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If you consider them as laws, then there must be independent definitions of force and mass but I don't think there's such definitions.

If you consider them as definitions, then why are they still called laws?

Qmechanic
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user5402
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    there must be independent definitions of force and mass but I don't think there's such definitions. Suppose I define force as what I measure with a spring scale, and I define mass as what I measure with a balance (which has been calibrated in the same location with a standard mass). –  Jul 06 '13 at 20:47
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    @BenCrowell I wish life is that simple. – user5402 Jul 06 '13 at 20:59
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    Just think of the laws as saying "one can assign a 'mass' to each object and a 'force' to each interaction so that ...". It is quite possible that with some laws of physics there would be no consistent assignments of 'masses' and 'forces' making Newton's laws hold. This means that Newton's laws are more than just definitions. For example, Newton's laws imply that an astronaut floating in space can't propel himself without throwing something. Can a definition imply anything? – Peter Shor Jul 06 '13 at 21:23
  • Conservative forces satisfy $m\ddot{\vec{x}}=-\vec{\nabla}V\left( x\right) $, so even in this special case an equation quantifying force isn't true by definition. A Lagrangian formulation makes this clearer with the on-shell concept. – J.G. Jun 22 '17 at 22:57
  • Related: https://physics.stackexchange.com/a/697000/226902 – Quillo Jun 04 '22 at 14:09
  • Many answers take as axiom that acceleration is indeed the "correct"/"natural" quantity to study. As for why, see the linked questions of https://physics.stackexchange.com/questions/90190/why-do-fundamental-physical-laws-involve-the-second-derivative, in particular https://physics.stackexchange.com/questions/4102/why-are-there-only-derivatives-to-the-first-order-in-the-lagrangian?noredirect=1&lq=1, and also an information-theoretic remark https://physics.stackexchange.com/questions/4102/why-are-there-only-derivatives-to-the-first-order-in-the-lagrangian?noredirect=1&lq=1#comment10908_4109 – D.R Nov 22 '22 at 07:33
  • They are fundamental laws https://www.physicskey.com/12/newton-s-laws-of-motion – Saral Mar 24 '24 at 15:42

6 Answers6

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In my view, standard statements of Newton's laws are usually overly concise, and this lack of detail causes confusion about what is a definition, and what is an empirical fact. To avoid this confusion, let's proceed in a systematic way that makes the distinctions between these definitions and empirical statements clear.

What follows certainly is not the original statement of the laws made by Newton himself; it is a modern interpretation meant to clarify the foundations of Newtonian mechanics. As a result, the laws will be presented out of order in the interest of logical clarity.

To start off, we note that the definitions of mass and force given below will require the concept of a local inertial frame. These are frames of reference in which when an object is isolated from all other matter, it's local acceleration is zero. It is an empirical fact that such frames exist, and we'll take this as the first law:

First Law. Local inertial reference frames exist.

How is this in any way related to the first law we know and love? Well, the way it is often stated, it basically says "if an object isn't interacting with anything, then it won't accelerate." Of course, this is not entirely correct since there are reference frames (non-inertial ones) in which this statement breaks down. You could then say, all right, all we need to do then is to qualify this statement of the first law by saying "provided we are making observations in an inertial frame, an object that doesn't interact with anything won't accelerate," but one could then object that this merely follows from the definition of inertial frames, so it has no physical content. However, going one step further, we see that it's not at all clear a priori that inertial frames even exist, so the assertion that they do exist does have (deep) physical content. In fact, it seems to me that this existence statement is kind of the essence of how the first law should be thought because it basically is saying that there are these special frames in the real world, and if your are observing an isolated object in one of these frames, then it won't accelerate just as Newton says. This version of the first law also avoids the usual criticism that the first law trivially follows from the second law.

Equipped with the first law as stated above, we can now define mass. In doing so, we'll find it useful to have another physical fact.

Third Law. If two objects, sufficiently isolated from interactions with other objects, are observed in a local inertial frame, then their accelerations will be opposite in direction, and the ratio of their accelerations will be constant.

How is this related to the usual statement of the third law? Well, thinking a bit "meta" here to use terms that we haven't defined yet, note that the way the third law is usually stated is "when objects interact in an inertial frame, they exert forces on each other that are equal in magnitude, but opposite in direction." If you couple this with the second law, then you obtain that the product of their respective masses and accelerations are equal up to sign; $m_1\mathbf a_1 = -m_2\mathbf a_2$. The statement of the third law given in this treatment is equivalent to this, but it's just a way of saying it that avoids referring to the concepts of force and mass which we have not yet defined.

