# Newton's Second Law

by pej.dgr@bigpond.com
Tags: newton
 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
 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
 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 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
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

 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 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).
 P: n/a On Nov 21, 10:26M-BM- pm, Tom Roberts 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.