The nature of force

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20 years 3 months ago #11278 by EBTX
Replied by EBTX on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Of course, this assumes that gravitons can collide with comets and accelerate them. But by extension, gravitons do not need to actually collide with matter ingredients in comets to create the appearance that they did, provided that some super-small entities create the appearance that they did collide. And so on, through an infinite range of scales<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I think I now understand completely your point of view, though I disagree with it at the deepest level. You've chased the problem down an infinite hole and decreed a solution which is identical the concept of "field". The only difference I can see is that "field" is considered a logical primary and no one frets over its lack of "parts".

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20 years 3 months ago #10139 by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
“Force” is defined as the time rate of change of momentum. So there could not exist a momentum change without a force acting<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Yes, force is classically defined as the rate of change of momentum because normally you are dealing with elastic collisions i.e. particles interacting via their static fields (e.g. gravitation, Coulomb force etc.). This is implicitly the assumption on which Newtonion physics rests, but for inelastic collisions (which are of a quantum 'mechanical' nature) this does not apply anymore (assuming that it does would be an unjustified generalization from Classical Mechanics). Of course, alongside an inelastic excitation of an atom by a charged particle there will be also some momentum transfer due to the Coulomb interaction force, but this is not what causes the excitation (and the associated change of momentum) in the first place. One also has to bear in mind that the Schrödinger Equation for an atom assumes the existence of a the Coulomb force field implicitly (in order to be able to define the potential energy), so there would not be any inelastic processes without a static force field anyway.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Let’s focus on why solar radiation pressure, which results from the collision of photons with bodies orbiting the Sun and is a major player these days in astronomy, is not a counterexample to your notion that all forces or recoils from collisions require a field to be present. Photons have no fields.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
'Photons' *are* fields as they are electromagnetic waves, but forget about light in this context because, as mentioned earlier, a consistent theory for radiation pressure does not exist (the corresponding observations can only be explained as a secondary effect due to the release of charged particles and subsequent elastic collisions on the surface of the material).

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In your view, can neutron stars (with no electrostatic interactions at the atomic level) collide, or not?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Obviously, neutrons can not (elastically) collide via an electrostatic field since they don't have one. If they have other static force fields then they can collide via this. If they don't have any static force fields at all, then they could 'collide' only inelastically (i.e. due to quantum 'mechanical' effects) but would not excert any force.



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20 years 3 months ago #10140 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />I think I now understand completely your point of view, though I disagree with it at the deepest level. You've chased the problem down an infinite hole and decreed a solution which is identical the concept of "field". The only difference I can see is that "field" is considered a logical primary and no one frets over its lack of "parts".<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">And that is a pity. Models that depend on axioms suspended from clouds (with no possible logical basis) such as "fields" without parts are very unsatisfying to a physicist because they invoke magic in that assumption. But such assumptions lacking in a physical basis are now so common in mathematics that few people even raise an eyebrow anymore.

If you find my explanation of contact distasteful, however did you get past Zeno's original paradox that there can be no motion? I see no logical way (i.e., without miracles) to resolve that paradox without the infinite one-to-one correspondence. Postulating smallest possible units of space and time creates numerous unresolvable paradoces to replace one resolvable one. -|Tom|-

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20 years 3 months ago #10141 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />Yes, force is classically defined as the rate of change of momentum because normally you are dealing with elastic collisions i.e. particles interacting via their static fields (e.g. gravitation, Coulomb force etc.). This is implicitly the assumption on which Newtonion physics rests, but for inelastic collisions (which are of a quantum 'mechanical' nature) this does not apply anymore (assuming that it does would be an unjustified generalization from Classical Mechanics).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">So you need to postulate that the quantum scale operates with different physical laws than the macroscopic scale. And you need to postulate that "inelastic collisions" are completely different processes than "elastic collisions", not just differing in degree of elastic deformation.

Neither Occam nor I have much appreciation for unneeded postulates, especially those that serve only a mathematical purpose but lack any justification on physics. (See below.)

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Of course, alongside an inelastic excitation of an atom by a charged particle there will be also some momentum transfer due to the Coulomb interaction force, but this is not what causes the excitation (and the associated change of momentum) in the first place.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">And what does?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">'Photons' *are* fields as they are electromagnetic waves, but forget about light in this context because, as mentioned earlier, a consistent theory for radiation pressure does not exist (the corresponding observations can only be explained as a secondary effect due to the release of charged particles and subsequent elastic collisions on the surface of the material).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That explains nothing about the source of the new momentum, which clearly arises from the electromagnetic wave, which carries a known momentum. Clearly, collision is the transfer vehicle, not "fields". My counterexample here to fields always being needed to transfer momentum stands unchallenged. And we are then back to collisions as the primary.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Obviously, neutrons can not (elastically) collide via an electrostatic field since they don't have one. If they have other static force fields then they can collide via this. If they don't have any static force fields at all, then they could 'collide' only inelastically (i.e. due to quantum 'mechanical' effects) but would not exert any force.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Please describe the details of this hypothetical forceless collision. So if a neutron star runs into a single neutron, the star will boumce back inelastically and leave the single neutron unaffected? But if not and momentum is transferred, then the change in momentum is a force by definition. I am struggling, but do not see a self-consistent model here than can avoid momentum transfers via simple, classical, elastic collisions, even for cases with too little elasticity to be observed.

