Gravity is done by force messengers -not by curved space !!

On Jun 13, 2:15 pm, Tom Roberts wrote:

physical laws are inherently coordinate-independent.

Yes, the laws of physics are inherently coordinate independent.
However, the mythical proper coordinate system is not the answer. In
fact, it is just stupid to run to this mythical proper coordinate
system for salvation. shrug

A coordinate-independent statement about this is that for a small pointlike
object its 4-momentum does NOT increase when falling in gravity.

What? The 4-momentum is a vector. Holding a vector unchanged would
require all the components of the vector unchanged as well. Do you
mean the amplitude? Either way, He challenges you on this one. The
field equations do not support what you are saying. That means what
you are claiming cannot be derived through or by the field equations.
shrug

But in relativity this is generalized to:
F = DP/d\tau
where F is the 4-force on a given small pointlike object, P is its 4-momentum,
and D/d\tau is the covariant derivative with respect to its proper time.

Well, excuse Me. The definition of a force under relativity is solely
of your own definition. Once again, the field equations do not
support that. The field equations do not address force at all. You
are making your own laws of physics. shrug

Note that for an object in freefall, in the presence of gravity, F=0 and
DP/d\tau=0 is the geodesic equation -- in the absence of applied forces, small
pointlike objects follow geodesic paths. Note that gravity is not a force in GR,
it appears in the connection buried inside that covariant derivative.

You are totally confused or just speaking with a forked tongue. You
have claimed your own mythical definition of force under relativity
and yet deny gravity ever being a force. shrug

[I rarely respond to you, and will resume not responding unless you
reply with something other than your usual nonsense and abuse.]

This is a convenient excuse. We have heard that one many times before
similar to the following.

“So, you don’t find God in your heart. That is because you have not
studied the Bible. Finding God cannot be done without studying the
Bible. Go read and study the Bible. Pay attention to the
scriptures. You are totally ignorant.”

shrug

Koobee Wublee wrote:
On Jun 13, 2:15 pm, Tom Roberts wrote:
A coordinate-independent statement about this is that for a small pointlike
object its 4-momentum does NOT increase when falling in gravity.

What? The 4-momentum is a vector.

Yes. You finally got something correct.


Holding a vector unchanged would
require all the components of the vector unchanged as well.

Ha. So a vector is not really a "single component", as you have claimed
elsewhere. The rest of us always knew that, and it's good to see you have
finally learned something.

In this case, your claim is not true except in locally-inertial coordinates
co-moving with the falling object. Yes, the statement that "an unchanging
4-vector has unchanging components" is a COORDINATE-DEPENDENT statement.

Consider a constant 3-vector in Euclidean space. Its components
with respect to rotating coordinates are NOT constant.

There is a major problem with attempting to discuss vectors via their
components: the components are INHERENTLY coordinate dependent, and aspects of
the coordinates affect the components of vectors. If one uses accelerated
coordinates, then the 4-momentum of a small pointlike freefalling object will
not have constant components, even though the underlying 4-momentum is unchanged
(see below for what that means).

That's why it is important to make coordinate-independent statements.

How does one know if a given set of coordinates are accelerating?
-- certain components of the connection are nonzero.


Do you
mean the amplitude?

The norm ("amplitude") of the object's 4-momentum is its mass, and thus is
constant and invariant.

I did indeed mean the 4-vector itself: by "unchanged" above I meant "unchanged
in the sense of being parallel propagated along the object's trajectory". In a
curved geometry, that is the closest meaning there is to the meaning in
Euclidean geometry.


The
field equations do not support what you are saying.

Yes they do. But you must LOOK IT UP in a textbook on GR, and you must
UNDERSTAND the book.


That means what
you are claiming cannot be derived through or by the field equations.

No. It merely means you do not understand GR.


But in relativity this is generalized to:
F = DP/d\tau
where F is the 4-force on a given small pointlike object, P is its 4-momentum,
and D/d\tau is the covariant derivative with respect to its proper time.

Well, excuse Me.

There is no excuse for you.


The definition of a force under relativity is solely
of your own definition. Once again, the field equations do not
support that. The field equations do not address force at all. You
are making your own laws of physics.

What I described here is not "mine" at all, it is discussed in every textbook on
GR, and is derived directly from the field equation given the condition that the
object in question is both pointlike and small enough that it does not
significantly affect the geometry. All you need to do is STUDY (and UNDERSTAND).


[I rarely respond to you, and will resume not responding unless you
reply with something other than your usual nonsense and abuse.]

This also applies to you.


