# What's wrong with this argument that Newton's second law implies all potentials are quadratic?

Newton's second law states:

$$F(\vec{x})=m\vec{\ddot{x}}$$

For $$\vec{x}$$ scaled by some arbitrary constant $$s$$, we obtain:

$$F(s\vec{x})=ms\vec{\ddot{x}} \Longleftrightarrow \frac{F(s\vec{x})}{s}=m\vec{\ddot{x}}$$

Which is clearly just $$F(\vec{x})$$! Therefore:

$$F(\vec{x})=\frac{F(s\vec{x})}{s}$$

for any $$s$$, which can only be satisfied by a quadratic potential. Therefore, if Newton's second law is to hold and be consistent, all potentials in the universe are quadratic! Is there a very obvious mistake here, or is this inconsistency related to the fact the classical mechanics is not a complete description of nature? Because this discrepancy does not seem to arise if we use Ehrenfest's theorem in QM.

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z May 11 '20 at 19:45
• This question is being discussed on Meta. – user258881 May 12 '20 at 16:39

While other answers are correct, they fail to address your specific issue. It looks like you are treating Newton's second law like it defines a single function, when it does not.

For example, in algebra if I say a function is $$f(x)=x^2 + 3$$, then I can "plug into" this function something like $$sx$$ so that $$f(sx)=(sx)^2+3$$ by how we defined the function.

This is not what Newton's second law is doing. $$F(x)=m\ddot x$$ is not a function saying "whatever I plug into the function $$F$$ I take its second derivative with respect to time and multiply it by $$m$$." So, your statement of $$F(sx)=ms\ddot x$$ is not correct. Newton's law is a differential equation, not a function definition. $$F(x)$$ is defined by the forces acting on our system, and Newton's second law then states that the acceleration is proportional to this force.

In order to deal with the kind of analysis you want to do, you have to be careful. It's a bit awkward to write $$F(\vec{x})=m\ddot{\vec{x}}$$ in the first place but you can write that as long as you understand what it means. It means that you are considering the force and acceleration both as fields because you're considering Newton's law at each point in space. So, a clearer way to write it is $$F(\vec{x})=m\ddot{\vec{x}}(\vec{x})$$

Edit $$1$$: Let me clarify the meaning of this expression a bit more clearly. As I said, I'm considering a particle at each point in space. So, $$\ddot{x}(x)$$ simply means the acceleration of the particle which is located at $$x$$. The $$x$$ is bracket is a label. For example, if I was writing down the Newton's second law for $$N$$ particles, I'd write $$F(x_i)=\ddot{x}_i$$ for $$i=1,2,...,N$$. Now, I put a particle at each coordinate point and the label $$i$$ is replaced with the coordinate label $$x$$. So, simply replacing $$i$$ with $$x$$ would get me $$F(x(x))=\ddot{x}(x)$$ where $$x$$ is a label just like $$i$$. Now, notice that $$F(x(x))$$ means the force at the position $$x$$ of a particle labeled by $$x$$. But the meaning of the coordinate labeling $$x$$, by definition, implies that the position $$x$$ of a particle labeled by $$x$$ would simply be $$x$$. Thus, I adopt a succinct notation for $$F(x(x))$$ and simply write $$F(x)$$. Thus, $$F(x(x))=\ddot{x}(x)$$ becomes $$F(x)=\ddot{x}(x)$$, which is the expression written above, except in vector notation.

Now, you can do the scaling game and write $$F(s\vec{x})=m\ddot{\vec{x}}(s\vec{x})$$

Now, you see that there is no reason to believe that $$\ddot{\vec{x}}(s\vec{x})=s\ddot{\vec{x}}(\vec{x})$$ in general. However, what you can do is try to see when this would be true. And if you do that, you can see that this would be true iff $$F(s\vec{x})=sF(\vec{x})$$

This is what you ultimately got. But this simply means that you've figured out the condition under which $$\ddot{\vec{x}}(s\vec{x})=s\ddot{\vec{x}}(\vec{x})$$ would be valid. Your mistake was that you assumed that $$\ddot{\vec{x}}(s\vec{x})=s\ddot{\vec{x}}(\vec{x})$$ is generically true (likely due to your confusing notation) and then concluded that $$F(s\vec{x})=sF(\vec{x})$$ should be true generically, which is not true because your implicit assumption is not generically true.

