer on your chair, and it shows 1g acceleration.
Put the accelerometer in the F1 car, it shows 1g horizontal
acceleration and 1g vertical acceleration, that is 1.41g.
Throw it out the window, and it shows zero acceleration
while in free fall.
> Now, having reviewd basic physics, you can make an accelerometer using
> a laser and measure position to a reference point. If you differentiate
> twice carefully you get acceleration. Digital encoders attached on
> motir shafts are used for this purpose all the time. You get a measure
> of acceleration and then multiply by mass to get the force.
Never before saw I anybody calling an instrument which measures
angular velocity an accelerometer. :-)
The accelerometers that are used all the time measure
proper acceleration.
> Remember, Force is not a directly measurable quantity but we measure
> only its affects.
Sure. There are indeed very few entities which can be measured directly.
But the external force acting on a mass is easily measured, although
indirectly.
Are you going to tell me that force is unmeasurable? :-)
> If you are clever enough and open minded you will understand the flaws
> of the Einstein thought experiment and also way although he was totally
> wrong and had misconceptions about basic physics it turns out the EP is
> valid locally.
Open minded? :-)
You do not understand GR, do you?
Both GR and Newtonian gravitation are consistent theories.
If they predicted the same for everything, they would be two alternative
interpretations of the same reality. In that case it would be a matter
of taste if you preferred to interpret gravitation as a force according
to Newton, or as a curvature of space-time according to GR.
But they don't predict the same for everything.
And in every case where there is a measurable difference between
the predictions of Newtonian gravitation and GR, GR has come out on top.
Newtonian gravitation is falsified.
GR is not - so far.
If it ever is, Newtonian gravitation is not an alternative.
A question:
If you are orbiting the Moon, what is your acceleration?
How would you measure it?
Paul
Paul B. Andersen
Yes, let's do that.
Let's use instruments.
What does an accelerometer measure?
>> [...]
>>
>> That is, the inertial mass and gravitatonal mass is one and the same
>> quality of the body. There is but one mass.
>> This is the equivalence principle on which GR rests.
>
> I am sure you don't understand GR.
>
>> | The second law of the mechanics is:
>> | F=m.a...........(1). And the law of the universal attraction
>> | is:
>> | G=kMm/r^2.....(2)
>>
>> Note that these are two different laws.
>> There is nothing in Newtonian mechanics that say the constant
>> of proportionality m in the former law and ditto in the latter law
>> have to be the same constant.
>
> Yeah, tell me about it. As one of the infinite solutions to the
> Einstein field equations was solved, we have the
> semi-proto-Schwarzschild metric described in the following spacetime
> equation.
>
> ds^2 = c^2 (1 + K / r) dt^2 - dr^2 / (1 + K / r)...
>
> Only through the Newtonian gravitational law (2) equating with the
> Euler-Lagrange equation associated with r which is basically (1) that
> you have the following
>
> K = - 2 G M / c^2
And how is this related to the fact that
there is nothing in Newtonian mechanics that say the constant
of proportionality m in the former law and ditto in the latter law
have to be the same constant?
>> Experiments indicate they are the same, but in Newtonian mechanics,
>> this is accidental, there is no law which say they have to be the same.
>
> True, but the standard interpretation to GR does not address this issue
> either. Only through proper interpretations to Noether's theorem,
> that the energy conservation is indeed a universal and fundamental
> phenomenon in which un-mistakenly identifies gravity as a force,
> Galileo's principle of Equivalence can be proven.
The equivalence principle is a postulate of GR.
I see.
You are the only one understanding GR. :-)
Are you also the only one understanding SR and the Lorentz transform?
In that case, maybe you could point out the error in the following:
In a frame of reference S, a photon with momentum p is moving
along the y -axis. The angle phi is thus pi/2
p
^
| phi
-----|---------------------------> x
A frame of reference S' is moving in the x direction of S
with the speed v.
p
^
| phi'
-----|---------------------------> x'
What is the angle phi' ?
The correct answer is phi' = arccos(-v/c)
\
\ phi'
----\-----------------------------------> x'
So we can conclude that if v and p are transverse in S,
they are NOT transverse in S', because the angle between
v and p is arccos(-v/c) in S'.
Conversely we can conclude that if v and p are transverse in S',
they are NOT transverse in S, because the angle between
v and p is arccos(v/c) in S.
We can sum it up thus:
If phi = pi/2, then phi' = arccos(-v/c)
if phi' = pi/2, then phi = arccos(v/c)
The general equation valid for any angle is:
cos(phi') = (cos(phi) - v/c)/(1 - (v/c)*cos(phi))
or:
cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
This is aberration.
Now, if the source is stationary in S, and the observer
is stationary in S' we can conclude:
If the velocity of the observer is transverse to the wave vector
in the source frame S, then phi = pi/2, and the observed Doppler shift
will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2) = f/sqrt(1 - v2/c2)
This is a blue shift
If the velocity of the source is transverse to the wave vector
in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
and the observed Doppler shift will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
f' = f*sqrt(1 - v2/c2)
This is a red shift.
Summed up:
----------------------------------------------------------------
# If the velocity of the observer is transverse to the wave vector
# in the source frame, the observed Doppler shift will be:
# f' = f/sqrt(1 - v2/c2)
#
# If the velocity of the source is transverse to the wave vector
# in the observer frame, the observed Doppler shift will be:
# f' = f*sqrt(1 - v2/c2)
----------------------------------------------------------------
Since this follows from the very mathematics of
the Lorentz transform which you claim to understand,
I will assume you know this is correct.
In the case you don't agree, I challenge you to show that
the above is NOT what the Lorentz transform predicts.
General talk about symmetry won't do.
You will have to do the math.
Paul
Koobee Wublee
> Koobee Wublee wrote:
> > True, but the standard interpretation to GR does not address this issue
> > either. Only through proper interpretations to Noether's theorem,
> > that the energy conservation is indeed a universal and fundamental
> > phenomenon in which un-mistakenly identifies gravity as a force,
> > Galileo's principle of Equivalence can be proven.
>
> The equivalence principle is a postulate of GR.
The principle of Equivalence nor Einstein's Equivalence Principle has
anything to do with GR. The principle of equivalence was first
proposed by Galileo.
> And how is this related to the fact that
> there is nothing in Newtonian mechanics that say the constant
> of proportionality m in the former law and ditto in the latter law
> have to be the same constant?
Newtonian's law of gravity stipulates on the principle of
Equivalence. The gravitational constant is added to show this.
> > Einstein Equivalence Principle is a dead end in the development of GR.
> > "Hey, look, mom. No hands. Me and Marcel [Grossmann] just jumped
> > out of the windows, and we expect to understand gravity." Yeah,
> > right! Einstein and Grossmann's work based on this principle went
> > nowhere. On the contrary, Newton came up with the law of gravity by
> > observing a falling apple. Einstein in a free fall could not do any
> > better than Newton observing objects in a free fall. As Dr. Roberts
> > has ingeniously pointed out, it would eventually meet the ground in a
> > tragic end.
> >
> > You may ask then where the Einstein field equations come from. Well,
> > the Einstein field equations can only be derived through the Lagrangian
> > Hilbert patched together like that monster by Doctor Frankenstein.
> > There is no other way, professor. Don't feel bad. Almost (near
> > 100%) of the experts in this field do not understand GR either. Dr.
> > Hartle, with great endorsement from professor Draper on a crusade to
> > teach the calculus of Variations applied to the geodesics first before
> > Riemannian differential geometry, may just spell doom for GR.
> > Currently, with Riemannian differential geometry, I can safely say not
> > too many really understand this subject. However, the peer pressure is
> > forcing them to see the nonexistence of the Emperor's new clothes
> > 'til this day. And this is the modern science. Very sad, indeed.
>
> I see.
> You are the only one understanding GR. :-)
Yes, I am glad that you are proud of me.
> Are you also the only one understanding SR and the Lorentz transform?
It appears to be the case as well. It is nice of you to point that out
as well. :-)
> In that case, maybe you could point out the error in the following:
>
> [...]
>
> This is a blue shift
>
> If the velocity of the source is transverse to the wave vector
> in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
> and the observed Doppler shift will be:
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
> f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
> f' = f*sqrt(1 - v2/c2)
> This is a red shift.
In this case, phi' means nothing. phi = pi / 2. Thus, cos(phi) = 0.
Then,
f' = f / sqrt(1 - v^2 / c^2)
It is still a blue shift. Both cases are indicating a blue shift
according to the Lorentz transform.
> Summed up:
> ----------------------------------------------------------------
> # If the velocity of the observer is transverse to the wave vector
> # in the source frame, the observed Doppler shift will be:
> # f' = f/sqrt(1 - v2/c2)
> #
> # If the velocity of the source is transverse to the wave vector
> # in the observer frame, the observed Doppler shift will be:
> # f' = f*sqrt(1 - v2/c2)
> ----------------------------------------------------------------
>
> Since this follows from the very mathematics of
> the Lorentz transform which you claim to understand,
> I will assume you know this is correct.
>
> In the case you don't agree, I challenge you to show that
> the above is NOT what the Lorentz transform predicts.
> General talk about symmetry won't do.
> You will have to do the math.
You still don't understand the Lorentz transform. That is because
you have failed to understand the symmetry therein. You don't
understand the principle of Relativity in which it is a property to the
Lornetz transform.
You still get an 'F' for failure.
Stamenin
floor was pushing him because the chest was pushing by the imaginary
being. Is this a not undertanding or a trick?
