Gregory L. Hansen wrote:
It gave me a new appreciation for the power of symmetries in determining
physical laws. I want to take another look at Newtonian gravity with
finite propagation, see if the same trick works with Galilean
transformations. Or arbitrary transformations, like the ones that almost
look like Lorentz transformations except that t'=t/sqrt(1-v^2/c^2),
dropping the term in x. But the discussion was pedagogically sparse. A
few articles by Poincare were referenced but our library doesn't have
those journals, and I think they're in French, one might be in Italian,
which wouldn't help me.


I do not want to stifle what looks as though it would be an interesting
investigation but finite propagation in a Newtonian system? Form general
properties of the Newtonian lagrangian we know for any Newtonian system the
force is derivable from a potential that depends only on distance (Landau -
Mechanics page 8). Indeed he states:

'The fact that the potential energy depends only on the positions of the
particles at a given instant shows that a change in the position of any
particle instaneously affects all other particles. We may say that
interactions are instanteously propagated. The necessity for interactions
in classical mechanics to be of this type is closely related to the premises
on which the subject is based, namely the absolute nature of time and
Galileo's relativity principle'

So this would seem to fundamentally rule out what your trying to do. Are
you sure a weak field gravity GR approximation (ie linerised GR) would not
be more suitable.

Of course in the spirit of making progress we should try all sorts of things
and what your trying to looks interesting. I am just not sure there is not
an inconsistency caused by the very foundations on which Newtonian mechanics
is built (actually the above looks like a pretty tight argument - over the
years I have learnt to trust Landau as a source).

Thanks
Bill

In article ,
Bill Hobba wrote:
Gregory L. Hansen wrote:

Of course in the spirit of making progress we should try all sorts of things
and what your trying to looks interesting. I am just not sure there is not
an inconsistency caused by the very foundations on which Newtonian mechanics
is built (actually the above looks like a pretty tight argument - over the
years I have learnt to trust Landau as a source).

Here's another one to ponder. In special relativity make the
transformation to rotation coordinates,

t' = t

x' = x cos wt - y sin wt

y' = x sin wt + y cos wt

z' = z

From that you can find a metric, connections, and geodesic equations. And
the geodesic equations will have terms interpretable as centrifugal and
Coriolis forces, which can be generalized in vector notation. Those
inertial terms come from g_00 and g_0i=g_i0.

I wanted to follow a similar program in Newtonian mechanics, since it's
the same change of coordinates, it's still just geometry. But Newtonian
spaces don't have g_00 or g_0i because time isn't a component of Newtonian
vectors. So that wonderful derivation doesn't seem available in Newtonian
mechanics, and the centrifugal and Coriolis terms have to be pounded
together by other means.

The geodesic equations tell you what a straight line is, I'd have figured
it would tell you what a straight line is even in the Newtonian rotating
frame. I really don't understand what went wrong there.

--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

(Gregory L. Hansen) wrote in message ...

Hi Greg

Here's another one to ponder. In special relativity make the
transformation to rotation coordinates,

t' = t

x' = x cos wt - y sin wt

y' = x sin wt + y cos wt

z' = z

From that you can find a metric, connections, and geodesic equations. And
the geodesic equations will have terms interpretable as centrifugal and
Coriolis forces, which can be generalized in vector notation. Those
inertial terms come from g_00 and g_0i=g_i0.

I wanted to follow a similar program in Newtonian mechanics, since it's
the same change of coordinates, it's still just geometry. But Newtonian
spaces don't have g_00 or g_0i because time isn't a component of Newtonian
vectors.

I think your tying one foot behind your back and entering an
ass kicking contest.

In Newtons Geometry you can still have 4D graphs, except
that these would be cartesian, so
g11 = g22 = g33 = g44 =1 and covariant and contravariant
components are equal.

So that wonderful derivation doesn't seem available in Newtonian
mechanics, and the centrifugal and Coriolis terms have to be pounded
together by other means.
The geodesic equations tell you what a straight line is, I'd have figured
it would tell you what a straight line is even in the Newtonian rotating
frame. I really don't understand what went wrong there.

I don't see what's wrong, if you permit Cartesian 4D.
Drop a ball on the x-axis.Set V4 = 1 and the geodesic
in this scenario is,

dV1/dx4 = (-1/2) [2*g14,4] = -g14,4

Acceleration A = dV1/dx4 , and for brevity,
dx4 =dt, V =V1, A=dV/dt

A = -g14,4 , and Newton finds g14 =-V.

If Newton has decided A = - GM/r^2 =A(g) then,

A(g) = &V/&t Eq. (PoE)

I think this is the Principle of Equivalence! which
shouldn't be too surprising as it is derived from a
geodesical equation. I think that because the
acceleration due to gravity A(g) is found to be
equivalent to the acceleration &V/&t, that would
be reproducible in an elevator.

Is this even close?
Ken S. Tucker

(Ken S. Tucker) wrote in message . com...
(Gregory L. Hansen) wrote in message ...

Hi Greg

Here's another one to ponder. In special relativity make the
transformation to rotation coordinates,

t' = t

x' = x cos wt - y sin wt

y' = x sin wt + y cos wt

z' = z

From that you can find a metric, connections, and geodesic equations. And
the geodesic equations will have terms interpretable as centrifugal and
Coriolis forces, which can be generalized in vector notation. Those
inertial terms come from g_00 and g_0i=g_i0.

I wanted to follow a similar program in Newtonian mechanics, since it's
the same change of coordinates, it's still just geometry. But Newtonian
spaces don't have g_00 or g_0i because time isn't a component of Newtonian
vectors.

I think your tying one foot behind your back and entering an
ass kicking contest.

In Newtons Geometry you can still have 4D graphs, except
that these would be cartesian, so
g11 = g22 = g33 = g44 =1 and covariant and contravariant
components are equal.

So that wonderful derivation doesn't seem available in Newtonian
mechanics, and the centrifugal and Coriolis terms have to be pounded
together by other means.
The geodesic equations tell you what a straight line is, I'd have figured
it would tell you what a straight line is even in the Newtonian rotating
frame. I really don't understand what went wrong there.

I don't see what's wrong, if you permit Cartesian 4D.

I was going to give a kneejerk response, then went away to think about
it. Galilean relativity doesn't have four-vectors, it has ordered
pairs of three-vectors and time, which are separately conserved under
transformations. A Cartesian metric clearly can't work in Galilean
relativity. E.g. in special relativity, for an observer at rest in
the lower-case frame observing something in the capital frame,

c^2 dT^2 = c^2 dt^2 - dx^2

dT^2 = dt^2 (1 - (dx/dt)^2/c^2)

dt = dT / sqrt(1 - v^2/c^2)

Time dilation. In a Cartesian metric,

k^2 dT^2 = k^2 dt^2 - dx^2

dt = dT / sqrt(1 + v^2/k^2)

Time contraction.

But consider k=0,

ds^2 = 0*dt^2 + dx^2 + dy^2 + dz^2

When you work the metric in special relativity for a rotating frame
you get

ds^2 = (-c^2 + w^2(X^2+Y^2)) dT^2 + dX^2 + dY^2 + dZ^2

+ 2 Y w dX dT - 2 X w dY dT

The only contribution from the dt^2 term in the metric is the -c^2 in
the first term. If you work the problem as before for the Galilean
metric I gave, you get exactly the same thing, but without the -c^2.
So the 0*dt^2 is acting as a place holder to say that something could
be there, even if there isn't anything there right now.

And then when you consider Galilean time intervals you need to
consider the separate metric

(ds_t)^2 = dt^2 + 0*(dx^2 + dy^2 + dz^2)

which I find kind of ugly. Besides the fact that the four-dimensional
version is clearly derived from another theory that's not Newtonian
mechanics from a guy that already knows what kind of answer he wants
to get.

The usual way of going to Newtonian mechanics is to let c-inf, or the
ratio v/c-0. Which works fine in the time,

dT^2 - (dX^2 + dY^2 + dZ^2)/c^2 = dt^2 - (dx^2 + dy^2 + dz^2)/c^2

- dT^2 = dt^2 as c-inf

But what can we say about the spatial part? We can ignore the spatial
part when the time part is much larger, but we can't ignore the time
part when the time part is much larger.

Maybe I'll try to understand chapters 4 and 5 of Goldstein in the
language of differential geometry. But I wanted to get a reply off to
you because I never know in advance how long my attention span will
hold out.

(Gregory L. Hansen) wrote in message . com...

(Ken S. Tucker) wrote in message . com...
(Gregory L. Hansen) wrote in message ...

Hi Greg

Here's another one to ponder. In special relativity make the
transformation to rotation coordinates,

t' = t

x' = x cos wt - y sin wt

y' = x sin wt + y cos wt

z' = z

From that you can find a metric, connections, and geodesic equations. And
the geodesic equations will have terms interpretable as centrifugal and
Coriolis forces, which can be generalized in vector notation. Those
inertial terms come from g_00 and g_0i=g_i0.

I wanted to follow a similar program in Newtonian mechanics, since it's
the same change of coordinates, it's still just geometry. But Newtonian
spaces don't have g_00 or g_0i because time isn't a component of Newtonian
vectors.

I think your tying one foot behind your back and entering an
ass kicking contest.

In Newtons Geometry you can still have 4D graphs, except
that these would be cartesian, so
g11 = g22 = g33 = g44 =1 and covariant and contravariant
components are equal.

So that wonderful derivation doesn't seem available in Newtonian
mechanics, and the centrifugal and Coriolis terms have to be pounded
together by other means.
The geodesic equations tell you what a straight line is, I'd have figured
it would tell you what a straight line is even in the Newtonian rotating
frame. I really don't understand what went wrong there.

