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Perhaps a trivial question, but it is something which I couldn't ever grasp ever since beginning physics. Why exactly should Newton's second law be linear in application of all the external forces?

For example, suppose I have a spring oscillating with a block (in absence of the gravity of Earth) then I add in gravity, then simply to find the new governing equation for acceleration I include gravity into the equation for the second law... but why? Why is there no cross effect between the two 'forces' acting on the block?


To be clear, I do not mean that force is linear in the sense of it being a linear function of time or a linear function of position. I know very well that forces could manifest whatever complicated function you can imagine, and in certain cases, as a series expansion.

My question asks about the linearity in terms of the application of different causes. As described by the example, let's add some more complication, let's say the Moon suddenly became 'massive enough' that its effect could be felt on our spring system i.e: the effect is no longer non-negligable for wherever I am on Earth conducting the experiment. Then the new governing equation of acceleration, again, would just be given by including the force of the Moon in my old force sum.

$$ a = \frac{\sum F_{ext} }{m} \to a'= \sum \frac{ \sum F_{ext} }{m} + \frac{F_{moon} }{m}$$

And, sure, a direct answer may be "because force is modelled by a vector and vectors add in so-and-so fashion", but, in physics we are trying to model what we see in the real world rather than impose our mathematical truths on it. Hence, again I ask, are there any deep reasons why in the real world, the forces (causes) of motion add up independently (without having a mixing effect)?

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    $\begingroup$ Force sums linearly because force is a vector by definition and the sum of vectors is the linear sum of their components. There's not much more why to it than that. $\endgroup$
    – g s
    Jun 26, 2021 at 21:25
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    $\begingroup$ Perhaps if you had a description where a cross-effect between forces was present, there would be a way to redefine the forces such that this term disappeared? $\endgroup$ Jun 27, 2021 at 0:44
  • $\begingroup$ The question is really about why is momentum linearly dependent to velocity. You just asked about the derivates both sides of $\boldsymbol{p} = m \boldsymbol{v}$ $\endgroup$
    – JAlex
    Jun 27, 2021 at 5:45
  • $\begingroup$ Conversely, constant force is an idealized and simplified model of how forces in the real world generally get applied. Constant force is only one case. There are higher-order derivatives like jerk, snap when objects collide e.g. golf club hitting ball. $\endgroup$
    – smci
    Jun 27, 2021 at 7:21
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    $\begingroup$ @gs There's nothing fundamental in your comment, and I would argue that it's incorrect. Force is not a vector by definition. That is a rather outrageous thing to say. We represent forces by vectors because we find that it works. One reason that we know it works is that the linear nature of forces is correctly described. Why forces add linearly is another question entirely $\endgroup$
    – garyp
    Jun 28, 2021 at 11:29

5 Answers 5

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Note: I have edited my original answer to take into account some comments and to add more generality, in particular stressing the superposition principle as a more general principle than the pair-wise additivity, the latter being a particular case of the former.

I wouldn't call linearity the property you are referring to. The proper name superposition principle. In the simplest case, the superposition principle coincides with the pair-wise additivity of the interactions, i.e., we assume that if the force on body $i$ due to body $j$ alone is ${\bf F}_{ij}$, and that due to body $k$ alone is ${\bf F}_{ik}$, the total force in $i$, due to the simultaneous presence of $j$ and $k$, is $$ {\bf F}_{i} = {\bf F}_{ij}+{\bf F}_{ik}. $$ Notice that in the separate case and in the combined case, each pair-wise force (${\bf F}_{ij}$ and ${\bf F}_{ik}$) is a function only of quantities of the corresponding pair. More in general, for an n-body system, $$ {\bf F}_{i} = \sum_{j=1;j\neq i}^n{\bf F}_{ij}. $$ Superposition (or pair-wise additivity) is definitely a mathematical property different from the bi-linearity of the vector sum. The former has to do with the functional dependence of the contributions to the force on the body parameters, the latter with the operations defined on these functions.

