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I am asking if there is any case in classical i.e., non-quantum, mechanics in which one cannot use Newton's second law $$\sum \mathbf{F}=\frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t},$$ to find the equations of motion of a system in classical mechanics (relativistic or not), and more general equations of motion are needed, just like in Quantum mechanics there are cases in which one cannot use the Schrödinger equation, e.g., for particles with spin.

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    $\begingroup$ Why can't one use Schrödinger equation for particles with spin? All time evolution in quantum mechanics (of isolated systems) is unitary, including those of particles with spin, and is described by Schrödinger equation. $\endgroup$ – Dvij D.C. Jun 1 at 21:57
  • $\begingroup$ @DvijD.C. Schrödinger equation is not enough to describe a spin $\frac12$ particle in a magnetic field. Pauli's equation is required. $\endgroup$ – GiorgioP Jun 1 at 22:48
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    $\begingroup$ @GiorgioP Pauli's equation is of the form $i\hbar\frac{d\psi}{dt}=H\psi$. If you restrict the form of the Hamiltonian in addition to it being Hermitian when you say "Schrödinger equation" then I agree but I don't think it is ever done in a modern context. $\endgroup$ – Dvij D.C. Jun 1 at 22:50
  • $\begingroup$ @DvijD.C. But with a two-component wave-function. Of course it depends on what one calls Schrödinger's or Newton's equation. In the Newtionian case, I would have hard time to call Newton's equation a second order differential equation involving matrices instead of vectors. $\endgroup$ – GiorgioP Jun 1 at 22:55
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    $\begingroup$ @GiorgioP It's still a vector in Hilbert space. Yes, it depends on what one calls the Schrödinger equation but the natural definition of Schrödinger equation is $i\hbar\frac{d\vert\psi\rangle}{dt}=\hat{H}\vert\psi\rangle$. If you want to focus on components, the position-basis wave-function is already an infinite-component wavefunction in itself. $\endgroup$ – Dvij D.C. Jun 1 at 23:02
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The simple answer is, no, Newton's second law is not violated anywhere in classical mechanics. But given that your question is a bit vague and includes the domain of relativity in classical mechanics, there is room for some elaboration.


The following statement is universally true.

Newton's second law is never violated in the dynamics of any system of particles in classical mechanics. The rate of change of momentum of a system of particles is always equal to the net force on the system.

However...

  • The more famous version of Newton's second law, i.e., $\sum \mathbf{F}_\text{ext}=\dfrac{d\mathbf{p}}{dt}$, is contingent upon the validity of Newton's third law for the forces internal to a system of particles. Not all forces between particles follow Newton's third law and thus, this form of Newton's second law is not always valid. However, this does not mean that Newton's second law is violated. $\sum \mathbf{F}=\dfrac{d\mathbf{p}}{dt}$ is still valid where $\sum\mathbf{F}=\sum\mathbf{F}_\text{ext}+\sum\mathbf{F}_\text{internal}$.

  • In special relativistic physics, Newton's second law $\sum \mathbf{F} =\dfrac{d\mathbf{p}}{dt}$ does not remain valid if we assign the usual meanings to symbols, i.e., if we take that $\mathbf{F}$ and $\mathbf{p}$ are non-relativistic three vectors. However, Newton's second law does remain valid in special relativistic physics as well if we are careful with what $\mathbf{F}$ and $\mathbf{p}$ represent. Namely, $\sum\mathbf{F}=\frac{d\mathbf{p}}{dt}(=m\gamma^3\mathbf{a}_{\parallel}+m\gamma\mathbf{a}_\perp)$ is relativistically correct where $\mathbf{F}$ is related to the Newtonian three vector force $\mathbf{f}$ via $\mathbf{F}=\mathbf{f}+(\gamma-1)\mathbf{v}\frac{\mathbf{v}\cdot\mathbf{f}}{v^2}$ and $\mathbf{p}=m\gamma\mathbf{v}$. A more manifestly relativistic way to write Newton's second law in special relativistic physics is $F^\alpha=\frac{dp^\alpha}{d\tau}$ where $\tau$ is the proper time along a particle's trajectory and $p^\alpha,F^\alpha$ are the four-momentum and four-force. In particular, $\{p^\alpha\}=(m\gamma,m\gamma\mathbf{v})$ and $\{F^\alpha\}=(\mathbf{v}\cdot\mathbf{F},\mathbf{F})$ where $\mathbf{F}$ is given in terms of Newtonian three force as described above.