>
> > 3) And the final case is when upon the material body acts an external
> > gravitational force. This is the most important case when the mass acts
> > in two ways, as an inertial mass according to the second law of the
> > mechanics, and as a gravitational mass according to the law of the
> > universal attraction.
>
> But the man in the chest standing on the ground knows but one thing.
> The floor is pushing him with a measurable external force.
> He can then conclude that he must be accelerating.
>
> There are no differences which can be measured locally.
right.
> > In this case are hidden the most of the not
> > understood phenomena, for example, why a man in falling at the earth do
> > not feel the gravitation.
>
> Not understood phenomena?
> According to GR, it is simple.
> A free falling body feels no force pushing him because
> no force is pushing him and he isn't accelerating.
> He is in inertial motion.
I have to recognize that the term "is in an inertial motion" is not
the best. Beter would be: can be considered as an inertial motion. A
free falling body do not feels the gravitational force of the earth
because upon it acts the inertial force of the body with which it trys
to opose to the gravitational force. There is an equilibrum between two
forces that act upon every atom of the body and this is a not
understood phenomenon.
> > And this case explains why are valid the laws
> > of the mechanics at the earth in spite of the fact that we have here
> > gravitational and inertial forces. The earth is rotating around the sun
> > and upon the earth acct gravitational and inertial forces but we do not
> > feel them because this forces act upon every atom of the material
> > bodies and upon us.
>
> Yes, of course that's the way Newtonian gravitation explains it.
> No need to explain what has been known for centuries.
I believe that the former explanation is a good negation for this
yours conclusion.
> > About these phenomena I give the following explanations:
> >
> >
> > The nature of the inertial force is not the same with the nature of
> > the gravitational force. One is exterior the other is interior in the
> > material body.
>
> Stop right there.
> There is no "inertial force", because it isn't really a force.
> You are probably referring to the reaction force to the external
> force that is accelerating the body. But this reaction force
> "act upon every atom of the material bodies and upon us", just
> like the Newtonian gravitational force does.
> So if the latter is "internal", so is the former.
I can conclude that you can't explain why a satelite of the earth
rotates permanently and why the earth does the same around the sun.
> > And this is an additional argument against the
> > acceptance of the replacing the force with a gravitational field. What
> > permitted to Einstein, to do this change is the fact, that this two
> > forces act upon every atom of the body's mass (m). But there is not
> > any reason to not make a difference between a force and a gravitational
> > field.
>
> Einstein did not remove inertia, he removed the gravitational force.
> According to GR there is no gravitational force.
So what else there do not exist?
Every thing is inconsistent, this is my oppinion.
>
> > By what have said we until now, results that all these curbed
> > motions can be considered in fact inertial motions. The material bodies being, the
> > cabin or the planets, all they are in a state of equilibrium without
> > any resultant force to be realized by man. These bodies are moving in
> > an empty space without braking forces and because of that their motion
> > is permanent. From this results, that the cosmos is in permanent motion
> > and that there aren't material bodies which are in a state of
> > staying. By this ones again we can say that not having such bodies, we
> > haven't any possibility to find out an absolute inertial system. But
> > the inertia of the material bodies creates inertial forces only when
> > the material bodies change their speed and their direction relatively
> > to the absolute inertial system.
>
> Interesting statement.
> Are you now arguing for GR? :-)
>
"all these curbed motions are in fact inertial motions" is
> according to GR, you know.
NO OF CAORSE NOT.
Mike
You seem not capable of learning. I explained to you that your usual
accelerometers measure reaction forces. In a free falling body, there
is no reaction force simple because it is FREE. But nothing says that
you must have immediate access to the reaction to have a force acting
on a body.
> > Now, having reviewd basic physics, you can make an accelerometer using
> > a laser and measure position to a reference point. If you differentiate
> > twice carefully you get acceleration. Digital encoders attached on
> > motir shafts are used for this purpose all the time. You get a measure
> > of acceleration and then multiply by mass to get the force.
>
> Never before saw I anybody calling an instrument which measures
> angular velocity an accelerometer. :-)
Digital encoders generate pulses. The pulses are added to measure
angular position. This can translate to linear position and you can
deduce from that linear acceleration if there is linear motion
involved. You never show one and you never used one so what do you
know? You car shaft rotates but your car moves linearly.
>
> The accelerometers that are used all the time measure
> proper acceleration.
You said nothing by that.
>
> > Remember, Force is not a directly measurable quantity but we measure
> > only its affects.
>
> Sure. There are indeed very few entities which can be measured directly.
> But the external force acting on a mass is easily measured, although
> indirectly.
I explained to you how to measure the force acting on a free falling
body. It is easy and accurate. That is if you understand Newton's law.
>
> Are you going to tell me that force is unmeasurable? :-)
Either you are stupid or trying hard to be one. I said force is not
directly measurable.
>
> > If you are clever enough and open minded you will understand the flaws
> > of the Einstein thought experiment and also way although he was totally
> > wrong and had misconceptions about basic physics it turns out the EP is
> > valid locally.
>
> Open minded? :-)
> You do not understand GR, do you?
Try harder.
>
> Both GR and Newtonian gravitation are consistent theories.
> If they predicted the same for everything, they would be two alternative
> interpretations of the same reality. In that case it would be a matter
> of taste if you preferred to interpret gravitation as a force according
> to Newton, or as a curvature of space-time according to GR.
> But they don't predict the same for everything.
> And in every case where there is a measurable difference between
> the predictions of Newtonian gravitation and GR, GR has come out on top.
I am tired to list at least 5 prediction of GR that have failed
experimental testing. One is gravitomagentic effects that turn out to
be million billion (yes) times larger than predicted by GR.
>
> Newtonian gravitation is falsified.
> GR is not - so far.
> If it ever is, Newtonian gravitation is not an alternative.
Everything that goes into deep space used Newtonian mechanics. It is
tested and glorified every day. can you write down the GR equations of
motion of a spring-mass system attached to a ceiling in a gravitational
field?
Say no, say you cannot and say that Newtonian mechanics does that as
your corroboirator Roberts will say.
>
> A question:
> If you are orbiting the Moon, what is your acceleration?
> How would you measure it?
I wonder if this is a question from your recent sophomore level
midterm?
Mike
>
> Paul
harry
Koobee, in this case the 'F' is all yours. Please take the above challenge
or leave it. The symmetry consists of the fact that the outcome is inverse
*for the inverse situation* no matter if one takes the source to be moving
and the observer in rest, the other way round, or even both moving. In the
case of purely "transverse Doppler", true (classical) Doppler is nil, and
the observed frequency shift is more appropriately called "time dilation".
You may see the symmetry as follows, slightly rephrasing the above.
Take one emitter-receiver transducer pair in each frame;
When the velocity of the "moving" transducer (f') is transverse to the wave
vectors in the "rest" frame S (with frequency f), the observed Doppler
shifts according to both parties will be:
f '/ f = sqrt(1 - v^2/c^2)
That corresponds to only time dilation as perceived in the "rest" frame, but
a combination of time dilation and true Doppler effect as perceived in the
"moving" frame.
Inversely, when the velocity of the "resting" transducer (f) is transverse
to the wave vectors in the "moving" frame S' (with frequency f'), the
observed Doppler shifts according to both parties will be:
f / f ' = sqrt(1 - v^2/c^2)
That corresponds to a combination of time dilation and true Doppler effect
as perceived in the "rest" frame, but only time dilation as perceived in
the "moving" frame.
In both cases, identical physical observations are made under identical
experimental conditions no matter who is claimed to be "in rest" - and
that's wat the LT are about.
Harald
Sorcerer
"harry" <harald.vanlin...@epfl.ch> wrote in message
news:116047...@sicinfo3.epfl.ch...
| In both cases, identical physical observations are made under identical
| experimental conditions no matter who is claimed to be "in rest" - and
| that's wat the LT are about.
|
| Harald
experimental
conditions, the entire crap is in your stupid imagination, you fuckin'
imbecile.
Koobee Wublee
> Koobee Wublee wrote:
> > [...]
> >
> > In this case, phi' means nothing. phi = pi / 2. Thus, cos(phi) = 0.
> > Then,
> >
> > f' = f / sqrt(1 - v^2 / c^2)
> >
> > It is still a blue shift. Both cases are indicating a blue shift
> > according to the Lorentz transform.
> >
> > [...]
>
> Koobee, in this case the 'F' is all yours. Please take the above challenge
> or leave it. The symmetry consists of the fact that the outcome is inverse
> *for the inverse situation* no matter if one takes the source to be moving
> and the observer in rest, the other way round, or even both moving. In the
> case of purely "transverse Doppler", true (classical) Doppler is nil, and
> the observed frequency shift is more appropriately called "time dilation".
> You may see the symmetry as follows, slightly rephrasing the above.
I have taken up the professor's challenge and found his derivation
false for the scenario where all motions are linear and parallel to
each. In this case, according to one of the two properties of the
Lorentz transform where the principle of Relativity is concerned, there
is no way to tell which one is moving and which one is stationary.
However, if the professor sticks to the rotating platform as the
experiment conducted by Mr. Thim, then he is correct about the second
phi as (phi = pi / 2 + v^2 / c^2). Mr. Thim's experiment is actually
a Sagnac lack-of-effect in disguise. Since we are basically discussing
about the Lorentz transform, the scenario about the linear motions
stands.