Edit $$2$$

I am considering the transformation $$x\to sx$$ to mean that it takes us from point $$x$$ to point $$sx$$ in the same units. So, if I'm writing Newton's law for the particle at position $$x=1$$ as $$F_1 = a_1$$, the transformation means that now I'm writing Newton's law for a different particle, one which is situated at $$x=s$$, and I'd write $$F_s=a_s$$. So nothing non-trivial is happening here. The assumption of the OP was that $$a_s=sa_1$$ which is a very non-trivial claim as it establishes a relation between accelerations of particles at different points. I simply point out the obvious that this is not true unless the forces at those positions are related in such a way to establish such a relation, i.e., unless $$F_s=sF_1$$.

• This notation definitely makes it more clear that I have committed the worst sin: abuse of notation, it all makes sense now! – Godzilla May 11 '20 at 19:29
• @miroslav351 Glad to see you're not at Newton's throat anymore ;) – Dvij D.C. May 11 '20 at 19:30
• I don't see what the expression $\ddot x(x)$ is supposed to mean. – Johnny Longsom May 11 '20 at 19:32
• I think Johnny Longsom objects to the notation: $\ddot{x}$ implies taking a second time derivative of a function of time, if the input is a vector (or the result of a vector valued scalar function), it doesn't really make sense. Maybe something like $\text{accel}(\vec x)$ is better; but note that $\text{accel}$, which gives you acceleration at a certain point in space, is not the same function as $\ddot x$ which gives you acceleration along a path parameterized by $t$ (I.e, it's not a field in space, but a field on a 1D path curving through space). – Filip Milovanović May 12 '20 at 3:55
• @JohnnyLongsom Usually, it just shows a link to move the discussion to chat (which I don't find here) but in any case, I created a room, I hope this is what you meant: chat.stackexchange.com/rooms/info/108589. I'll be online on and off but looking forward to your input, thanks! :) – Dvij D.C. May 28 '20 at 13:00

What you found here is not a inconsistency of Newton's mechanics, but a symmetry of the harmonic oscillator. Consider for simplicity a point particle in $$\mathbb{R}^n$$. The force can be considered as a function $$F:\mathbb{R}^n\rightarrow\mathbb{R}^n$$ taking the position of the particle as argument. Newton's law states that a physical trajectory $$\gamma:\mathbb{R}\rightarrow\mathbb{R}^n$$ of a point-particle of mass $$m$$ satisfies the equation $$F(\gamma(t))=\ddot\gamma(t)$$ for all times $$t\in\mathbb{R}$$.

Now concerning your question, you observed that if we take a physical trajectory $$\gamma$$ and scale it by a real number $$s\in\mathbb{R}$$, this satisfies Newton's law only if $$F$$ is linear. However, this is no inconsistency of Newton's mechanics, since scaling of a physical trajectory in general does not give you a new physical trajectory. Instead, the proper interpretation of what you found here is that this Kind of scaling symmetry is a characteristic of the harmonic oscillator (quadratic potential).

As a conclution, you assumed, that scaling a physical trajectory gives a new physical trajectory, which is not true in general. What you found is that this symmetry is a property of linear forces/quadratic potentials. I hope this could help you! Cheers!

• This answer (like several of your other ones) is very clear, and your bold phrase clearly pinpoints the assumption OP made. (I recognized what the issue was, but didn't know how exactly to put it into words) +1 – peek-a-boo May 24 '20 at 21:24

For $$\vec x$$ scaled by some arbitrary constant $$𝑠$$, we obtain: $$F(s\vec{x})=ms\vec{\ddot{x}} \Longleftrightarrow \frac{F(s\vec{x})}{s}=m\vec{\ddot{x}}$$

This is not true. Remember that $$\vec x$$ and $$\vec F(\vec x)$$ represent something physical; $$\vec x$$ represents the position and $$\vec F(\vec x)$$ represents the net force as a function of position. Once you've multiplied $$\vec x$$ by $$s$$, it no longer holds that meaning, so $$\vec F(s\vec x)$$ no longer represents the net force as a function of position, unless $$\vec F$$ happens to be linear (in which case it is proportional to force). Here, you implicitly assume that $$\vec F$$ is linear, and then show that that is consistent with it being linear; this is your discrepancy.