Now, we use the third law to define mass. Let two objects $O_0$ and $O_1$ be given, and suppose that they are being observed from a local inertial frame. By the third law above, the ratio of their accelerations is some constant $c_{01}$; \begin{align} \frac{a_0}{a_1} = c_{01} \end{align} We define object $O_0$ to have mass $m_0$ (whatever value we wish, like 1 for example if we want the reference object to be our unit mass), and we define the mass of $O_1$ to be \begin{align} m_1=-c_{01}m_0 \end{align} In this way, every object's mass is defined in terms of the reference mass.

We are now ready to define force. Suppose that we observe an object $O$ of mass $m$ from a local inertial frame, and suppose that it is not isolated; it is exposed to some interaction $I$ to which we would like to associate a "force." We observe that in the presence of only this interaction, the mass $m$ accelerates, and we define the force $\mathbf F_{I}$ exerted by $I$ on $O$ to be the product of the object's mass and its observed acceleration $\mathbf a$; \begin{align} \mathbf F_{I} \equiv m\mathbf a \end{align} In other words, we are defining the force exerted by a single interaction $I$ on some object of mass $m$ as the mass times acceleration that a given object would have if it were exposed only to that interaction in a local inertial frame.

Second Law. If an object $O$ of mass $m$ in a local inertial frame simultaneously experiences interactions $I_1, \dots, I_N$, and if $\mathbf F_{I_i}$ is the force that would be exerted on $O$ by $I_i$ if it were the only interaction, then the acceleration $\mathbf a$ of $O$ will satisfy the following equation: \begin{align} \mathbf F_{I_1} + \cdots +\mathbf F_{I_N} = m \mathbf a \end{align}