> That corresponds to a combination of time dilation and true Doppler effect
> as perceived in the "rest" frame, but only time dilation as perceived in
> the "moving" frame.
> In both cases, identical physical observations are made under identical
> experimental conditions no matter who is claimed to be "in rest" - and
> that's wat the LT are about.
Whoever does not see the symmetry does not understand the principle of
Relativity. Since the principle of Relativity is a property of the
Lorentz transform, anyone who does not understand the principle of
Relativity cannot have understood the Lorentz transform.
So, you get an "F" as well. Better go back and understand the
principle of Relativity. For reference, this principle was identified
by Galileo almost 4 hundred years ago. You, the professor, and most of
the others are 400 years behind in science. What's up with that?
Paul B. Andersen
> Paul B. Andersen wrote:
>> Koobee Wublee wrote:
>> Are you also the only one understanding SR and the Lorentz transform?
>
> It appears to be the case as well. It is nice of you to point that out
> as well. :-)
>
>> In that case, maybe you could point out the error in the following:
>>
>> [...]
>>
>> This is a blue shift
>>
>> If the velocity of the source is transverse to the wave vector
>> in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
>> and the observed Doppler shift will be:
>> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
>> f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
>> f' = f*sqrt(1 - v2/c2)
>> This is a red shift.
>
> In this case, phi' means nothing. phi = pi / 2. Thus, cos(phi) = 0.
> Then,
>
> f' = f / sqrt(1 - v^2 / c^2)
>
> It is still a blue shift. Both cases are indicating a blue shift
> according to the Lorentz transform.
You are babbling. May I remind you:
Paul B. Andersen wrote August 19 :
| I challenge you to prove that you understand
| the very mathematics of the Lorentz transform
| by solving the following problem:
|
| In a frame of reference S, a photon with momentum p is moving
| along the y -axis. The angle phi is thus pi/2
|
| p
| ^
| | phi
| -----|---------------------------> x
|
| A frame of reference S' is moving in the x direction of S
| with the speed v.
|
| p
| ^
| | phi'
| -----|---------------------------> x'
|
| What is the angle phi' ?
|
| Put up, or shut up.
You responded:
| OK, since you insist. However, I do reserve the right to make my life
| similar that is for me to set the velocity of x as observed by x'
| parallel to both x and x' axes. The magnitude of the velocity is
| positive going from left to right. I am also defining this angle as 0
| when your angle = pi / 2. Also defining clockwise phi as positive, we
| have
|
| phi' = - cos-1(v / c)
Later corrected to.
| From my definition of phi', I actually meant
|
| phi' = - sin-1(v / c)
This is correct from your definition of the angle.
With my definition of the angle,
the correct answer is phi' = arccos(-v/c)
We agree about what the actual angle is.
Don't you forget it, and don't talk nonsense.
So we better star over from here:
\
\ phi'
----\-----------------------------------> x'
So we can conclude that if v and p are transverse in S,
they are NOT transverse in S', because the angle between
v and p is arccos(-v/c) in S'.
Conversely we can conclude that if v and p are transverse in S',
they are NOT transverse in S, because the angle between
v and p is arccos(v/c) in S.
We can sum it up thus:
If phi = pi/2, then phi' = arccos(-v/c)
if phi' = pi/2, then phi = arccos(v/c)
The general equation valid for any angle is:
cos(phi') = (cos(phi) - v/c)/(1 - (v/c)*cos(phi))
or:
cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
This is aberration.
Now, if the source is stationary in S, and the observer
is stationary in S' we can conclude:
If the velocity of the observer is transverse to the wave vector
in the source frame S, then phi = pi/2, and the observed Doppler shift
will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2) = f/sqrt(1 - v2/c2)
This is a blue shift
If the velocity of the source is transverse to the wave vector
in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
and the observed Doppler shift will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
f' = f*sqrt(1 - v2/c2)
This is a red shift.
Summed up:
----------------------------------------------------------------
# If the velocity of the observer is transverse to the wave vector
# in the source frame, the observed Doppler shift will be:
# f' = f/sqrt(1 - v2/c2)
#
# If the velocity of the source is transverse to the wave vector
# in the observer frame, the observed Doppler shift will be:
# f' = f*sqrt(1 - v2/c2)
----------------------------------------------------------------
Since this follows from the very mathematics of
the Lorentz transform which you claim to understand,
I will assume you know this is correct.
In the case you don't agree, I challenge you to show that
the above is NOT what the Lorentz transform predicts.
General talk about symmetry won't do.
You will have to do the math.
Paul
Koobee Wublee
| Paul B. Anderson bragged about his own mistake again:
> Koobee Wublee wrote:
> | I challenge you to prove that you understand
> | the very mathematics of the Lorentz transform
> | by solving the following problem:
> |
> | In a frame of reference S, a photon with momentum p is moving
> | along the y -axis. The angle phi is thus pi/2
> |
> | p
> | ^
> | | phi
> | -----|---------------------------> x
> |
> | A frame of reference S' is moving in the x direction of S
> | with the speed v.
> |
> | p
> | ^
> | | phi'
> | -----|---------------------------> x'
> |
> | What is the angle phi' ?
> |
> | Put up, or shut up.
>
> | [...]
> |
> > phi' = - sin[^]-1(v / c)
>
> This is correct from your definition of the angle.
> With my definition of the angle,
> the correct answer is phi' = arccos(-v/c)
>
> [...]
>
> \
> \ phi'
> ----\-----------------------------------> x'
>
> So we can conclude that if v and p are transverse in S,
> they are NOT transverse in S', because the angle between
> v and p is arccos(-v/c) in S'.
Yes. For reference, this is the scenario of S1 to S'.
> Conversely we can conclude that if v and p are transverse in S',
> they are NOT transverse in S, because the angle between
> v and p is arccos(v/c) in S.
No. For reference, this is the scenario of S' to S2.
The scenarios of S1 to S' and of S' back to S2 are completely
independent of each other.
The situation when S1 and S' are traveling in parallel to each other
is the same as when S' and S2 are traveling in parallel to each
other. This is not the scenario of Mr. Thim's setup.
In the scenario of S1 to S', we have
E1' = (E1 - v * p1) / sqrt(1 - v^2 / c^2)
Where
** E1' = h f'
** E1 = h f
** p1 = Momentum vector of the photon as observed by S
** v = Velocity vector of S' as observed by S1
Further more,
** | p1 | = h f1 / c
In a transverse case, we have
f1' = f1 / sqrt(1 - v^2 / c^2)
Where
** v * p1 = 0
Notice we don't give a damn about the angle of the dot product, v'
* p1', which is related to your phi'.
Now, in the scenario of S' to S2, we have
E2 = (E2' - v' * p2') / sqrt(1 - v'^2 / c^2)
Where
** E2 = h f2
** E2' = h f2'
** p2' = Momentum vector of the photon as observed by S'
** v' = Velocity vector of S2 as observed by S'
Further more,
** | p2' | = h f2' / c
S1 and S2 are at rest relative to each other. Thus, more
simplification,
** v' = - v
In a transverse case, we have
f2 = f2' / sqrt(1 - v'^2 / c^2) = f2' / sqrt(1 - v^2 / c^2)
Where
** v' * p2' = 0
Again, notice we don't give a damn about the angle of the dot
product, v * p2, which is related to your phi.
In a nutshell, we have the frequency of the photon emitted by S1 and
observed by S' as follows.
f1' = f1 / sqrt(1 - v^2 / c^2)
This indicates a blue shift.
Continue with the situation from S' to S2, we have the frequency of
the photon emitted by S' as f2' and received by S2 as f2.
f2 = f2' / sqrt(1 - v^2 / c^2)
This also indicates a blue shift.
Although f1' does not have to be f2', you probably would demand
(f1' = f2'). This is OK. This is a special case. In such as
case, we have
f2 = f1' / sqrt(1 - v^2 / c^2) = f1 / (1 - v^2 / c^2)
So, as a photon is emitted in S1, absorbed by S', re-emitted by S',
and at last absorbed by S2 (S1 and S2 are at rest relative to each
other), the observed photon by S2 will be blue shifted from the
original S1. The photon gains energy by doing so. The velocity of
S' will decrease and change direction to compensate for the energy
gain in the photon.
This is the prediction of the Lorentz transform.
> [...]
The rest of garbage snipped due to the fundamental blunder by the
professor as pointed out in the above derivation.
In Mr. Thim's experimental setup, the rotating apparatus (S' in our
terminology) is actually a conductor. Thus, special relativity does
not apply. However, if he were to actually add sensors all along the
edge of S' and transmitter transmitting exactly frequency of the
received photon, then the situation of S' going to S2 is very
different from what I have derived above. Instead, we have
v' * p2' = E2' v'^2 / c^2
Where
** E2 = (E2' - v' * p2') / sqrt(1 - v'^2 / c^2)
This is because only at that certain angle, the emitted photon from the
rotating S' can reach S2. Then, we have
f2 = f2' sqrt(1 - v'^2 / c^2)
The overall observed frequency by S2 as emitted by S1 is
f2 = f1
Just as you have said, but with a very different reason. This is
another one of the situations of the Sagnac non-effect.
Now, put up or shut up.
harry
"Sorcerer" <Headm...@hogwarts.physics_b> wrote in message
news:BhLWg.107743$aP3.1...@fe3.news.blueyonder.co.uk...
:)))
harry
> harry wrote:
>> Koobee Wublee wrote:
>
>> > [...]
>> >
>> > In this case, phi' means nothing. phi = pi / 2. Thus, cos(phi) = 0.