• I think OP does not assume that $F$ is linear, this is what OP concludes from the assumtion that each multiple of a solution of the EoM is again a solution. This is assuming a symmetry which is only satisfied for a linear force $F$. – Johnny Longsom May 11 '20 at 19:15
• @JohnnyLongsom the statement $F(s \vec x)=ms\vec{\ddot x}$ assumes that $F$ is linear. I said "implicitly assume" because the OP did not realise that this assumption was made. – Sandejo May 11 '20 at 19:19
• @JohnnyLongsom, when the OP writes an equation that holds only if $F$ is linear in $\vec{x}$, and then claims that this implies that $F$ must be linear in $\vec{X}$, the OP is guilty of "begging the question", i.e., assuming that which is to be concluded. – Alfred Centauri May 11 '20 at 19:19
• As far as I understood, OP got the equation $F(sx)=sF(x)$ as a consequence of the assumption that from $F(x)=m\ddot x$ follows $F(xs)=ms\ddot x$. This is different than assuming that $F$ is linear, it is assuming a symmetry of the space of trajectories. And the linearity of $F$ is a consequence of this symmetry. – Johnny Longsom May 11 '20 at 19:25
• @JohnnyLongsom, I see your point. – Alfred Centauri May 11 '20 at 20:49

It looks to me like you attempted an illegal change of variables. You can't just substitute $$s\vec{x}$$ for $$\vec{x}$$.

Remember that the equation $$F(\vec{x}) = m \vec{\ddot{x}}$$ isn't supposed to hold for all possible time-varying quantities $$\vec{x}$$. It's a particular assertion about $$\vec{x}$$, which is true for some time-varying quantities $$\vec{x}$$ and false for other time-varying quantities $$\vec{x}$$. Newton's second law asserts that the equation is true if $$\vec{x}$$ is the position of a particle in a force field defined by the function $$F$$.

So if you want to do a change of variables similar to the one that you did, you'll have to define a new quantity $$\vec{y} = \frac{\vec{x}}{s}$$, and then you can substitute $$s\vec{y}$$ for $$\vec{x}$$.

From there, you can conclude that

$$F(s\vec{y})=ms\vec{\ddot{y}} \Longleftrightarrow \frac{F(s\vec{y})}{s}=m\vec{\ddot{y}} = \frac{m}{s}\vec{\ddot{x}},$$

and the rightmost expression here is clearly just $$\frac{F(\vec{x})}{s}$$. Therefore:

$$\frac{F(\vec{x})}{s} = \frac{F(s\vec{y})}{s}.$$

Of course, this isn't a contradiction at all. It's a tautology, since $$\vec{x} = s \vec{y}$$ by the definition of $$\vec{y}$$.

Newton's second law does not state $$\vec F(\vec x) = m\ddot{\vec x}\ .$$ It states $$\vec F = m\ddot{\vec x}\ .$$ That is, the force $$\vec F$$ is, in general, not a function of the position. (Note that the position does not even appear on the right hand side!) If you want, for a given trajectory $$\vec x(t)$$, it is a function of the function $$\vec x(t)$$ (since then, you can compute $$\ddot{\vec x}(t)$$).

But generally, you simply can't write $$\vec F(\vec x)$$ on the left hand side, and thus, whatever you derive from your first equation suffers from this incorrect starting point.

• Of course force is a function of position. It takes a function $x$ and maps it to $m\ddot{x}$. Position does appear on the right-hand side. – jonathan May 25 '20 at 21:57
• @jonathan So you are telling me that if you see a photograph of a car, you know how fast it was and how much it accelerated? -- The force is not a function of the position, but a function of the function $x(t)$, which is precisely what I write. $F(x)$ takes a number $x$ as an argument, not a function. – Norbert Schuch May 26 '20 at 11:11
• Apologies, I misread what you wrote. Force is a function of the position function $x$, as you have explained. I think that really all of this confusion stems from abuse of notation (for example, denoting the position function by $x$ but also denoting a specific position by $x$). This would be remedied if Physicists were more careful about where the domains and ranges of their function are, and explicitly defining the objects they're talking about. – jonathan May 26 '20 at 18:01

I think that most of the answers fail to correctly convey what's going on. The equation you write is almost true, the only "sin" you committed is to assume that the force $$F$$ is not affected by the rescaling of the variable $$\vec{x}$$.

When you say "$$\vec{x}$$ scaled by an arbitrary constant," you are actually meaning: "define the variable $$\vec{y}$$ to be $$\vec{x}/s$$." So $$\vec{x}$$ might be the position of the particle that I measure in kilometers and $$\vec{y}$$ the position that you measure in miles. And, accordingly, we have two different functions for the force: the force expressed in kilometers and the one expressed in miles.

