#1
Nov2008, 05:00 AM

P: n/a

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 


#2
Nov2108, 05:00 AM

P: n/a

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 Newtonâ€™s second law to dp/dt = >dpâ€™/dt + F, where pâ€™ is the particleâ€™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 


#3
Nov2208, 05:00 AM

P: n/a

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 locallystraight 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 4second flight it travels 4 lightseconds 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 3momentum, t is the time coordinate (of an inertial frame), and F is the total applied force (a 3vector in which gravity appears). In GR this becomes (applies only to a test particle): DP/d\tau = F where P is the particle's 4momentum, \tau is the particle's proper time, F is the total applied 4force (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 


#4
Nov2208, 05:00 AM

P: n/a

Newton's Second Law
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 


#5
Nov2308, 05:00 AM

P: n/a

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 em fields (important caveat!). The contravariant 4momentum is p^a = m dx^a/dT, where m is the invariant rest mass and T is the proper time. The covariant 4momentum 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 4force by F_a = dV/dx^a. Clearly, this equation has the form that you suggest. However, if an em field,described by 4potential A^a(x^u), is present, then things look rather more messy! (although possibly they still simplify in terms of the em 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). 


#6
Dec608, 05:00 AM

P: n/a

On Nov 21, 10:26MBM 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 timedependent and/or positiondependent.  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 nonzero only if the total momentum of the two is nonzero. 


#7
Dec708, 05:00 AM

P: n/a

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. > 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 backandforth 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 selfinconsistent 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 > nonzero only if the total momentum of the two is nonzero. 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 nonzero"). Your attempts at descriptions are woefully inadequate, and disagree with simple, basic observations of billiard balls  let me use your definitions of words in singlequotes: 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 thencurrent theories and experiments. You will probably find that your new paradigm is unnecessary. 


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