I don't see what's wrong, if you permit Cartesian 4D.

I was going to give a kneejerk response, then went away to think about
it. Galilean relativity doesn't have four-vectors, it has ordered
pairs of three-vectors and time, which are separately conserved under
transformations. A Cartesian metric clearly can't work in Galilean
relativity. E.g. in special relativity, for an observer at rest in
the lower-case frame observing something in the capital frame,

c^2 dT^2 = c^2 dt^2 - dx^2

dT^2 = dt^2 (1 - (dx/dt)^2/c^2)

dt = dT / sqrt(1 - v^2/c^2)

Time dilation. In a Cartesian metric,

k^2 dT^2 = k^2 dt^2 - dx^2

dt = dT / sqrt(1 + v^2/k^2)

Time contraction.

But consider k=0,

ds^2 = 0*dt^2 + dx^2 + dy^2 + dz^2

When you work the metric in special relativity for a rotating frame
you get

ds^2 = (-c^2 + w^2(X^2+Y^2)) dT^2 + dX^2 + dY^2 + dZ^2

+ 2 Y w dX dT - 2 X w dY dT

The only contribution from the dt^2 term in the metric is the -c^2 in
the first term. If you work the problem as before for the Galilean
metric I gave, you get exactly the same thing, but without the -c^2.
So the 0*dt^2 is acting as a place holder to say that something could
be there, even if there isn't anything there right now.

And then when you consider Galilean time intervals you need to
consider the separate metric

(ds_t)^2 = dt^2 + 0*(dx^2 + dy^2 + dz^2)

which I find kind of ugly. Besides the fact that the four-dimensional
version is clearly derived from another theory that's not Newtonian
mechanics from a guy that already knows what kind of answer he wants
to get.

The usual way of going to Newtonian mechanics is to let c-inf, or the
ratio v/c-0. Which works fine in the time,

dT^2 - (dX^2 + dY^2 + dZ^2)/c^2 = dt^2 - (dx^2 + dy^2 + dz^2)/c^2

- dT^2 = dt^2 as c-inf

But what can we say about the spatial part? We can ignore the spatial
part when the time part is much larger, but we can't ignore the time
part when the time part is much larger.

As I pointed out in my previous post, Newton could have
concievable found (unit vector t) dot (unit vector x) is non zero
and called this g_14 if he could use the geodesic formula.

From this, the Lorentz transform follows, but the constant c,
still needs to be postulated from empirical evidence. This came
from Maxwell's Equations and MMX.

Maybe I'll try to understand chapters 4 and 5 of Goldstein in the
language of differential geometry. But I wanted to get a reply off to
you because I never know in advance how long my attention span will
hold out.

Good idea. If you get something interesting, hope you would take
a moment to post your ideas.
Ken S. Tucker

(Ken S. Tucker) wrote in message . com...
(Gregory L. Hansen) wrote in message . com...

As I pointed out in my previous post, Newton could have
concievable found (unit vector t) dot (unit vector x) is non zero
and called this g_14 if he could use the geodesic formula.

From this, the Lorentz transform follows, but the constant c,
still needs to be postulated from empirical evidence. This came
from Maxwell's Equations and MMX.

Maybe I'll try to understand chapters 4 and 5 of Goldstein in the
language of differential geometry. But I wanted to get a reply off to
you because I never know in advance how long my attention span will
hold out.

Good idea. If you get something interesting, hope you would take
a moment to post your ideas.
Ken S. Tucker

The geodesic equations give you inertial forces, the terms that aren't
the accelerations. But those depends on velocities. One velocity
you're guaranteed of, especially in Newtonian mechanics, is the
passage of time. And in fact, if you take as your metric

ds^2 = 0*dt^2 + dx^2 + dy^2 + dz^2

when you put something in g_00, (and consider a particle at rest) it
amounts to the potential, or phi/2. And by potential I mean exactly
what's meant in Newton's law,

dp/dt = - grad phi

And that's old news for a weak gravitational potential in general
relativity, but I was just kind of surprised to see it pop out in
Newtonian mechanics that way. If you have a time-dependent potential
you get a rate of time flow that can vary according to position, I
suppose that should be artificially discarded.

And then consider the ol' plane wave from, e.g., quantum mechanics.

psi = A * exp i(p.x - Et)

What does that look like to you? Looks to me a lot like a
four-momentum multiplied by a four-position.

It's like a four-dimensional description really turns out to be the
most natural approach even if you don't have Einstein on the mind.

Interestingly, in Goldstein's suggested reading of chapter 5, he
mentions Felix Klein's book on the mathematical theory of the top,
published around 1896. That and Sommerfeld's four-volume work on tops
used a four dimensional formalism for mathematical convenience, years
before Minkowski used it in special relativity, although no physical
significance was given to the time dimension. Maybe I'll try to look
up Klein's book.

But it's amazing how short the ol' attention span lasts when you hit
Euler angles.