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Classical Fields and Newton's 2nd Postulate of Motion

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dr_k
#19
Feb7-13, 03:33 PM
P: 69
Quote Quote by DaleSpam View Post
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.
Bear with me here, [itex]m_I[/itex] is a new definition that denotes one of the two intrinsic properties of matter, [itex]m_I[/itex] and [itex]q_I[/itex], such that the "new" [itex]m_I[/itex] is not defined to be "the proportionality between net force and acceleration", but part of a scalar function [itex]f(q_I, m_I)[/itex] that expands/contracts the acceleration in Newton's 2nd Postulate. A scalar function with a first dominant term whose current interpretation is "the proportionality between force and acceleration". This new [itex]m_I[/itex] may or may not be equivalent to [itex]m_{grav}[/itex].

For those who design langrangians to see what the action spits out, it may be an interesting exercise.

This is just a thought experiment that changes axiomatic definitions and postulates, to see how/if things change.
DaleSpam
#20
Feb7-13, 03:37 PM
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Quote Quote by dr_k View Post
Bear with me here, [itex]m_I[/itex] is a new definition
Then don't call it "inertial mass", because that is already defined and is f(q,m). I wouldn't even use a subscript "I" for it, since people will naturally assume that you mean "inertial mass".

You should think about how you would measure your third mass.
dr_k
#21
Feb7-13, 04:24 PM
P: 69
Quote Quote by DaleSpam View Post
Then don't call it "inertial mass", because that is already defined and is f(q,m). I wouldn't even use a subscript "I" for it, since people will naturally assume that you mean "inertial mass".

You should think about how you would measure your third mass.
Thanks for your input.

I wouldn't interpret there being 3 masses, just 2. Let [itex]m_I[/itex] and [itex]q_I[/itex] stand for the intrinsic properties of matter, mass and charge, for an isolated test particle. They are part of the scalar function [itex]f(q_I,m_I)[/itex] that expands/contracts the acceleration in Newton's 2nd Postulate of Motion. [itex]m_{grav}[/itex], on the other hand, is defined by Newton's Postulate of Gravity, and [itex]q_{Coulomb}[/itex] is defined by Coulomb's Electrostatic Postulate. The series expansion of [itex]f(q_I,m_I)[/itex] gives, to first order, a term [itex]m_I[/itex] which is currently called the "inertial mass", where [itex]m_{inertial}\equiv m_I[/itex].

This is my fault, for being so vague w/ my words. I appreciate your input. Thanks.
PAllen
#22
Feb7-13, 05:44 PM
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Quote Quote by dr_k View Post
Thanks for your input.

I wouldn't interpret there being 3 masses, just 2. Let [itex]m_I[/itex] and [itex]q_I[/itex] stand for the intrinsic properties of matter, mass and charge, for an isolated test particle. They are part of the scalar function [itex]f(q_I,m_I)[/itex] that expands/contracts the acceleration in Newton's 2nd Postulate of Motion. [itex]m_{grav}[/itex], on the other hand, is defined by Newton's Postulate of Gravity, and [itex]q_{Coulomb}[/itex] is defined by Coulomb's Electrostatic Postulate. The series expansion of [itex]f(q_I,m_I)[/itex] gives, to first order, a term [itex]m_I[/itex] which is currently called the "inertial mass", where [itex]m_{inertial}\equiv m_I[/itex].

This is my fault, for being so vague w/ my words. I appreciate your input. Thanks.
You don't seem to have responded to my description of the GR take on this (this is the relativity forum, not the classical physics forum). In GR, it is already true that a charged particle of mass m contributes differently as a gravitational source than a neutral particle (where m is the the inertial mass defined by force and proper acceleration).

So, if you take m to be gravitational source mass, and f(m,q) to be <= m(grav), then GR already incorporates your idea, in a way. If you want to make f(m,q) >= m(grav), then it appears to me your concept is inherently counter factual, given the strength of evidence for GR.
DaleSpam
#23
Feb7-13, 06:09 PM
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Quote Quote by dr_k View Post
The series expansion of [itex]f(q_I,m_I)[/itex] gives, to first order, a term [itex]m_I[/itex] which is currently called the "inertial mass", where [itex]m_{inertial}\equiv m_I[/itex].
No, what is currently called the "inertial mass" is your [itex]f(q_I,m_I)[/itex]. This is not just to first order, this is exact. The DEFINITION of inertial mass, m, is m=F/a.

You are positing a third mass, "intrinsic mass", which is related to inertial mass, m, by [itex]m=f(q_I,m_I)[/itex].

Do you see that now? I don't know how I can be more clear.

You can measure the gravitational mass using a balance scale. You can then measure the inertial mass by dropping the object and measuring the acceleration. I cannot think of a way to measure the intrinsic mass.
dr_k
#24
Feb7-13, 06:33 PM
P: 69
Quote Quote by PAllen View Post
You don't seem to have responded to my description of the GR take on this (this is the relativity forum, not the classical physics forum). In GR, it is already true that a charged particle of mass m contributes differently as a gravitational source than a neutral particle (where m is the the inertial mass defined by force and proper acceleration).

So, if you take m to be gravitational source mass, and f(m,q) to be < m, then GR already incorporates your idea, in a way. If you want to make f(m,q) >= m(grav), then it appears to me your concept is inherently counter factual, give the strength of evidence for GR.
Dale and PAllen,

I placed my thread in this forum to get expert GR opinions; I appreciate your input. GR was not my specialty. I will ponder your recent comments. Thanks again.
DaleSpam
#25
Feb8-13, 09:46 AM
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You are welcome. I would strongly encourage you to think about how to measure your "intrinsic mass". Unless you can come up with some independent way to measure it then all you have is [itex]m=f(q_I,m_I)[/itex] which is kind of one equation in two unknowns (f and [itex]m_I[/itex]). You simply won't have enough information to do it.
atyy
#26
Feb8-13, 11:10 AM
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P: 8,374
As DaleSpam said, there is inertial mass u, gravitational charge m, and electrical charge q. I don't believe there is any mathematical necessity in classical physics for u to be proportional to m or q. In Newtonian mechanics, the equivalence principle is put in by hand u=m. In GR the equivalence principle is also put in by hand using minimal coupling. In quantum mechanics, there is apparently Weinberg's low energy theorem for relativistic spin 2 particles in which the equivalence principle is derived.

So for each particle the force laws should go something like:

Gm1m2/r2 + Gm1m3/r2 + ... + Kq1q2/r2 + Kq1q3/r2 + ... = u1a1

I haven't thought it through, but I wonder if DaleSpams's concern about the number of equations for arbitrary ui for each particle can be answered with enough particles and particle configurations?
DaleSpam
#27
Feb9-13, 07:03 AM
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Quote Quote by atyy View Post
I wonder if DaleSpams's concern about the number of equations for arbitrary ui for each particle can be answered with enough particles and particle configurations?
I considered that, but I can't think of how you would determine an unknown function of an unknown parameter regardless of the amount of data. I think if you know the function then you can fit the parameter with sufficient data, and if you know the parameter you can approximate the function as close as you like with sufficient data, but I don't see how you can do both even with an infinite amount of data.
atyy
#28
Feb9-13, 11:18 AM
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Quote Quote by DaleSpam View Post
I considered that, but I can't think of how you would determine an unknown function of an unknown parameter regardless of the amount of data. I think if you know the function then you can fit the parameter with sufficient data, and if you know the parameter you can approximate the function as close as you like with sufficient data, but I don't see how you can do both even with an infinite amount of data.
Yes, that doesn't seem possible for an arbitrary function since that would be (assuming analyticity) an infinite number of Taylor coefficients.


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