tommyedison Posted December 10, 2004 Report Share Posted December 10, 2004 For nuclear fusion reactions to occur, two positively charged nuclei must collide. According to Coulomb's law for electrostatic repulsion, when the particles are just about to collide, the repulsion would be infinite. If the repulsion is infinite, how can the positively charged nuclei collide at all? And do physicists use special equations to measure the amount of speed required for two +ve particles to collide?Does gravity play any role? What am I missing? Thanks in advance for any help. Quote Link to comment Share on other sites More sharing options...

punk Posted December 10, 2004 Report Share Posted December 10, 2004 Coulomb's law breaks down for very short distances and very high energies. Or in other words: Coulomb's law doesn't apply in the situation of nuclear fusion. Every physical "law" should be stated with a domain of validity. If you are discussing any situation outside of that domain, the law does not apply. Put another way, every physical law should be thought of as an approximation to a still more general law. In the case of Coulomb's law, contemporary physics would view it as a mean field approximation to Quantum Electrodynamics, which is itself an approximation to Quantum Electro-weak Theory. To really look at nuclear fusion you would have to go to the Standard Model (Electro-weak Theory plus Quantum Chromodynamics). Quote Link to comment Share on other sites More sharing options...

Scientist Posted December 10, 2004 Report Share Posted December 10, 2004 For nuclear fusion reactions to occur, two positively charged nuclei must collide. According to Coulomb's law for electrostatic repulsion, when the particles are just about to collide, the repulsion would be infinite. If the repulsion is infinite, how can the positively charged nuclei collide at all? And do physicists use special equations to measure the amount of speed required for two +ve particles to collide?Does gravity play any role? What am I missing? Thanks in advance for any help. Iam sure Stephen will come here and give you a much better explanation than I will, but I never post so I thought I would give it a try . First, there is another force involved other than the electromagnetic force called the strong nuclear force. This is the force that holds atoms together, and it is much stronger than the electromagnetic force (but it only works up to a certain distance... past that distance it essentially "turns off.") For fusion to occur, the charged particle must get into the region where the strong nuclear force dominates over the electromagnetic force. The particle can do this by having a great deal of kinetic energy (from its temperature) but there is also a quantum effect called "tunneling" that can allow particles with less than this kinetic energy to reach the region where the strong force dominates. This happens in the Sun. If there were no quantum effects, then the temperature required in the center of the Sun for the fusion we see would be 1,000 times greater than the temperature we actually know is there. The "tunneling" effect explains why the temperature can be cooler than we would otherwise expect. Does this explain at all what you were asking? I wanted to show a picture of the potential of a nucleus as a function of distance so you could see where the strong force starts to dominate, but I can't find one online right now! I noticed that punk posted while I was writing this. He is right in that a totally correct analysis of fusion would require the most advanced theories, but the explanation I gave you is what is usually used when we talk about it, and the calculations you get using it are pretty accurate at the level we are looking at. So, now I just have to wait for Stephen to come and tell me all the things that are wrong with what I said Quote Link to comment Share on other sites More sharing options...

ewv Posted December 11, 2004 Report Share Posted December 11, 2004 ...when the particles are just about to collide, the repulsion would be infinite.Â If the repulsion is infinite, how can the positively charged nuclei collide at all? Coulomb's law breaks down for very short distances and very high energies. Or in other words: Coulomb's law doesn't apply in the situation of nuclear fusion.Â Every physical "law" should be stated with a domain of validity. If you are discussing any situation outside of that domain, the law does not apply.There is no such thing as an actual infinity. Whenever you find one "predicted" by any theory it necessarily means you are beyond its domain of validity. A singularity can tell you a lot based on the rate of increase in its vicinity, but it should not be interpreted to predict infinity at the singular point (or values too close to it). Put another way, every physical law should be thought of as an approximation to a still more general law. In the case of Coulomb's law, contemporary physics would view it as a mean field approximation to Quantum Electrodynamics, which is itself an approximation to Quantum Electro-weak Theory.Speaking philosophically, it is not true that every law is an approximation to a better one, which would imply an infinite regress (and a strict ordering of all comparable theories in all respects). Better to recognize that every theory has a finite domain of validity, with a finite degree of precision -- then go on from there to characterize what aspects of what theory may be regarded as an approximation to what aspects of another in terms of explanation or precision. There is no such thing as an "ideal" "infinite precision" that renders all physical laws otherwise "approximate". Correct theories are exact within their domain of validity and to the limits of their known degree of precision; "exactness" is not synonomous with an unattainable actual infinite, which has no identity and cannot exist (and about which punk didn't claim otherwise). Referring to theories as "approximate" may be meant in only a loose way or may be specified more definitively, but be careful when using the term "approximation" this way; it has a long record of epistemological abuse. Quote Link to comment Share on other sites More sharing options...