[... further nonsense and inability to read]


Tom Roberts

Aristarchus proved the curvature of space, locally of course,
because when is gravitation not local?... he coordinated
with a friend at the same longitude,
different lattitude, taking the shadows
of two gnomons at noon & trigonometry.

Gauss proved the curvature, surveying Allsace-Lorraine
with his theodolite for the goment of France.

but, you can also prove it with a hanging chain,
a catenary, which has those cool, funicular properties.

On Jun 14, 7:17 pm, Tom Roberts wrote:
Koobee Wublee wrote:

What? The 4-momentum is a vector.

Yes. You finally got something correct.

shrug

Holding a vector unchanged would
require all the components of the vector unchanged as well.

Ha. So a vector is not really a "single component", as you have claimed
elsewhere. The rest of us always knew that, and it's good to see you have
finally learned something.

First of all, let’s write down what a position vector is as an example
in both the common spherically symmetric polar and the linearly
rectangular coordinate systems.

[s] = r r = x x + y y + z z

Where all symbols should be self-explanatory

So, yes, a vector has several components, but there is no way in hell
that you can write a matrix and call that a vector. Thus,
effectively, the position vector [s] is a single element. Its number
of components depends on your choice of coordinate system as the two
examples brought up. Thus, a vector is a scalar in exorcism to the
common myth that a vector is not a scalar. A vector can also serve as
an integration constant. Such an example can be seen in the solution
to Maxwell’s equations in free space where these vector integration
constants are the amplitudes to the electric and magnetic field wave
equations. shrug

In this case, your claim is not true except in locally-inertial coordinates
co-moving with the falling object. Yes, the statement that "an unchanging
4-vector has unchanging components" is a COORDINATE-DEPENDENT statement.

Consider a constant 3-vector in Euclidean space. Its components
with respect to rotating coordinates are NOT constant.

Word salad again? shrug

There is a major problem with attempting to discuss vectors via their
components: the components are INHERENTLY coordinate dependent, and aspects of
the coordinates affect the components of vectors.

It is exactly opposite of what you are saying. When He wrote [s] as
the position vector, it is what it is. It is a scalar. Remember a
vector represents the geometry as well. So, it must be invariant.
Its ‘components’ can vary depending on your choice of coordinate
system. Is (r = x)? Is (r = y)? Is (r = z)? Is (r = x)? Is
(r = y)? Is (r = z)? These components are clearly
different. shrug

If one uses accelerated
coordinates, then the 4-momentum of a small pointlike freefalling object will
not have constant components, even though the underlying 4-momentum is unchanged
(see below for what that means).

That's why it is important to make coordinate-independent statements.

How does one know if a given set of coordinates are accelerating?
-- certain components of the connection are nonzero.

Do you
mean the amplitude?

The norm ("amplitude") of the object's 4-momentum is its mass, and thus is
constant and invariant.

I did indeed mean the 4-vector itself: by "unchanged" above I meant "unchanged
in the sense of being parallel propagated along the object's trajectory". In a
curved geometry, that is the closest meaning there is to the meaning in
Euclidean geometry.

Writing the momentum vector as follows, He supposes that the
derivative of it can vanish.

[p] = P_0 p_0 + P_1 p_1 + P_2 p_2 + ... = P_i p_i

Where all symbols should be self-explanatory

However, this momentum is inappropriate. There is no concept of what
you call momentum under GR, namely the field equations. Please try
not to confuse the so-called energy momentum tensor with this
momentum. After all, all practical applications require this energy
momentum to be null, no? shrug

The
field equations do not support what you are saying.

Yes they do. But you must LOOK IT UP in a textbook on GR, and you must
UNDERSTAND the book.

Well, He sees that you are pushed to the wall, and this is your last
resort. Well, you are the one who is oblivious of how the Ricci
tensor is derived. Most newer textbooks are clueless. An example is
the book Mrs. Kennie Sucker has been staring at in the past four
decades, and he still cannot figure out a Lagrangian is. shrug

The Ricci tensor is a 4x4 matrix in the sense of spacetime, and it is
derived by contracting the 4x4x4x4 matrix of the Riemann tensor. That
was a task with no justification --- a mindless mathematical exercise
by Levi-Civita. shrug

The Riemann tensor was derived by Ricci by sorting through the double
covariant derivative of the difference in two adjacent coordinates.
Of course, there are more than one way to arrange this sorting. Ricci
merely found one. Field equations built out of different such sorting
would be very different. However, they all give the same results when
the metric is diagonal, and this cannot be construed as the Riemann
tensor is invariant. shrug