>> > Then,
>> >
>> > f' = f / sqrt(1 - v^2 / c^2)
>> >
>> > It is still a blue shift. Both cases are indicating a blue shift
>> > according to the Lorentz transform.
>> >
>> > [...]
>>
>> Koobee, in this case the 'F' is all yours. Please take the above
>> challenge
>> or leave it. The symmetry consists of the fact that the outcome is
>> inverse
>> *for the inverse situation* no matter if one takes the source to be
>> moving
>> and the observer in rest, the other way round, or even both moving. In
>> the
>> case of purely "transverse Doppler", true (classical) Doppler is nil, and
>> the observed frequency shift is more appropriately called "time
>> dilation".
>> You may see the symmetry as follows, slightly rephrasing the above.
Regretfully, here you deleted an essential part of my posting...
> I have taken up the professor's challenge and found his derivation
> false for the scenario where all motions are linear and parallel to
> each.
Indeed the part that you deleted was about exactly that case.
> In this case, according to one of the two properties of the
> Lorentz transform where the principle of Relativity is concerned, there
> is no way to tell which one is moving and which one is stationary.
That is exact, as I tried to explain to you in the deleted part.
> However, if the professor sticks to the rotating platform as the
> experiment conducted by Mr. Thim, then he is correct about the second
> phi as (phi = pi / 2 + v^2 / c^2). Mr. Thim's experiment is actually
> a Sagnac lack-of-effect in disguise. Since we are basically discussing
> about the Lorentz transform, the scenario about the linear motions
> stands.
We seem to agree that we're not talking about that case.
>> That corresponds to a combination of time dilation and true Doppler
>> effect
>> as perceived in the "rest" frame, but only time dilation as perceived in
>> the "moving" frame.
>> In both cases, identical physical observations are made under identical
>> experimental conditions no matter who is claimed to be "in rest" - and
>> that's wat the LT are about.
>
> Whoever does not see the symmetry does not understand the principle of
> Relativity. Since the principle of Relativity is a property of the
> Lorentz transform, anyone who does not understand the principle of
> Relativity cannot have understood the Lorentz transform.
We also agree on that!
> So, you get an "F" as well. Better go back and understand the
> principle of Relativity. For reference, this principle was identified
> by Galileo almost 4 hundred years ago. You, the professor, and most of
> the others are 400 years behind in science. What's up with that?
Harald
Paul B. Andersen
I can only interpret your "no" to mean that you claim
my statement in front of that "no" to be wrong.
Let's settle this before we continue.
The axes of S' and S are aligned, that is the x'-axis of S'
is parallel to the x-axis of S, ditto for the y and z axes.
The origo of S' is moving in the positive x-direction in S
with the speed v, and consequently the origo of S is moving
in the negative x'-direction in S' with the speed v.
The Lorentz transform from S to S' is:
t' = (t - vx/c^2)/sqrt(1-v^2/c^2)
x' = (x - vt)/sqrt(1-v^2/c^2)
y' = y
z' = z
The Lorentz transform from S' to S is:
t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
According to this transformation, the direction of the velocity
of a light beam (or the velocity/momentum of a photon)
transforms like this:
From S to S':
-------------
^
| phi
-----|---------------------------> x
\
\ phi'
----\-----------------------------------> x'
The general equation relating phi and phi' is:
cos(phi') = (cos(phi) - v/c)/(1 - (v/c)*cos(phi))
If phi = pi/2, phi' = arccos(-v/c)
So we can conclude that if v and p are transverse in S,
they are NOT transverse in S', because the angle between
v and p is arccos(-v/c) in S'.
This far, we agree.
You do however NOT agree to the following:
From S' to S
============
^
| phi'
-----|---------------------------> x'
/
/ phi
----/----------------------------> x
The general equation relating phi' and phi is:
cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
If phi' = pi/2, phi = arccos(v/c)
So we can conclude that if v and p are transverse in S',
they are NOT transverse in S, because the angle between
v and p is arccos(v/c) in S.
You claimed this to be wrong.
Please specify exactly what's wrong.
We will go on when this is settled.
There is no point in applying the LT on Thims's experiment
before we agree how the velocity of light transforms
according to SR.
Paul
harry
> Paul B. Andersen wrote:
>> Koobee Wublee wrote:
No additonal math is needed to show that you make a logical mistake:
>> If the velocity of the source is transverse to the wave vector
>> in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
>> and the observed Doppler shift will be:
>> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
>> f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
>> f' = f*sqrt(1 - v2/c2)
>> This is a red shift.
>
> In this case, phi' means nothing. phi = pi / 2. Thus, cos(phi) = 0.
> Then,
>
> f' = f / sqrt(1 - v^2 / c^2)
>
> It is still a blue shift. Both cases are indicating a blue shift
> according to the Lorentz transform.
One can't change the physical situation and claim that such has no effect on
the measurement. As I pointed out (in the part you snipped), when S and S'
switch place, also f and f' must switch place. It's a simple case of
substitution.
Harald
Sorcerer
| I can only interpret your "no" to mean that you claim
| my statement in front of that "no" to be wrong.
|
| Let's settle this before we continue.
|
| The axes of S' and S are aligned, that is the x'-axis of S'
| is parallel to the x-axis of S, ditto for the y and z axes.
| The origo of S' is moving in the positive x-direction in S
| with the speed v, and consequently the origo of S is moving
| in the negative x'-direction in S' with the speed v.
(but velocity -v)
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img44.gif
| The Lorentz transform from S to S' is:
| t' = (t - vx/c^2)/sqrt(1-v^2/c^2)
| x' = (x - vt)/sqrt(1-v^2/c^2)
| y' = y
| z' = z
|
| The Lorentz transform from S' to S is:
Oops!
The ANDERSEN transform from S' to S is:
| t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)
| x = (x' + vt')/sqrt(1-v^2/c^2)
| y = y'
| z = z'
HAHAHAHA!
Hilarious!
The faster S travels, the longer it takes to get there!
Well, you'd expect that..
If ship A approaches ship B at speed c, then
ship A's clock has stopped relative to ship B,
(because moving clocks run slow)
hence ship B's clock continues on forever relative to
ship A as ship B approaches ship A while standing still.
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img43.gif
"Let's settle this before we continue."
HAHAHAHAHA!
Androcles
Sorcerer
f' * sqrt(1 - v^2 / c^2) = f
wooden block = 1/( Haradl's addled brain)
It's a simple case of substitution. :)))
No additional math is needed to show that Andersen make a stupid error:
The origo of S' is moving in the positive x-direction in S
with the speed v, and consequently the origo of S is moving
in the negative x'-direction in S' with the speed v.
(and the VELOCITY -v)
The Andersen transform from S' to S is:
t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)
^^
Andersen negated x' but did not negate v.
It's a simple case of substitution. :)))
x = (x' + vt')/sqrt(1-v^2/c^2)
^ ^
Andersen negated v but did not negate x'.
It's a simple case of substitution. :)))
wooden block = 1/( Andersen's troll brain)
It's a simple case of substitution. :)))
( Andersen's troll brain) = ( Haradl's addled brain)
It's a simple case of substitution. :)))
FUCKWITS! :)))
Androcles.
Koobee Wublee
> "Koobee Wublee" <koobee...@gmail.com> wrote in message
> > f' = f / sqrt(1 - v^2 / c^2)
> >
> > It is still a blue shift. Both cases are indicating a blue shift
> > according to the Lorentz transform.
>
> One can't change the physical situation and claim that such has no effect on
> the measurement.
I have failed to understand what you are referring to.
> As I pointed out (in the part you snipped), when S and S'
> switch place, also f and f' must switch place.
That is the case according to common sense yes.
> It's a simple case of substitution.
However, we are discussing physics obeying the Lorentz transform. You
have claimed the validity of the Lorentz transform. You must apply it.
You must not creative special cases for the Lorentz transform.
Koobee Wublee
> Koobee Wublee wrote:
> The axes of S' and S are aligned, that is the x'-axis of S'
> is parallel to the x-axis of S, ditto for the y and z axes.
> The origo of S' is moving in the positive x-direction in S
> with the speed v, and consequently the origo of S is moving
> in the negative x'-direction in S' with the speed v.
>
> The Lorentz transform from S to S' is:
> t' = (t - vx/c^2)/sqrt(1-v^2/c^2)
> x' = (x - vt)/sqrt(1-v^2/c^2)
> y' = y
> z' = z
For reference, the above set of equations is equation 1.
> The Lorentz transform from S' to S is:
> t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)
> x = (x' + vt')/sqrt(1-v^2/c^2)
> y = y'
> z = z'
For reference, the above set of equations is equation 2.
> According to this transformation, the direction of the velocity
> of a light beam (or the velocity/momentum of a photon)
> transforms like this:
>
> From S to S':
> -------------
>
>
> ^
> | phi
> -----|---------------------------> x
>
>
> \
> \ phi'
> ----\-----------------------------------> x'
>
> The general equation relating phi and phi' is:
> cos(phi') = (cos(phi) - v/c)/(1 - (v/c)*cos(phi))
For reference, the above is equation 3.
> If phi = pi/2, phi' = arccos(-v/c)
>
> So we can conclude that if v and p are transverse in S,
> they are NOT transverse in S', because the angle between
> v and p is arccos(-v/c) in S'.
>
> This far, we agree.
Yes, but you have not derived equation 3 from either or both equations
1 and 2. Your equations 1 and 2 are not setup to give you equation 3.
