Re: Is F=ma a definition or not?

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 01/13/05 

Date: Thu, 13 Jan 2005 10:07:54 +0100

mmeron@cars3.uchicago.edu wrote:
> In article <cs2slr$9jp$2@news.urz.uni-heidelberg.de>, Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> writes:
>
>>mmeron@cars3.uchicago.edu wrote:
>>
>>>In article <efb319e0.0501120125.6e034a8a@posting.google.com>, jaganlee@gmail.com (Lee, Jagan) writes:
>>>
>>>
>>>>The definition of force seems to be F=ma.
>>>>Newton's the first and the third law can be deduced from the second
>>>>law, F=ma.
>>>>If Newton's the second law, F=ma is just a definition, can it be told
>>>>as a physics or science? I think it's just only a math or philosophy.
>>>
>>>
>>>Nope.
>>
>>Well, what *is* the definition of "force" then?
>>
>
> Well, I'm not even sure to what extent it is defined and to what
> extent it is a basic, intuitively understood notion.

Well, all I've seen so far seems to suggest the second.

> Perhaps the most
> basic notion in this context is "interaction" and we observe that not
> all interactions (even if of the same type) have same magnitude of
> effects. So we can ascribe to them some notion of "strength of
> interaction" and that's what the force is.

That's still quite vague and not a definition, IMO.

> Consider the simple case of an elastic force, i.e. F = kx.

Which holds only in some special cases and is even itself
called a law (Hooke's law).

> Is there
> any mention of acceleration or mass, in there? Of course, you can say
> that if and when this force is applied to a mass m, and acceleration
> will result. Sure. But, we can talk about this force as is, without
> reference to any acceleration. That's what it is a valuable concept.

I don't dispute that it's a valuable concept.

> Would all we can say about a force is that it is the product of mass
> and acceleration than it would've been, physically, a useless concept,
> having no predictive value.
>
> F = ma is an equation, not an identity. A big difference. An
> identity is a statement which is always true. An equation is a
> statement which is nearly always false. That's what makes is valuable
> since when you know that "entity X satisfies an equation E", this
> statement carries information with it. It narrows the set of possible
> values for X from, apriori, all imaginable values, to values in the
> small subset which satisfies E. On the other hand, an identity
> carries no information. To illustrate by a simple example, the
> statement
>
> X = X
>
> is an identity, always true, and tells you nothing about X. On the
> other hand the statement
>
> X - 3 = 0
>
> Is nearly always false. In fact, it is false for any imaginable value
> of X, with the exception of the single value 3. Therefore, if you
> happen to know that X satisfies this equation, you're in possession
> of information uniquely identifying X.

Oh, *that* you meant with "nearly always false". You confused
me quite a bit at first - F = ma nearly always false sounded
quite strange! ;-)

> So, what is F = ma. Well, we consider an interaction between a test object
> Ob and an external "interaction source" Ex. Ex may be characterized
> by some set of physical parameters X. Similarly the test object Ob
> may be characterized by some set of intrinsic parameters Y, as well as
> kinematic parameters, i.e. location z, velocity z'=v, acceleration z''
> = a etc.
>
> Now, we postulate that it is possible to
>
> 1) *Consistently* assign to the interaction a strength F, which may depend
> on the parameter sets X and Y, possibly also on z, z', but not on
> higher derivatives of z.
>
> 2) *Consistently* assign to the object a value m = m(Y) (meaning, it
> depends on the properties of the object but *not* on the interaction).
>
> Such that the relationship
>
> F(X,Y,z,z') = m(Y)z''
>
> holds.
>
> Note that this is *not* an identity (which a definition would've
> been). It cannot be an identity since on each side of the expression
> there are variables present which are absent from the other side.
> Since, apriori, any of these variables can have any value, both sides
> of the relationship do not (again, apriori) have to be equal. So,
> this is an equation, not an identity, and the finding that the
> assignments above can be carried in such way that for all observed
> motions the relation holds, carries an enormous amount of information.

Sounds sensible.

Thanks!

Bye,
Bjoern

Relevant Pages

  • Re: Is F=ma a definition or not?
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