joshphysics
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    Great answer! I have a few queries. Firstly, why do we just "define" the force to be the mass x acceleration. That seems a bit arbitrary. We can define anything we want, but that doesn't necessarily mean it'll help us do physics. Obviously, this definition does, but I don't know why other than I suppose through observation? Secondly, what do you mean inertial frames may not exist a priori? Never heard that, but would love to read more on it. Lastly, in the first sentence of your "Third Law" bit, where is the implication there that there is an acceleration? Or is it just saying if there is? – sangstar Jun 15 '17 at 17:30
  • @sangstar 1. We make that definition because from observation we know that it will lead to the Second Law I wrote down, and that Law can be used to make predictions. 2. Inertial frames are a particular kind of thing. Our universe could have been such that when you isolated objects they'd move in some complicated way instead of at constant velocity. 3. I'd say it doesn't matter -- can there ever not be an acceleration? – joshphysics Jun 15 '17 at 17:39
  • I see! For 3., yes the case in which an acceleration is not present is possible here. Or at least I'm confident that's the case. – sangstar Jun 15 '17 at 17:42
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    @sangstar Why so confident? All objects will gravitationally interact. – joshphysics Jun 15 '17 at 17:45
  • @joshphysics Gah! Conventionally, I assume it's negligible, but you're definitely correct. – sangstar Jun 15 '17 at 17:48
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    @joshphysics Wait, if there is never not an acceleration on an object since gravitational fields have infinite ranges, how can local inertial frames exist? – sangstar Jun 16 '17 at 07:35
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    @sangstar that's a great question. In fact, we know thanks to Einstein that global inertial frames do not actually exist. We get around this by using local inertial frames, in which the first law holds as accurately as you want provided you only do experiments in a small region of spacetime. – Javier Jun 22 '17 at 17:08
  • @sangstar Infinite range of gravitational force isn't an issue for local inertial frames at all. Consider a uniform gravitational field filling all of space and we can even have global inertial frames - these will be the frames freely falling in the gravitational field. Similarly, even when the fields are non-uniform, if we take indefinitely small region then we can say the gravitational fields are uniform within those small regions - allowing us to have local inertial frames in those regions. As before, these will be the local frames falling freely in the locally uniform gravitational field. –  Jun 22 '17 at 23:06
  • @Dvij so in the reference frame of a free falling object, such a global inertial frame can exist in the presence of a gravitational force? But what if I applied a force in the opposite direction of that "inert yet free falling" object? Surely, it wouldn't accelerate in that applied direction at full force even if our reference frame considered the object in free fall - it's still accelerating regardless of our opinion of reference, and an opposing opposite force will reduce it - or have the opposite force be reduced? – sangstar Jun 23 '17 at 00:05
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    This formulation has a loophole. Consider objects $O_0, O_1, O_2$ with given accelerations ratios $c_{01}, c_{02}, c_{21}$ and define mass as $m_1 = c_{01} m_0, , m_2 = c_{02} m_0$. Now, however, there is no guarantee that $m_1 = c_{21} m_2$ and the definition of mass is thus non-unique. The definition of mass can be unique only if a transitivity postulate $c_{ij}/c_{ik} = c_{kj}$ holds for any three objects $O_i, O_j, O_k$. – Void Jun 25 '17 at 14:27
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    @Void That's an interesting observation, but I don't see how this is a problem if we all agree to use $O_0$ as our reference mass. If we want the ability to use another mass, say $O_1$, to measure masses after we have measured it against $O_0$, then I agree that the procedure outlined above would not guarantee that we would get consistent results, and we would need to add your transitivity postulate, but that seems unnecessary if we want a simple, predictive theory that works in principle, even if in practice it's hard to work with because we always have to use $O_0$ to measure all masses. – joshphysics Jun 25 '17 at 18:25
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    @joshphysics Let me give a more concrete example to demonstrate the problem. We take a situation where $c_{01}=1$ and thus if we take either $O_0$ or $O_1$ as the referential object which weighs our unit $1 kg$ we get $m_0 =m_1 = 1kg$. But now your postulates allow for the possibility that $c_{12} \neq c_{02}$, let's say $c_{12} = 2$ and $c_{02} = 1$. This means that if we take $O_0$ as the referential object for mass definitions, we get $m^{(0)}_2 = m_1 = m_0 = 1kg$ and if we take $O_1$ as the reference, we get $m^{(1)}_2 = 2m_1 = 2 m_0 = 2kg$. – Void Jun 25 '17 at 19:05
  • @Void Oh yes I agree. Recognizing the observation you just made was the entire point of my last reply. However, just because the laws/definitions as stated don't rule out this inconsistency, doesn't mean that there will actually be an inconsistency when you apply them -- I believe your transitivity law is in fact true. In other words, I agree that if we want to know for certain that using different reference masses won't get us into trouble, we need to know transitivity holds, but not knowing/stating this doesn't mean we will get inconsistent results when we apply the laws as stated. – joshphysics Jun 25 '17 at 19:17
  • @joshphysics Just to clarify. The definitions here, they don't tell us what mass and Force are. But they do tell us how mass, acceleration and force are related to each other, right? Also, the validity of these definitions comes from the fact that they are consistent with the observations we make, am I correct? – R004 Nov 23 '17 at 03:05
  • @joshphysics "the assertion that inertial frames do exist does have a (deep) physical content." Could you elaborate on this? – R004 Nov 23 '17 at 04:42
  • @Void I think that in order to retain the correct Newton's laws it is in fact necessary to assume that $$c_{12} = \frac{c_{02}}{c_{01}}.$$ If this is not done then it can be shown that for two interacting masses, where none of them are the reference mass $m_0$ we get that (using the second law) $$ c_{12} \equiv \frac{a_1}{a_2} = \frac{F_1/m_1}{F_2/m_2} = \frac{F_1}{F_2} \frac{c_{02}}{c_{01}}.$$
    So that if the relation I stated above does not hold then it is not certain that the forces acting on two interacting bodies are of equal magnitude.
    – JezuzStardust Apr 05 '18 at 07:32
  • @joshphysics Actually, not assuming the transitivity rule as was suggested above will give rise to inconsistencies. If two equal point masses interact and we do not require the transitivity rule, then we may have $a_1/a_2 = c_{12} \neq 1$ and this leads to an spontaneous acceleration of the center of mass. I am not sure how to fix this, but I think that the transitivity should be taken as an additional physical law (since it is actually an empirical result that this holds). So this has to either be added into the third law, or taken as an additional law. – JezuzStardust Apr 06 '18 at 08:44
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    It's atrocious and totally unacceptable that this either isn't taught or isn't taught enough. How many go into physics and/or a discipline which features strong use of physics (e.g. engineering) and yet don't manage to get a rectified understanding of some of these most elementary principles of basic mechanics, or get it much later than they should? It makes one want to distrust a lot of what one gets out of the school/education system. If this is jixed up, what else might be as well? – The_Sympathizer Sep 02 '18 at 16:03
  • @JezuzStardust : Having a transitivity law though doesn't seem too bad - it's not too different from how that temperature is defined in thermodynamics. You need "Newton's zeroth law" :) – The_Sympathizer Sep 02 '18 at 16:04
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    What is an "inertial frame"? Generally inertial frames are defined in the context of force, i.e., a frame in which objects under no net force will have constant velocity. But your definition must be independent of force to not be circular. Is it defined in terms of interactions? If so, then what's an interaction? – Bridgeburners Jun 20 '19 at 19:36
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    @joshphysics I just want to take a moment to congratulate you for such an amazing answer! I am looking it after a year, and it still gives me enormous satisfaction! – Atom Jun 20 '20 at 14:45
  • @Void I agree with you that unless $c_{jk} = m_k/m_j$ is not postulated, one can’t, just from Josh’s third law, derive the common statement of the third law (i.e., $m_j\mathbf{a}_j=-m_k\mathbf{a}_k$ for two (possibly) interacting particles isolated from rest of the world and being observed in a local intertial reference). – Atom Jun 20 '20 at 17:29
  • @joshphysics Do you agree with my latest comment (addressed to Void)? – Atom Jun 20 '20 at 17:31
  • @joshphysics can you please provide any references on the last one 'the second law' while I could find all the points mentioned with a rigorous treatment in a book called Desloge Mechanics. it along with others fail to mention that the principle of superposition of force is a postulate. – Gaurang Agrawal Nov 12 '21 at 03:38
  • @joshphysics also thanks for this brilliant answer. This one enamored me so much that I had to pick my undergrad classical mech. based on the fact that the book must contain all this. strangely enough, tho the author does define force as the rate of change in momentum of a particle after defining mass, and the third law like the way here and says that for a system of particles force is 'given' by this certain vector sum. He didn't mention it as one of the postulates of classical mech. like he did for other postulates like the existence of inertial ref. frame – Gaurang Agrawal Nov 12 '21 at 03:41
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    The book Analytical Mechanics by Marmo and Fasano takes a similar approach while introducing Newton's laws. – Apoorv Potnis Dec 14 '21 at 13:03
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    @ApoorvPotnis Thanks for that reference. I was unaware of that book, and it seems like it contains a nice combination of physical and mathematical precision. – joshphysics Dec 14 '21 at 23:38
  • Please check my answer to this question bellow, so you can see the difference in approaching this subject. I claim that this answer, though greatly explained, misses the point of the question. – Mihailo_Serbia Jun 27 '22 at 10:33
  • If someone is interested in more difficult subject with similar way of thinking, please check out this https://physics.stackexchange.com/questions/488770/where-does-the-metric-directly-manifest-in-physics/715790#715790 – Mihailo_Serbia Jul 13 '22 at 09:37
  • I am now obsessed with this topic and amazed that 40 years after being awarded a physics degree from a fancy university this is the first time I am contemplating this. Was vaguely aware of critiques of how we think of Newton’s laws (eg absolute space, bucket paradox) going back ages including Leibniz, Mach, Einstein, etc but not that it touched on fundamental notions of force and mass. Given this long history, no paradigm will be perfect but @joshphysics’s among the most satisfying. – user175324 Jul 19 '22 at 16:23
  • Now two questions
    1. Not all objects “interact” at Newtonian scales. Doesn’t that require different reference mass objects for each type of “interaction” to measure the mass of an object. If my object has no charge, is not ferromagnetic and is small enough that there is no macro gravitational “interaction” then I’m left with various macro “contact” forces (arising from quantum EM interactions). Does this make transitivity issue more complicated?