punk Posted December 11, 2004 Report Share Posted December 11, 2004 The conventional view in science is that if any physical "law" has sufficient observational data to be held as "valid", then some new more general theory which is intended to replace it should be able to derive the old "law" in some appropriate limit. So one test of General Relativity is the ability to derive Newton's Law of Gravity in the appropriate non-relativistic, low energy limit. In that sense the old "law" was an "approximation" of the more general theory. I dont intend to imply an infinite regress to approximations, although I do admit it was a rather loose way of speaking. At any given time we will always have a finite set of data points in a limited range from which to construct a theory. Given this, we should always assume that the best theory we have is not the grand theory of everything (because we would now need an infinite amount of data points to prove this). So we are always in the position of supposing that there is a more general theory out there than the one we currently have. So we always suppose the current best theory is an approximation to something else. Of course this leaves open the chance that we might find the theory of everything, but we'll never be totally sure that we have it. But this is more of an epistemic question and far beyond the original topic here. Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 11, 2004 Report Share Posted December 11, 2004 Iam sure Stephen will come here and give you a much better explanation than I will, No need. You do just fine. I usually don't read ahead, but I am glad I did. but I never post so I thought I would give it a tryÂ . I hope you will post more often. And, for the record, I agree with the overall sense of the points made by ewv. Quote Link to comment Share on other sites More sharing options...

ewv Posted December 11, 2004 Report Share Posted December 11, 2004 ... we are always in the position of supposing that there is a more general theory out there than the one we currently have. So we always suppose the current best theory is an approximation to something else."Out" where? There is no theory "out" anywhere not known by some consciousness. Until that occurs, there is no "something else" to approximate. Only the facts that give rise to a theory, or which may give rise to one in the future, exist. Also, because there is no such thing as an "infinite number" of data points (or anything else), any notions of anything that would require that for "validation" are also meaningless, invalid concepts. What would a "grand theory of everything" even mean? Omniscience? That, too, would be "infinite" and therefore nonsensical. All knowledge is definite and finite -- and therefore necessarily "limited" -- in every respect in order to have identity. Saying that knowledge is "limited" is not a disparagement or characterization of it as merely "approximate": "limitation" of knowledge is necessary for its identity and therefore for it to exist at all. Those are some of the problems implied by too loose a characterization of theories in terms of "approximation", even though there is an element of truth behind it if properly interpreted, and unspecified ideas of a "theory of everything" as a kind of ideal limiting case thought of as making sense of existing, finite knowledge. (In terms of the widest possible abstractions, the philosophical principles of existence and identity are as broad as you can get, but by themselves don't provide a way to deduce details.) Quote Link to comment Share on other sites More sharing options...