Superposition, or the more specific pair-wise additivity of the forces, is often used in Newtonian mechanics and it was taken for granted by Netwon, but it is not a necessary condition. Indeed it is quite easy to provide examples of more complicated forms of force law. Even more important, although rarely stressed in the textbooks, the most accurate models of the effective forces among atoms or molecules in condensed matter are certainly not pair-wise additive (see the comment at the end).

Probably the most simple example of a force that is not pair-wise additive is the force between two neutral but polarizable particles, say $1$ and $2$. If only these two particles are present the mutual forces are zero: ${\bf F}_{12}=0$ and ${\bf F}_{21}=0$. However, if we introduce a third, charged particle say number $3$, both the original particles get an induced electric dipole and, in addition to the dipole-charge interactions with the charged body, ${\bf F}_{21}\neq 0$ and ${\bf F}_{12} \neq 0$, due to the dipole-dipole interaction.

A couple of final comments are in order:

  1. the formal structure of Newtonian mechanics is able to accommodate non-pairwise forces without problems. It is only the expression of the total force on each particle that is more complex. It should be clear that pair-wise non-additivity does not break the second-law relation between total force and acceleration. Simply put, there is nothing like the additivity of the accelerations due to the presence of different external bodies. This has nothing to do with the vector character of accelerations and forces, of course.
  2. if the example of the polarizable particles in the presence or absence of a charge seems too artificial, one should remember that the effective interactions among atoms in condensed matter are always originating from an operation of partial trace over electronic degrees of freedom. An example is the well-known Born-Oppenheimer approximation where the interatomic interaction energy contains a many-body term (i.e. non-pairwise interaction) corresponding to the ground state energy of the electrons in the presence of fixed nuclei.
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    $\begingroup$ I don't get the difference between pair-wise additivity and linearity. By varying F_ij, you can prove that the effect of 0 force is 0, and the effect of n times the same force is n times the effect of the force. How is that not linearity? $\endgroup$ Jun 27, 2021 at 11:02
  • $\begingroup$ @EricDuminil Look at the example in the second paragraph of the original question and at the example with polarizable bodies. In the first case, there are two forces that add and are individually the same in the presence or in the absence of the other. In my example, the final force is again a sum of forces, but it is not the sum of the forces in the two separate situations. That's the reason for using a term different from linear. Forces are vectors and their sum is a bilinear operator. However, in general, it is not true that each force contribution depends only on the state of 2 bodies. $\endgroup$ Jun 27, 2021 at 12:03
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    $\begingroup$ Pair-wise additivity is one of the features of linear functions/operators/systems. Linearity is the appropriate mathematical term so long as it's clear that it's referring to it's interaction with other instances of the same type (rather than it's polynomial order). $\endgroup$ Jun 27, 2021 at 16:47
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    $\begingroup$ Pair-wise additivity is the consequence of linearity, it's not one {or,instead of} the other. $\endgroup$
    – Soleil
    Jun 27, 2021 at 18:42
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    $\begingroup$ Presumably 'linearity' here refers to the formal mathematical definition of the term, whereby components can be added and subtracted (your pair-wise linearity). It is in this sense that Fourier transforms are 'linear operations', although they're clearly not the linear-as-in-straight-line definition we all heard in high school. $\endgroup$ Jun 28, 2021 at 4:54
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Essentially because, since Newton postulated said linear relationship in order to explain the observations made by himself and his predecessors Galileo, etc., it has proven to be accurate (to be an accurate fit; to be a correct hypothesis).

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@GiorgioP has covered much of the ground, but I would like to add a few remarks:

