My physics text says that force is "an interaction between two bodies or a body and its environment."

When an object undergoes acceleration we explain it with a force. But we don't measure force, we measure the change in acceleration. Is force just a helpful construct to be a placeholder in the equation F = ma?

Is force defined as the quantity mass times acceleration, or is it really "an interaction"?

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    $\begingroup$ Can you be more specific on what a you consider a "real thing" or "an interaction?" $\endgroup$ – BMS Jan 16 '14 at 4:28
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    $\begingroup$ What would it mean to be "a real thing" or to be "not a real thing"? $\endgroup$ – dmckee --- ex-moderator kitten Jan 16 '14 at 4:34
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    $\begingroup$ @ user 37464, +1, yes it is a good question. Force is just a concept defined by m and a. So force, in a sense, is just a concept for convenience. $\endgroup$ – wonderich Jan 16 '14 at 5:12
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70186/2451 and links therein. $\endgroup$ – Qmechanic Jan 21 '14 at 22:46

Newton's 2nd law: $\vec{F}=m \vec{a}$ is simply a definition of force. (So it is actually NOT a law experimentally/theoretically derived, unlike Newtonian gravitational force $\vec{F}_{\text{ (source or cause)}}=(GMm/r^2) \hat{r}$. ps. However, notice that the first $\vec{F}_{\text{(consequence)}}=m \vec{a}$ is the consequence/result of the force, while the later [take $\vec{F}_{\text{ (source or cause)}}=GMm/r^2$ as an example] is the source/origin/cause of the force. These two concepts, though related, are rather different. Importatnly, if one does not define the force in Newton's 2nd law, one has trouble later to talk about Newtonian gravity.)

It is easy to understand this from the measurement viewpoint: While $\vec{a}$ can be measured by (space) rulers and (time) clock; and the mass $m$ can be measure by a scale by comparing to a standard mass unit. So we now have well-defined the force $\vec{F}_{\text{(consequence)}}$ by measurable quantities.

(The consequence/result of the) Force is just a concept defined by m and a. However, this is an important useful concept. This force concept can be used to connect to the source or cause of the force: such as gravity, such as classical E & M, the Lorentz force law on e charge and currents - see also this post. Such as a potential $U(x)$, causes $F=-dU(x)/dx$. Also, such as thermodynamics pressure on an area. Or simply external forces by human/animals. etc.

$\bullet$ So the concept of force is very useful - connecting different source or cause of the motion $\sum_i \vec{F_i}_{\text{ (source or cause)}}$ together to the consequence/result $\vec{F}_{\text{(consequence)}}=m \vec{a}$ of the motion with acceleration $\vec{a}$, or, more precisely to the momentum $\vec{P}$ changes: $\vec{F}_{\text{(consequence)}}=d\vec{P}/dt$.

We may summarize as: $$ \text{ (source: E&M, gravity, thermal, external)}\to\vec{F}_{\text{ (source or cause)}}\\ =\vec{F}_{\text{(consequence)}}=m \text{(inertia)}\vec{a} \text{(motion in the spacetime)} $$

Be aware that, to me, the Newton's 2nd law is all about this statement: $$ \vec{F}_{\text{ (source or cause)}}=\vec{F}_{\text{(consequence)}} $$ While philosophically you can say the force define the interactions - the cause. The $\vec{a}$ is the consequence. It is also about the causality.

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    $\begingroup$ From F (source or cause): it defines the interaction. But for F (consequence): it defines its consequence - spacetime motion. These two viewpoints are on the two sides of the causality. $\endgroup$ – wonderich Jan 16 '14 at 5:44
  • $\begingroup$ ps. notice someone votes down earlier, please leave comments below what you disagree/dislike. I take what has been said above as a personal belief. Many thanks! :) $\endgroup$ – wonderich Jan 16 '14 at 17:45
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    $\begingroup$ I think that this answer is a bit confusing. $F=ma$ is not a definition of force. It is an experimental fact: the evolution of a system is fully determined by the initial positions and velocities. $\endgroup$ – pppqqq Jan 22 '14 at 14:57
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    $\begingroup$ I did not down vote, but I found your answer quite confused concerning Force cause / Force effect that seems to be a crucial distinction in your mind. You should try to clarify it. My answer is maybe too technical (after all I am a mathematical physicists) for the OP, I wrote it since I hope that some other, more experienced or mathematically minded, people could be however interested, though the OP could not be content with it. $\endgroup$ – Valter Moretti Jan 22 '14 at 15:30
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    $\begingroup$ I would argue that there is content beyond mere definition in saying that the set of interactions between objects add linearly, and their sum will give you the acceleration. Newton's second law probably works more cleanly as a definition of mass. $\endgroup$ – Jerry Schirmer Jan 22 '14 at 15:47

Here is my opinion. In my view Newtonian dynamics works as follows.

