[Enthalpy]

Hello dear friends!

Would you know a table of the energy of last ionization for varied atoms please?

I'm throwing thoughts at the relativistic correction. Up to know (engineer level) I've heard that the electron's kinetic energy injects a correction on the electron's inertia, often expressed as a modified mass. I've not heard that the electrostatic energy injects any correction on the inertia - and though, the electrostatic energy is twice as strong as the kinetic one, of opposite sign, and every energy is supposed to have inertia.

To try to check that, I compare the ionization energy for the last electron with varied numbers of protons. Besides the known Z² law, more protons give the electron a bigger kinetic and electrostatic energy as compared with the electron's mass, which makes the relativistic correction more important and visible. Begun there:
http://www.scienceforums.net/topic/85377-relativistic-corrections-to-hydrogen-like-atoms/
where I see the correction due to the kinetic energy, minus a few % that I haven't tried to attribute yet - in progress. Maybe this small difference is meaningful, maybe not, and it looks linear, as opposed to the higher terms of m/m₀.

My source of data up to now was Webelements
http://www.webelements.com/hydrogen/atoms.html
which doesn't state clearly its source; it could be the Crc Handbook of Chemistry and Physics.

In the next step, I'll exploit the Hdbk of Chem & Phys now at hand. Its source is:
C.E.Moore, Ionization Potentials and Ionization Limits Derived from the Analysis of Optical Spectra (1970).

You get it: the kind of data I seek is

• Experimental of course
• 0.1% accurate
• With nuclei heavier than Ca if possible
• And from an other source, to cross-check.

Thank you!

[Enthalpy]

This curve relates:
- The deviation from Z² of the last ionization energy, for nuclei of varied Z;
- With the kinetic energy of the electron over its rest mass.

The ionization energy is from C.E.Moore:
Ionization Potentials and Ionization Limits Derived from the Analysis of Optical Spectra (1970)
obtained over the CRC Handbook of Chemistry and Physics, 72nd edition
section "Atomic, Molecular and Optical Physics", chapter "Ionization Potential of Neutral and Ionized Atoms".
The data I got previously over Webelements.com is uniformly 23ppm bigger and must have the same source.

The only correction ("pinned") included so far is the electron-to-nucleus mass ratio. For instance the nucleus' size is not included.

I see on this curve that
- The kinetic energy contributes to the deviation: 1% kinetic energy vs rest mass increases the ionization energy by 0.46%, nearly the 50% expected from the increased mass+energy.
- The electrostatic energy doesn't contribute much to the inertia. I expected it would, reducing it.
- The experimental points fit a straight line very neatly, suggesting that the 0.50 and even the 0.46 are significant and don't result from higher powers of E/mc².

Comments are welcome of course, additional datasources too.

Marc Schaefer, aka Enthalpy

[Enthalpy]

This is how I believe(d) to understand the energy and mass of the electrostatic interaction:

• The electrostatic field E has an energy per volume unit ε|E|²/2. It is observed and measured at radio waves. Mainstream science: "a capacitor stores the energy in the vacuum between the electrodes".
• The field of a lone charged particle is energy, so to say an "auto-action", computed by integrating the squared field over the volume. Mainstream science: virtual particle pairs and dressed particles were invented to keep this energy finite at sizeless particles like electrons.
• When two charged particles interact, their fields add as vectors, and the integral over the volume of the variation of energy density versus the individual ones give the interaction energy -Zq²/(4πεR) here. An interaction is then the superposition of both "auto-actions" made nonlinear because the field is squared.
• See the drawing: in the hydrogen-like atom, the individual fields reinforce between the nucleus and the electron's possible position, increasing the mass density, but weaken elsewhere, including far from the symmetry axis. The peak of reinforcment is stronger, but the weakening over a bigger volume wins somehow to reduce the energy globally.
• The electrostatic energy has a mass related by c² that we obseve and measure. It increases the mass per nucleon, as atoms heavier than iron show. This mass acts equally as a weight and as inertia, as is observed at atoms.

I had expected that:

• The attraction between the nucleus and the electron makes the atom lighter.
• The orientation, extension, strength of this mass change (versus the sum of both auto-actions) follows the possible positions of the electron.
• The mass of the interaction contributes to the electron's inertia, the contribution of each volume element being weighed by |x|²/|R|².

