# $F = ma$ not applicable at atomic dimensions?

Is the following logic valid?

$F = ma$ does not hold at atomic dimensions because it is a continuous function. But since we know from the Einstein-Plank relation $E = nh\nu$, energy is quantized and not continuous. Therefore, $F = ma$ does not apply at very low quantum numbers.

• $F=ma$ is not valid on the quantum scale because the position of an object is not well-defined - much less its second derivative. Energy is not always quantized in quantum mechanics. – J. Murray Feb 25 '18 at 16:42
• You seem to have mis-understood the Einstein-Planck relations The energy in the EM field at frequency $\nu$ is quantized in the way you indicate. But the field can contain contributions from any frequency, so the energy in the EM field can take on any value. – dmckee Feb 25 '18 at 16:44
• BTW. We have MathJax running on the site, so you can write math in a LaTeX-alike way. For instance writing $E = n h \nu between single dollar-signs would result in$E = nh\nu$. – dmckee Feb 25 '18 at 16:46 • Thanks much for the excellent feedback. You say "position not well-defined" ... do you say this because of Heisenberg Uncertainty Principle? – ChemEGuy Feb 25 '18 at 16:54 • Position is not well-defined because the state of a typical quantum mechanical system does not correspond to a single position in space. – J. Murray Feb 25 '18 at 17:02 ## 3 Answers We do use concepts that you would recognize from a course in Newtonian physics in quantum mechanics, but the formulation is based on the Hamiltonian (or Lagrangian for quantum field theories) formulation of mechanics. In those formalisms the central role is played by energy rather than by force. So when you are, say, solving the Schrödinger equation $$\left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r})\right] \Psi(\vec{r},t) = i\hbar\frac{\partial}{\partial t} \Psi(\vec{r},t)$$ the potential function$V(\vec{r})$shows up explicitly. Similarly the formalism for working out differential scattering cross-sections in quantum mechanics make explicit use of a potential function. You can, of course, compute a force law from the potential—and in may cases they are familiar one Coulomb's and Hooke's laws—but we don't use those functions directly. • We also have$\frac{\mathrm{d}}{\mathrm{d}t}\left\langle \hat{\boldsymbol{p}}\right\rangle =-\left\langle \boldsymbol{\nabla}V\right\rangle $. – J.G. Feb 25 '18 at 18:17 • @J.G. That is probably worth of an answer in and of itself. – dmckee Feb 25 '18 at 18:41 Following dmckee's suggestion I've expanded a comment of mine into an "answer", albeit one I still consider more a long comment than anything else. Classically, Newton's second law in a conservative field can be written as$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{p}=-\boldsymbol{\nabla}V(\mathrm{x})$. Quantum-mechanically this is upgraded with some averages of operators, viz.$\frac{\mathrm{d}}{\mathrm{d}t}\langle\hat{\mathbf{p}}\rangle=-\langle\boldsymbol{\nabla}V(\hat{\mathrm{x}})\rangle$. We can add a suitable function to$-\boldsymbol{\nabla}V\$ in either case to incorporate non-conservative forces.

As you sign ChemEguy I will suppose physics is not your stronger suit.

F=ma does not hold at atomic dimensions because it is a continuous function.

At atomic dimensions quantum mechanics reigns, it does not hold as such because it is an emergent classical formula from the underlying quantum mechanical interactions. At this level fundamental forces are defined as exchanges of momentum using the formalism of feynman diagrams. The link expands on this.

But since we know from the Einstein-Plank relation E=nhν, energy is quantized and not continuous.

As answered in the comments, energy may or may not be quantized, depending on the solutions of the boundary value problems using quantum mechanical equations. Bound states are quantized.

Therefore, F=ma does not apply at very low quantum numbers.

It has a different manifestation at dimensions commensurate with h, the Planck constant. Quantum numbers are the numbers distinguishing elementary particles from each other, and are not immediately connected with exchange forces; various interactions conserve or not quantum numbers, and it will depend on the exchange forces whether they are conserved or not.

## protected by Qmechanic♦Feb 25 '18 at 20:05

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