On Fri, 22 Oct 2004 22:56:09 GMT, "Bill Hobba" wrote:


"Paul Draper" wrote in message
. com...
"eleaticus" wrote in message
...
In various ways matters of frames versus equations have arisen.

A. Consider the distance D=vt equation, about the movement of an
automobile
on the street outside your home. What frames are involved? We can say
three: the frame of the observer (you), the street, and the frame of
whatever is moving. Yes, your frame and the auto's is the same. Say that
the
D=0=x location is the nearest corner to you.

Should we wish to transform the equation to the viewpoint of the
Goodyear
blimp overhead, moving at velocity u with respect to you and the street,
are
we required to ignore the fact that the equation is about the auto
versus
the street corner and pretend it is about the auto and you?

There's a big difference. The 'observer' - meaning the stationary
frame -
is an imaginary construct in its general usage.

If it is the auto versus you the equation is about imaginary observers
and
physical autos, not about physical autos and the street they are on.

If it is auto vs observer then the transform must be about the auto vs
the
blimp observer and D=vt becomes D'=(u-v)t.

If it is auto vs street corner, a correct transform from a physical
viewpoint must retain the physical facts of the situation: a car moving
relative to a physical object, the street corner, at velocity v.

This important difference is extremely imporant in the next case

B. Consider the old kinetic energy equation, K = .5mv^2. Here we have
only
two frames in evidence. The imaginary observer and the non-inertial
moving
object.

As it stands, per my description, the equation doesn't even have the
physical standing of the D=vt equation in which I posited a street
(corner)
as well as the imaginary observer, and the car.

Indeed, the equation is in effect meaningless, who cares what the
supposed
kinetic energy is from the viewpoint of any of the infinities of such
observers throughout the universe. There is only one 'object' for which
the
equation has real meaning: an impact object, an object the moving mass
hits.
Yes, as it stands the frame of the impact object is at rest with respect
to
the imaginary observer's frame.

Innyhoo. that's the third frame, and the question arises: are we somehow
bound to ignore the only kind of frame with respect to which the
equation
has real meaning? What is the law of physics that says we must play the
ass
and say it is the imaginary observer to which this object's kinetic
energy
is relevant.

The question has import when we decide to refer the physical facts to
another frame. If K=.5mv^2 is at all really about the imaginary observer
then surely it is also really about the second imaginary observer, and
should transform to K=.5m(u-v)^2.

But if K=.5mv^2 is about meaningful physical fact then the v terms is
relative to such an impact object and the transform should preserve the
relative velocity of the moving object versus the impact object for
which
the equation is meaningful.

Summary.

What are equations about in physics, imaginary observers, or real (ie
meaningful) physical objects, fields?

eleaticus

It's an excellent question, one that has to be answered carefully.

There is a distinction between an equation and a physical law.
Since your title indicates frames, let me remind you that physical
laws are what remain invariant in different inertial frames. This is
not to say that equations remain invariant. You point out KE, which is
not even invariant under a Galilean transformation, let alone Lorentz
transformations. But the *conservation* of energy and momentum (the
laws) are invariant.

Just because there is an equation associated with it doesn't make it a
physical law.

Very true Paul. The interpretation of the POR requires one to be careful
what one means by physical law. For example some have said it has no
content - it is in fact a definition of physical law. Such is obviously
false as it is required for the derivation of the Lorentz transformations so
must have content. Exactly what counts as physical law must be examined on
a case by case basis. For example suppose an aether did exist and had
detectable consequences. Obviously the laws that govern how objects
interact with the aether would be the same in any frame so exactly why would
it violate the POR? The answer has to do with the symmetry properties of an
inertial frame - an inertial frame is isotropic. The existence of a
detectable aether wind breaks isotropy so we are not really dealing with an
inertial frame. ...

In what observable or measurable sense???

Some people claim that the CBMR provides a way to tell one inertial frame
from another in violation of the POR

What 'measurable' violation???

- so what is going on with it? The answer is the CBMR can be screened out -

Yeah, and so can sound by use of a acoutically damper screen. That doesn't
make the air dissapaer. What an absolutely lousy illogicial argument! Just
because Tom says it doesn't improve it validity...

it is an object contained in a frame and can be gotten rid of. The aether,
being, for example, the medium that light is supposed to undulate in, can
not be screened out, its very existence is necessary for the laws of nature
as we understand them to exist eg we know that light is EM radiation
described by Maxwell's equations - so if an aether did exist it would be
an intrinsic part of the frame otherwise Maxwell's equations would go
down the tube.

As eluded to above, waves can be screened in a medium and yet the medium
remains. Nuff said...

Paul Stowe

"Paul Stowe" wrote in message
...
On Fri, 22 Oct 2004 22:56:09 GMT, "Bill Hobba"
wrote:


"Paul Draper" wrote in message
. com...
"eleaticus" wrote in message
...
In various ways matters of frames versus equations have arisen.

A. Consider the distance D=vt equation, about the movement of an
automobile
on the street outside your home. What frames are involved? We can
say
three: the frame of the observer (you), the street, and the frame of
whatever is moving. Yes, your frame and the auto's is the same. Say
that
the
D=0=x location is the nearest corner to you.

Should we wish to transform the equation to the viewpoint of the
Goodyear
blimp overhead, moving at velocity u with respect to you and the
street,
are
we required to ignore the fact that the equation is about the auto
versus
the street corner and pretend it is about the auto and you?

There's a big difference. The 'observer' - meaning the stationary
frame -
is an imaginary construct in its general usage.

If it is the auto versus you the equation is about imaginary
observers
and
physical autos, not about physical autos and the street they are on.

If it is auto vs observer then the transform must be about the auto
vs
the
blimp observer and D=vt becomes D'=(u-v)t.

If it is auto vs street corner, a correct transform from a physical
viewpoint must retain the physical facts of the situation: a car
moving
relative to a physical object, the street corner, at velocity v.

This important difference is extremely imporant in the next case

B. Consider the old kinetic energy equation, K = .5mv^2. Here we
have
only
two frames in evidence. The imaginary observer and the non-inertial
moving
object.

As it stands, per my description, the equation doesn't even have the
physical standing of the D=vt equation in which I posited a street
(corner)
as well as the imaginary observer, and the car.

Indeed, the equation is in effect meaningless, who cares what the
supposed
kinetic energy is from the viewpoint of any of the infinities of such
observers throughout the universe. There is only one 'object' for
which
the
equation has real meaning: an impact object, an object the moving
mass
hits.
Yes, as it stands the frame of the impact object is at rest with
respect
to
the imaginary observer's frame.

Innyhoo. that's the third frame, and the question arises: are we
somehow
bound to ignore the only kind of frame with respect to which the
equation
has real meaning? What is the law of physics that says we must play
the
ass
and say it is the imaginary observer to which this object's kinetic
energy
is relevant.

The question has import when we decide to refer the physical facts to
another frame. If K=.5mv^2 is at all really about the imaginary
observer
then surely it is also really about the second imaginary observer,
and
should transform to K=.5m(u-v)^2.

But if K=.5mv^2 is about meaningful physical fact then the v terms is
relative to such an impact object and the transform should preserve
the
relative velocity of the moving object versus the impact object for
which
the equation is meaningful.

Summary.

What are equations about in physics, imaginary observers, or real (ie
meaningful) physical objects, fields?

eleaticus

It's an excellent question, one that has to be answered carefully.

There is a distinction between an equation and a physical law.
Since your title indicates frames, let me remind you that physical
laws are what remain invariant in different inertial frames. This is
not to say that equations remain invariant. You point out KE, which is
not even invariant under a Galilean transformation, let alone Lorentz
transformations. But the *conservation* of energy and momentum (the
laws) are invariant.

Just because there is an equation associated with it doesn't make it a
physical law.

Very true Paul. The interpretation of the POR requires one to be
careful
what one means by physical law. For example some have said it has no
content - it is in fact a definition of physical law. Such is obviously
false as it is required for the derivation of the Lorentz
transformations so
must have content. Exactly what counts as physical law must be examined
on
a case by case basis. For example suppose an aether did exist and had
detectable consequences. Obviously the laws that govern how objects
interact with the aether would be the same in any frame so exactly why
would
it violate the POR? The answer has to do with the symmetry properties
of an
inertial frame - an inertial frame is isotropic. The existence of a
detectable aether wind breaks isotropy so we are not really dealing with
an
inertial frame. ...

In what observable or measurable sense???

Please see my supposition - 'For example suppose an aether did exist and had
detectable consequences.'. An example would be a positive result on the
Trouton-Noble experiment.


Some people claim that the CBMR provides a way to tell one inertial
frame
from another in violation of the POR

What 'measurable' violation???]

The frequency of the CBMR radiation woul be different is the direction of
motion - thus isotropy would be broken. But because the CBMR can be removed
(screened out) frames do not have to contain it thus its existence doe not
violate the POR - unlike an aether which could not be removed.


- so what is going on with it? The answer is the CBMR can be screened
out -

Yeah, and so can sound by use of a acoutically damper screen. That
doesn't
make the air dissapaer. What an absolutely lousy illogicial argument!
Just
because Tom says it doesn't improve it validity...

So you are claiming the existence of air is an intrinsic property of frames
attached to the earth? Yea right.