Though, the graph of 10 November 2015 shows an increase of the electron's inertia as the nucleus' charge increases, and fitting rather well (0.46 versus 0.50) the sole kinetic energy of the electron. The electrostatic attraction, a negative energy twice as strong as the kinetic energy, would reduce the electron's inertia.

Until I put some figures on what I had expected, here are qualitative thoughts:

• If the nucleus and the electron concentrated equally the electrostatic interaction energy that is twice as strong as the kinetic energy, we would observe no change at all in the electron's inertia.
• The interaction is globally a negative (change of) energy. This is the integral of the (change of) energy density over the volume: the increase between the nucleus and the electron, the decrease elsewhere that outweighs it.
• The increase is nearer to the nucleus, the decrease is farther away. So not only is the decrease stronger over the volume, its effect on the electron's inertia, weighed by the squared distance, must be a stronger reduction.

Marc Schaefer, aka Enthalpy

[Enthalpy]

Preliminary results (see the drawing).

• Taking the permittivity as 1, the energy density W of the electric field E is traditionally W=E²/2. W_i, the energy density of the interaction, defined as the result of both charges together minus the result of each separately, is just the scalar product of both fields.
• This interaction is zero where the individual fields are perpendicular, that is, at the sphere with poles at the charges. It's positive within and negative outside the sphere.
• The interaction energy density is proportional to both charges. Its distribution is independent of them, here of the atomic number.
• For the hypothetic contribution of the electrostatic interaction energy to the equivalent electron mass, each volume element must be weighted according to its position.
  
  • The contribution usable with both azimuthal directions of delta(psi) is weighted by x²+y², after proper reorientation of the arbitrary z direction around the x axis.
  • The contribution usable with the radial directions of delta(psi) is weighted by x²+y²+z².
  • Instead of the usual scalar in vacuum, this hypothetic contribution to the mass would be a second-order tensor like in many solids.
  • For the 1s orbitals considered here, the radial component matters.
• The energy density of the interaction is infinite near each charge but it compensates before and after the charge along the x axis, and compensates to the second order of y and z near the x axis, so its integral over a small volume around a charge is finite and small. A simple scheme of numerical integration suffices.
• The interaction energy is the same in identical volumes around both charges.
• Along the x axis, the nucleus' field varies as x^-2 and the weight of its hypothetic mass contribution as x², so the contributions from 0<x<1 and from 1<x<2 compensate - and they nearly compensate near the x axis.

A piece of software to evaluate the hypothetic contribution to the mass is in progress.
Marc Schaefer, aka Enthalpy

[Irlanur]

To be honest I didn't read everything you wrote so far, but maybe I can still give some input.

-First of all I think it makes sense in the framwork of Quantum Chemistry to work with the rest mass and express the relativistic momentum and energy with it.

-Then: do you want a "relativistic correction" or a full relativistic framework? I ask that because it makes a conceptual difference if you use the relativistic parts of a Hamiltonian in a variational or a perturbative approach. Relativistic Hamiltonians are NOT variationally stable!

-What I haven't found yet in your texts are retardation effects. This is what changes the electromagnetic interaction energy in a relativistic framework (besides that you have spin...).

[Enthalpy]

Hi Irlanur and all, thanks for your interest!

I'd be happy with a relativistic correction! No bigger ambition than that... The one I've seen up to now is the electron's kinetic energy making a correction to the rest mass, and this one fits the observation well.

What bothers me is that the electrostatic energy is twice as big as the kinetic energy and it has a mass too, as we see at heavy nuclei. There, the repulsion among protons adds energy to the nucleus, enough so to change the mass per nucleon by a very observable amount - the mass of the electric field. In hydrogen-like ions, I expect the mass of the electric interaction to enlight the electron, but this is not seen experimentally. I try to evaluate how big the discrepancy is.

Would the spin make a big difference for a 1s orbital? I hope not. I've neglected it. The ionization energy table I have doesn't mention neither, despite giving more decimal places than I need.