The covariant derivative was DEFINED by Ricci from the geodesic
equations derived by Christoffel. There are two ways to re-arrange
these connection coefficients. Christoffel chose to represent the
geodesic equations in the more symmetric form, and that was how Ricci
DEFINED his covariant derivative from. There is another re-
arrangement to these connection coefficients. Grossmann found this,
and He discovered it independently 100 years later. Taking the double
covariant derivatives of these two arrangements of connection
coefficients would yield completely different curvature tensors in
general. These two forms of covariant derivatives only agree with
each other when the metric is diagonal. shrug

At the end of the day, both the Riemann and the Ricci tensors do not
have anything to do with momentum. So, defining gravitation as taking
the covariant derivative of the momentum is just a conjecture on your
own part attempting to play god by creating your own laws of physics
in your make-belief world. shrug

That means what
you are claiming cannot be derived through or by the field equations.

No. It merely means you do not understand GR.

Hmmm... You are the one who does not understand GR and thinks you
understand GR through these mythical word salads of yours. shrug

But in relativity this is generalized to:
F = DP/d\tau
where F is the 4-force on a given small pointlike object, P is its 4-momentum,
and D/d\tau is the covariant derivative with respect to its proper time.

Well, excuse Me.

There is no excuse for you.

You need to follow your own advice. Study the subject. Pay more
attention to the mathematics involved because GR is all about the
mathematics and NOT ABOUT THE WORD SALAD created by you. Try not to
spread mysticism through word salad that cannot be backed up by sound
mathematics. shrug

The definition of a force under relativity is solely
of your own definition. Once again, the field equations do not
support that. The field equations do not address force at all. You
are making your own laws of physics.

What I described here is not "mine" at all, it is discussed in every textbook on
GR, and is derived directly from the field equation given the condition that the
object in question is both pointlike and small enough that it does not
significantly affect the geometry. All you need to do is STUDY (and UNDERSTAND).

The field equations do not address force. GR does not address force
although you can argue to construct the force through Newton’s second
law. In this case, you have to address the momentum issue. Good luck
on that because GR does not address the momentum. Momentum involves
this relativistic mass. It can only be properly derived through the
geodesic equations. Try not to use circular reasoning to justify the
concept of momentum under GR. shrug

[I rarely respond to you, and will resume not responding unless you
reply with something other than your usual nonsense and abuse.]

This also applies to you.

You are free to run with your tail between your legs, or better yet,
you can follow PD’s shoes and file a complaint with Caltech bitching
about how He has ripped you to pieces in debates on a subject that you
as a self-styled physicist suppose to be an expert in. shrug

[... further nonsense and inability to read]

As He has said many times over, the Dark Side of physics has no
future. Sooner or later, the truth will prevail. It is just a matter
of time. shrug

“Hey, look, Mom. The emperor has no clothes on!” shrug

1treePetrifiedForestLane wrote:
Aristarchus proved the curvature of space, locally of course,
because when is gravitation not local?... he coordinated
with a friend at the same longitude,
different lattitude, taking the shadows
of two gnomons at noon & trigonometry.

He proved the curvature of THE SURFACE OF THE EARTH, not of "space". Indeed, in
his computation he implicitly assumed Euclidean space, which is flat.


Gauss proved the curvature, surveying Allsace-Lorraine
with his theodolite for the goment of France.

Again, OF THE SURFACE OF THE EARTH, not "space". He also used Euclidean space in
his computation.


but, you can also prove it with a hanging chain,
a catenary, which has those cool, funicular properties.

That is curvature OF THE CHAIN, not "space". And one also applies Euclidean
geometry.

From GR, in all these case we know that the curvature of space-TIME, as applied
to each physical measurement, is much smaller than the measurement resolutions.

Yet the curvature of space-TIME is easily measured by dropping
a rock. That's because the curvature is MUCH larger in a
space-time plane than in the space-space planes of your examples.


Tom Roberts

On 6/15/11 8:13 AM, Tom Roberts wrote:
1treePetrifiedForestLane wrote:
Aristarchus proved the curvature of space, locally of course,
because when is gravitation not local?... he coordinated
with a friend at the same longitude,
different lattitude, taking the shadows
of two gnomons at noon & trigonometry.


Quoting Carl Sagan--who is likely overly optimistic about the
accuracy of Eratosthenes (not Aristarchus) observations:


There was once a time when our little planet seemed immense, when
it was the only world we could explore, its true size was first
worked out in a simple and ingenious way by a man who lived here
in Egypt in the third century B.C.

