Relativistic corrections to hydrogen-like atoms

[Enthalpy]

Hello you all!

Models for the hydrogen atom and hydrogen-like often tell that the relativistic correction acts on the electron's mass as a consequence of the kinetic energy. Though, the kinetic energy equals half the (negative) electrostatic energy, so if the electron carried half of the electrostatic energy, the effect would cancel out the kinetic energy. What do you think?

To check that, I took the energy of last ionisation for varied elements so the relativistic effect varies, there
http://www.webelements.com/hydrogen/atoms.html
hoping the data is measured and accurate enough, since the effect is at most 1%, and tinkered a spreadsheet
RelativisticHydrogen.zip
(expand, open with Gnumeric, Excel or equivalent).

The first correction was to compensate by 1+m/M the effect of light nuclei to approach the Z² ionisation energy law of immobile nuclei. Seems to work well.

Then I already get a relative excess of the ionisation energy with Z that is nearly half the proportion of kinetic energy added to the electron's rest mass, which would be consistent with the kinetic energy alone acting on the electron's inertia.

Though, I observe less than 50% of the relative kinetic energy acting on the ionisation energy, something like 46%. The fit is less perfect at hydrogen and helium (fewer digits in the data as well), so I observe the variation referred to lithium, where the very clean fit suggests that the 46% are meaningful despite they stretch the data's accuracy.

So: what woud be the next correction after the kinetic energy added to the rest mass?

• Part of the electrostatic energy subtracting from the electron's mass? I feel normal to attribute far less than half of it to the electron, but haven't yet checked how much.
• Other causes?

Thank you!

[ajb]

What have you done?

 

You should expand the relativistic expression for the kinetic energy using a Taylor series to get corrections that involve the mass. The first correction is something like p^4/m^3 (ignoring the physical constants and numerical factors). You can then first order pertubation theory to get corrections to the energy spectra.

 

Then you should also consider spin-orbit coupling effects, again you can use first order perturbation theory.

[Enthalpy]

Thanks!

 

You imagine things waaaay more subtle than I've done! I'm a coarse engineer, you know: when a 16mm diameter screw breaks I put a 24mm one.

 

I've corrected the energy as the mass, that is in a first-order approximation, and only to compare with the departure from the Z² law. At Scandium (I have no data beyond) the correction is 1.2% so I expect the second order around 0.01% which indeed approaches the 0,46 vs 0,50 of 1.2%. Only one magnitude below, it would be the next necessary refinement.

 

Is the mass (or mass + kinetic energy) correction applied once for all positions of the electron, or individually for each position? (Which would already be too subtle for me).

 

Have you ever seen the inertia of the electrostatic energy injected in the hydrogen-like atom?

 

I check only 1s orbitals here, that is, the last ionisation of every atom. Would there be any significant coupling with the spin?

[ajb]

Is the mass (or mass + kinetic energy) correction applied once for all positions of the electron, or individually for each position? (Which would already be too subtle for me).

 

You write the classical kinetic energy, either the non-relativistic one or the corrected one, in terms of the momentum and the mass. The position variable does not appear here, it will appear in the potential energy.

 

You then let the momentum become an operator by applying the rules of quantum mechanics.

 

Have you ever seen the inertia of the electrostatic energy injected in the hydrogen-like atom?

No.

 

I check only 1s orbitals here, that is, the last ionisation of every atom. Would there be any significant coupling with the spin?

I cannot recall details. Look up relativistic corrections and spin-orbit coupling.

[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.

 

Here's the spreadsheet:

RelativisticHydrogen2.zip

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 eps*|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/(4pi*eps*R) 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

[Sensei]

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

You calculate it like electrician.

Not like quantum physicist.

Start with mass-energy of photon, then go to pair-production of electron-positron pair,

then go to pair-production of proton-antiproton,

and how proton joins with electron, to bound Hydrogen (this release energy- photon),

then to fusion,

and further...

Like I described here:

http://www.scienceforums.net/topic/92274-say-you-believe-or-else/#entry893302

 

From quantum-physics to classic physics, and electronics, not reverse.

 

I had expected that:

• The attraction between the nucleus and the electron makes the atom lighter.

Did you? I would say it's pretty standard knowledge..

Free proton has mass 938272046 eV/c²

Free electron has mass 510998.928 eV/c²

Bound together ₁¹H has 938272046 eV + 510998.928 eV - 13.6 eV = 938783031.328 eV (/c² if you want mass)

Divide by 931494061 eV and you have value in a.m.u.

938783031.328 eV / 931494061 eV = ~ 1.007825 u

(and you have isotope mass from f.e. here https://en.wikipedia.org/wiki/Hydrogen_atom right panel)

(you can't compare with the main Hydrogen website, because they include Deuterium in calcs: like ( mass of isotope*abundance + mass of other isotope * abundance ) = average mass etc)

 

 

If you add "missing energy" 13.6 eV, you will separate proton and electron, and make atom ionized.

