Weighing Electrons
Don Herbison-Evans
(donherbisonevans@outlook.com)

No-one has ever actually weighed electrons. The fundamental axiom to Einstein's theory of General Relativity is the principle of Equivalence: that the inertial mass of all particles is identical to their gravitational mass. This has been proved by experiment of the particles composing the nuclei of atoms. It has not been proved for electrons. If for example: if it could be shown experimentally that electrons were, for example, not subject to gravity, with zero gravitational mass, Einstein's axiom would be proved wrong, and cosmological theory would need complete revision. It has only been possible hitherto to measure the inertial mass of electrons. No-one previously has been able to measure their gravitational mass, to assess the validity of Einstein's assumption. The experiments described here would be able to this. Two methods come to mind.

1. Gravimetric Chemistry
For the general case, consider the conversion of molecule S with molecular mass 'S' Daltons, into a molecule T with molecular mass 'T' Daltons. Let 's' be the number of electrons in S, and 't' be the number electrons in T. Let 'e' be the mass of an electron in Daltons. Bearing in mind that, generally, atomic masses are determined by Mass Spectrometry, and so are inertial masses, then if electrons do have gravitational weight: 1 gram of S will be produce T/S grams of T. But if electrons have no gravitational weight: (S-se) grams of S will produce (T-te) grams of T. So then 1 gram of S will produce (T-te)/(S-se) grams of T if electrons have no weight

If x is the difference in weight of product T from 1 gram of S

x = (T-te)/(S-se) - T/S
   = [S(T-te) - T(S-se)]/[S.(S-se)]
   = (sTe - tSe)/[S.(S-se)]
   ≈ e.(sT - tS)/(S2)             : equation 1
   ≈ e.(s/S - t/T).(T/S)             : equation 2
where e = 548 micrograms Although electrons only contribute about 548 micrograms per Gram Dalton, balances are now commercially available that can measure 2.5 grams to an accuracy of 1 microgram, such as the Sartorius Cubis Ultramicro Balance MSA.

Electrons constitute the highest proportion of the mass of molecules that are composed of atoms of the lowest atomic number, so it is tempting to use the lightest elements for this experiment. However measurements based on chemical reactions can be complicated by changes in isotopic ratios for molecules containing elements with more than one naturally ocurring stable isotope. The lightest elements that have only one stable isope, and so avoid this fractionation, are Be and F.

Be -> BeF2

So let us consider the reaction:

Be + F2 -> BeF2

The principle is to find out what weight of BeF2 is produced from a weighed amount of Be.

With this reaction: we have

Be -> BeF2 :-
S = 9.012182 ≈ 9
s = 4
T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47
t = 22
sT ≈ 4 x 47 = 188
tS ≈ 22 x 9 = 198
S2 ≈ 81
x ≈ e.[188 - 198)]/81 = -e.[10/81] ≈ -0.123e ≈ -0.123x548 ≈ -68 micrograms per gram Be
So 1 gram of Be should produce approximately
either 47.009042/9.012182 = 5.216167
or (47.009042 + 22x0.000548)/(9.012182 + 4x0.000548)
= (47.009042 + 0.012056)/(9.012182 + 0.002192)
= 47.021098/9.014374 = 5.216236 grams of BeF2
depending on whether electrons are weightless or heavy.

BeH2 -> BeF2

Now let us consider the reaction:

BeH2 + 2HF -> BeF2 + 2H2 With this reaction: we have BeH2 -> BeF2 :-
S = 9.012182 + 2x1.007825 = 9.012182 + 2.015650 = 11.027832 = ≈ 11
s = 6
T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47
t = 22
sT ≈ 6 x 47 = 282
tS ≈ 22 x 11 = 242
S2 ≈ 121
x ≈ e.[282 - 242)]/121 = e.[40/121] ≈ 0.331e ≈ 0.331x548 ≈ 181 micrograms / gram BeH2
So 1 gram of BeH2 should produce approximately
either 47.009042/11.027832 = 4.262764
or (47.009042 + 22x0.000548)/(11.027832 + 6x0.000548)
= (47.009042 + 0.012056?)/(11.027832 + 0.003288?)
= 47.021098/11.031120 = 4.262586 grams of BeF2
depending on whether electrons are weightless or heavy.

BeD2 -> BeF2

Because natural Hydrogen contains some Deuterium, let us consider the reaction :

BeD2 + 2HF -> BeF2 + (say 2HD) With this reaction: we have BeD2 -> BeF2 :-
S = 9.012182 + 2x2.014102 = 9.012182 + 4.028204 = 13.040386 = ≈ 13
s = 6
T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47
t = 22
sT ≈ 6 x 47 = 282
tS ≈ 22 x 13 = 286
S2 ≈ 169
x ≈ e.[282 - 286)]/169
= -e.[4/169] ≈ -0.024e ≈ -0.024x548 ≈ -13 micrograms / gram BeD2
So 1 gram of BeD2 should produce approximately either 47.009042/11.027832 = 4.262764 or
(47.009042 + 22x0.000548)/(11.027832 + 6x0.000548)
= (47.009042 + 0.012056)/(11.027832 + 0.003288)
= 47.021098/11.031120 = 4.262586 grams of BeF2
depending on whether electrons are weightless or heavy.