  • In general relativity, Newton's second law looks substantially different. It reads $F^\alpha=\frac{Dp^\alpha}{D\tau}$ where $F^\alpha,p^\alpha$ are four-force and four-momentum of the particle in some arbitrary coordinate system and $\frac{Dp^\alpha}{D\tau}=\frac{dp^\alpha}{d\tau}+\Gamma^\alpha_{\mu\nu}p^\mu p^\nu$ where $\Gamma^\alpha_{\mu\nu}$ are the Christoffel symbols and capture the gravitational field as observed in the given coordinate system. $\tau$ is, as usual, an affine parameter along the trajectory of a particle that can be chosen to be the proper time for timelike trajectories.

  • Finally, Newton's second law is obviously only valid for a system of particles. It doesn't describe the dynamics of fields. For example, Maxwell equations that describe the dynamics of electromagnetic fields cannot be interpreted as a manifestation of Newton's second law. This is not a "violation" of Newton's second law because Newton's second law doesn't intend to describe the dynamics of systems that cannot be described as a system of particles. Of course, it does successfully describe the dynamics of a system of particles interacting with fields.

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Yes. If you are on a rotational frame then Newton's second law fails.

There is a more general equation describing such systems:

$$\mathbf{a} = \frac{\mathbf{F}}{m} - 2\boldsymbol\Omega\times\frac{d\mathbf{r} }{dt}-\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf{r})-\frac{d\boldsymbol\Omega}{dt}\times\mathbf{r},$$ where $\boldsymbol\Omega$ is the angular velocity pseudo-vector. Note that if $\boldsymbol\Omega = 0$ you recover Newton's law: $\mathbf{a}=\mathbf{F}/m$.

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    $\begingroup$ I don't think it might count as a failure of Newton's laws, the Corolis and centrifugal terms are fictitious forces, and Newton's law is of the form $\mathbf{F}_\mathrm{ficticious}+\sum \mathbf{F}=m\mathbf{a}$. $\endgroup$ – Don Al Jun 2 at 3:10
  • $\begingroup$ @Don Al I thinks it does count. After all, ficticious forces are introduced because Newton's laws fail there. $\endgroup$ – Javi Jun 2 at 3:27
  • $\begingroup$ @Javi Fictitious forces are there only because of the coordinate system used. $\endgroup$ – Don Al Jun 2 at 5:29
  • $\begingroup$ @Don Al I agree with you, but that is not what we are discussing here, I believe. $\endgroup$ – Javi Jun 2 at 5:43
  • $\begingroup$ Your answer is at odds with the question. The intention of the question is: is there some unexpected case where the second law doesn't hold good? Now, as we know the second law is valid for motion with respect to an inertial coordinate system. That restriction is so basic that it goes without saying. It's something that everybody knows. Theory of motion uses motion with respect to an inertial coordinate system, always. There is no exception to the following rule: every equation of motion that is valid uses an inertial coordinate system as reference. $\endgroup$ – Cleonis Jun 2 at 16:56
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A full answer would imply that you specify, more than you did, what is the meaning of symbols and in particular of $\bf F$. In my answer I'll assume that by $\bf \sum F$ is for the resulting total force acting on a point particle, due to the action of other bodies, and/or to the choice of a non-inertial frame. If ${\bf p}$ represents the momentum of one point particle, even staying at the level of classical mechanics, the answer is yes.

Usually, non-relativistic variable mass systems cannot be described by $\frac{{\mathrm{d}}{\bf p}}{\mathrm{d}t}= {\bf \sum F}$. They require the more general form $$ m(t) \frac{{\mathrm{d}}{\bf v}}{\mathrm{d}t} = {\bf \sum F} + {\bf u }\frac{{\mathrm{d}}{m}}{\mathrm{d}t} , $$ where $m(t)$ is the time varying mass, and $\bf u$ is the relative velocity of escaping or incident mass relative to the point like body. It is always possible to put such an equation in the form $$ \frac{{\mathrm{d}}{\bf p}}{\mathrm{d}t}= {\bf \sum F} + \left({\bf u} + \frac{ {\bf p}}{m}\right)\frac{{\mathrm{d}}{m}}{\mathrm{d}t}. $$ Therefore, we have to add a new term to the total force acting on the particle. Even if the new term could be formally interpreted as an additional force, like the spin case mentioned in the question, it more than a formal change, since it introduces two additional quantities necessary to specify the dynamical state ( ${\bf u}(t)$ and $m(t)$).

I think that the situation is very close to the spin case where the presence of an additional degree of freedom (the spin) forces a major change in the spinless Schrödinger equation in the presence of a magnetic field.

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