H + 13.6 eV -> p⁺ + e^-

Providing more energy accelerate newly freed proton and freed electron.

 

Similarly,

if you have Deuterium:

Free proton has mass 938272046 eV/c²

Free neutron has mass 939565413 eV/c²

938272046 eV/c² + 939565413 eV/c² - 2.22 MeV = approximate mass-energy of Deuterium nucleus.

D⁺ + 2.22 MeV -> p⁺ + n⁰

In other words Deuterium nucleus "miss" 2.22 MeV

To separate proton from neutron bound together in Deuterium, provide missing energy (any way).

This way neutron generators are made.

https://en.wikipedia.org/wiki/Neutron_source

They just need source of energy with K.E. >= 2.22 MeV

Can be used high energy photons, high energy alpha particles, etc.

Edited February 14, 2016 by Sensei

[swansont]

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.

Evidence?

[Enthalpy]

You calculate it like electrician.

Not like quantum physicist.

Start with mass-energy of photon, then go to pair-production of electron-positron pair,

then go to pair-production of proton-antiproton,

and how proton joins with electron, to bound Hydrogen (this release energy- photon),

then to fusion,

and further...

Like I described here:

http://www.scienceforums.net/topic/92274-say-you-believe-or-else/#entry893302

 

From quantum-physics to classic physics, and electronics, not reverse.

 

Sensei, don't be ridiculous. One year ago you believed enegy was quantized in multiples of h.

Evidence?

 

I suppose you agree that the mass defect of atoms is maximum around iron, and that heavier atoms have a smaller mass defect, and eventually become unstable beyond lead, because the repulsion of the protons overcomes the strong force.

 

It is this electrostatic repulsion that adds mass to these heavier atoms. The strong force, commonly modelled as a short-range attraction between nucleons, gives a mass defect per nucleon approximately constant for atoms not too light, where most nucleons have a neighbour. It just gets stronger as the volume of heavier nuclei increases faster than their "surface", but slowly.

 

This is exactly the "Coulomb term" of the standard "Liquid drop model"

https://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Coulomb_term

 

As an expert for Relativity, you didn't ask about "acts equally as a weight and as inertia", did you?

https://en.wikipedia.org/wiki/Equivalence_principle#Experiments

This has been tested to <10^-13, while the electrostatic force changes the mass per nucleon by 10^-3 between iron and lead. Any difference would be known.

Edited February 15, 2016 by Enthalpy

[swansont]

 

I suppose you agree that the mass defect of atoms is maximum around iron, and that heavier atoms have a smaller mass defect, and eventually become unstable beyond lead, because the repulsion of the protons overcomes the strong force.

 

It is this electrostatic repulsion that adds mass to these heavier atoms. The strong force, commonly modelled as a short-range attraction between nucleons, gives a mass defect per nucleon approximately constant for atoms not too light, where most nucleons have a neighbour. It just gets stronger as the volume of heavier nuclei increases faster than their "surface", but slowly.

 

This is exactly the "Coulomb term" of the standard "Liquid drop model"

https://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Coulomb_term

 

 

No, not that. You implied electrostatic attraction between the electron and nucleus makes the atom more massive. I want evidence this is true.

[Enthalpy]

Preliminary results.

 

1. 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.
2. 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.
3. The interaction energy density is proportional to both charges. Its distribution is independent of them, here of the atomic number.
4. For the hypothetic contribution of the electrostatic interaction energy to the equivalent electron mass, each volume element must be weighted according to its position.
   4.1. 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.
   4.2. The contribution usable with the radial directions of delta(psi) is weighted by x²+y²+z².
   4.3. Instead of the usual scalar in vacuum, this hypothetic contribution to the mass would be a second-order tensor like in many solids.
   4.4. For the 1s orbitals considered here, the radial component matters.
5. 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.
6. The interaction energy is the same in identical volumes around both charges.
7. 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

 

(Hi Swansont, thanks for your interest, answering soon)

[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 reproduced Sensei's attempt with 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 like this:

 

 

A first result is that I get the right interaction energy of 1/(4pi). 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

[swansont]

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

 

But you didn't say that. You said "It increases the mass per nucleon". If you agree that it reduces the mass then this is not an issue.

[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.

 

EnergyOnly.zip

 

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

Marc Schaefer, aka Enthalpy

Edited March 18, 2016 by Enthalpy

[Enthalpy]

I've checked what corrections are usually applied to hydrogenoids (one electron around a nucleus), and they fit well the dataset from CE 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 <Psi|p⁴|Psi> 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 (click for full scale):

• 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, so 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. Expand, open with GnuMeric or Excel.

RelativisticHydrogen5.zip

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