Working within an inertial reference frame, interactions are described by forces, that have to be thought of as functions ot time and position and velocity of the point of matter on which they are applied: $$\vec{F}=\vec{F}(t, \vec{x}, \vec{v}) \qquad (1)$$ Above, $\vec{x}$ denotes the position of the point of matter and $\vec{v}$ is its velocity. Every interaction has its own functional form that distinguishes this from other interactions.

In case the interaction is completely due to another similar point of matter with position $\vec y$ and velocity $\vec u$, the functional form of the force acting on $\vec x$ has to include these parameters and the dependence form $t$ does not appear (time homogeneity in inertial reference frames)

$$\vec{F}=\vec{F}(\vec{x}, \vec{v}, \vec{y}, \vec{u})\qquad (2)$$

and the third Newton's law must hold, concerning the couple of forces $\vec F$ and $\vec F'$ acting on $\vec y$:

$$\vec{F}(\vec{x}, \vec{v}, \vec{y}, \vec{u})= -\vec{F}'(\vec{y}, \vec{u}, \vec{x}, \vec{v})\:.\qquad (3)$$

In (1) the appearance of $t$ in the functional formula of $\vec F$ represents an external system imposing that force on the point at $\vec x$. It is assumed that the evolution of that system is given. Forces due to strength fields are of the said type: Think, in particular, of forces due to known fields ${\vec E}(t,\vec{x})$ and ${\vec B}(t,\vec{x})$ ($^{*}$).

One of the major tasks of '700-'800 physics was finding out the functional form of forces associated to the known different interactions. Gravitational force, Coulomb force, Lorentz force were discovered and described in accordance to this paradigm.

Why the mathematical structure (1) is relevant in physics?

This is because, if the function $\vec{F}$ is sufficiently smooth, the 2nd law of dynamics, intepreted as a differential equation:

$$\frac{d^2\vec x}{dt^2} = \frac{1}{m}\vec{F}\left(t, \vec{x}, \frac{d\vec{x}}{dt}\right) \qquad (4)$$

admits a unique solution for given initial data $\vec{x}(0)$, $\frac{d\vec x}{dt}|_{t=0}$, as established by a celebrated theorem due to various mathematicians ($^{\dagger}$). That is nothing but the mathematical representation of the mechanical determinism.

I would like to stress that, for systems of many points,

  • (a) assuming (3) for each couple of forces

as well as

  • (b) supposing the validity of superposition principle of different forces acting on a point of mass,

one ends up with a system of differential equations with the same property of (4): Existence ad uniqueness of its solutions are always guaranteed for given initial data.

Regarding the notion of mass, i.e. the constant $m$ associated with the point of matter in (4), it is worth stressing that it does not depend on the type of interaction, namely, on the particular functional form of the force $\vec{F}$ acting on the point. In this sense the mass is a property of the point of matter.

The approach I have (so quickly) illustrated does not work, as is well known, due to various reasons (all modern physics, essentially, arises from that failure). There is, however, in classical physics a class of very familiar interactions that cannot be completely encompassed within that paradigm. These are reactive forces due to geometric constraints. We know that they are of electrical nature, however their functional form is practically unknown and it is replaced by some geometrical information about the constraint.

Think of a point constrained to belong to a given curve and also subjected to a force of known functional form $\vec{F}$. The reaction $\vec \phi$ acting on the matter point, due to the constraint, is an unknown of the dynamical problem: $$m\frac{d^2\vec x}{dt^2} = \vec{\phi}+ \vec{F}\left(t, \vec{x}, \frac{d\vec{x}}{dt}\right) \qquad (5)$$

The Lagrangian approach to mechanics represents, in my opinion, one of the most powerful attempt to deal with these situations, for a certain class of forces due to constraints (including smooth constrains, but also forces due to the rigidity constraint and other physically relevant cases).

Obviously one of the most relevant and devastating problems with Newtonian paradigm was the failure of the third principle (3) in the presence of electromagnetic forces with moving charges in dynamical regime.

So referring to your questions:

(Q1) Is force just a helpful construct to be a placeholder in the equation $F = ma$?

(A1) No. It is instead a much more sophisticated theoretical notion, because it is a function of the kinematic status of the system and every interaction has its own particular functional form.

(Q2) Is force defined as the quantity mass times acceleration, or is it really "an interaction"?

(A2) It is not defined as the quantity mass times acceleration. $F=ma$ is a (differential) equation where, barring a few (but physically relevant) cases $F$ is known while the motion is unknown. Furthermore the term $F$ showing up in $F=ma$ is the sum of all forces acting on the matter point due to all interactions. Also for this reason $F=ma$ cannot be considered the definition of $F$. Each single force appearing in the RHS of $F=ma$ is the mathematical description of a corresponding interaction.

