Classical Fields and Newton's 2nd Postulate of Motion


by dr_k
Tags: classical, fields, motion, newton, postulate
dr_k
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#1
Feb4-13, 09:40 PM
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Although I retired from active physics research almost 2 decades ago, there's a question that has annoyed/intrigued me for almost 40 years...

In Classical Field Theory, matter has 2 intrinsic properties, namely mass and charge. Given Newton's 2nd Postulate, [tex]{\bf F}_{net} = m {\bf a},[/tex] I've always wondered why the expansion/contraction, m, of the acceleration [tex]{\bf a}[/tex] isn't a scalar function of mass and charge, i.e. [tex]{\bf F}_{net} = f(m,q) {\bf a}[/tex] where f(m,q) = m + (negligible terms) except in extreme/rare situations that are experimental outliers, e.g. extreme cosmological physical situations. Although Newton's 2nd Postulate is...an a priori postulate, an educated guess that agrees w/ common everyday classical empirical phenomena, is there any mathematical basis, from Classical Field theory, that forces [tex]f(m,q) = m[/tex] precisely?
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elfmotat
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#2
Feb4-13, 09:48 PM
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Well, that's what we observe from experiment. There's nothing to suggest deviations from [itex]f(m,q)=m[/itex].
dr_k
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#3
Feb4-13, 10:02 PM
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Quote Quote by elfmotat View Post
Well, that's what we observe from experiment. There's nothing to suggest deviations from [itex]f(m,q)=m[/itex].
I agree w/ your assertion. Newton's 2nd Postulate agrees wonderfully w/ all our observed phenomena...so far. My query is very specific though. Is there a mathematical line of logic, from Classical Field Theory and/or or GR, that forces the scalar f(m,q) to precisely be m?...Not a proof of [itex]{\bf F}_{net} = m {\bf a}[/itex], which is impossible since it's an axiomatic postulate.

It's conceivable that there are extreme cosmological phenomena that might possibly disagree w/ f(m,q) = m.

:)

atyy
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#4
Feb5-13, 10:00 AM
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Classical Fields and Newton's 2nd Postulate of Motion


How is q defined?
martinbn
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#5
Feb5-13, 10:13 AM
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I would say that the two intrinsic parameters are gravitational mass and charge. They would appear on the left side of the equation, where the force is, since thy characterize interactions. What appears on the right side is inertial mass, which characterizes another property, and charge has nothing to do with that property.
dr_k
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#6
Feb5-13, 11:19 AM
P: 69
aty and martin,

For lack of a more vivid imagination, I would call q, in f(m,q), the "inertial charge" analogous to the "inertial mass" m in f(m,q).

f(m,q) = m + (negligible terms), except in instances perhaps not yet encountered, would imply the definition of "inertial mass" would be modified into a definition where "inertial mass and charge" would have relationships to both Newton's Postulate of Gravity and Coulomb's Electrostatic Postulate.

Simply, isolate a single point test particle, with classical intrinsic properties mass and charge, which can experience fields set up by external source charges and masses, and ask yourself, why is the expansion/contraction of acceleration in Newton's 2nd Postulate of Motion, [itex]{\bf F}_{net} = m {\bf a}[/itex], not affected by charge? Newton's 2nd Postulate and Coulomb's Electrostatic Postulate are separated by ~ 125 years, so I wouldn't expect Newton to postulate a f(m,q), but still, it's an annoying question that still puzzles me. Is is possible that f(m,q) = m + (neglible terms) where we have yet to see situations where the (negligible terms) could be measured?

I posed the query here on the off chance that a GR/Classical Field specialist might know a mathematical proof that forces f(m,q) to precisely equal the inertial m.

Thanks for your time.
atyy
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#7
Feb5-13, 11:33 AM
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Following martinbn's lead, Newton's second law is F=ka, where k is not necessarily q or m. In Newtonian gravity k is proportional to m, an expression of the principle of equivalence.

The principle of equivalence can be derived from the quantum field theory of a relativistic massless spin 2 field, eg. section 2.2.2 of http://arxiv.org/abs/1007.0435
dr_k
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#8
Feb5-13, 11:52 AM
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Quote Quote by atyy View Post
Following martinbn's lead, Newton's second law is F=ka, where k is not necessarily q or m. In Newtonian gravity k is proportional to m, an expression of the principle of equivalence.

The principle of equivalence can be derived from the quantum field theory of a relativistic massless spin 2 field, eg. section 2.2.2 of http://arxiv.org/abs/1007.0435
aty,
Thanks for your input, I'll read that paper. I could be wrong, but I see a little bit of circular logic here: the principal of equivalence uses the 2nd Postulate [itex]{\bf F}_{net}=m{\bf a}[/itex] as an a priori assumption to show acceleration is independent of the role of test mass versus source mass, i.e. inertial versus gravitational mass. If f(m,q) actually is a function of the test charge, or "inertial charge", then the principal of equivalence would also need to be modified to include the "inertial charge".

I don't mean to waste everyone's time, and there's no doubt that all observed data currently agrees with f(m,q) = m; I suppose it's just a thought experiment that continues to bother me.