    2. How does framework account for fact that objects held in fixed relation can behave as a single object at a COM (ie why do masses add?)

    – user175324 Jul 19 '22 at 16:38
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    In case it's useful, the textbook by R.D. Gregory, Classical Mechanics (CUP, 2006), follows essentially this approach. Notably, the statement of the third law fully includes the consistency requirement. – Emilio Pisanty Oct 31 '22 at 23:47
  • (On the other hand, it also goes for the strong statement of the law of action and reaction (i.e. requiring the forces to be along the line between the two particles), which is clearly bunk $-$ simple counterexamples can be constructed with two interacting dipolar magnets.) – Emilio Pisanty Oct 31 '22 at 23:48
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joshphysics's answer is excellent, and a perfectly good logical ordering of concepts, in which force is defined in terms of mass. I personally prefer a slightly different logical ordering (which of course ends up being equivalent), in which mass is defined in terms of force:

First law: Local inertial reference frames exist.

I can't improve on joshphysics's excellent explanation here.

Second law: Every object's mass exists, and is independent of the force applied to it.

We define a "force" $F_i$ to be a physical influence arising from a repeatable experimental setup. ($i$ is just a label, not a vector component.) For example, we could consider a single rubber band, stretched by a fixed amount, to which we connect a series of different "test objects." This defines a force $F_1$ which is not a vector quantity (hence the lack of bold script), but instead a label for a particular experimental setup. Or we could consider the gravitational pull $F_2$ from Jupiter on various "test objects" when it is at a particular location and distance relative to the test object. A given force $F_i$ acting on a given test object $o_j$ will impart on it a measurable acceleration vector ${\bf a}(F_i, o_j)$.

Now we find three nontrivial empirical results:

(i) If forces $F_1$ and $F_2$ induce accelerations ${\bf a}_1$ and ${\bf a}_2$ in an object when applied individually, then they induce acceleration ${\bf a}_1 + {\bf a}_2$ in the object when applied simultaneously.

(ii) A given force $F_i$ accelerates all test objects in the same direction (although with different magnitudes). In other words, $${\bf a}(F_i, o_j) \parallel {\bf a}(F_i, o_{j'})$$ for all $i$, $j$, and $j'$.