Do you really understand the Lorentz transform, the equations 1 and 2?
You then, wrote down the energy transformation equation as equation 4
below.
f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
Since (phi = pi / 2), the above equation becomes
f' = f / sqrt(1 - v^2 / c^2)
> You do however NOT agree to the following:
>
> From S' to S
> ============
>
> ^
> | phi'
> -----|---------------------------> x'
>
>
> /
> / phi
> ----/----------------------------> x
>
> The general equation relating phi' and phi is:
> cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
>
> If phi' = pi/2, phi = arccos(v/c)
>
> So we can conclude that if v and p are transverse in S',
> they are NOT transverse in S, because the angle between
> v and p is arccos(v/c) in S.
>
> You claimed this to be wrong.
I actually don't see you are wrong here. What I meant previously was
your the next step in which you are about to take is wrong --- the step
on transformation of energy. Your mistake was to recycle equation 4.
The equation of you mistake is
f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
It should be the following related to equation 2.
f = f'(1 + v cos(phi') / c) / sqrt(1 - v^2 / c^2)
Since (phi' = pi / 2), the above equation becomes
f = f' / sqrt(1 - v^2 / c^2)
Now, you need to go back to address my previous post.
Paul B. Andersen
Every word I wrote above was true, wasn't it?
>>> Now, having reviewd basic physics, you can make an accelerometer using
>>> a laser and measure position to a reference point. If you differentiate
>>> twice carefully you get acceleration. Digital encoders attached on
>>> motir shafts are used for this purpose all the time. You get a measure
>>> of acceleration and then multiply by mass to get the force.
>> Never before saw I anybody calling an instrument which measures
>> angular velocity an accelerometer. :-)
>
> Digital encoders generate pulses. The pulses are added to measure
> angular position. This can translate to linear position and you can
> deduce from that linear acceleration if there is linear motion
> involved. You never show one and you never used one so what do you
> know? You car shaft rotates but your car moves linearly.
Quite.
You can also throw your log in the water, measure the time
between the knots, and calculate the ship's acceleration.
Same principle.
So that's how the acceleration in inertial navigation
systems is measured, isn't it?
>> The accelerometers that are used all the time measure
>> proper acceleration.
>
> You said nothing by that.
Didn't I by that say that the instrument that is called
an "accelerometer" measures proper acceleration?
>>> Remember, Force is not a directly measurable quantity but we measure
>>> only its affects.
>> Sure. There are indeed very few entities which can be measured directly.
>> But the external force acting on a mass is easily measured, although
>> indirectly.
>
> I explained to you how to measure the force acting on a free falling
> body. It is easy and accurate. That is if you understand Newton's law.
By using digital encoders? :-)
>> Are you going to tell me that force is unmeasurable? :-)
>
> Either you are stupid or trying hard to be one. I said force is not
> directly measurable.
>
>>> If you are clever enough and open minded you will understand the flaws
>>> of the Einstein thought experiment and also way although he was totally
>>> wrong and had misconceptions about basic physics it turns out the EP is
>>> valid locally.
>> Open minded? :-)
>> You do not understand GR, do you?
>
> Try harder.
It was a silly question.
You are stuck in the 19th century, so the answer is obvious.
>> Both GR and Newtonian gravitation are consistent theories.
>> If they predicted the same for everything, they would be two alternative
>> interpretations of the same reality. In that case it would be a matter
>> of taste if you preferred to interpret gravitation as a force according
>> to Newton, or as a curvature of space-time according to GR.
>> But they don't predict the same for everything.
>> And in every case where there is a measurable difference between
>> the predictions of Newtonian gravitation and GR, GR has come out on top.
>
> I am tired to list at least 5 prediction of GR that have failed
> experimental testing. One is gravitomagentic effects that turn out to
> be million billion (yes) times larger than predicted by GR.
>
>> Newtonian gravitation is falsified.
>> GR is not - so far.
>> If it ever is, Newtonian gravitation is not an alternative.
>
> Everything that goes into deep space used Newtonian mechanics. It is
> tested and glorified every day. can you write down the GR equations of
> motion of a spring-mass system attached to a ceiling in a gravitational
> field?
>
> Say no, say you cannot and say that Newtonian mechanics does that as
> your corroboirator Roberts will say.
You are living in your own fantasy world where GR is falsified
and Newtonian gravitation is not, right? :-)
>> A question:
>> If you are orbiting the Moon, what is your acceleration?
>> How would you measure it?
>
> I wonder if this is a question from your recent sophomore level
> midterm?
Quite, it is from my homework.
Should I answer that we can measure it by using a digital encoder?
Paul
Koobee Wublee
> "Koobee Wublee" <koobee...@gmail.com> wrote in message
> Regretfully, here you deleted an essential part of my posting...
>
> > I have taken up the professor's challenge and found his derivation
> > false for the scenario where all motions are linear and parallel to
> > each.
>
> Indeed the part that you deleted was about exactly that case.
>
> > In this case, according to one of the two properties of the
> > Lorentz transform where the principle of Relativity is concerned, there
> > is no way to tell which one is moving and which one is stationary.
>
> That is exact, as I tried to explain to you in the deleted part.
You started out disagree with me. Since the discussion between I and
Professor Andersen leaves no grey area, you have to agree with the
professor as well, and that is how I understand it.
Ok, in the part I snipped, you wrote:
> Take one emitter-receiver transducer pair in each frame;
>
> When the velocity of the "moving" transducer (f') is transverse to the wave
> vectors in the "rest" frame S (with frequency f), the observed Doppler
> shifts according to both parties will be:
> f '/ f = sqrt(1 - v^2/c^2)
>
> That corresponds to only time dilation as perceived in the "rest" frame, but
> a combination of time dilation and true Doppler effect as perceived in the
> "moving" frame.
>
> Inversely, when the velocity of the "resting" transducer (f) is transverse
> to the wave vectors in the "moving" frame S' (with frequency f'), the
> observed Doppler shifts according to both parties will be:
> f / f ' = sqrt(1 - v^2/c^2)
>
> That corresponds to a combination of time dilation and true Doppler effect
> as perceived in the "rest" frame, but only time dilation as perceived in
> the "moving" frame.
Your mathematics actually agrees with me but not your interpretations
in the beginning of your post. That confused me. In fact, what you
wrote did not make much sense to me. Furthermore, you mentioned about
the 'true' Doppler shift as if there is a 'false' Doppler
shift. This further confuses me. I have no intention of being dragged
down discussing true and false Doppler shifts. Thus, I had to snip it
all. After all, you have been a champion of Lorentz and thus the
Lorentz transform for quite sometime.
You further wrote:
> In both cases, identical physical observations are made under identical
> experimental conditions no matter who is claimed to be "in rest" - and
> that's wat the LT are about.
You seem to agree with me. I have no intention of being dragged down
discussing agreeing with me, disagreeing with me, agreeing with me,
disagreeing with me... Thus, again, I had to snip it and just comment
on the very first paragraph you wrote which he had disagreed with me.
Yes, regrettably, I have to snip the rest of your current reply because
I still cannot make out what you are talking about disagreeing with me
and agreeing with me at the same time. It's nothing personal.
Please don't feel so rejected. Hey, my kids are singing 'ABCDEFG
HIJKLMNO PQRSTU VWXY and Z. Now, I know my ABC. Come..." Give it a
try. Maybe you will feel better.
Paul B. Andersen
It is trivially simple to derive equation 3 from either one of 1 and 2.
> You then, wrote down the energy transformation equation as equation 4
> below.
>
> f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
>
> Since (phi = pi / 2), the above equation becomes
>
> f' = f / sqrt(1 - v^2 / c^2)
So if the source is stationary in S and the observer in S',
we can say:
If v and p are transverse in the source frame S (phi = pi/2)
then the observer will see:
f' = f / sqrt(1 - v^2 / c^2)
which is a blue shift.
>> You do however NOT agree to the following:
>>
>> From S' to S
>> ============
>>
>> ^
>> | phi'
>> -----|---------------------------> x'
>>
>>
>> /
>> / phi
>> ----/----------------------------> x
>>
>> The general equation relating phi' and phi is:
>> cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
>>
>> If phi' = pi/2, phi = arccos(v/c)
>>
>> So we can conclude that if v and p are transverse in S',
>> they are NOT transverse in S, because the angle between
>> v and p is arccos(v/c) in S.
>>
>> You claimed this to be wrong.
>
> I actually don't see you are wrong here. What I meant previously was
> your the next step in which you are about to take is wrong --- the step
> on transformation of energy. Your mistake was to recycle equation 4.
> The equation of you mistake is
>
> f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
>
> It should be the following related to equation 2.
>
> f = f'(1 + v cos(phi') / c) / sqrt(1 - v^2 / c^2)
Right.
f' = f sqrt(1 - v^2 / c^2)/(1 + v cos(phi') / c)
> Since (phi' = pi / 2), the above equation becomes
>
> f = f' / sqrt(1 - v^2 / c^2)
Right.
f' = f sqrt(1 - v^2 / c^2)
So if the source is stationary in S and the observer in S',
we can say:
If v and p are transverse in the observer frame S' (phi' = pi/2)
then the observer will see:
f' = f * sqrt(1 - v^2 / c^2)
which is a red shift.
> Now, you need to go back to address my previous post.
I will rather go back to my previous post.