Bill


it is an object contained in a frame and can be gotten rid of. The
aether,
being, for example, the medium that light is supposed to undulate in,
can
not be screened out, its very existence is necessary for the laws of
nature
as we understand them to exist eg we know that light is EM radiation
described by Maxwell's equations - so if an aether did exist it would be
an intrinsic part of the frame otherwise Maxwell's equations would go
down the tube.

As eluded to above, waves can be screened in a medium and yet the medium
remains. Nuff said...

Paul Stowe

Paul Draper wrote:

"eleaticus" wrote in message ...
In various ways matters of frames versus equations have arisen.

A. Consider the distance D=vt equation, about the movement of an automobile
on the street outside your home. What frames are involved? We can say
three: the frame of the observer (you), the street, and the frame of
whatever is moving. Yes, your frame and the auto's is the same. Say that the
D=0=x location is the nearest corner to you.

Should we wish to transform the equation to the viewpoint of the Goodyear
blimp overhead, moving at velocity u with respect to you and the street, are
we required to ignore the fact that the equation is about the auto versus
the street corner and pretend it is about the auto and you?

There's a big difference. The 'observer' - meaning the stationary frame -
is an imaginary construct in its general usage.

If it is the auto versus you the equation is about imaginary observers and
physical autos, not about physical autos and the street they are on.

If it is auto vs observer then the transform must be about the auto vs the
blimp observer and D=vt becomes D'=(u-v)t.

If it is auto vs street corner, a correct transform from a physical
viewpoint must retain the physical facts of the situation: a car moving
relative to a physical object, the street corner, at velocity v.

This important difference is extremely imporant in the next case

B. Consider the old kinetic energy equation, K = .5mv^2. Here we have only
two frames in evidence. The imaginary observer and the non-inertial moving
object.

As it stands, per my description, the equation doesn't even have the
physical standing of the D=vt equation in which I posited a street (corner)
as well as the imaginary observer, and the car.

Indeed, the equation is in effect meaningless, who cares what the supposed
kinetic energy is from the viewpoint of any of the infinities of such
observers throughout the universe. There is only one 'object' for which the
equation has real meaning: an impact object, an object the moving mass hits.
Yes, as it stands the frame of the impact object is at rest with respect to
the imaginary observer's frame.

Innyhoo. that's the third frame, and the question arises: are we somehow
bound to ignore the only kind of frame with respect to which the equation
has real meaning? What is the law of physics that says we must play the ass
and say it is the imaginary observer to which this object's kinetic energy
is relevant.

The question has import when we decide to refer the physical facts to
another frame. If K=.5mv^2 is at all really about the imaginary observer
then surely it is also really about the second imaginary observer, and
should transform to K=.5m(u-v)^2.

But if K=.5mv^2 is about meaningful physical fact then the v terms is
relative to such an impact object and the transform should preserve the
relative velocity of the moving object versus the impact object for which
the equation is meaningful.

Summary.

What are equations about in physics, imaginary observers, or real (ie
meaningful) physical objects, fields?

eleaticus

It's an excellent question, one that has to be answered carefully.

There is a distinction between an equation and a physical law.
Since your title indicates frames, let me remind you that physical
laws are what remain invariant in different inertial frames. This is
not to say that equations remain invariant. You point out KE, which is
not even invariant under a Galilean transformation, let alone Lorentz
transformations. But the *conservation* of energy and momentum (the
laws) are invariant.

Just because there is an equation associated with it doesn't make it a
physical law.

PD

Your answer is pearls before swine. Eleaticus, who is such a gutless
piece of **** that he won't even put his name on his postings, has
had this pointed out to him for years. Until recently, he didn't
even respond to replies to his postings. Do you really think
that you're going to educate him/her?

Help the people who have legitimate questions. Don't waste time
on cranks.

John Anderson

"Bill Hobba" wrote in message
...

"eleaticus" wrote in message
. ..
Is the v in KE physically meaningful wrt the arbitrary, imaginary
observer?

No one claimed it was not.

They should. It is only by doing so that the one 'experiment' is falsified
into more than one experiment.

Saying the laws of physics are the same implies
the same experiment done under the same conditions will give the same
results - not that you have the same experiment when you go to a different
frame. Shoot a bullet at a stationary target in an inertial frame and you
see the bullet move towards the target. Go to a frame where the bullet is
at rest and you see the target move toward the bullet. Two different
experiments. What the POR says is the laws of physics that govern the two
situations are the same eg F=ma is still true, momentum and energy is
still
conserved.

You switch from KE to F, from v to a. Strange that staying with v would
nullify your point.

Is the v in KE physically meaningful wrt to an object of impact?

Yes it is meaningful - but because it is not the same in different
inertial
frames it can not be considered a law of nature ie it is not one of those
things that govern mechanics regardless of which frame you are in.