jedymastyr Posted December 11, 2004 Report Share Posted December 11, 2004 For nuclear fusion reactions to occur, two positively charged nuclei must collide. According to Coulomb's law for electrostatic repulsion, when the particles are just about to collide, the repulsion would be infinite. If the repulsion is infinite, how can the positively charged nuclei collide at all? And do physicists use special equations to measure the amount of speed required for two +ve particles to collide?Does gravity play any role? What am I missing? Thanks in advance for any help. I don't know much about advanced physics (strong/weak nuclear forces), but even in the more basic E&M that engineers learn in college this isn't a problem (if two positively charged nuclei must collide). Take two protons, for example: If you look at the spherical shell theorem 2, you'll see that a spherical particle acts (with respect to Coulomb's Law) as if it is a point charge at the center of the sphere. I'm pretty sure this (shell theorem) directly comes from Gauss' Law (from which Coulomb's Law is derived). [i seem to remember my mechanics teacher explaining how Newton invented calculus (partially, perhaps?) in order to calculate the force of gravity on an object due to a large spherical object (such as the Earth) by adding up all the components of gravity from small portions of the Earth's mass. If I remember correctly, he invented calculus, and then proved it with geometry, afterward. I think, due to the similarity between the gravity/electrostatics equations, this applies similarly to electric forces. So, I think Newton was the one that discovered it was the same as the point charge at the center of Earth equal to the Earth's mass.] Anyway, back to the problem at hand--if you have two particles, which take up some amount of space, the electrostatic force would only be infinite if their centers corresponded to the same location in space. Since this is obviously not possible, it is impossible to get any infinite force requirements out of this. If I'm wrong about any of this, feel free to correct me. I have the feeling that it's not that significant anyway, though, since as has already been mentioned there are much stronger forces at that range (nevertheless, it's important to know that it's not an infinite force required to pull them in). Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 11, 2004 Report Share Posted December 11, 2004 The conventional view in science is that if any physical "law" has sufficient observational data to be held as "valid", then some new more general theory which is intended to replace it should be able to derive the old "law" in some appropriate limit.Â So one test of General Relativity is the ability to derive Newton's Law of Gravity in the appropriate non-relativistic, low energy limit.Â In that sense the old "law" was an "approximation" of the more general theory. I do not think of that as a proper perspective. Clearly the predictions of a more general theory must reduce to the correct predictions of the prior theory, but sharing a mathematical formalism as a limit is not necessarily an approximation of the more general theory, at least not in any meaningful sense. A globally-flat spacetime with small perturbations may be an approximation to general relativity, but only in the sense of the limit process starting from general relativity. It is not simply the reduction to a shared mathematical Newtonian limit that matters; afterall, special relativity and the Lorentz ether theory share the same mathematical base, but one could hardly consider Lorentz' theory as the local limit of GR, as can be done for SR. And, as far as the physical theory itself is concerned, in what manner can the instantaneous propagation of gravity of Newton's law be considered an approximation for changes in the gravitational field propagating at light speed, in GR? So we always suppose the current best theory is an approximation to something else. But since Newtonian gravitation and general relativity are as disparate as two physical systems could be, the use of "approximation" obscurs the only really important fact that links the two -- that is that general relativity expands the context in which Newtonian gravitation properly applies, not as the latter being an approximation to the former, but as an expansion of the Newtonian domain of applicability by a fundamentally different theory. Quote Link to comment Share on other sites More sharing options...

tommyedison Posted December 11, 2004 Author Report Share Posted December 11, 2004 Thank you for the replies. If you look at the spherical shell theorem 2, you'll see that a spherical particle acts (with respect to Coulomb's Law) as if it is a point charge at the center of the sphere. I'm pretty sure this (shell theorem) directly comes from Gauss' Law (from which Coulomb's Law is derived). [i seem to remember my mechanics teacher explaining how Newton invented calculus (partially, perhaps?) in order to calculate the force of gravity on an object due to a large spherical object (such as the Earth) by adding up all the components of gravity from small portions of the Earth's mass. If I remember correctly, he invented calculus, and then proved it with geometry, afterward. I think, due to the similarity between the gravity/electrostatics equations, this applies similarly to electric forces. So, I think Newton was the one that discovered it was the same as the point charge at the center of Earth equal to the Earth's mass.]Anyway, back to the problem at hand--if you have two particles, which take up some amount of space, the electrostatic force would only be infinite if their centers corresponded to the same location in space. Since this is obviously not possible, it is impossible to get any infinite force requirements out of this. I did think of this but assuming the diameter of one proton to be 1 fm (which is I think around the average diameter of nucleus), the repulsion acceleration was coming out to be in orders of 29. I thought this is too high for protons at temperatures of order 8 or 9 to collide. -------------------------------------------- Incidentally, I have finished Resnick and Halliday: Fundamentals of Physics. Could you please recommend more advanced books. I am mostly interested in electromagnetism and quantum mechanics. Quote Link to comment Share on other sites More sharing options...

