Return to Answer

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects. Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes. To deny the vanishing, one would have to reject the rotational symmetry.

Let me wrap by saying that the quadrupole moment may always be interpreted as some "elliptical shape" of the object or particle. This ellipsoid would be stretched along some axes and shrunk along other axes. However, the electron's spin-up and spin-down state really pick the same preferred axis in space – the sign doesn't matter for the quadrupole – so they can't have different values of the quadrupole moment. In other words, the quadrupole moment doesn't depend on the spin, and because the spin is the only rotational-symmetry-breaking quantum number that the electron has, the quadrupole moment has to be zero. (A Pauli-matrix-free proof.)

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects (and similarly, of course, for all spin-0 objects). Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes. To deny the vanishing, one would have to reject the rotational symmetry.

Let me wrap by saying that the quadrupole moment may always be interpreted as some "elliptical shape" of the object or particle. This ellipsoid would be stretched along some axes and shrunk along other axes. However, the electron's spin-up and spin-down state really pick the same preferred axis in space – the sign doesn't matter for the quadrupole – so they can't have different values of the quadrupole moment. In other words, the quadrupole moment doesn't depend on the spin, and because the spin is the only rotational-symmetry-breaking quantum number that the electron has, the quadrupole moment has to be zero. (A Pauli-matrix-free proof.)

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects. Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes.

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects. Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes. To deny the vanishing, one would have to reject the rotational symmetry.

Let me wrap by saying that the quadrupole moment may always be interpreted as some "elliptical shape" of the object or particle. This ellipsoid would be stretched along some axes and shrunk along other axes. However, the electron's spin-up and spin-down state really pick the same preferred axis in space – the sign doesn't matter for the quadrupole – so they can't have different values of the quadrupole moment. In other words, the quadrupole moment doesn't depend on the spin, and because the spin is the only rotational-symmetry-breaking quantum number that the electron has, the quadrupole moment has to be zero. (A Pauli-matrix-free proof.)

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects. Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes.

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$ Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij}) $$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects. Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes.