Newton's Second Law

pej.dgr@bigpond.com

According to general relativity a test particle senses the curvature
of space ( a function of the density of all matter in the universe)
and responds to it with a continuous change in momentum. The only
forces that act on such a particle are contact forces that constrain
its motion. We should therefore revise Newton’s second law to dp/dt =
dp’/dt + F, where p’ is the particle’s momentum in a state of free
fall.

Is that a fair statement? If not how should it read?

Phil Gardner

Oh No

Thus spake pej.dgr@bigpond.com
>According to general relativity a test particle senses the curvature
>of space ( a function of the density of all matter in the universe)
>and responds to it with a continuous change in momentum. The only
>forces that act on such a particle are contact forces that constrain
>its motion. We should therefore revise Newton’s second law to dp/dt =
>dp’/dt + F, where p’ is the particle’s momentum in a state of free
>fall.
>
>Is that a fair statement? If not how should it read?
>
Not really. In general relativity a particle does not sense the
curvature of space and, in the absence of an active force, travels in a
locally straight line during each part of its motion. The curvature of
space means that a line which is everywhere locally straight does not
end up looking straight when viewed on the large scale, and is called a
geodesic.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Tom Roberts

pej.dgr@bigpond.com wrote:
> According to general relativity a test particle senses the curvature
> of space ( a function of the density of all matter in the universe)
> and responds to it with a continuous change in momentum.

This is nowhere near true. Perhaps you should study GR before attempting
to make assertions about it.

In GR a test particle moves in a locally-straight line when no external
forces act on it. Note that gravity is not a force in GR. Even though
the particle's trajectory is locally straight, on larger scales the
trajectory can appear to be curved, due to the curvature of spacetime,
which we call "gravity". For a thrown baseball, its trajectory is
accurately straight at the scale of nanoseconds (feet), but not at the
scale of seconds (a billion feet).

That thrown baseball follows a geodesic path through spaceTIME,
and in Cartesian coordinates fixed to the earth's surface is
nearly a straight line -- during a 4-second flight it travels
4 light-seconds along the time axis (4 billion feet), but
the curvature of the path you watch is less than 200 feet.
So its geodesic differs from a straight line by about 0.05
parts per million. The earth's gravity is quite weak!

> The only
> forces that act on such a particle are contact forces that constrain
> its motion.

This is also not true. Electromagnetic forces are quite common, and are
not "contact". Indeed, "contact forces" are really electromagnetic.

> We should therefore revise Newton's second law to dp/dt =
> dp'/dt + F, where p' is the particle's momentum in a state of free
> fall.

Again no. In Newtonian physics one has:

dp/dt = F where p is an object's 3-momentum, t is the
time coordinate (of an inertial frame), and
F is the total applied force (a 3-vector in
which gravity appears).

In GR this becomes (applies only to a test particle):

DP/d\tau = F where P is the particle's 4-momentum, \tau is the
particle's proper time, F is the total applied
4-force (no gravity), and D/d\tau is the covariant
derivative (gravity appears in the connection
inside it).

This is considerably more complicated than Newtonian physics....

Tom Roberts

Oh No

Thus spake Tom Roberts <tjroberts137@sbcglobal.net>
>> The only
>> forces that act on such a particle are contact forces that constrain
>> its motion.
>
>This is also not true. Electromagnetic forces are quite common, and are
>not "contact".
>Indeed, "contact forces" are really electromagnetic.
>
Good post, but might make a couple of points. It is not clear here
whether you mean that em forces are contact or not, and I think this is
very much a matter of view point. I regard them as contact forces, where
the contact is actually made with photons acting as carriers for the
force. Others might say that contact is not made, and that the force is
transmitted by photons.

Also, the most dominant force which might be described as "contact" is
the Pauli exclusion principle. This is the force which prevents one
solid object from merging into another solid object.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

a student

On Nov 20, 4:13*am, pej....@bigpond.com wrote:
> According to general relativity a test particle senses the curvature
> of space ( a function of the density of all matter in the universe)
> and responds to it with a continuous change in momentum. *The only
> forces that act on such a particle are contact forces that constrain
> its motion. *We should therefore revise Newton’s second law to dp/dt =
> dp’/dt + F, where p’ is the particle’s momentum in a state of free
> fall.
>
> Is that a fair statement? *If not how should it read?

Despite some of the other posts in reply, one can rewrite Newton's
second law in a form quite similar to the one you suggest, at least
when there are NO e-m fields (important caveat!).

The contravariant 4-momentum is
p^a = m dx^a/dT,
where m is the invariant rest mass and T is the proper time. The
covariant 4-momentum is
p_a = g_ab p^b ,
where g_ab is the metric tensor.

It is not difficult (although slightly painful) to show that the
geodesic
equation (i.e., motion with no external forces) can be written as
dp_a/dT = (1/2m) (dg^uv/dx^a) p_u p_v . (*)
Let us denote the righthand side of this equation by dp'_a/dT. One
may similarly show that if the particle moves under a scalar
potential V(x^u), that the above form of the geodesic equation is
replaced by the more general equation
dp_a/dT = dp'_a/dT + F_a, (**)
where one defines the 4-force by F_a = dV/dx^a. Clearly, this
equation has the form that you suggest.