ewv Posted December 11, 2004 Report Share Posted December 11, 2004 (edited) Basically, yes. For any inverse square law, a uniform spherical shell or solid has the same field as the corresponding source regarded mathematically as concentrated at a point at the center, as can be shown using either Newton's calculus or geometry. But looking at the structure inside and very close to an electron is outside the realm of applicability for an inverse square law alone. An equation with a singularity does not literally apply either at the singular point or very near to it. Computation of an "infinity" at a point or an enormous value very nearby is a guarantee that you have "gone too far", but is not the only thing you have to know about the domain of validity and the local requirements to invoke a more advanced theory taking other phenomena into account. I'll leave it to Stephen and the others who know a lot more than I do about the advanced physics of this to elaborate on details. [This was for For jedymastyr 12:40 AM EST, a few more posts intervened.] Edited December 11, 2004 by ewv Quote Link to comment Share on other sites More sharing options...

ewv Posted December 11, 2004 Report Share Posted December 11, 2004 I have finished Resnick and Halliday: Fundamentals of Physics. Could you please recommend more advanced books. I am mostly interested in electromagnetism and quantum mechanics. Resnick and Halliday is very good as a start - but best in the form of the more detailed two volume version written for scientists and engineeers, so you might do that first. An advanced classic on E&M is Jackson, but you will probably want more between that and where you are now. Good candidates are Shadowitz (inexpensive Dover) and Marion. For QM there is a lot to cover, including a prior classical study of more advanced analytical mechanics (Marion; Goldstein; and Landau & Liftshitz are good candidates) and some of the historical background on modern physics. QM itself has a lot of controversial interpretations not to be taken at face value, but not yet fully explained properly either. At some point you will want to read the classic 3 vol Feynman Lectures, written as undergraduate texts for mechanics through QM, but actually much better for insights once you already know more; you are ready for some of that now, following Resnick & Halliday. Feynman's non-technical QED is also very interesting. Quote Link to comment Share on other sites More sharing options...

jedymastyr Posted December 11, 2004 Report Share Posted December 11, 2004 Basically, yes...but is not the only thing you have to know [This was for For jedymastyr 12:40 AM EST, a few more posts intervened.] Just for clarification (I don't think I said it clearly enough before)-- If the electromagnetic force were infinity, it wouldn't matter that there are other forces acting (nuclear strong/weak)--an infinite force would have to be overcome. So even though there are other (stronger) forces acting, there still aren't any "infinite" forces. I just wanted to show that there weren't any infinite electric forces the strong force had to overcome in order to hold the protons together. Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 11, 2004 Report Share Posted December 11, 2004 Basically, yes. For any inverse square law, a uniform spherical shell or solid has the same field as the corresponding source regarded mathematically as concentrated at a point at the center, as can be shown using either Newton's calculus or geometry.Â But looking at the structure inside and very close to an electron is outside the realm of applicability for an inverse square law alone.Â An equation with a singularity does not literally apply either at the singular point or very near to it.Â Computation of an "infinity" at a point or an enormous value very nearby is a guarantee that you have "gone too far" ... But I think it important to underscore that not all mathematical infinities imply a physical infinity (which is a good thing since we know that physical infinities cannot exist). Nor are all mathematical infinities necessarily best described as a singularity. Granted that the Coulomb solution of Maxwell's equations has a singularity at r = 0, but the "singularity" at r = 2M for the Schwarzschild geometry, where the g_rr component becomes infinite at the horizon, is an artifact of the coordinates by which the line element is expressed. If you properly calculate the components of the Riemann curvature tensor, which is a measure of the tidal forces felt by an observer along the radius r, none of the components become infinite at the horizon. So the physical implication is that an observer can pass the Schwarzschild horizon, experiencing only finite tidal forces, despite the "singularity." Of course, this is no longer the case for the central singularity of the black hole, at r = 0, where the mathematics does imply a physical singularity. And, as I am sure you well-know, there are a multitude of techniques by which we deal with mathematical singularities, not all of which implies we have "gone too far." Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 11, 2004 Report Share Posted December 11, 2004 I did think of this but assuming the diameter of one proton to be 1 fm The current (2004) best determination of the charge radius of the proton is 0.870 fm, +/- 0.008 fm, as an average. The precise determination of this value is of great importance to quantum chromodynamics. If you are really interested in this stuff, the Particle Data Group makes their 1109 page Review of Particle Physics available. This enormous compilation lists the averaged properties of gauge bosons, leptons, quarks, mesons, and baryons and it has succinct summaries of the current state of various particle physics topics. See their web site here. Quote Link to comment Share on other sites More sharing options...