Paul B. Andersen wrote previously:
| Now, if the source is stationary in S, and the observer
| is stationary in S' we can conclude:
|
| If the velocity of the observer is transverse to the wave vector
| in the source frame S, then phi = pi/2, and the observed Doppler shift
| will be:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2) = f/sqrt(1 - v^2/c^2)
| This is a blue shift
|
| If the velocity of the source is transverse to the wave vector
| in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
| and the observed Doppler shift will be:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
| f' = f*(1 - (v/c)^2)/sqrt(1 - v^2/c^2)
| f' = f*sqrt(1 - v^2/c^2)
| This is a red shift.
You called this gibberish.
Now you have confirmed that it is correct.
Can we now consider this for settled, and apply it
on Thim's paper, or are you yet again going to say that
what is correct, is wrong, and then demonstate that it is correct?
Paul
harry
> harry wrote:
>> "Koobee Wublee" <koobee...@gmail.com> wrote in message
>
>> > f' = f / sqrt(1 - v^2 / c^2)
>> >
>> > It is still a blue shift. Both cases are indicating a blue shift
>> > according to the Lorentz transform.
>>
>> One can't change the physical situation and claim that such has no effect
>> on
>> the measurement.
>
> I have failed to understand what you are referring to.
>
>> As I pointed out (in the part you snipped), when S and S'
>> switch place, also f and f' must switch place.
>
> That is the case according to common sense yes.
I merely rewrote Andersen's math in such a way to make crystal clear that
his math actually agrees with that.
>> It's a simple case of substitution.
>
> However, we are discussing physics obeying the Lorentz transform. You
> have claimed the validity of the Lorentz transform. You must apply it.
> You must not creative special cases for the Lorentz transform.
I'll try to remove the confusion in my other, parallel reply.
harry
Hopefully your confusion is not caused by a typo or mix-up of me. We'll see
below!
I intended to rewrite Paul's math in such a way that the symmetry is more
evident, and starting with the standard "time dilation" case. Apparently I
succeeded in that, but you dispute that the math is the same and I could
have messed up the explanation. Here follows a check.
Paul:
Summed up:
----------------------------------------------------------------
# If the velocity of the observer is transverse to the wave vector
# in the source frame, the observed Doppler shift will be:
# f' = f/sqrt(1 - v2/c2) [1a]
#
# If the velocity of the source is transverse to the wave vector
# in the observer frame, the observed Doppler shift will be:
# f' = f*sqrt(1 - v2/c2) [2a]
----------------------------------------------------------------
Harald:
----------------------------------
When the velocity of the "moving" transducer (f') is transverse to the wave
vectors
in the "rest" frame S (with frequency f), the observed Doppler shifts
according to both parties will be:
f '/ f = sqrt(1 - v^2/c^2) [2b]
[...]
Inversely, when the velocity of the "resting" transducer (f) is transverse
to the wave vectors
in the "moving" frame S' (with frequency f'), the observed Doppler shifts
according to both parties will be:
f / f ' = sqrt(1 - v^2/c^2) [1b]
-------------------------------------
That looks the same to me; I merely formulated it in a slightly more precise
and general way and rearranged the terms.
I hope you can agree that [1a] = [1b], and [2a] = [2b].
> In fact, what you
> wrote did not make much sense to me. Furthermore, you mentioned about
> the 'true' Doppler shift as if there is a 'false' Doppler
> shift. This further confuses me. I have no intention of being dragged
> down discussing true and false Doppler shifts. Thus, I had to snip it
> all. After all, you have been a champion of Lorentz and thus the
> Lorentz transform for quite sometime.
With "true" Doppler I mean "classical" Doppler. "Relativistic Doppler" is
just a correction of the classical Doppler effect for time dilation, as
Einstein also explained in 1907. The term "relativistic Doppler" can be
misleading, as it is in fact a combination of two physically different
effects: Doppler and time dilation.
Consequetly, "transverse Doppler" is IMO a misnomer, since it is
relativistic Doppler *without* the Doppler effect: in the frame in which
"transverse" is true, one only has to account for time dilation as the
distance is momentaneously constant.
But all that is not needed to verify if I made an error with rearranging
Paul's math or not.
Regards,
Harald
Koobee Wublee
The above equation is how S' observes a photon sent by S which is a
blue shift.
The above equation is how S observes a photon sent by S' which is
also a blue shift.
> Right.
>
> f' = f sqrt(1 - v^2 / c^2)
The above equation is what frequency of the photon sent by S' as it
is observed by S which indicates a red shift in the emitted frequency.
> So if the source is stationary in S and the observer in S',
No, it does not matter if S or S' is moving. All transverse
observations are blue shifted.
> we can say:
> If v and p are transverse in the observer frame S' (phi' = pi/2)
> then the observer will see:
> f' = f * sqrt(1 - v^2 / c^2)
> which is a red shift.
You are grossly confused. The above equation indicates the emitted
frequency not the observed frequency. The emitted frequency relative
to the observed frequency is red shifted. The observed frequency
relative to the emitted frequency is again also a blue shift.
> > Now, you need to go back to address my previous post.
>
> I will rather go back to my previous post.
>
> Paul B. Andersen wrote previously:
> | Now, if the source is stationary in S, and the observer
> | is stationary in S' we can conclude:
> |
> | If the velocity of the observer is transverse to the wave vector
> | in the source frame S, then phi = pi/2, and the observed Doppler shift
> | will be:
> | f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2) = f/sqrt(1 - v^2/c^2)
> | This is a blue shift
> |
> | If the velocity of the source is transverse to the wave vector
> | in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
> | and the observed Doppler shift will be:
> | f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
> | f' = f*(1 - (v/c)^2)/sqrt(1 - v^2/c^2)
> | f' = f*sqrt(1 - v^2/c^2)
> | This is a red shift.
>
> You called this gibberish.
Yes, a magic trick, sign of a swindler or a con-artist. You can fool
harry but not me.
> Now you have confirmed that it is correct.
No, I have confirmed you are wrong.
> Can we now consider this for settled, and apply it
> on Thim's paper, or are you yet again going to say that
> what is correct, is wrong, and then demonstate that it is correct?
Yes, it is settled that you are wrong.
Paul B. Andersen
How the hell is it possible to calculate the correct
equation and still claim them to be wrong?
Look. Let us repeat what you have stated is correct.
This is the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S.
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
That is, when p and v are transverse in the source frame S,
then phi = pi/2 and the equation above yields:
f' = f/sqrt(1 - v^2/c^2)
But if v and p are transverse in the observer frame S',
then phi' = pi/2 and cos(phi) = v/c
Paul B. Andersen wrote;
| If phi' = pi/2, phi = arccos(v/c)
|
| So we can conclude that if v and p are transverse in S',
| they are NOT transverse in S, because the angle between
| v and p is arccos(v/c) in S.
|
| You claimed this to be wrong.
Koobee Wublee responded:
| I actually don't see you are wrong here.
That is because it is correct!
So get this into your head, and don't repeat that
it is wrong in every second posting!
Read carefully:
| If v and p are transverse in the observer frame S',
| they are NOT transverse in the source frame S,
| because the angle between v and p is arccos(v/c) in S.
You have admitted that this is correct!
And when cos(phi) = v/c, the Doppler equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
yields:
f' = f*sqrt(1 - v^2/c^2)
which is a red shift
========================================================
So when v and p are transverse in the observer frame,
a red shift is observed!
========================================================
You cannot keep contradicting yourself in every second posting,
so unless you admit that this is correct, you are a crank!
Paul
Koobee Wublee
No. Your equation relating the frequency is wrong.
> Look. Let us repeat what you have stated is correct.
>
> This is the equation for the Doppler shift of a signal
> emitted from a source in S, and observed by an observer in S.
>
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
>
> That is, when p and v are transverse in the source frame S,
> then phi = pi/2 and the equation above yields:
> f' = f/sqrt(1 - v^2/c^2)
>
> But if v and p are transverse in the observer frame S',
> then phi' = pi/2 and cos(phi) = v/c
You should write p' instead of p to indicate the momentum of the
photon as observed by S'.
> Paul B. Andersen wrote;
> | If phi' = pi/2, phi = arccos(v/c)
> |
> | So we can conclude that if v and p are transverse in S',
> | they are NOT transverse in S, because the angle between
> | v and p is arccos(v/c) in S.
> |
> | You claimed this to be wrong.
>
> Koobee Wublee responded:
> | I actually don't see you are wrong here.
>
> That is because it is correct!
> So get this into your head, and don't repeat that
> it is wrong in every second posting!
The issue is not the above.
> Read carefully:
> | If v and p are transverse in the observer frame S',
> | they are NOT transverse in the source frame S,
> | because the angle between v and p is arccos(v/c) in S.
>
> You have admitted that this is correct!
>
> And when cos(phi) = v/c, the Doppler equation:
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
The equation above is wrong. It should be
f = f' (1 - v cos(phi') / c) / sqrt(1 - v^2 / c^2)
> yields:
> f' = f*sqrt(1 - v^2/c^2)
> which is a red shift
Wrong here. The observed frequency by S in this case should be
f = f' / sqrt(1 - v^2 / c^2)
Which is always a blue shift.
> ========================================================
> So when v and p are transverse in the observer frame,
> a red shift is observed!
> ========================================================
========================================================
No, any transverse Doppler shift according to the Lorentz transform is
observed to be blue shifted as the mathematics indicates.