(iii) Suppose we have two different forces $F_1$ and $F_2$ (e.g. two rubber bands of different stiffness) and two different test objects $o_A$ and $o_B$. The following equality always holds:

$$\frac{|{\bf a}(F_1, o_A)|}{|{\bf a}(F_1, o_B)|} = \frac{|{\bf a}(F_2, o_A)|}{|{\bf a}(F_2, o_B)|}.$$

This suggests a natural way to systematically quantify the effects of the various forces. First take a particular test object $O$ and assign to it an arbitrary scalar quantity $m_O$ called its "mass." Don't worry about the physical significance of this quantity yet. Note that only this one particular object has a well-defined "mass" at this stage. Now apply all of your different forces to the object $O$. Each force $F_i$ will induce some acceleration ${\bf a}(F_i, O)$ on $O$. Now assign to each force $F_i$ a vector quantity $${\bf F}_i := m_O\, {\bf a}(F_i, O)$$ which "records" its action on the test object $O$. Note that Newton's second law is trivially true only for the particular test object $O$. Also note that changing the value of $m_O$ simply dilates all the force vectors by the same amount, so you might as well just choose mass units in which it has the numerical value of $1$. The empirical observation (ii) above can now be rephrased as

(ii') For all forces $F_i$ and test objects $o_j$, $${\bf F}_i \parallel {\bf a}(F_i, o_j).$$

We can therefore define a scalar quantity $m_{(i,j)}$, which depends both on the applied force and on the test object, such that $${\bf F}_i = m_{(i,j)} {\bf a}(F_i, o_j).$$

This justifies the first claim of the Second Law, that every object's mass exists. Recall from the definition of the force vector that $$m_O {\bf a}(F_i, O) = m_{(i,j)} {\bf a}(F_i, o_j),$$ so only the ratio $m_{(i,j)} / m_O$ is physically measurable, as mentioned above.

If we let $o_B$ be the test object $O$, then empirical observation (iii) above can be rearranged to $m_{(1,A)} = m_{(2,A)}$ for all test objects $o_A$, justifying the second claim of the Second Law that an object's mass does not depend on the external force applied to it.

Finally, the facts that (a) induced accelerations add as vectors and (b) an object's mass does not depend on the applied force, together imply that applied forces add as vectors as well.

Third law: When one object exerts a force on a second object, the second object simultaneously exerts a force equal in magnitude and opposite in direction on the first object.

We already defined the force vector ${\bf F}$ above, so this is clearly a nontrivial empirical observation rather than a definition.