Again, thanks for your time.
martinbn
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#9
Feb5-13, 12:45 PM
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dr_k, I don't understand why you need to introduce inertial charge. Inertial mass, by definition, is that coefficient in front of the acceleration.
stevendaryl
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#10
Feb5-13, 12:54 PM
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Quote Quote by dr_k View Post
For lack of a more vivid imagination, I would call q, in f(m,q), the "inertial charge" analogous to the "inertial mass" m in f(m,q).
I'm not sure what the analogy is. Mass really plays 3 roles in physics: (1) It acts as a source of a gravitational field. (2) It is affected by a gravitational field. (3) It acts as a "resistance" to acceleration (because of F=ma, acceleration is inversely proportional to mass, when F is held constant). Newton's third law (equal and opposite forces) implies that the masses associated with (1) and (2) are equal, and the equivalence principle (or alternatively, the universality of freefall) insures that the masses associated with (2) and (3) are proportional (if not equal). It's number (3) that is called "inertial mass".

In the case of charge, you still have (1) and (2): charge can be the source of an electric field, and it is affected by an electric field. But I don't see what could correspond to (3). So I don't see the role for an "inertial charge".
dr_k
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#11
Feb5-13, 01:12 PM
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Quote Quote by martinbn View Post
dr_k, I don't understand why you need to introduce inertial charge. Inertial mass, by definition, is that coefficient in front of the acceleration.
Good Afternoon Martin and Steven,

Suppose I postulate that Newton's 2nd Postulate be modified, such that [itex]{\bf F}_{net} = f(m,q) {\bf a}[/itex]. Then I postulate that f(m,q) is a series where the first term is "inertial mass" m, and the second term is a function of the test particle's inertial mass m and charge q, which I will call "inertial charge" for lack of a better term. This second term in the series, and all subsequent terms, have never been noticed since they are so small they fall within the current experimental error bars for [itex]m_{grav} = m_{inertial}+/- error[/itex].

Perhaps the "inertial charge", the intrinsic property of the isolated test charge, needs to be extremely large to be noticed in terrestrial experiments. One can attain extremely large [itex]{\bf E}[/itex]-fields, on the "pointy bits" of isolated conductors in electrostatic equilibrium, with relatively small q, since they occur at extremely small radii of curvature, but perhaps an extremely large test charge concentration, w/o dielectric breakdown of the surrounding medium, required to show the existence of the second term in f(m,q), has never been attained...yet.
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#12
Feb5-13, 01:33 PM
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Quote Quote by dr_k View Post
Suppose I postulate that Newton's 2nd Postulate be modified, such that [itex]{\bf F}_{net} = f(m,q) {\bf a}[/itex]. Then I postulate that f(m,q) is a series where the first term is "inertial mass" m, and the second term is a function of the test particle's inertial mass m and charge q, which I will call "inertial charge" for lack of a better term. This second term in the series, and all subsequent terms, have never been noticed since they are so small they fall within the current experimental error bars for [itex]m_{grav} = m_{inertial}+/- error[/itex].
That's certainly possible - the error bars are tight indeed but they're not zero.

I'm inclined to reject any unobserved higher-order terms, but that's an aesthetic preference not a logically reasoned one:
1) GR as a theory provides a very elegant explanation for the complete equality of gravitational and inertial mass.
2) Absent experimental data, introducing those hypothetical higher-order terms neither simplifies any problem for me nor advances my understanding.

So what I have is pretty and it works for me, so I'm inclined to stick with it as long as keeps on working, there are no hard counterexamples, and no one has come up with anything prettier.
dr_k
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#13
Feb5-13, 01:49 PM
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Quote Quote by Nugatory View Post
That's certainly possible - the error bars are tight indeed but they're not zero.

I'm inclined to reject any unobserved higher-order terms, but that's an aesthetic preference not a logically reasoned one:
1) GR as a theory provides a very elegant explanation for the complete equality of gravitational and inertial mass.
2) Absent experimental data, introducing those hypothetical higher-order terms neither simplifies any problem for me nor advances my understanding.

So what I have is pretty and it works for me, so I'm inclined to stick with it as long as keeps on working, there are no hard counterexamples, and no one has come up with anything prettier.
Your comments are rational, and precise. My query is only conjecture, one that has been w/ me for decades...just something, for me, to think about...I thought I would run it by the Big Brains that I've encountered here. Aside: I've run this by many a physics compatriot over the years and received the same response. It is an aesthetic avenue, on my part, since I don't understand why Newton's 2nd Postulate, experimentally, doesn't incorporate the second intrinsic property of classical matter.