========================================================
I gave you a most thorough and professional derivation over yours, and
you have not even read it.
http://groups.google.com/group/sci.physics.relativity/msg/de06b6a03ec2e430?hl=en&
> You cannot keep contradicting yourself in every second posting,
> so unless you admit that this is correct, you are a crank!
I have taken your challenge and pointed out exactly where you have
failed. If you insist on these trollish behaviors, I have no time to
discuss with a professor such as yourself who blunders at the most
basic math and physics.
Paul B. Andersen
It is trivially simple to derive equation 3 from either one of 1 and 2.
> You then, wrote down the energy transformation equation as equation 4
> below.
>
> f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
>
> Since (phi = pi / 2), the above equation becomes
>
> f' = f / sqrt(1 - v^2 / c^2)
So if the source is stationary in S and the observer in S',
we can say:
If v and p are transverse in the source frame S (phi = pi/2)
then the observer will see:
f' = f / sqrt(1 - v^2 / c^2)
which is a blue shift.
>> You do however NOT agree to the following:
>>
>> From S' to S
>> ============
>>
>> ^
>> | phi'
>> -----|---------------------------> x'
>>
>>
>> /
>> / phi
>> ----/----------------------------> x
>>
>> The general equation relating phi' and phi is:
>> cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))
>>
>> If phi' = pi/2, phi = arccos(v/c)
>>
>> So we can conclude that if v and p are transverse in S',
>> they are NOT transverse in S, because the angle between
>> v and p is arccos(v/c) in S.
>>
>> You claimed this to be wrong.
>
> I actually don't see you are wrong here. What I meant previously was
> your the next step in which you are about to take is wrong --- the step
> on transformation of energy. Your mistake was to recycle equation 4.
> The equation of you mistake is
>
> f' = f (1 - v cos(phi) / c) / sqrt(1 - v^2 / c^2)
>
> It should be the following related to equation 2.
>
> f = f'(1 + v cos(phi') / c) / sqrt(1 - v^2 / c^2)
or:
f' = f sqrt(1 - v^2 / c^2)/(1 + v cos(phi') / c)
> Since (phi' = pi / 2), the above equation becomes
>
> f = f' / sqrt(1 - v^2 / c^2)
Right.
f' = f sqrt(1 - v^2 / c^2)
So if the source is stationary in S and the observer in S',
we can say:
If v and p are transverse in the observer frame S' (phi' = pi/2)
then the observer will see:
f' = f * sqrt(1 - v^2 / c^2)
which is a red shift.
> Now, you need to go back to address my previous post.
I will rather go back to my previous post.
Paul B. Andersen wrote:
| Now, if the source is stationary in S, and the observer
| is stationary in S' we can conclude:
|
| If the velocity of the observer is transverse to the wave vector
| in the source frame S, then phi = pi/2, and the observed Doppler shift
| will be:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2) = f/sqrt(1 - v2/c2)
| This is a blue shift
|
| If the velocity of the source is transverse to the wave vector
| in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
| and the observed Doppler shift will be:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
| f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
| f' = f*sqrt(1 - v2/c2)
| This is a red shift.
You called this gibberish.
Now you have confirmed that it is correct.
Can we now consider this for settled, and apply it
on Thim's paper, or are you yet again going to say that
what is correct, is wrong, and then demonstate that it is correct?
Paul
Paul B. Andersen
> Paul B. Andersen wrote:
>> Look. Let us repeat what you have stated is correct.
>>
>> This is the equation for the Doppler shift of a signal
>>
>> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
>>
>> That is, when p and v are transverse in the source frame S,
>> then phi = pi/2 and the equation above yields:
>> f' = f/sqrt(1 - v^2/c^2)
>>
>> But if v and p are transverse in the observer frame S',
>> then phi' = pi/2 and cos(phi) = v/c
>
> You should write p' instead of p to indicate the momentum of the
> photon as observed by S'.
Sure.
But the important point is that the equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
is correct!
>> Paul B. Andersen wrote;
>> | If phi' = pi/2, phi = arccos(v/c)
>> |
>> | So we can conclude that if v and p are transverse in S',
>> | they are NOT transverse in S, because the angle between
>> | v and p is arccos(v/c) in S.
>> |
>> | You claimed this to be wrong.
>>
>> Koobee Wublee responded:
>> | I actually don't see you are wrong here.
>>
>> That is because it is correct!
>> So get this into your head, and don't repeat that
>> it is wrong in every second posting!
>
> The issue is not the above.
>
>> Read carefully:
>> | If v and p are transverse in the observer frame S',
>> | they are NOT transverse in the source frame S,
>> | because the angle between v and p is arccos(v/c) in S.
>>
>> And when cos(phi) = v/c, the Doppler equation:
>> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
>
> The equation above is wrong. It should be
Say - are you drunk, or what?
This is still the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S':
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
It is as correct now as it was above,
so what the hell are you talking about?
>
> f = f' (1 - v cos(phi') / c) / sqrt(1 - v^2 / c^2)
>
>> yields:
>> f' = f*sqrt(1 - v^2/c^2)
>> which is a red shift
>
> Wrong here. The observed frequency by S in this case should be
>
> f = f' / sqrt(1 - v^2 / c^2)
>
> Which is always a blue shift.
Sober up, man.
From whence did you get the idea that the observer is in S?
The observer is still in S', we never changed that.
---------------------------
Can't you read?
I even said:
Read carefully:
| If v and p are transverse in the observer frame S',..
##################
Why the hell don't you read what I write and respond to that
in stead of responding to something I never wrote?
Read the following again - carefully.
Please point out exactly what is wrong in _what I have written_!
Read carefully:
===============
This is the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S':
-------------------------------------------------------------
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
That is, when p and v are transverse in the source frame S,
then phi = pi/2 and the equation above yields:
f' = f/sqrt(1 - v^2/c^2)
That is, the observer in S' will observe a blue shift.
-----------------
But if v and p' are transverse in the observer frame S',
then phi' = pi/2 and cos(phi) = v/c
and the Doppler equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
yields:
f' = f*sqrt(1 - v^2/c^2)
That is, the observer in S' will observe a red shift.
------------------
You cannot make this wrong by pointing out that something
else is wrong!
Paul
Koobee Wublee
> Koobee Wublee wrote:
> > Paul B. Andersen wrote:
> >> Look. Let us repeat what you have stated is correct.
> >>
> >> This is the equation for the Doppler shift of a signal
> >> emitted from a source in S, and observed by an observer in S'.
> >>
> >> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
> >>
> >> That is, when p and v are transverse in the source frame S,
> >> then phi = pi/2 and the equation above yields:
> >> f' = f/sqrt(1 - v^2/c^2)
> >>
> >> But if v and p are transverse in the observer frame S',
> >> then phi' = pi/2 and cos(phi) = v/c
NO, WE DON'T CARE WHAT ANGLE V AND P FORM IN THE OBSERVER FRAME S'.
The other case was from S' to S. Now, are you talking about from
S' S, again?
If that is the case then the answer should be the same in both cases.
If you have a different answer, that means you have not understood the
Lorentz transform.
And then, your (phi = pi / 2) and (cos(phi) = 0). This is the
definition of transverse. It is transverse in the emitter frame.
> > You should write p' instead of p to indicate the momentum of the
> > photon as observed by S'.
>
> Sure.
> But the important point is that the equation:
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
> is correct!
f' = f / sqrt(1 - v^2 / c^2)
Because (phi = pi / 2).
This is again a blue shift.
phi = pi / 2
> > f = f' (1 - v cos(phi') / c) / sqrt(1 - v^2 / c^2)
> >
> >> yields:
> >> f' = f*sqrt(1 - v^2/c^2)
> >> which is a red shift
> >
> > Wrong here. The observed frequency by S in this case should be
> >
> > f = f' / sqrt(1 - v^2 / c^2)
> >
> > Which is always a blue shift.
>
> Sober up, man.
> From whence did you get the idea that the observer is in S?
> The observer is still in S', we never changed that.
> ---------------------------
>
> Can't you read?
> I even said:
> Read carefully:
> | If v and p are transverse in the observer frame S',..
> ##################
v and p are never transverse in the observer frame. In transverse
Doppler, v * p = 0.
> Why the hell don't you read what I write and respond to that
> in stead of responding to something I never wrote?
>
> Read the following again - carefully.
> Please point out exactly what is wrong in _what I have written_!
>
> Read carefully:
> ===============
>
> This is the equation for the Doppler shift of a signal
> emitted from a source in S, and observed by an observer in S':
> -------------------------------------------------------------
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
>
> That is, when p and v are transverse in the source frame S,
> then phi = pi/2 and the equation above yields:
> f' = f/sqrt(1 - v^2/c^2)
> That is, the observer in S' will observe a blue shift.
> -----------------
>
> But if v and p' are transverse in the observer frame S',
Then, this is not the case of a transverse Doppler effect.
> then phi' = pi/2 and cos(phi) = v/c
> and the Doppler equation:
> f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
> yields:
> f' = f*sqrt(1 - v^2/c^2)
> That is, the observer in S' will observe a red shift.
I apologize for not reading your very deliberately misleading scenarios
very carefully. In fact, your two scenarios are exactly the same (both
from S' to S), and since the laws of physics must be the same, both
scenarios should conclude with the same answer.
It is also my fault to expect a more sophisticated discussion out of
you. That is why I keep the notions of the situation from S to S'
and then from S' back to S.
> ------------------
>
> You cannot make this wrong by pointing out that something
> else is wrong!
Your mistake is to violate the second postulate of SR.
In a nutshell, we have the two scenarios below.