tparker
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    Truly fantastic answer! Maybe seems more so to me as this is exactly how I arrange the content of the Newton's laws for myself. It is frustrating how textbooks (or any other book) completely ignores all of this and turns towards applications. I would like to point out that this way of thinking is a bit advantageous over the otherwise excellent explanation by @joshphysics because of the fact that this way isolates the third law from the more fundamental definitions of mass and force. The third law is pretty much always violated and I would not like it to be the basis of my definitions ;-). –  Jun 22 '17 at 22:58
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    There is a problem with this answer if one takes into account electrical charge. Consider the two experimental setups where $F_1$ corresponds to placing an object at a certain distance from Jupiter (assumed electrically neutral), and $F_2$ corresponds to placing an object at a certain distance from a positively charged object with the same mass as Jupiter. Then rule (iii) breaks down if you use objects $o_A$ and $o_B$ with the same mass but different charge. – Tob Ernack Feb 26 '19 at 18:13
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    In fact, for gravitational forces the ratio $\frac{|\mathbf{a}(F_1, o_A)|}{|\mathbf{a}(F_1, o_B)|}$ is always equal to $1$ since the acceleration is independent of the mass of the object. This is not true for other forces in general. I think the problem is that the forces are not properties of the experimental setup only, but also depend on properties of the test objects themselves, and there is more than one degree of freedom because we have both mass and charge. – Tob Ernack Feb 26 '19 at 18:16
  • What this means is that in your procedure for defining mass, you still have a good definition for $m_{(i, j)}$, but in fact this quantity depends on the applied force. – Tob Ernack Feb 26 '19 at 18:42
  • @DvijMankad as for your comments about the Third Law being a definition rather than empirical in joshphysics' treatment, I am not completely sure, but perhaps one could change his treatment by refraining from claiming that accelerations are in opposite directions while defining mass. Instead of going straight to the Third Law, one could simply define mass using the empirical observation that ratio of magnitudes of the accelerations is constant. The Second Law is then defined as he did, and the Third Law (that forces have opposite directions) is an empirical fact. – Tob Ernack Feb 26 '19 at 18:59
  • Of course, this is still implicitly assuming that forces have equal magnitudes, which is also part of the Third Law. I am not sure how to address this at the moment. – Tob Ernack Feb 26 '19 at 19:04
  • @TobErnack No, I agree with you--I guess I misrepresented my point. I am not saying that the third law becomes a definition in Josh's treatment. I realize that it still remains a perfectly empirical physical law. I was pointing out precisely what you point in your second comment that in Josh's treatment, the definition of mass is built upon the empirical facts contained in the third law. [...] –  Feb 26 '19 at 19:21
  • [...] But since the existence of mass is more robust than the third law--and can be deduced independent of the third law (as tparker did), I just prefer to represent the discussion in a way where the definition of mass is independent of the validity of the third law. –  Feb 26 '19 at 19:21
  • @DvijMankad ah I see then. Do you see a way to resolve the issues I mentioned above? I also would like this approach to work, but I think there might be some complications. Perhaps we could restrict to experimental setups where the force is independent on the test object (such as rubber bands or springs) but this seems a bit strange because it requires knowing which experimental setups have this property a priori. – Tob Ernack Feb 26 '19 at 19:25
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    @TobErnack You make a very fair point. As you point out, the issues occur when the ratio of the acceleration that an object feels to the acceleration that the test object feels depends fundamentally on properties intrinsic to the object. Newton's laws were primarily formulated to describe macroscopic contact forces, which are (to an excellent approximation) independent of the object's internal properties. In this context, everything works pretty well. – tparker Feb 27 '19 at 02:16
  • @TobErnack I don't think there's any complete clean way to present this is full generality that always works. I think you just have to embrace the messiness and first define mass in terms of contact-type forces, and then address electromagnetic interactions separately. – tparker Feb 27 '19 at 02:19
  • @TobErnack If you're only considering electromagnetic forces then things actually become simpler, because then (I think) only the charge-to-mass ratio is uniquely defined. If you're only considering gravitational forces then things become even simpler, because you can directly define an object's mass in terms of the acceleration that it induces on any other object a certain distance away. These subtleties only come up when you need to combine different kinds of forces together. – tparker Feb 27 '19 at 02:22
  • @TobErnack The fundamental issue is that when you first explain Newton's laws to a beginner, you're implicitly claiming that they work for a general force without specifying the nature of the force's source. This setup is clearly fundamentally a bit ill-defined, because whether it works will depend a bit on exactly what you consider to be a force. If you have a system where all the interactions are completely specifyied by deterministic laws - e.g. one described by the Lagrangian or Hamiltonian formalisms - then it's arguably best to forget about the notion of forces entirely and work ... – tparker Feb 27 '19 at 02:27
  • ... directly with trajectories in configuration or phase space. – tparker Feb 27 '19 at 02:27
  • Term "force" is introduced in the Second Law without being defined. Not cool! joshphysics's answer is the way. – Esmailian Jun 03 '21 at 17:09
  • Addendum: The book Analytical Mechanics by Marmo and Fasano takes a similar approach while introducing Newton's laws. – Apoorv Potnis Dec 14 '21 at 13:03
  • Amazing answer! Although, I have a doubt... All the logics are placed beautifully but they somehow don't indicate that the quantity m is the object's mass... They only indicate that whatever this quantity is, it's object dependent... So how do we come to this conclusion that m is a measure of "quantity of matter"... And for that matter, is there really any rigorous idea as to what "quantity of matter" really means? – user266637 Dec 28 '23 at 14:26
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First, I want to say I find your question excellent! It is very important, for anyone who wants to call himself a physicist, to know the answer on your question.

EVERY PHYSICAL QUANTITY must be defined through operation of measurement OR through mathematical relations to other physical quantities which are already defined through operations of measurements. Meaning, we must know how to measure a physical quantity (directly or indirectly).

For example, we define velocity as time derivative of position vector, and this makes sense only if we know how to measure time and length.

Time is "defined" as measurement of specific clock (which has some specific properties in every way independent of time - we cannot say our specific clock, which we want to use as instrument for measuring time, must have properties of ticking after same TIME interval). We call one tick of our specific clock one second. Then, duration of some process we are observing is measured by counting ticking of our clock. N ticking means process lasted N seconds. Of course, if that process did not occur at the same place, we must use more than one same (i.e. having same properties) specific clock. We must use two clocks, but then clocks must be synchronized (by some defined procedure e.g. using light signals). I just want to add that what I said does not mean that every laboratory should have same specific clocks. We just defined time that way. Once we have done it, we then use some other clock and compare it with our specific clock. If their ticking match we can also use other clock for measuring time and so on.

Length is defined similarly. We take some stick which we call one meter. That stick cannot have properties of being constant length (i.e. rigid) because we want to define length using that stick (we do not want circular definitions), so we want that our stick have some specific properties independent of length (we want it to be at the same pressure, temperature etc.). Then length of some object is how much our specific sticks we have between ending points of that object (we must know how we attach our sticks to each other i.e. what is straight line and we also must know simultaneously where ending points are, but I do not want to talk further about spacetime). Suppose we have N sticks we say length is N meter long. Once we defined procedure we can use some other sticks or methods for measuring length as long as they give same results as our specific stick (which we can check by comparison).