In Alexandria, at that time, there lived a man named Eratosthenes.
One of his envious contemporaries called him Beta, the second
letter of the Greek alphabet, because, he said, Eratosthenes was
second best in the world in everything, but it seems clear that in
many fields Eratosthenes was Alpha: he was an astronomer,
historian, geographer, philosopher, poet, theater critic and
mathematician.

He was also the chief librarian of the great library of Alexandria
and one day while reading a papyrus book in the library he came
upon a curious account: far to the south, he read, at the frontier
outpost of Syene something notable could be seen on the longest
day of the year. On June 21 the shadows of a temple column or a
vertical stick would grow shorter as noon approached and as the
hours crept towards midday the sun's rays would slither down the
sides of a deep well which, on other days would remain in shadow
and then precisely at noon columns would cast no shadows and the
sun would shine directly down into the water of the well. At that
moment the sun was exactly overhead.

It was an observation that someone else might easily have ignored:
sticks, shadows, reflections in wells, the position of the sun,
simple everyday matters, so what possible importance might they
be.

But Eratosthenes was a scientist and his contemplations of
these homely matters changed the world, in a way made the world.
Because Eratosthenes had the presence of mind to experiment to
actually ask whether back here near Alexandria a stick cast a
shadow near noon on June the twenty first, and it tums up sticks
do. An overly skeptical person might have said that the report
from Syene was an error but an absolutely straightforward
observation why would anyone lie on such a trivial matter?

Eratosthenes asked himself how it could be that at the same moment
a stick in Syene would cast no shadow and a stick in Alexandria,
eight hundred kilometers to the north, would cast a very definite
shadow.

If the shadow at Syene is of a certain lenght and the shadow at
Alexandria is the same lenght that also makes sense on a flat
Earth, but how could it be, Eratosthenes asked, that at the same
instant there was no shadow at Syene and a very substantial shadow
at Alexandria.

The only answer was that the surface of the earth is curved, not
only that, but the greater the curvature the bigger the difference
in the length of the shadows. The sun is so far away that its rays
are parallel when they reach the Earth. Sticks at different angles
to the sun's rays will cast shadows of different lengths for the
observed difference in the shadow length the distance between
Alexandria and Syene had to be about 7 degrees along the surface
of Earth, by that I mean if you would imagine these sticks
extending all the way down to the center of the Earth they would
there intersect at an angle of about 7 degrees, well seven degrees
is something like a fiftieth of the full circumference of the
Earth 360 degrees

Eratosthenes knew the distance between Alexandria and Syene, he
knew it was eight hundred kilometers, why? because he hired a man
to pace out the entire distance so that he could perform the
calculation. I'm talking about now 800 kilometers times fifty is
forty thousand kilometers so that must be the circumference of the
Earth, that's how far it is to go once around the earth, that's
the right answer.

Eratosthenes's only tools were sticks, eyes, feet, and brains plus
a zest for experiment. With those tools he correctly deduced the
circumference of the Earth to high precision with an error of only
a few percent, that's pretty good figuring for twenty-two hundred
years ago.

Le 15/06/11 15:13, Tom Roberts a écrit :
1treePetrifiedForestLane wrote:
Aristarchus proved the curvature of space, locally of course,
because when is gravitation not local?... he coordinated
with a friend at the same longitude,
different lattitude, taking the shadows
of two gnomons at noon & trigonometry.

He proved the curvature of THE SURFACE OF THE EARTH, not of "space".
Indeed, in his computation he implicitly assumed Euclidean space, which
is flat.

Please don't bother those people with FACTS.... They do not care about
facts, and the difference between SPACE and the surface of the earth
is beyond their mental abilities.

it applies a line-of-sight, yes, which is dependent
upon the classical problem of the brachistochrone/tautochrone,
a.k.a. the "ray" of light in a medium with increasing index
of refraction.

I am frequently barfing over the "rubbersheet model,"
showing Eaaarth weighing "down South" in a depression,
with satellites rolling around the norther side
of this infinite surface ... a catenary-of-rotation?

also, I get a regular emitic from Minkowski's silly slogans
about phase-space ... but I'm something of an electonics guy, so,
you can see how I might have already barfed over your screen.

* * * * Yet the curvature of space-TIME is easily measured by dropping
* * * * a rock. That's because the curvature is MUCH larger in a
* * * * space-time plane than in the space-space planes of your examples.

I was going to say, "(or Eratosthenes?)"

wow, YP ...
does *any* thing have relevance,
other than your crappy math and/or English?

a scalar can be considered as the "zero-vector" with a nonzero scalar
part;
that is, just a point with a magnitude, viz temperature or IQ -- tee-
hee.