Combining these two experiments, with BeH2 and BeD2, assuming the proportion D/H in natural Hydrogen is about 1/5000, if we did the experiment with natural Hydrogen we would get a differnece of approximately

[4999 x (181) + 1 x (-13)]/5000 = 180.96 micrograms / gram BeH2 so the experiment would be fine with natural BeH2

General Considerations

One would of course work with arbitrary but accurately weighed amounts of the reactants and scale the results accordingly. Limiting the effects of impurities and interactions with the atmosphere and reaction vessels to values well below a few micrograms would be a challenge, particularly as metallic Be in air becomes covered in a thin oxide layer, and F2 attacks nearly everything.

From equation 2, we appear to need to maximise the difference s/S - t/T, while minimising T/S. Hydrogen has s/S ≈ 1, but gases are hard to work with and weigh accurately, Fortunately hydrides are solid and could be used to advantage. Other lighter elements have s/S ratios near 0.5, and the heavier elements have values approaching 92/238 ≈ 0.39. So we might try to drop the t/T ratio by using a reaction that converts between heavy elements to light elements, for example: reacting BeI2 with F2, and evaporating off the iodine and iodine fluorides produced, would give

BeI2 -> BeF2: S = 263, s = 110, T = 47, t = 22.
sT ≈ 110 x 47 = 5,170
tS ≈ 22 x 263 = 5,786
S2 ≈ 69,169
x ≈ e.(5170 - 5786)/69169 = -e.616/69169 ≈ -0.008906e
≈ -4.88 micrograms per gram of BeI2
This appears to be a worse result than before, but the electron weight difference for 1 gram of the product BeF2 -4.88 x (263/47) ≈ -27.3 micrograms which is an improvement over -12.9 micrograms from the reaction Be -> BeF2. However, the spontaneous inflammability of BeI2 in air might make this experiment even more difficult and hazardous than Be -> BeF2.

One might alternatively consider other reactions involving the other stable mono-isotopic elements:

  Element  
  Atomic  
Number
  Approx  
Atomic
Mass
Be
4
9
F
9
19
Na
11
23
Al
13
27
P
15
31
Sc
21
45
V
23
51
Mn
25
55
Co
27
58
As
33
75
Y
39
89
Nb
41
93
Rh
45
103
I
53
127
Cs
55
133
Pr
59
141
Tb
65
159
Ho
67
165
Tm
69
169
Au
79
197
Bi
83
209

As can be seen, only four of these are non-metals, which severely limits our options. However, as we have seen: it is also possible to use compounds with multi-isotopic elements although the weights would be complicated by the isotopic ratios, which would need to be measured and allowed for.

2. Beam sag
Consider a vacuum tube L metres long with an electron gun at one end accelerating an electron beam by some voltage V striking a phosphorescent screen at the other end. If electrons have a gravitational weight proportional to their inertial mass: the beam will sag due to the electrons falling by gravity as they fly from one end of the tube to the other.

The amount of sag can be calculated by noting that the electrons will have an energy of

Ve = (1/2)mv2 joules, where V is in volts
e is the electron charge ≈ 2 x 10-19 coulombs
m is the electron mass ≈ 10-30 kilograms
so that v2 = 2Ve/m
v = (2Ve/m)1/2 metres/second

So the time of flight travelling L metres will be

t = L/v seconds during which it will sag s = (1/2)gt2 metres where g ≈ 10 metres/second2
so that s = (1/2)gL2/v2 = (1/2)gL2/(2Ve/m) = gmL2/(4Ve) If V = 1 Volt, and L = 1 metre: s ≈ 10 x 10-30/(4 x 2 x 10-19) ≈ 10-11 metres Sadly this appears to be too small to detect by conventional methods.

The result varies as the square of the tube length and inversely with the accelerating voltage. The minimum voltage will be determined by the thermal distribution of electron velocities as they leave the cathode of the electron gun.

If the cathode temperature is T Kelvin, the electrons will have a range of velocities about a median which is the equivalent of a voltage Vc where

eVc = k.T where k = Boltzmann's constant = 1.3806488 x 10-23 J/K so that Vc = kT/e For a cathode temperature of say 500 K Vc ≈ 1.4 x 10-23 x 500/(2 x 10-19) ≈ 0.035 Volt So if we dropped the voltage to say 0.1 Volt and lengthened the tube to 1 Kilometre, the sag of the electron beam would become s ≈ 10-11.106/0.1 = 10-4 m = 0.1 mm The curvature of the electron beam by the magnetic field of the earth can be minimised by shielding it with a one kilometre long mu-metal tube aligned along the local field lines of force.

Conclusion

Both methods described here are at the limit of feasibility. The Gravimetric Chemistry method is probably feasible with available equipment and materials, but would require very pure reactants, ideally 99.9999% pure, very careful handling, and possibly the use of inert atmospheres. Determining the weight of electrons by the Beam Sag would require the building of extensive and expensive apparatus. The result of either experiment may be that Einstein's Principle of Equivalence is true. Then nothing would change. But it may be that the result would disprove the Principle, which might then lead to an understanding of how to reconcile the two main incompatible current theories of theoretical physics: Quantum Theory and Relativity Theory. That tantalising posibility may make it worthwhile to undertake one of these experiments.

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written 26 August 2014, revised 1 May 2020