1. S emits a photon to be observed by S'.
2. S emits a photon to be observed by S'.
Both scenarios are identical. In a transverse Doppler shift, (v * p =
0) in S (emitter frame). S' (observer frame) would always observers
a blue shift according to the Lorentz transform.
> Paul
Your application of mathematical methodology is way below par. You
should not teach period. Androcles is so right about you.
Paul B. Andersen
This is too stupid to be meant seriously.
The scenario phi = pi/2 and the scenario phi = arccos(v/c)
are obviously not the same, and the equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
should obviously not give the same answer for two different angles.
> It is also my fault to expect a more sophisticated discussion out of
> you. That is why I keep the notions of the situation from S to S'
> and then from S' back to S.
I am not interested in a "more sophisticated discussion"
about something different from the issue.
I am only interested in getting you to point out exactly
what is wrong in MY scenario, which never was anything else
than the above.
THIS is still the issue:
My original statement in my original posting:
Paul B. Andersen wrote September 18:
| There is a moving transmitter, and
| a stationary receiver. In this case, the angle between
| the direction of wave propagation and the velocity of
| the transmitter is pi/2 _in the rest frame of the receiver_.
| The angle phi' in the rest frame of the receiver relates
| to the angle phi in the rest frame of the transmitter thus:
| cos(phi') = (cos(phi)-v/c)/(1 - (v/c)cos(phi))
| combining this with:
| f' = f*(1-(v/c)cos(phi))/sqrt(1-v^2/c^2)
| yield:
| f' = f*sqrt(1-v^2/c^2)/(1+(v/c)cos(phi'))
| which, when phi' = pi/2 simplifies to:
| f' = f*sqrt(1-v^2/c^2)
Koobee Wublee responded:
| Your analysis would violate the very basic principle of Relativity in
| which you are secretly wishing for an absolute frame of reference to
| bail you out of this one. You cannot change the rules of mathematics
| associated with the Lorentz Transformations (plural indicating
| symmetrical reverse transform).
Ever since, I have tried to make you point out what was wrong,
but all you have done so far, is to confirm most of what I wrote
above, and you have failed to point out a single concrete error.
But still, you keep hand waving like this:
> Your mistake is to violate the second postulate of SR.
So here we go for the umpteenth time:
Please point out _exactly_ where the following is wrong.
Exactly where is the violation of the second postulate?
In which line(s) is there an error?
===================================
Read again:
This is the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S':
-------------------------------------------------------------
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
That is, when p and v are transverse in the source frame S,
then phi = pi/2 and the equation above yields:
f' = f/sqrt(1 - v2/c2)
That is, the observer in S' will observe a blue shift.
-----------------
But if v and p' are transverse in the observer frame S',
then phi' = pi/2 and cos(phi) = v/c
and the Doppler equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
yields:
f' = f*sqrt(1 - v2/c2)
That is, the observer in S' will observe a red shift.
------------------
Is THIS right, or is it wrong?
Paul
Koobee Wublee
further.
You have the following scenarios.
1. S emits a photon to be observed by S'. S' observes a Doppler
shift which is BLUE.
2. S emits a photon to be observed by S'. S' observes a Doppler
shift which is RED.
Both scenarios are identical and yet observations are different.
Your students should be able to help you to resolve these
inconsistencies of yours. There is no need to waste any more of my
time.
Paul B. Andersen
> You need to resolve your inconsistencies first before we can go
> further.
Don't you think this is getting ridiculous? :-)
> You have the following scenarios.
This is the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S':
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
It is valid in both scenarios.
> 1. S emits a photon to be observed by S'. S' observes a Doppler
> shift which is BLUE.
Quite.
If p and v are transverse in the source frame S,
phi = pi/2 and the equation above yields:
f' = f/sqrt(1 - v2/c2)
The observer in S' will observe a blue shift.
> 2. S emits a photon to be observed by S'. S' observes a Doppler
> shift which is RED.
Quite.
If v and p' are transverse in the observer frame S',
then phi' = pi/2 and phi = arccos(v/c)
and the Doppler equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
yields:
f' = f*sqrt(1 - v2/c2)
The observer in S' will observe a red shift.
Where are the inconsistencies?
> Both scenarios are identical and yet observations are different.
Isn't it a bit too stupid to claim that the scenarios
are identical when the angle phi is different? :-)
> Your students should be able to help you to resolve these
> inconsistencies of yours. There is no need to waste any more of my
> time.
My students understand that when you insert two different
angles phi into the equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
then you get two different answers.
One would expect it to be impossible to miss that,
so how come you fail to understand it?
I think you have realized that I am right, but won't admit it.
The hallmark of a crank is that he will never admit an error.
You are a crank.
Paul
Koobee Wublee
> Koobee Wublee wrote:
> > You need to resolve your inconsistencies first before we can go
> > further.
>
> Don't you think this is getting ridiculous? :-)
Yes, absolutely. It is indeed very ridiculous to discuss about your
own inconsistencies. I am afraid you have to work that one out
yourself. Go and have fun with yourself. :-)
> [...]
>
> Isn't it a bit too stupid to claim that the scenarios
> are identical when the angle phi is different? :-)
The angles are all the same because the scenarios are the same.
Remember that
> > You have the following scenarios.
> >
> > 1. S emits a photon to be observed by S'. S' observes a Doppler
> > shift which is BLUE.
> >
> > 2. S emits a photon to be observed by S'. S' observes a Doppler
> > shift which is RED.
> >
> > Both scenarios are identical and yet observations are different.
> > Your students should be able to help you to resolve these
> > inconsistencies of yours. There is no need to waste any more of my
> > time.
>
> My students [...]
Your students just don't have the gut to tell you that you are wrong
and that you are self inconsistent. You are on your own. Good luck.
> The hallmark of a crank is that he will never admit an error.
It begins to look that way, yes. You are a crank, and you do not
acknowledge an inconsistent error in yourself.
> You are a crank.
Day are getting shorter in Norway, and you are fantasizing. Good bye.
Sorcerer
| Koobee Wublee wrote:
| > You need to resolve your inconsistencies first before we can go
| > further.
|
| Don't you think this is getting ridiculous? :-)
Of course you are!
Anyone that doesn't know what the three cube roots of 8 are
and thinks cross-product means multiply is wide open to ridicule.
Hilarious, yes?
Paul B. Andersen
> Paul B. Andersen wrote:
>> Koobee Wublee wrote:
>
>>> You need to resolve your inconsistencies first before we can go
>>> further.
>> Don't you think this is getting ridiculous? :-)
>
> Yes, absolutely. It is indeed very ridiculous to discuss about your
> own inconsistencies. I am afraid you have to work that one out
> yourself. Go and have fun with yourself. :-)
>
>> [...]
Is there any particular reason why you snipped this?
| This is the equation for the Doppler shift of a signal
| emitted from a source in S, and observed by an observer in S':
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
| It is valid in both scenarios.
|
|| 1. S emits a photon to be observed by S'. S' observes a Doppler
|| shift which is BLUE.
|
| Quite.
| If p and v are transverse in the source frame S,
| phi = pi/2 and the equation above yields:
| f' = f/sqrt(1 - v2/c2)
| The observer in S' will observe a blue shift.
|
|| 2. S emits a photon to be observed by S'. S' observes a Doppler
|| shift which is RED.
|
| Quite.
| If v and p' are transverse in the observer frame S',
| then phi' = pi/2 and phi = arccos(v/c)
| and the Doppler equation:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
| yields:
| f' = f*sqrt(1 - v2/c2)
| The observer in S' will observe a red shift.
>> Isn't it a bit too stupid to claim that the scenarios
>> are identical when the angle phi is different? :-)
>
> The angles are all the same because the scenarios are the same.
> Remember that
So pi/2 and arcos(v/c) where v > 0 are the same angle? :-)
You know better of course.
So why do you pretend otherwise?
>>> You have the following scenarios.
>>>
>>> 1. S emits a photon to be observed by S'.
.. and p and v are transverse in the source frame S,
so phi = pi/2
>>> S' observes a Doppler shift which is BLUE.
>>>
>>> 2. S emits a photon to be observed by S'.
.. and v and p' are transverse in the observer frame S',
so phi = arccos(v/c),
>>> S' observes a Dopplershift which is RED.
>>>
>>> Both scenarios are identical and yet observations are different.
.."are too, are too!"
Are you also rolling on the floor, screaming and kicking? :-)
>> > Your students should be able to help you to resolve these
>> > inconsistencies of yours. There is no need to waste any more of my
>> > time.
>>
>> My students [...]
Is there any particular reason why you snipped this?
| My students understand that when you insert two different
| angles phi into the equation:
| f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v2/c2)
| then you get two different answers.
|
| One would expect it to be impossible to miss that,
| so how come you fail to understand it?
> Your students just don't have the gut to tell you that you are wrong
> and that you are self inconsistent. You are on your own. Good luck.
I still think it is impossible to miss that the scenarios are different.
So you haven't really missed it.
So why don't you admit it?
Too embarrassing? :-)
>> The hallmark of a crank is that he will never admit an error.
>
> It begins to look that way, yes. You are a crank, and you do not
> acknowledge an inconsistent error in yourself.
>
>> You are a crank.
>
> Day are getting shorter in Norway, and you are fantasizing. Good bye.
Have a good time, and keep claiming that pi/2
is equal to arccos(v/c) even when v > 0.
That's proper behaviour by a crank! :-)
Paul