Martin, here's a picture that might help to explain my inadequate words:

http://i45.tinypic.com/rid79t.jpg
dr_k
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#14
Feb6-13, 05:24 PM
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Quote Quote by dr_k View Post
Martin, here's a picture that might help to explain my inadequate words:

http://i45.tinypic.com/rid79t.jpg
Arrghhh. This picture is one where I made some assumptions. A more general picture is:
http://i48.tinypic.com/hs6ssw.jpg
PAllen
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#15
Feb7-13, 01:40 AM
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Quote Quote by dr_k View Post
Arrghhh. This picture is one where I made some assumptions. A more general picture is:
http://i48.tinypic.com/hs6ssw.jpg
If I understand the gist of this, then, in the context of classical (geometric) GR, where gravity is not a force at all, the suggestion amounts to saying that for a charged particle, the mass you use in 4-momentum (which leads to force), and mass you put in the stress energy tensor (acting as source of gravity) are slightly different. It is not immediately clear to me what implications that would have. Note that for a charged particle, EM field already causes it to contribute differently as a gravitational source than a neutral particle of the same mass. So are you suggesting an additional difference?

(I interpret you suggestion as saying inertial mass is your f(m,q), and gravitational mass is m).
DaleSpam
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#16
Feb7-13, 08:38 AM
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Quote Quote by dr_k View Post
is there any mathematical basis, from Classical Field theory, that forces [tex]f(m,q) = m[/tex] precisely?
Hmm, this is an interesting question.

Let's suppose that F=f(m,q)a. Then let M=f(m,q). So we would have F=Ma. Then all of the things that used to depend on m now depend on M. So I don't think that there is any difference between F=ma and F=f(m,q)a.

In other words, we measure mass and charge for each fundamental particle. If F=f(m,q)a then the mass we would measure and put in our tables would be M. But we wouldn't be able to detect any difference between that and a universe where the masses in the tables were m and F=ma held.
dr_k
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#17
Feb7-13, 10:36 AM
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Quote Quote by PAllen View Post
(I interpret you suggestion as saying inertial mass is your f(m,q), and gravitational mass is m).
PAllen and Dale,

The most general picture that I can draw is http://i48.tinypic.com/hs6ssw.jpg ,

where Newton's 2nd Postulate is [itex]{\bf F}_{net} = f(q_I,m_I) {\bf a}[/itex] , here [itex]m_I[/itex] is the inertial mass and [itex]q_I[/itex] is the "inertial charge", Coulomb's Electrostatic Postulate is [itex]{\bf F}_{q_{Coulomb}} = q_{Coulomb} {\bf E}[/itex], and Newton's Postulate of Gravity is [itex]{\bf F}_{m_{grav}} = m_{grav}~{\bf g}[/itex].

[itex]{\bf g}[/itex] and [itex]{\bf E}[/itex] are fields, created by external [itex]m'[/itex] s and [itex]q'[/itex] s.

Then I postulate that the expansion/contraction of the acceleration in Newton's 2nd Postulate is dominated by the first term, the inertial mass, [itex] f(q_I,m_I) = f_0 + f_1 +... = m_I + f_1(q_I,m_I) + ...[/itex], such that

[itex] {\bf g} = (\frac{m_I}{m_{grav}} + \frac{f_1(q_I,m_I)}{m_{grav}} + ...) {\bf a} - \frac{q_{Coulomb}}{m_{grav}} {\bf E}[/itex]

At this point I can make different assumptions, does [itex]m_I = m_{grav}[/itex] and/or does [itex]q_I = q_{Coulomb}[/itex]?

For example, if we assume [itex]m_{grav} = m_I \equiv m[/itex], and [itex]q_{Coulomb} = q_I \equiv q[/itex], in the absence of an external [itex]{\bf E}[/itex], we'd get

[itex] {\bf g} = (1 + \frac{f_1(q,m)}{m} + ...){\bf a}[/itex].

This conjecture of mine has always been based upon the idea that perhaps the higher order terms in [itex]f(q_I,m_I)[/itex] haven't been experimentally noticed, whether [itex]m_I = m_{grav}[/itex] and/or [itex]q_I = q_{Coulomb}[/itex] or not, since perhaps the charge required to notice the next term, [itex]{f_1(q_I,m_I)}[/itex], is extraordinarily large. For example, to get a mere 1C of charge, on an isolated conducting shell in electrostatic equilibrium, would require a radius of ~55m w/o atmospheric dielectric breakdown. [itex]f_1(q_I,m_I)[/itex] would, of course, have physical units of mass, so its functional dependence on q is complete conjecture, except that it hasn't been noticed yet.

Again, everyone, thanks for humouring this old man in his dotage. :)
DaleSpam
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#18
Feb7-13, 03:00 PM
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Quote Quote by dr_k View Post
where Newton's 2nd Postulate is [itex]{\bf F}_{net} = f(q_I,m_I) {\bf a}[/itex] , here [itex]m_I[/itex] is the inertial mass
This doesn't work. By DEFINITION inertial mass is the proportionality between force and acceleration: [itex]F=m_I a[/itex] (by definition of inertial mass). So if you also have [itex]F=f(q_I,m_I)a[/itex] (by postulate) then you immediately get [itex]f(q_I,m_I)=m_I[/itex], which is not what you want.

You can have [itex]f(q_I,m_{grav})[/itex] and still have an interesting question, but whatever m and q you put in there, by definition [itex]m_I=f(q,m)[/itex]

It seems to me that your question reduces to asking about the equivalence of gravitational and inertial mass.


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