ewv Posted December 11, 2004 Report Share Posted December 11, 2004 ... there are a multitude of techniques by which we deal with mathematical singularities, not all of which implies we have "gone too far."Yes, not "going too far" in the sense of having them, but in getting too close for meaningful pointwise evaluation of a function that becomes unbounded. The singularities themselves are often necessary to properly characterize the local and global behavior of a function, both physically and mathematically. Even when not meaningful as the value at a point, they still meaningfully contribute to the values of improper integrals, defining poles of a transform, local rates of growth measured by coefficients, etc. They are essential to mathematics, not demons of physics implying an actual infinity! Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 12, 2004 Report Share Posted December 12, 2004 Yes, not "going too far" in the sense of having them, but in getting too close for meaningful pointwise evaluation of a function that becomes unbounded. Sometimes that is not always clear. See, for instance, the ingenious gr-qc/0101011 at the e-Print Archive. The singularities themselves are often necessary to properly characterize the local and global behavior of a function, both physically and mathematically.Â Even when not meaningful as the value at a point, they still meaningfully contribute to the values of improper integrals, defining poles of a transform, local rates of growth measured by coefficients, etc.Â They are essential to mathematics, not demons of physics implying an actual infinity! I wish more people understood this. Too often mathematical singularities are rejected out of hand by those who do not know proper mathematics. Concepts of method are not the same things as physical reality. Quote Link to comment Share on other sites More sharing options...

punk Posted December 13, 2004 Report Share Posted December 13, 2004 Perhaps it would be better to say that where a theory predicts a measurable infinity then one has approached the end of the range of validity to the theory. One view of mathematical physics is that it aims to predict measurements, and it is only these which need to be justified against the structure of the actual world. Anything else which occurs in the process of making the prediction can be taken as purely computational in nature and should not be justified against the physical world. That is that these things can be thought of as elaborate versions of the "carrying the digit" as we were taught in elementary school where when adding 14 to 16 we do the following: 1 16 + 14 ______ 30 Thus the carried one has no physical meaning it is just an artifact of the algorithm. Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 13, 2004 Report Share Posted December 13, 2004 Perhaps it would be better to say that where a theory predicts a measurable infinity then one has approached the end of the range of validity to the theory. If by "theory" you mean the principles which explain deterministic causal behavior, then no, your above way of saying things is most certainly not "better." A better way to say it in that case is that the theory is wrong, because actual infinities do not exist in the physical world. If, on the other hand, by "theory" you really mean the mathematical formalism (which seems to be what you mean), then your way of saying it is not "better," just that that case refers to a different fact, namely that not all solutions to mathematical formulas necessarily pertain to physical reality. One view of mathematical physics is that it aims to predict measurements, and it is only these which need to be justified against the structure of the actual world. And that view (a prevalent one) would be terribly wrong. Correspondence with experimental fact is a necesary condition of any valid theory, but to actually rise to the level of being a theory it must also provide consistent, non-contradictory causal explanations of the experimental facts. Anything else is just mathematics, not a theory of physics. (Not that there is anything wrong per se with "just mathematics," as long as we do not pretend that such mathematics be more than it actually is.) Quote Link to comment Share on other sites More sharing options...