However, if an e-m field,described by 4-potential A^a(x^u), is
present,
then things look rather more messy! (although possibly they still
simplify in terms of the e-m momentum?) - the equation of motion can
be
written as
dp_a/dT = (del/del x^a) sqrt[ g^uv (p_u - eA_u)(p_u - eA_v) ] +
F_a,
where the sqrt term is also equal to the rest mass m - I wouldn't
want to try and sort this out just for the sake of it!

BTW, it is not clear what you mean by 'contact forces'. If you mean
collisions, then this is tricky to deal with in GR (eg, there are no
rigid bodies).

pej.dgr@bigpond.com

On Nov 21, 10:26M-BM- pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> pej....@bigpond.com wrote:
>
> (Snip)
> > The only forces that act on such a particle are contact forces that constrain
> > its motion.
>
> This is also not true. Electromagnetic forces are quite common, and are
> not "contact".
>
> (Snip)
>
> Tom Roberts

I would argue that:

The (total, net) force on a particle or body is best defined by F =
del E, where E is its total energy and the derivatives are with
respect to the particle's coordinates.

This definition is consistent with the results of every experiment
that can be made in which the force is measured (by a macroscopic
measuring device) rather than postulated as a consequence of some
observed acceleration.

In any elastic interaction between two identical particles or bodies,
the total momentum of which is zero, symmetry considerations require
their total energies to be equal and constant.

[[Mod. note -- Your definition of force looks quite similar to what's
usually called the "principle of virtual work". But the conclusion
you reach in the immediately preceding paragraph is wrong: the total
energies may well be time-dependent and/or position-dependent.
-- jt]]

It must therefore follow that there is no force between them, however
large their accelerations are.

The electromagnetic force between two colliding electrons is therefore
non-zero only if the total momentum of the two is non-zero.

Tom Roberts

pej.dgr@bigpond.com wrote:
> The (total, net) force on a particle or body is best defined by F =
> del E, where E is its total energy and the derivatives are with
> respect to the particle's coordinates.

This simply does not work for magnetic forces, for which the energy of a
moving charged particle does not change (the force is perpendicular to
its velocity).

But you attempt to introduce new definitions of words.
This is just one instance of the many PUNs you use.
Such unacknowledged PUNs destroy your arguments.
(More on this below.)


> This definition is consistent with the results of every experiment
> that can be made in which the force is measured (by a macroscopic
> measuring device) rather than postulated as a consequence of some
> observed acceleration.

Then you should not use the words "energy" and "force", as your
definitions are unusual, and differ in significant ways from the normal
meanings of those words. You confuse both your readers and yourself with
such PUNs: You are (implicitly) using both the normal definitions and
your unusual ones in various places of your discussion, and thus obtain
nonsense. Such PUNs destroy any argument you attempt to make.

Also you seem to not understand the role of science.
Science is an attempt to model the world; in Newtonian
mechanics, when an acceleration is observed, a force
must be responsible. There is no problem "postulating"
electromagnetic and gravitational forces, even though
they cannot be directly measured, because they are
utterly necessary for the model to describe actual
observations of the world.


If you want to use your definitions, you cannot rely on ANY of the
normal relationships, and must derive things from scratch using YOUR
definitions, and your definitions ONLY -- you need to start from scratch
defining YOUR OWN mechanics. Good luck!

This is a general problem for people like yourself attempting to
redefine standard words -- most people find it impossible to stick to
their new definitions. You certainly did not write consistently using
JUST your own definitions -- you slip back-and-forth between
conventional meanings and your peculiar meanings in just about every
sentence you write. Words are important, as are their definitions,
because they are what we use to THINK with (and also to communicate).
Words with slippery definitions are useless -- you need to clean up your
act before you can hope to write or think anything sensible in your new
paradigm. So you need to avoid the usual words and make up your own
words, to be able to prevent conventional meanings from slipping in. I'm
pretty sure that when you do so, you will find your new paradigm to be
either self-inconsistent or useless (applicable to no real phenomena).


> In any elastic interaction between two identical particles or bodies,
> the total momentum of which is zero, symmetry considerations require
> their total energies to be equal and constant.
> It must therefore follow that there is no force between them, however
> large their accelerations are.

So those accelerations come from magic? With conventional words there is
no problem, but using your meanings there is an enormous hole....


> The electromagnetic force between two colliding electrons is therefore
> non-zero only if the total momentum of the two is non-zero.

You use far too many PUNs, and obtain nonsense (with your definitions,
"electromagnetic force" is an oxymoron -- it can never be directly
measured, even if "total momentum is non-zero"). Your attempts at
descriptions are woefully inadequate, and disagree with simple, basic
observations of billiard balls -- let me use your definitions of words
in single-quotes: one can measure the 'forces' between colliding
billiard balls (strain gauges on their surfaces), and yet you claim
those 'forces' are zero (from "symmetry considerations").

I repeat: you need to develop YOUR OWN mechanics from scratch, using new
words with YOUR unusual definitions, and ONLY your definitions.



And a suggestion: learn standard physics first. History shows that
nobody has ever made a significant contribution to the field without
being familiar with then-current theories and experiments. You will
probably find that your new paradigm is unnecessary.