I also do not see a mechanism for rebound if the collision was truly inelastic. What physical process propels the colliding entity backwards in not the elastic deformation of the target body springing back toward its original shape? -|Tom|-

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20 years 3 months ago #10143 by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
So you need to postulate that the quantum scale operates with different physical laws than the macroscopic scale. And you need to postulate that "inelastic collisions" are completely different processes than "elastic collisions", not just differing in degree of elastic deformation.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I do not need to postulate that inelastic collisions are fundamentally different from elastic collisions as it is self-evident (inelastic collisions do not exist in Classical Mechanics). On the contrary, it would be a (rather unreasonable) postulate if you assume that the principles of Classical Mechanics apply identically also in Quantum 'Mechanics'.
This has nothing to do as such with the scale (as elastic collisions occur also on the atomic scale) but with the completely different nature of the physical processes involved.


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Of course, alongside an inelastic excitation of an atom by a charged particle there will be also some momentum transfer due to the Coulomb interaction force, but this is not what causes the excitation (and the associated change of momentum) in the first place.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">And what does?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The existence of a resonant physical mechanism that enables certain kind of particles with a certain kinetic energy to excite atomic transitions (and lose their kinetic energy in the process). The Schrödinger Equation gives the cross section for this (basically it is determined by the dipole integral involving the wave functions of the free and bound states of the particle).


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Please describe the details of this hypothetical forceless collision. So if a neutron star runs into a single neutron, the star will boumce back inelastically and leave the single neutron unaffected? But if not and momentum is transferred, then the change in momentum is a force by definition. I am struggling, but do not see a self-consistent model here than can avoid momentum transfers via simple, classical, elastic collisions, even for cases with too little elasticity to be observed.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I am not very familiar with nuclear physics and don't know what kind of inelastic collisions (or elastic collisions for that matter) exist here, but if the collision is truly inelastic (in the sense described by me above), then a neutron would simply be stopped and its energy converted into a non-mechanical 'energy'-form (e.g. radiation). In this case no momentum (and hence no force) would be transferred as there is no mechanical contact like for elastic collisions (which are due to the static force field).


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I also do not see a mechanism for rebound if the collision was truly inelastic. What physical process propels the colliding entity backwards in not the elastic deformation of the target body springing back toward its original shape?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">For truly inelastic collisions there is obviously no rebound (like for elastic collisions) because the relative speed of the two particles is zero after the 'collision'.



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20 years 3 months ago #11281 by EBTX
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<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">And that is a pity. Models that depend on axioms suspended from clouds (with no possible logical basis) such as "fields" without parts are very unsatisfying to a physicist because they invoke magic in that assumption. But such assumptions lacking in a physical basis are now so common in mathematics that few people even raise an eyebrow anymore.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I couldn't agree more about current models having no connection to actual human experience. But I don't see the field concept as divorced from experience ... at least it wasn't in Faraday's experience. In the case of neutral bodies the concept of collision is obviously paramount but when looking at the less common charged (or magnetic) body experience there is no obvious collision element. So Faraday invented the field concept.

There was nothing else to do ... and ... more importantly, no reason to postulate something unseen to fulfill a collision requirement. It is not that the field cannot, in principle, be reduced to particles as MM suggests ... it's that it is logically unnecessary to do so ... unless ... one can detect (directly and unambiguously) the particles of which it is composed. Until then, all the particles which compose the "field" can be tied up in a bag and ... I should just examine the bag itself as the primary existential element.

I see the field concept as being composed of only a few parameters. These are properties which denote ... distance, direction and a tension analog as in twists or compression/stretching ... all of which have experiential analogs in the real world (like a block of rubber). A field, in my view, can have no attribute without a clear physical analog. For instance, there can be no Higgs field which exists solely to assign mass to particles. My question would then be ... How ?? in terms of motion, expansion/contraction, rotation, etc.

Thus, there are only a few conceivable things that can happen in 3-space. An object can move relative to another object laterally or toward/away. It might rotate on an axis (with some complications). It might expand or contract like a balloon (again with some complications). It might disappear and reappear elsewhere (not necessarily observed, just conceivable). If we add a field concept we also have the above mentioned properties.

Beyond that, there is nothing else that can happen in our universe ... unless we go to "floating abstractions" of which there are an infinite number. We need only make up a word to identify the phenomenon then we have phlogiston and P-branes all over the place. We would then be putting "experiential pieces" into a bag labeled "P-brane" and accepting that as an existential primary which I reject. Our difference is that you have only particles in your model. I have fields as well but they are limited to experiential properties.

I say that you do not have enough stuff to build the observed universe. You say that I have too much. But my view is no more open-ended than yours. I can't in principle add anything more either. ;o)

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