ewv Posted December 13, 2004 Report Share Posted December 13, 2004 Perhaps it would be better to say that where a theory predicts a measurable infinity then one has approached the end of the range of validity to the theory.When the value at a point representing a quantitative factual attribute is deduced to be infinity, i.e, indefinitely unbounded and therefore meaningless, you have already gone beyond the pointwise range of validity, not “approaching” it (which is probably not what you meant). There will be a finite interval around the point of singularity where the values are so large that the whole neighborhood is outside the valid domain. There is no point in the mathematical continuum infinitely “next” to a singularity (or any other point), there is no point where a finite evaluation can itself be “next to infinity” (there is no such thing as either an infinite magnitude nor a finite value being next to the indefinitely large), and there may be other reasons why a region may be outside the valid domain because of, and/or aside from, an excessive size of the magnitudes evaluated, depending on the nature of the theory and the context of its validation. (Also, speaking more carefully, the theory does not “predict infinity”, the mathematics would lead to that if applied outside the valid domain of the theory. This is the same issue as the often heard expression that a “theory breaks down” in some region: the theory does not literally break. It does not apply there to begin with; if anything is “broken” it is an improper application or tentative interpretation in a realm of uncertainty. Normally that is understood as a kind of shorthand in a manner of speaking, but often theories are improperly treated as floating abstractions manipulated as independent “models” in consciousness floating over reality and available for later “connections” to facts on a hit or miss basis, thus causing, in this “model mentality”, the theory to “break”.) One view of mathematical physics is that it aims to predict measurements, and it is only these which need to be justified against the structure of the actual world. Anything else which occurs in the process of making the prediction can be taken as purely computational in nature and should not be justified against the physical world.The first part of that is a narrow, pragmatist view of the aim of a theory, which must also provide explanatory value even though some of its components may be mathematically methodological. Not every parameter in a theory need apply directly and by itself to a single physical magnitude, but it must be related at least in some indirect way at some level of abstraction. Otherwise parts of the theory would be totally disconnected from reality with no meaning. There can be no floating abstractions bridging “voids” of knowledge. Knowledge is interconnected and completely integrated. “Concept of method” does not mean philosophical pragmatism. <tt> <i>1</i> 16 <u>+14</u> 30</tt> Thus the carried one has no physical meaning it is just an artifact of the algorithm. The “<i>1</i>” that is carried means one unit of “tens”: 6 + 4 is one “ten” and no “ones”. It is indirectly related in that way, through a larger unit consisting of ten of the original units, to whatever is measured by the 16, 14 and 30. Counting the “carried” units of tens (omitting the total “ones” summinng to less than 10) as part of the addition process serves a methodological purpose in the algorithm, but still has an ultimate “physical” meaning: It is a very simple example of an abstract concept of method that is related indirectly to the facts through a hierarchical chain of abstractions (in essence, the abstract unit is ten of the original units, together with a rule for what is being counted by it in accordance with the purpose of the algorithm). Usually, such connections are not seen that simply because they are based on a vast accumulation of prior concepts and methods, and are much more abstract and remote. Quote Link to comment Share on other sites More sharing options...

punk Posted December 13, 2004 Report Share Posted December 13, 2004 To clarify (which I think you already figured out), I was using "theory" to refer to the mathematical formalism alone. So my version of "theory" would be a mathematical formalism which produces predictions which can be compared to physical measurement. And of course a good theory makes predictions very close to what is measured. You seem to prefer the idea that a "theory" is a mathemtical formalism which makes predictions which can be compared to physical measurement but which also has a clear interpretation. And my guess is that an interpretation should map structures within the theory (beyond the predictions) to structures in the physical world, and that these structures should meet certain criteria such as obeying causality. Please correct me if I misunderstood. My view is that the interpretation part cannot be applied to the theory itself but can be applied to the set of predictions (which are assumed to in principle be comparable to some physical measurement). So if we take the set of all predictions we can find structures which should correspond to structures in the physical world, but structures within the formalism which do not show up in the predictions have no meaning in the interpretation. Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 13, 2004 Report Share Posted December 13, 2004 You seem to prefer the idea that a "theory" is a mathemtical formalism which makes predictions which can be compared to physical measurement but which also has a clear interpretation. First, the use of "prefer" is too loose a word for that which I have described as an essential aspect of the very meaning of "theory." Second, "interpretation" is too vague as it is currently used; I am referring to the ideas and principles which are intimately connected -- of necessity -- with the mathematical formalism, and that which makes the mathematics sensible. And my guess is that an interpretation should map structures within the theory (beyond the predictions) to structures in the physical world, and that these structures should meet certain criteria such as obeying causality. I am not wont to use such terminology as "map" or 'structures" in this context. To provide a mapping is like providing a function to associate elements in sets, and in this case, one "set" being ideas and the other "set" being physical reality. The epistemological significance -- the relation between the conceptual realm and physical concretes -- is much more clearly defined and explicated than can be captured by "mapping." I refer you to Ayn Rand's Introduction to Objectivist Epistemology for the details. Quote Link to comment Share on other sites More sharing options...