LAWS OF PHYSICS are mathematical relations between physical quantities and we discover them by method of observations (empirically). Law is correct if our experiment says so. If I cannot experimentally (I neglect here technology problems) check some mathematical statement, then that statement is nothing more than mathematical expression, it is not a physical law.

So, mass, as physical quantity, is defined through measurement. We have some specific weighing scale and some specific object which we call one kilogram. We put other object we want to measure on the one plate of scale and counting how much our specific objects we must put on the other plate so out weighing scale is balanced. We counted N, so our object has mass of N kilograms. We can check that mass is additive quantity i.e. if we put two same objects we see that mass is 2N kilograms etc. We can measure mass by using different apparatus as long as they give same result as our first device (which we used for definition of mass).

Same story is applied when we want to measure force. We define one Newton, procedure of measuring etc. We check that force is vector, find some other ways to measure force (they only need to match our first way).

Momentum is defined as product of mass and velocity and measured indirectly.

Now we know how mass and force are measured we can further explore properties of them i.e. we can now look for some law (mathematical relations) connecting quantities of mass and force. And we found through observations that F=ma and now we can interpret mass as measure of inertia of body and force as how much we would push or pull some body, but that is not definition of mass and force. If we defined force as F=ma, then this relation is not a physical law and we do not know nothing yet about force expect that it is calculated as product of mass and acceleration. Of course, we defined mass and force so they would be related somehow because we experience this Newton law on daily basis and we have already knew some properties which we want force and mass to have.

"The development of physics is progressive, and as the theories of external world become crystallised, we often tend to replace the elementary physical quantities defined through operations of measurement by theoretical quantities believed to have a more fundamental şignificance in the external world. Thus the vis viva mvv, which is immediately determinable by experiment, becomes replaced by a generalised energy, virtually defined by having a property of conservation; and our problem becomes inverted - we have not to discover the properties of thing which we have recognized in nature, but to discover how to recognize in nature a thing whose properties we have assigned." - Arthur Stanley Eddington - Mathematical theory of relativity

Conservation of momentum then becomes experimentally provable. If we defined mass through conservation of momentum (by measuring ratio of accelerations of two isolated body and calling one body 1kg), then we cannot to check if conservation of momentum is true, bacause it would not be a law, but a definition of mass.

NEWTON LAWS ARE LAWS!

The first Newton law is most complicated, because it is hard to know if our system really is inertial or not (general theory of relativity beautifully explains this problem). But we can, as Newton originally did, says that distant stars are inertial system and every system in uniform motion relative to them is also inertial and second and third laws are correct in them.

Answer of "joshphysics" is logically precise, but physically wrong.

  • In my opinion, this answer is very close. I completely agree with the first part on measurement. Just needs more illustration on how we just knew that $F=ma$. – Khalid T. Salem Jun 22 '17 at 16:27
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    Important question is if there is something you do not agree with (Do you think I wrote something incorrect?). I did not say we JUST knew that F=ma, we discovered it experimentally, but you are maybe right that I should have written more about that. – Mihailo_Serbia Jul 02 '17 at 14:27
  • No I don't think there's something wrong with your answer, I liked it. – Khalid T. Salem Jul 02 '17 at 15:10
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To understand what Newton's three Laws really are, one needs to consider the notion of momentum. Momentum $\vec{p}$ of a point particle is the product of its mass $m$ (which will be defined implicitly later) and its instantaneous velocity $\vec{V}$, so $\vec{p}:=m\vec{V}$. Also, $m \in \mathbb{R}_+$ mass units and $ m:=const $ (reasons are so that $ m$ characterises a particle and does not make vectors $\vec{V}$ and $\vec{p}$ point in a different directions). One also needs to consider the Law of Conservation of a Linear Momentum, which is the consequence of space translation symmetry (contrary to a pupular belief that it is the consequence of Newton's Laws).

Now, let's talk about the Newton's Laws:

Newton's first and third laws: consequence of the Law of Conservation of a Linear Momentum, nothing more.

Newton's second law: a definition of a force, $\sum \vec{F}:=\dot{\vec{p}}$ (which also yields the familiar $\sum \vec{F}=m\vec{a}$)

Remark: a question about measuring masses of point particles may arise, so here is the answer. Consider a system of two point particles moving along the $ x $-axis towards each other. Law of Conservation of Linear Momentum states:

\begin{align}m_1 \left |\vec{V}_{11} \right | - m_2 \left |\vec{V}_{21} \right | = m_2 \left |\vec{V}_{22} \right |-m_1 \left |\vec{V}_{12} \right |\end{align}

Defining $ m_1 $, for example, to be equal to one unit of mass, it is possible to calculate $ m_2 $ (measuring the values of the velocities of the particles before and after the collision is a standard procedure that can be carried out).