There is nothing about KE=.5mv that would make it unlawful to consider the
velocity to be wrt to the impact object rather than the imaginary observer,
but acting as if it is wrt the observer makes it nonsense. Only if it is wrt
to an impact object does it make consistent sense and science.

WRT the object, the equation is obviously invariant, the point of the whole
post!

Remember
saying the laws of nature are the same is not saying you have exactly the
same experiment.

Certainly not! IFF you are foolish enough to take an equation about physical
fact that is valid only wrt the interaction between two physical objects and
say it is about a physical object vs an imaginary object.

eleaticus

wrote in message ...

Your answer is pearls before swine. Eleaticus, who is such a gutless
piece of **** that he won't even put his name on his postings, has
had this pointed out to him for years. Until recently, he didn't
even respond to replies to his postings. Do you really think
that you're going to educate him/her?

Help the people who have legitimate questions. Don't waste time
on cranks.

To quote myself:

Focus well on negative 'responses'. Are they vicious ranting? Are the
replies actually responsive? Do they rant about gravity, or how Relativity
is proved correct a million times each day, or some other 'we are proved
right' rave that doesn't deal in details about the debunking done here? It
is typically General Relativity or items about the energy and mass of moving
objects that are being waved at you, and such items are completely
irrelevant to coordinate transformations and invariance..

Just ask them for a list of all the observations that have been made of the
shortening (contraction) of moving objects that Special Relativity says
always occurs.

eleaticus



John Anderson

"eleaticus" wrote in message
news

"Bill Hobba" wrote in message
...

"eleaticus" wrote in message
. ..
Is the v in KE physically meaningful wrt the arbitrary, imaginary
observer?

No one claimed it was not.

They should. It is only by doing so that the one 'experiment' is falsified
into more than one experiment.

Your point being?


Saying the laws of physics are the same implies
the same experiment done under the same conditions will give the same
results - not that you have the same experiment when you go to a
different
frame. Shoot a bullet at a stationary target in an inertial frame and
you
see the bullet move towards the target. Go to a frame where the bullet
is
at rest and you see the target move toward the bullet. Two different
experiments. What the POR says is the laws of physics that govern the
two
situations are the same eg F=ma is still true, momentum and energy is
still
conserved.

You switch from KE to F, from v to a. Strange that staying with v would
nullify your point.

I give examples of laws of nature - the value of KE is not - the form of the
equation is.


Is the v in KE physically meaningful wrt to an object of impact?

Yes it is meaningful - but because it is not the same in different
inertial
frames it can not be considered a law of nature ie it is not one of
those
things that govern mechanics regardless of which frame you are in.

There is nothing about KE=.5mv that would make it unlawful to consider the
velocity to be wrt to the impact object rather than the imaginary
observer,
but acting as if it is wrt the observer makes it nonsense.

That makes no sense. Except for direction a particular KE (for a given
mass) implies a particular velocity which can change between frames.

Only if it is wrt
to an impact object does it make consistent sense and science.

I have no idea what you are trying to say - as usual you use words in ways
that make no sense.


WRT the object, the equation is obviously invariant, the point of the
whole
post!

The above makes no sense - as far as I can give any meaning to what you say
wrt to the object means in a frame that the object is at rest hence has zero
KE.


Remember
saying the laws of nature are the same is not saying you have exactly
the
same experiment.

Certainly not! IFF you are foolish enough to take an equation about
physical
fact that is valid only wrt the interaction between two physical objects
and
say it is about a physical object vs an imaginary object.

What can I say - the above makes no sense just like the rest of your
rubbish.

Bill


eleaticus

"eleaticus" wrote in message .. .

"Martin Hogbin" wrote in message
...


Are you trying to learn some physics at last? Or will
you 'educate' us all?

You seem to be missing the fact that y'all (SR-cult cretins) opt for the
idiot's side of the question.

Just a tiny flicker of hope.

Martin Hogbin

"Martin Hogbin" wrote in message
...


Just a tiny flicker of hope.

Not well fanned by avoiding the principled questions in the thread.


Martin Hogbin

eleaticus

"eleaticus" wrote in message . ..

"Martin Hogbin" wrote in message
...


Just a tiny flicker of hope.

Not well fanned by avoiding the principled questions in the thread.

You never replied to
http://groups.google.com/groups?&thr...t01.boi.hp.com
So you are stupid or you are a coward.
Probably both.

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

"eleaticus" wrote in message
. ..

"Martin Hogbin" wrote in message
...


Just a tiny flicker of hope.

Not well fanned by avoiding the principled questions in the thread.

You never replied to
http://groups.google.com/groups?&thr...t01.boi.hp.com
So you are stupid or you are a coward.
Probably both.

I should follow slime to their slime pool? When we have both the newsgroup
and email before us?

eleaticus

Dirk Vdm