punk Posted December 13, 2004 Report Share Posted December 13, 2004 A mathematical formalism by definition has no requirement for an interpretation. That's what a "formalism" is. It is simply a set of symbols with a set of rules by which one can go from one set of symbols to another set of symbols. Also a mathematical formalism in physics can describe multiple physical situations depending on the "interpretation" given the symbols (I'm sorry, I'm not sure what word you would prefer instead of "interpretation"). So the same differential equation can describe many different physical models. Moreover there have been cases where the same formalism for the same physical situation has been subject to various interpretations from various physicists. Let's take Maxwell's Equations. We no longer view them using the interpretation Maxwell used in developing them. Or Schroedinger's Equation where many of the arguments he used to get there are now considered faulty. The formalism stands since it corresponds with observation where various interpretations come and go. Quote Link to comment Share on other sites More sharing options...

stephen_speicher Posted December 14, 2004 Report Share Posted December 14, 2004 A mathematical formalism by definition has no requirement for an interpretation.Â That's what a "formalism" is.Â It is simply a set of symbols with a set of rules by which one can go from one set of symbols to another set of symbols. If the mathematics corresponds to physical reality then the formalism leads to valid experimental predictions. If we want to understand the nature of the phenomena then we need a theory, a set of ideas and principles which provide causal explanations. Also a mathematical formalism in physics can describe multiple physical situations depending on the "interpretation" given the symbolsThen, at most, only one theory which leads to and incorporates the mathematical formalism can be correct. (I'm sorry, I'm not sure what word you would prefer instead of "interpretation").Â So the same differential equation can describe many different physical models. The same form of differential equation can describe different phenomena, and different physical theories can explain that different phenomena. Moreover there have been cases where the same formalism for the same physical situation has been subject to various interpretations from various physicists.Yes, that is true, as I have stated and detailed on this forum many times. But, the fact remains that, at most, only one of those theories is a correct explanation of the given physical situation. Let's take Maxwell's Equations.Â We no longer view them using the interpretation Maxwell used in developing them.Â Or Schroedinger's Equation where many of the arguments he used to get there are now considered faulty.Â The formalism stands since it corresponds with observation where various interpretations come and go. One final time: the fact remains that, at most, only one of these "interpretations" is correct. Quote Link to comment Share on other sites More sharing options...

punk Posted December 14, 2004 Report Share Posted December 14, 2004 But since: 1. all of our theories are only valid in certain domains 2. when a theory is replaced by a new theory in a wider domain, that new theory often provides a radically different understanding of the phenomenon that quite often that the conceptual understanding of the old theory was wrong, and the conceptual understanding in the new theory doesn't really apply to the old theory (take Coulomb's Law versus QED), that likely for all our theories except for the latest bleeding edge ones there is no correct conceptual understanding of the formalism. And it is quite likely that the conceptual understanding of the bleeding edge theories is likely wrong as well. So we are left with zero correct interpretations for physical theories in reality and at most one correct interpretation in principle. I understand where a conceptual understanding is required to do theoretical physics, given that quite a bit of theoretical physics is not rigorous mathematics, so the theorist has to go with concepts rather than rigorous methodology. It just doesn't seem philosophically required in specifying what a physical theory is. I dont see where you need more than a formalism and data it compares well to. And where we can find structures within the data that is fine to consider, but structures which occur in the theory but not in the data, I have trouble granting meaning to. But I am interested in understanding your position. I'm not being deliberately contrary. Quote Link to comment Share on other sites More sharing options...

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