Retardation effects... Do you mean propagation time? That's a slippery track! For charged articles with a constant speed - not the case in an atom - the electric field points to the present direction of the particle, not to its position one propagation time ago; an analytic solution has been derived from Maxwell's equations. For the gravitaion field of the Earth and the Moon too, and this is easy to proove. This is a discussion theme in itself, I still have to give more thoughts to it.

[Irlanur]

By retardation I mean that that changes in electromagnetic fields propagate with the speed of light only. So for the interaction energy of two moving particles you cannot just use the static picture.

Maybe have a look at this also:
https://en.wikipedia.org/wiki/Breit_equation

[Enthalpy]

Why I say that the attration between the nucleus and the electron makes the atom lighter:

• It's my understanding that every energy is mass. When considering the composite object (here the atom), the interaction of the components appears as a rest mass.
• For heavy nuclei, the big energy of the electrostatic repulsion between the protons is observed at the atom's mass. The same must happen with protons and electrons. The attraction would enlighten the hydrogen atom by 2*13.6eV and the kinetic energy make it heavier by 13.6eV.
• I have attempted to compare the rest mass of the proton, the electron and the hydrogen atom, but have big doubts about it.

Proton, electron and hydrogen atom masses from Nist, here in 10^-9 amu = 0.931eV:
http://physics.nist.gov/cuu/Constants/index.html (2014 Codata)
http://www.nist.gov/pml/data/comp.cfm

1 007 276 466.879 +-0.091 proton
0 000 548 579.909 +-0.000 016 electron
1 007 825 032.23(9) hydrogen atom
-------------------
           14.6*10^-9 amu = 13.6eV lighter

This looks too good... I suspect that the atom's mass is derived from the proton's one or the other way, because their uncertainty is the same but the measures would use different methods. Then, the 13.6eV difference would prove only that other people understand it as I do.

I would like to spread these 2*13.6eV defect over components of the atom. It's the query of this thread, because difficulties arise. Experimental data shows the electron heavier by 13.6eV (and more with more protons in the nucleus), consistently with the kinetic energy alone and with very little effect of the electric interaction that would make the electron lighter.

If the electron bore all the interaction energy, it would be 13.6eV lighter - no.
Splitting the interaction 50/50 among the nucleus and the electron would keep the electron's rest mass - no.
My current attempt is to localize the mass where the electric energy is said to be: in the vacuum between the charges, and try to evaluate the effect on an equivalent electron mass.

I've written a piece of software that integrates over space the change of the electric energy density due to the interaction. It uses three meshes as on the drawing.

A first result is that I get the right interaction energy of 1/(4π). This is consistent with the usual interpretation that charges have a self-energy before any interaction modifies the energy density when charges are near. Standard physics wants the electric field to contribute to the electron's rest mass, prior to any electric interaction, and this rest mass is used to compute hydrogen's spectrum, so I feel legitimate to incorporate the interaction's mass somewhere too, and partly at the electron.

As opposed, my attempt to weigh the local change in energy density by x²+y² or x²+y²+z² fails, both computationally - it can't converge at big distance - and conceptually - among others, the nucleus' field doesn't depend on the electron's possible locations but is a component of the change in energy density. I have to meditate this longer.

Marc Schaefer, aka Enthalpy

[Enthalpy]

Here's already a software's simplified version that computes the energy of electrostatic interaction but no inertia. It finds the correct interaction energy by comparing the individual fields and their sum, as well as the points in the 02 March 2016 message related with energy.

While this is no proof, it is consistent with the idea that charged particles have an electric "self-energy" prior to interactions.

To run the program, remove the .txt extension and launch the htm in Firefox where it takes few seconds. Windows Script Host and IE6 would be 1000 times slower.

Marc Schaefer, aka Enthalpy

[Enthalpy]

I've checked what corrections are usually applied to hydrogenoids (one electron around a nucleus), and they fit well the dataset from C.E. Moore, with no inertia from the electrostatic interaction. Here for the energy E of last ionization, where n=1 l=0.