  • what's wrong in considering the conservation of linear momentum as a consequence of Newton Laws? If you assume $F= \dot p$, which is true in classical mechanics, 3rd law and conservation of linear momentum are completely equivalent. – pppqqq Jul 08 '13 at 16:36
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    Conservation of linear momentum is fundamentally the consequence of space translation symmetry, Newton's first and second laws are special cases. – Constantine Jul 08 '13 at 17:18
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I think the answer by Joshphysics is very good. In particular the statement that asserting existence is a key element.

The idea is to restate the laws of motion in such a way that the question law versus definition issue becomes clearer.
In analogy with thermodynamics I will state a 'law zero'; a law that comes before the historical 'First law'.
As with Joshphysics's answer the following treatment is for the Newtonian domain.

Law zero:
(Assertion of existence)
There exists opposition to change of an object's velocity. This opposition to change of velocity is called 'inertia'.

First law:
(The uniformity law)
The opposition to change of velocity is uniform in all positions in space and in all spatial directions.

Second law:
(The acceleration law)
The change of velocity is proportional to the exerted force, and inversely proportional to the mass.

The above statements are not definitions.
For comparison, the zero point of the Celcius scale is a definition; it is interexchangeable with another definition of zero point of temperature scale. The laws of motion are not exchangeable for other statements.

The concept of force is also applicable in statics, hence Force can also be defined in the context of a static case (compression), and then we check for consistency with Force defined in terms of dynamics. As we know: we find consistency.

For mass things are more interesting. Mass is in fact defined by the laws of motion. Trivial example: if you would use the volume of an object as a measure of its mass the second law would not apply universally. It's the law of motion that singles out what an object's mass is: precisely that property for which the second law holds good.

The lesson is that if you would insist that any statement is either a physics law, or a definition, you would totally bog yourself down.

Our physics laws are both: they are statements about inherent properties of Nature, and they define the concepts that the laws are valid for.




Additional remarks:

The first and second law together are sufficient to imply the historical third law. This can be recognized in the following way:

Let object A and object B both be floating in space, not attached to any larger mass.
From an abstract point of view it might be argued: there is a difference between:
Case 1: object A exerting a force upon object B, but B not on A
Case 2: object A and object B exerting a force upon each other.
According to the laws of motion the above distinction is moot. Observationally the two cases are identical, making it meaningless to distinguish between them on an abstract level.

Assume for argument sake that object A exerts a attracting force upon object B, but B not upon A. Both A and B are floating in space. The leverage that object A has to pull object B towards itself is A's own inertia. A has no other leverage, A is not attached to any larger mass. A can pull B closer to itself if and only if A is itself in acceleration towards B. There is no scenario, no observation, where Case 1 and Case 2 are distinguishable, hence Case 1 and Case 2 must be regarded as one and the same case.



The first law and second law together are sufficient to imply the superposition of forces.

Cleonis
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    How "The first law and second law together are sufficient to imply the superposition of forces"? You can't tell if a force will disrupt the source of another force. You have to assume the superposition of forces. – Shing Jun 23 '17 at 14:15
  • You forgot to add the only one very important thing of the Newton's laws: the above are only valid in inertial frames (and not elsewhere), where such frames are defined as existing by the first law (I don't really understand what you instead state the first law to be). – gented Jul 02 '17 at 17:06
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Newton's Law are in addition to laws of force and mass.

Newton's law of mass, changes in mass are caused in proportion to changes in density and changes in amount of matter (this might be paraphrased too badly).

Force Laws (there are many, ones for gravity, ones for springs, etc.)

Newton's third law of motion constrains what force laws you consider (effectively you only use/consider force laws that conserve momentum).

Newton's second law of motion turns these force laws into predictions about motion, thus allowing the force laws to be tested, not just eliminated for violating conservation of momentum. This works because he postulates that we can test force laws by using calculus and then looking at the prediction from solutions to second order differential equations.

Newton's first law of motion then excludes certain solutions that the second law allowed. I'm not saying that historically Newton knew this, but it is possible (see Nonuniqueness in the solutions of Newton’s equation of motion by Abhishek Dhar Am. J. Phys. 61, 58 (1993); http://dx.doi.org/10.1119/1.17411 ) to have solutions to F=ma that violate Newton's first law. So adding the first law says to throw out those solutions.

In summary: the third law constrains the forces to consider, the second makes predictions so you can test the force laws, and the first constrains the (too many?) solutions that the second law allows. They all have a purpose, they all do something.

And you need to first have laws of mass and/or laws of forces before any of Newton's laws of motion mean anything.

Timaeus
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