Added comments in particular after some remarks by Idear

  • In the absence of forces, the motion of a point of matter is, in a sense, the natural one: constant velocity (possibly vanishing velocity) in every inertial reference frame. In this view, forces can be seen as the causes of deviations from the natural motion. Acceleration is therefore viewed here as the effect of forces.

  • The functional form of $\vec{F}$, in principle, could include derivatives of higher order, as $\frac{d^3\vec{x}}{dt^3}$. It actually happens for some physical systems, like the model of a charged particle taking the EM radiation self-acceleration into account. These functional forms, though possible in principle, generally violate the requirement of a differential equation in normal form: $$\frac{d^n\vec{x}}{dt^n}= G(t, \mbox{derivatives of order $< n$})\qquad (6)$$ In this case, in general, there is not an existence and uniqueness theorem for the solutions of the said differential equation with natural initial conditions: $\vec{x}(0),\ldots, \frac{d^{n-1}\vec{x}}{dt^{n-1}}|_{t=0}$. Physically speaking determinism could fail. (The heuristic procedure outlined in the footnote($^{\dagger}$) cannot be implemented, in general, if the form (6) of the differential equation is not valid.)

  • Do the notion of force encompass all (macroscopic) interactions? That is a very difficult issue for several reasons, in particular for the following one. Suppose to describe physics within a fixed inertial reference frame. Consider a set of $N$ point of matter very far from each other and from the other bodies in the Universe. We know that no force acts on them and their motion has constant velocity. In particular, the relative velocity of one of them is constant with respect to any other point of the set. This is a very particular situation. How is it possible? What is the interaction acting on all those points constraining them to move with a so peculiar (relative) motion? This interaction cannot be described in terms of forces by definition. This interaction is just the one which defines inertial frames. As is well known this problem was tackled by Mach and gave rise to the so called Mach's principle that, in turn, played a crucial role in developing General Relativity by A. Einstein.


($^{*}$) Homogeneity of space, isotropy of space and invariance under Galileian transformations valid for inertial reference frames, impose further severe constrains on the functional form of (2). For instance only the difference $\vec{x}-\vec{y}$ and $\vec{v}-\vec{u}$ are allowed to show up in the explicit expression of $\vec{F}(\vec{x}, \vec{y}, \vec{v}, \vec{u})$.

($^{\dagger}$) It is sufficient that $\vec F$ is continuous in all variables jointly and is locally Lipschitz in the variable $(\vec{x}, \vec{v})$. $C^1$ jointly in all variables is largely enough. However there is a nice heuristic argument (I think due to Newton) suggesting the existence of a solution for given initial conditions. Inserting initial data in the RHS of (4), the LHS produces $\frac{d^2\vec x}{dt^2}|_{t=0}$. The procedure can be iterated ad libitum, since the functional form of $\vec{F}$ is known so that one can compute all derivatives in $t$ of the RHS of (4), obtaining this way all derivatives $\frac{d^n\vec x}{dt^n}|_{t=0}$, for $n=0,1,2,3,\ldots$. These coeffcients give rise to the (formal) Taylor expansion of a (hopefully analytic) solution of the differential equation.

  • $\begingroup$ Nice answer. About the third law, I've also found some interesting things in physics.stackexchange.com/questions/12122/… . $\endgroup$ – pppqqq Jan 22 '14 at 15:30
  • $\begingroup$ I am not sure I agree with you entirely. The concept of causality is not mentioned. And the definition of measurement is not emphasized and clarified. Everything a physicist does should better based on something measurable. I am a theoretical physics, but caring much more experimental facts. $\endgroup$ – wonderich Jan 22 '14 at 21:11
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    $\begingroup$ It is also possible to have $d^nx/dt^n$ for $n\geq3$ in your functional. How should one argue this does not appear? In principle, this can appear in complicated fluid dynamical system and in semiclassical limit of quantum system (such as low energy ground state of solid state band model - certain e.g. topological Chern insulators). So in a sense, one need to define the consequence (the output of determinism) to be velocity changes = i.e. acceleration. So this is a definition of force. While the cause (the input of determinism) can be confirmed by the experimental measurement. $\endgroup$ – wonderich Jan 22 '14 at 21:30
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    $\begingroup$ i.e. the cause (the input of determinism) can be confirmed/tested by the experimental measurement (such as Newton gravity, Lorentz force, or even the semiclassical force in the solid state band model). $\endgroup$ – wonderich Jan 22 '14 at 21:32
  • $\begingroup$ But I still agree this answer is nice and valuable. Am willing to upvote +1 still. $\endgroup$ – wonderich Jan 22 '14 at 21:37

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