• The relativistic correction to the mass depends on the local kinetic energy hence on the distance to the nucleus. That's the <Ψ|p⁴|Ψ> in Wiki's explanation of the perturbation method. Then, it increases the ionization energy by 2.5×E/mc² rather than 1×.
• No mention of spin-orbit coupling has popped up for a spherical orbital, not even a quadratic effect.
• The Darwin term reduces the ionization energy by 2×E/mc².
• The Lamb shift has no analytical expression nor excellent prediction beyond hydrogen. Subtracting a Z² extrapolation from hydrogen would be exaggerated, and the fraction of Z² drops nonlinearly with Z.

The diagram shows:

• On X axis, the electron's ionization energy (taken as its kinetic energy) divided by its rest mass-energy;
  
• On Y axis, the relative deviation from a Z² extrapolation from hydrogen. All hydrogenoids here are pinned: the consequence of the nucleus' "movement" is removed by computation;
• The measures by C.E.Moore and their fit by a straight line that would be a Z² relative, or Z⁴ absolute, deviation;
• The contribution by relativistic mass and Darwin, following both Z² and already a nice fit of the measures;
• And the biggest possible contribution of Lamb shift to relativistic mass and Darwin. I've applied Z² on the measured hydrogen Lamb shift (×2³ for 1s instead of 2s).

Picking some amount of Lamb shift would just fit the experiment.

Some sources:
https://en.wikipedia.org/wiki/Fine_structure
https://en.wikipedia.org/wiki/Lamb_shift
http://cua.mit.edu/8.421_S06/Chapter3.pdf
http://crunch.ikp.physik.tu-darmstadt.de/nhc/pages/lectures/rhiseminar06-07/djapo.pdf
Djapo.pdf gives on p34 a list of many contributions, where I don't see the inertia of the electrostatic interaction.

So the mystery remains for me. We measure a mass for the electrostatic interaction of protons in heavy nuclei; the possible distributions of this mass should depend on the possible positions of the electron; but its inertia, which would be about as big as the relativistic mass correction, is unwanted to fit experimental data.

The spreadsheet follows. Remove the .txt, expand, open with GnuMeric or Excel.

Marc Schaefer, aka Enthalpy

[Enthalpy]

Here's a situation similar to the hydrogenoid atom, but simpler as it involves no quantum mechanics.

Two particles collide head-on and rebound elastically, for instance a proton and a positron with speed relativistic but not enough to make new particles. Their common centre of mass is immobile as is a first observer who sees them reverse their speed.

A second observer has a constant speed u, say perpendicular to the particles' path. He sees both particles having a speed component -u before, during and after the rebound. The punched screens make it more dramatic: the immobile observer sees the particles come back through the holes, hence the moving observer too, and the screens have kept their speed -u.

The particles' momentum along u is u times their relativistic mass before and after the collision. We wish this momentum component to be constant over the collision, so the moving observer must attribute to the particles a mass that is constant over the collision - even when the transverse speed gets smaller or zero as the kinetic energy converts into electrostatic energy. That's consistent with the mass of heavy nuclei, where an observer external to the nucleus weighs the protons' electrostatic repulsion. By the way, this increase doesn't depend on the speed u, which can be small.

Though, the immobile observer computes the collision with a particle mass depending on the speed only, not on the electrostatic energy. Worse, the moving observer too computes the transverse speeds during the rebound using no mass contribution from the electrostatic interaction.

So the mass of the electrostatic interaction depends on the observer, or worse, on his purpose, yuk. Possibly like: the repulsion energy makes particles heavier except for the acceleration that results from this interaction. O good.

Spread the electrostatic contribution to the mass in the vacuum where the fields of both particles interfere, rather than on the particles? But why wouldn't that mass slow the particles' acceleration due to the repulsion? The interference of the fields moves with the particles.

Uncomfortable too: the lighter particle, which carries the biggest increase of relativistic mass since the centre of mass is immobile, also carries the biggest increase of electrostatic mass, despite both particles experience the same electrostatic potential, including the slope and curvature. So the electrostatic contribution doesn't depend on local field quantities, but on the particle's history through the field, or possibly on the rest mass.

I didn't consider the magnetic induction here, despite charges move. Nor the radiation, despite charges accelerate (but identical particles would radiate little).

Did I botch something? Would someone kindly shed light on this mess?

Marc Schaefer, aka Enthalpy