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

"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?

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

Martin Hogbin

"Martin Hogbin" wrote in message
...

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

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.

Martin Hogbin

eleaticus

"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.

No they are not. The observer's frame and the auto's frame are
different frames.

There is no frame for "the street", only observational frames.

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?

No. But we are required to establish that all velocities are relative
(thus, the "Principle of Relativity").

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

[snip]

....and it goes downhill from there.

"Titan Point" wrote in message
om...
"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.

No they are not. The observer's frame and the auto's frame are
different frames.

Yes, that was 'only' a bad and strange 'typo' , as is evident from
everything else herein.

There is no frame for "the street", only observational frames.

Too bad that a street can't be an imaginary observer even though no actual
observer is actually required.

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?

No. But we are required to establish that all velocities are relative
(thus, the "Principle of Relativity").

Ooops. Almost missed you agreeing with the obvious.

But what are the consequences? Must we nonetheless transform the equation as
if the auto-street (corner) erlationship did not exist?

Yes, or no.

Please be explicit.

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

[snip]

...and it goes downhill from there.

Straight down to the part where your admission above burns your ass, being
about an equation where your admission acknowledges improper - certainly
unnecessary (mis)usage by your intellectual kin.

So, you are on the idiot's side of the question en re k = .5mv^2?

eleaticus

eleaticus

"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

"Paul Draper" wrote in message
om...
It's an excellent question, one that has to be answered carefully.

Thanks for your thoughtful post (obviously thoughtful, given that sentence.
lol)

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.

Actually, the/a point of the piece is that KE is fully meaningful only wrt
to a target/impact object, so it is indeed in that regard invariant. Think
about it.

Forced choice:

Is the v in KE physically meaningful wrt the arbitrary, imaginary observer?
Is the v in KE physically meaningful wrt to an object of impact?

eleaticus


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

PD

"Paul Draper" wrote in message om...
"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.

That was my first answer to another raving lunatic a few years ago.
His name was, still is, and will always be, Ancrocles.

Dirk Vdm

"Paul Draper" wrote in message
om...
"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. Some people claim that the CBMR provides a way to tell one
inertial frame from another in violation of the POR - so what is going on
with it? The answer is the CBMR can be screened out - 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.

Thanks
Bill


PD

"eleaticus" wrote in message
. ..

"Paul Draper" wrote in message
om...
It's an excellent question, one that has to be answered carefully.

Thanks for your thoughtful post (obviously thoughtful, given that
sentence.
lol)

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.

Actually, the/a point of the piece is that KE is fully meaningful only
wrt
to a target/impact object, so it is indeed in that regard invariant.
Think
about it.

Forced choice:

Is the v in KE physically meaningful wrt the arbitrary, imaginary
observer?

No one claimed it was not. 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.

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. Remember
saying the laws of nature are the same is not saying you have exactly the
same experiment. Understanding the POR requires care in what is meant by a
law of nature. People who have studied classical mechanics know this quite
well - different frames have differ experiments but those experiments are
all governed by Newton's laws which are the same. From his 1905 paper
Einstein stated:

'They suggest rather that, as has already been shown to the first order of
small quantities, the same laws of electrodynamics and optics will be valid
for all frames of reference for which the equations of mechanics hold good.
We will raise this conjecture (the purport of which will hereafter be called
the ``Principle of Relativity'') to the status of a postulate'

Einstein assumes the reader is familiar enough with mechanics to understand
what is being said ie what is meant by law of nature.

Bill


eleaticus


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

PD