  • Addition of forces is not exclusive to Newtonian mechanics — it is also the case in theoretical mechanics, relativity, quantum mechanics, etc. Although some of these disciplines do not explicitly use the concept of forces, the interactions remain pairwise (although this qualifier might have to be made more precise in the QFT context).
  • The pairwise nature of interactions is, of course, an experimental fact. However, there are deeper reasons for it: the fundamental equations of physics are linear, reflecting the symmetries of the universe. Had we had non-additive forces in the Newton's laws, it would be hard (or impossible) to make these laws invariant with respect to translations and rotations in space. More importantly, it would be hard to make them consistent with the relativity principle (Newton's first law). We would have then very different physics indeed.
  • Higher level physics theories (elasticity, hydrodynamics, macroscopic electrodynamics) do use various kinds of non-linear interactions. These are usually empirical ones, which are reducible to linear interactions when analyzed on a microscopic level.
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    $\begingroup$ +1 since this also points the requirement of consistency of forces in different static transformations of frames in space. $\endgroup$
    – Aditya
    Jun 27, 2021 at 7:57
  • $\begingroup$ GR doesnt have it. $\endgroup$
    – lalala
    Jun 28, 2021 at 6:44
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If you think of force in terms of potential: $$F=-dU/ds$$ Where U is the potential, and s the displacement. If the potentials are linear, that is, if we can calculate the total potential by adding the various partial potentials that our object is experiencing, then the overall force will also be the linear sum of partial forces.

With that context out of the way, there really is no reason a-priori to expect our potentials to add linearly like this. Indeed, in General Relativity they do not! However, these deviations from linearity are only significant in very strong gravitational fields. For weak-fields, potentials are so close to linear you can barely measure the difference.

So, linearity is a weak-field effect. Now, the natural follow-up question here is: why are potentials (and forces) linear in the weak-field limit? For any smooth, continuous mathematical function $F(x)$, if we know $F(x_0)$ we can approximate $F(x_0+h)$ via: $$F(x_0+h)=F(x_0)+h\times dF(x_0)/dx$$ That is, we can linearly extrapolate from any point to calculate nearby points. How large $h$ can be before we need to consider non-linear effects will depend on the function, and is something we can determine experimentally. An example of this is Ohm's Law. For small currents, $I$: $$V=IR$$ But if our current is large enough that linear relationship will break down and non-linear effects will kick in.

So, to answer your question: no, forces do not add linearly (in general), but for most practical purposes the deviation from linearity is so small we can ignore it.

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  • $\begingroup$ Hmm I'm a noob in GR, but doesn't there exist in non conservative forces in it? This may cause an incomplete analysis as not all forces can be realized as spatial derivative of some potential $\endgroup$ Jun 28, 2021 at 9:14
  • $\begingroup$ @Buraian In GR you can't use simple vector addition either, as the geometry is non-Euclidian. You could introduce new linear virtual forces if you want, it is a mathematical model after all, but linearity as a weak-field limit comes up in other areas of physics so I'd tend to favour that approach. $\endgroup$ Jun 29, 2021 at 3:11
  • $\begingroup$ To put that another way, nonconservative forces are the forces where the vectors a+b are not the linear sum of their components. $\endgroup$ Jun 29, 2021 at 3:19
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    $\begingroup$ This is the answer I want to accept, but I am a bit suprised it wasn't upvoted much. The point you brought up seems quite significant to me $\endgroup$ Jul 4, 2021 at 11:13
  • $\begingroup$ @Buraian: I suppose it is because many people don't read much past the first answer if it answers their own questions. In your case, my impressions is that you were asking the deeper question about why linear superposition works in the first place, which I suspect is something many people simply take for granted. Physics is deep, and there are many nuances new learners overlook. $\endgroup$ Jul 4, 2021 at 23:29
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The truth is that the addition isn't really linear. Linearity just happens to be a very good approximation in every day situations.

In gravity, the reason that linearity is not exact is that the gravitational field also gravitates (in GR, see https://en.wikipedia.org/wiki/Stress%E2%80%93energy%E2%80%93momentum_pseudotensor), which makes it that the gravitational field from two objects is not exactly a superposition of the gravitational fields of the two objects. The reason that the approximation is good is that in most situations, the gravitational field is so weak that its energy is too small to gravitate very much and can be ignored.

Similarly, in electromagnetism, charged particles polarize the vacuum (through creation of virtual particle-antiparticle pairs, see https://en.wikipedia.org/wiki/Vacuum_polarization), and I think (but I could be wrong about this) that the field from two charged particles is not merely the sum of the fields of the individual particles because of this effect. But again this is a tiny effect and can usually be ignored at low energy, so superposition is a good approximation.

Why is superposition so often a good approximation? I'm afraid I don't have a good answer to that at the moment but I think it's a good question.

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