So if we apply Newton's law in the US we would have ($$\vec{y}$$ is in miles) $$F_{\mathrm{mi}}(\vec{y}) = m \vec{\ddot{y}}\,.$$ Whereas if we apply it in any other country that adopted the metric system ($$\vec{x}$$ is in kilometers) $$F_{\mathrm{km}}(\vec{x}) = m \vec{\ddot{x}}\,.$$ Again: note that $$F_{\mathrm{mi}}$$ and $$F_{\mathrm{km}}$$ need not be the same function. Then the equation should be stated as $$F_{\mathrm{km}}(\vec{x}) = m \vec{\ddot{x}}\quad \Longrightarrow\quad F_{\mathrm{km}}(s\vec{y}) = m s\vec{\ddot{y}}\,.$$

This together with the first equation amounts to the following equality $$F_{\mathrm{km}}(s \vec{y}) = s\, F_{\mathrm{mi}}(\vec{y})\,.$$ So you just discovered that, under a change of scale, the force is not a scalar quantity, but it's a quantity that transform with a weight of one. This happens in many many other instances in physics and it's important because it tells you what kind of object the force is. The angular momentum, for instance, would transform differently.

The problems I have with the other answers are

• It's true that generally $$F$$ is not a function of the position, but it could be. And it's certainly not true that when $$F$$ is a function of the position, then it must be linear.

• It doesn't matter that there is a derivative, if $$s$$ is a constant, it goes through. The argument wouldn't work for $$s$$ a function of time, but we are not asking about that.

• It's obviously true that $$\ddot{x} \to s \ddot{x}$$ under a rescaling. That's a property of the derivatives: derivatives are linear. Let's not get confused here.

• $$\ddot{x}(sx)$$ also doesn't really have any meaning. $$\ddot{x}$$ does not depend on the position. Or at least it does but in a circular way because we are using the differential equation $$F=m\ddot{x}$$ to say what's the value of $$\ddot{x}$$ at a given point. Therefore $$F(x)=\ddot{x}(x)$$ would be an empty statement.

• One always needs to make a distinction between the $$F$$'s before and after the transformation. And the transformation it's not a tautology but it's an important property of the force.

• The accepted answer is kind of correct but confusing: what's the difference between an equation and a definition? The equal sign should behave properly in both cases.

Of course I agree with the answer about the harmonic oscillator. There the viewpoint is different, so let me address that. @Johnny Longsom is making a rescaling of the variables without changing the units! So it's not looking at the same system with imperial or metric. It's looking at different systems which are related to each other by one being the scaled up version of the other. The conclusion of the post is correct: that's a symmetry of the system!

And you see: to discover that it was a symmetry we needed to know how $$F$$ transformed. So, every time you see a new animal in physics, ask it how it transforms.

• +1: This is a clarifying answer and a good summary of how different answers are addressing the question. Your answer is quite similar to the answer by Tanner Swett but more explicitly clarified IMO. So, as applied to a single particle, there are two ways to address the question. One is yours which considers the scaling as a change of units and the other is Johnny Longsom's which considers scaling of a trajectory in the same units. – Dvij D.C. May 15 '20 at 1:31
• I'd like to clarify my answer: I am looking at $F(x)=ma(x)$ as a set of equations applied over the whole of space at once. In other words, I'm talking about multiple particles at once. So $a(x)$ has the meaning that it represents the acceleration of the particle at position $x$. $x$ written as an argument is a label if you wish. If we only had $N$ particles, I'd write $N$ equations as $F(x_i) = a_i$. Now I put a particle at each point and the discrete label $i$ is replaced by the coordinate label $x$. – Dvij D.C. May 15 '20 at 1:34
• A more explicit version of my equation would be $F(x(x))=a(x)$ But x(x) is trivially $x$ by definition (because the force at the position of a particle at position $x$ is just force at $x$) and thus I write it simply as $F(x)=a(x)$. I should have clarified this all more explicitly in my answer I guess. – Dvij D.C. May 15 '20 at 1:39

In the right side of the equation $$a$$ is a function of time: $$a = a(t) = \frac{d^2x}{dt^2}$$ So the same is true for the left side. $$F = F(t)$$

It is perfectly possible for example to have different values of acceleration (and force) for the same $$x$$ in different times, like a car in an race circuit. So it can not in general be a function of $$x$$.

And if we want to know $$\mathbf F(st) = m\mathbf a(st)$$, of course $$\mathbf a(st) \ne s\mathbf a(t)$$ in general.

