Coulomb's law fails when the distance $r$ between to point charges vanishes. As $r\to 0$, the electric field between two point changes increases without limit. I should be bothered about the limit $r\to 0$ only if that limit is a physical limit. For example, does anyone ever directly have to use Coulomb's law to determine the force between two overlapping electrons? Even in quantum mechanics, the singularity of the Coulomb potential at $r=0$ do not pose any problem. Hydrogen atom exists and is solvable. So my question is are there situations in physics where $r\to 0$ can be a physical limit and one needs to worry (particularly, classical physics)?
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asked Jan 13, 2019 at 12:21
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$\begingroup$ The hydrogen atom model in QM has an angular momentum barrier. The singularity is still there. $\endgroup$– user196418Jan 13, 2019 at 12:32
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$\begingroup$ @ggcg Of course, the singularity is there. The point I'm trying to make is that $r=0$ is not a physical domain. Hence, we need not worry about the potential becoming infinity at $r=0$. In H atom the electron cannot occupy the $r=0$ location which makes the $r=0$ is not at all a physical limit. $\endgroup$– SolidificationJan 13, 2019 at 12:38
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$\begingroup$ @mithusengupta123 What if this were the case of two protons in the nucleus? That makes it possible for particles to exist within $r \to 0$ . $\endgroup$– KarthikDec 24, 2019 at 8:10
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2 Answers
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No. Nobody has been able to do an experiment that distinguishes an electron from a true point particle. That is to say, if it has a non-zero radius, it is smaller than we can measure.
By overlapping electrons, you likely mean electrons with overlapping wave functions. This does not mean the electrons occupy the same extended regions of space. It means they are likely to be found at separate points in the same region.
Protons do have a finite radius, but this is misleading. Protons contain quarks. The location of the quarks is limited by the Uncertainty Principal. The radius of the proton is more or less the uncertainty in position of the quarks. The quarks themselves have no known size.
E&M works very well on the smallest scales we can measure.
answered Jan 13, 2019 at 12:46
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Let us separate the classical 1/r singularities from the quantum potentials used in quantum mechanical equations.
In classical physics, there are no point particles by definition of particle, which has to have a mass. So there will be no possibility of two charged particles to interact at r=0. Their mass will not allow it.
In quantum mechanics the classical 1/r potentials are used to derive the wave functions that describe the quantum mechanical particles. These wave functions with the boundary conditions imposed will give $Ψ*Ψ$ as the probability to find the quntum mechanical particle at $(0,0,0)$. For some systems this is a nonzero value, which means that the orbitals pass through the pole of the potential, but the interaction is only through the quantum mechanical solutions, not directly particle with the potential. Here are the hydrogen orbitals, and some do pass through $(0,0,0)$.
This overlap is how electron capture works in nuclear interactions.
In general one can say that quantum mechanics with its probabilistic nature takes care of classical singularities, which is how effective quantization of gravity is used in the Big Bang model.