The claim is essentially that Newton's second law will change if we change the units in which we measure distance. However, in this case the units of force will change as well. In other words, the correct rescaling is $$\mathbf{x} \rightarrow s\mathbf{x}, \mathbf{F} \rightarrow s\mathbf{F}$$, so that, regraless of $$s$$: $$m\ddot{\mathbf{x}}=\mathbf{F}.$$

Since in retrospect I realise this question can be taken in several different ways, I am posting another reply myself, which combines some of the best answers so far and could hopefully help somebody else understand why this obviously wrong physical reasoning occurs mathematically.

As is clear at this point, the original issue was bad notation and there are at least four different ways in which this question can be understood (and resolved). From now on I am going to refer to the current position as $$\vec{x}_0$$ and the trajectory as a function of this starting position and time as $$\vec{x}(t)$$, such that $$\vec{x}(0)=\vec{x}_0$$, in order to avoid confusion. In all cases the other initial condition $$\vec{\dot{x}}(0)=\vec{v}_0$$ will not be given much consideration, but we still have to be aware of it.

1. First scenario: as correctly pointed out, Newton's second law does not imply that the force comes from a potential and/or is a function of the current position. In this case the force is simply a definition and the question doesn't hold anymore, since the force does not have to be a function of the trajectory:

$$\vec{F}(t)=m\vec{\ddot{x}}(t)$$

1. Second scenario: the force comes from a potential, such that $$\vec{F}(\vec{x}(t))=-\left.\frac{\partial U(\vec{x}')}{\partial \vec{x}'}\right|_{\vec{x}'=\vec{x}(t)}$$ and we make a coordinate transformation of the whole trajectory $$\vec{x}(t)=s\vec{y}(t)$$ for some scaling factor $$s$$. In this case:$$\vec{F}(\vec{x}(t))=m\vec{\ddot{x}}(t) \Leftrightarrow \vec{F}(s\vec{y}(t))=ms\vec{\ddot{y}}(t), \qquad\vec{F}(\vec{x}(t))=\vec{F}(s\vec{y}(t))$$ This is a simple change in variables and there is not much else to be found here - certainly no contradictions anyway.

2. Third scenario: the force comes from a potential and we rescale the initial condition of the differential equation, so that $$\vec{\tilde{x}}(0)=s\vec{x}_0$$. In general, this trajectory will be completely different to $$\vec{x}(t)$$ and the only way $$\vec{\tilde{x}}(t)=s\vec{x}(t)$$ is if the force corresponds to a quadratic potential.

3. Fourth scenario: the force comes from a potential and we rescale the solution $$x(t)$$ of the differential equation, such that $$\vec{\tilde{x}}(t)=s\vec{x}(t)$$. In general, this rescaled function no longer satisfies the original differential equation and nothing else can be said. The only way the new rescaled trajectory satisfies the original differential equation is if we had a harmonic potential.

Note that the difference between 3. and 4. is subtle: the new trajectory $$\vec{\tilde{x}}(t)$$ in 3. satisfies Newton's second law but its shape is in general different to $$\vec{x}(t)$$, while the $$\vec{\tilde{x}}(t)$$ in 4. doesn't in general satisfy Newton's second law (and doesn't have to) and its shape is the same as $$\vec{x}(t)$$. In both 3. and 4. the trajectories can only degenerate for a harmonic potential (although one might have to fiddle with the velocity initial condition to achieve that), which is what my original post "proved", although it certainly isn't clear whether I proved 3. or 4. (possibly neither) due to my bad notation!

I'm not sure where you think there is an inconsistency? All you're saying by your final equation, is that the Force @ position $$\vec{x}$$, = Force @ scaled position $$s\vec{x}$$ divided by the scaling, which is completely true? In the same way $$sF(\vec{x}) = F(s\vec{x}) \tag{1}$$ if $$F$$ is linear.

The use of $$F$$ is just to denote your calculating the Force, $$F(\vec{x})\neq F(s\vec{x})$$

Perhaps it is clearer to you to denote the force on position $$\vec{x}$$ as $$F_x$$ and the force on scaled position $$s\vec{x}$$ as $$F_{sx}$$.

• That's only true if $F$ is linear. – Sandejo May 11 '20 at 19:02
• But this equation can only be true if $F(x)$ is a linear function, no? – Godzilla May 11 '20 at 19:02
• Yes, that's true, but you've wrote it as linear, i.e. your second equation stating that $F(sx) = ms\ddot{x}$ – SamuraiMelon May 11 '20 at 19:23