In PSE here electrons are added to a sphere and gravitational modifications are expected.

My question is:
Is there any experiment that show that a negatively charged object is source of a stronger gravitational field than the same uncharged object?

In other words:
The active gravitational mass of an electron is equal to his passive gravitational mass by experiment?

added, for clarification:
from here active/passive grav mass:
- active gravitational mass: establishes the field
- passive gravitational mass: responds to the field
- by experiment: a carefully designed setup that evidences how electrons interact in the presence of a gravitational field. Does it exists already? If no, is it doable? etc.

  • $\begingroup$ ""active gravitational mass of an electron is equal to his passive gravitational mass"" Please, what do you mean with this active and passive? I only know about mass. And I doubt that anybody ever has measured the gravity of an electron, its just too small compared to Coulombs forces. $\endgroup$
    – Georg
    Mar 9, 2011 at 18:06
  • 2
    $\begingroup$ The identity of an object surely includes its charge, so if the charge of object A is different than the charge of object B, it follows that A and B "cannot be the same". There is no one-to-one map between charge and uncharged objects, certainly not one that would universally preserve the mass. So your question makes no sense. Also, there are no "passive" and "active" masses. If you ask why the mass given by the "strength of gravitational field" is the same as the mass "how much an object responds to an external g. field", it's because of momentum conservation. $\endgroup$ Mar 9, 2011 at 18:11
  • $\begingroup$ And if you were asking why the inertial mass is the same as the gravitational (active or passive) mass, it's because of the equivalence principle. It's been tested by showing the same acceleration of all objects in the gravitational fields. Also, if you ask why electrostatic potential energy influences mass, uranium bombs are an example because the fission energy comes primarily from a reduced positive electrostatic potential energy after the big uranium nucleus decays. $\endgroup$ Mar 9, 2011 at 18:12
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    $\begingroup$ I think the question cannot be answered before clarifying what is "active gravitational mass", and "the same uncharged objects", both of which are highly problematic and probably cannot be made well-defined. People already started speculating what the PO might mean, maybe it is up to him to clarify. $\endgroup$
    – user566
    Mar 9, 2011 at 20:52
  • $\begingroup$ Comment to the question (v3): OP writes: [...] Is there any experiment that shows that a negatively charged object is source of a stronger gravitational field than the same uncharged object? In other words: The active gravitational mass of an electron is equal to his passive gravitational mass [...] How can an electron be stripped of its charge to compare? Phys.SE deals only with mainstream physics. $\endgroup$
    – Qmechanic
    Jul 27, 2014 at 13:38

4 Answers 4


I think you'll be unsatisfied with an answer about the gravitational field of the electron--to my knowledge, no one has tested anything involving the gravitational field created by microscopic particles. The closest we've come is tests of gravitational redshift involving scattering microscopic particles in the external field of the Earth.

There, is, however, a known solution to Einstein's equation and Maxwell's equations that represents a black hole with a nonzero charge, known as the Reissner-Nordström metric, given as:

$$ds^{2} = - f dt^{2} + \frac{dr^{2}}{f} + r^{2} d \Omega^{2}$$

Where $f=1-\frac{2M}{r} + \frac{q^{2}}{r^{2}}$. In the context of this metric, there is a difference in the gravitational field induced by the charge from what it would be without the charge--you get the $r^{2}$ variation in what becomes the gravitational potential function for timelike geodesics. This is, in principle, measurable (and a failure to measure it would be a contradiction of either Maxwell's equation or General Relativity, which are both stringently tested--the former moreso than the latter).

One, however, must be careful with what one means by 'active gravitational mass' of a system like this--the ADM Mass of this system is still $M$, and is not modified by $Q$, even if particles near the black hole feel different forces due to the presence of the charged particle.

Finally, as a bit of a interesting aside, it should be noted that the presence of a charge moves the location of the horizon to the location $r_{\pm}=M \pm \sqrt{M^{2}-Q^{2}}$, so there is no horizon at all if $Q>M$. It turns out that if you put the values for the electron mass and charge into this equation, you will find that it predicts that the electron should be a naked singularity.


The question was motivated because I have a suspicion that the electron does not participate to the source of a gravitational field, and eventually not even responds to such a field.

In 1908 Milikan measured the charge on a single electron. The charge-to-electron mass ratio $q/m_{e}$ was calculated by Thomson in 1897 using the angular momentum and the deflection due to a perpendicular magnetic field. I think that this value reflects the inertial mass.

I found this old (1964) doc Gravitational and resonance experiments on very low-energy free electrons by Fairbank, and this entry Experimental Comparison of the Gravitational Force on Freely Falling Electrons and Metallic Electrons by WittBorn and Fairbank, 1967, with the abstract:

A free-fall technique has been used to measure the net vertical component of force on electrons in a vacuum enclosed by a copper tube. This force was shown to be less than 0.09mg, where m is the inertial mass of the electron and g is 980 cm/sec2. This supports the contention that gravity induces an electric field outside a metal surface, of magnitude and direction such that the gravitational force on electrons is cancelled.

It seems that the issue remains unsettled. - Docs of 1992 and 2007 - (Tests of the weak equivalence principle for charged particles in space)

Fairbank, as everyone else back then, and even now, believed that the electron must participate in gravity, contrary to my suspicion, preferring to imagine the existence of an imaginary induced electric force, possibly because in the 50s and 60s existed some hype about possible effects relating electricity and gravity.

Experimentation is central to advancement of Physics, and the solution of this unsettled issue may prove important in the genesis of a sucessful theory encompassing both the so called particles and the gravitation field.

I can understand that performing the test of Fairbank under microgravity conditions can discriminate if the electron responds to the gravity field.

To test if the electron has an active gravitational mass a possible test that may work ('I do not know the tecnological limits...') could be done using a collimated beam of slow neutrons with a path traversing between the plates of a very large high-voltage capacitor. The possible deflection, or not, of the beam may prove very interesting.

quoting from CERN: "AGAIN and AGAIN: This is an unsolved experimental problem!!!", "IMHO THIS REMAINS AN UNSOLVED PROBLEM WARRANTING EXPERIMENTAL ATTENTON" review and literature inside - 2015

  • $\begingroup$ A test with electric field and ultra cold neutrons was done now Nature-Realization of a gravity-resonance-spectroscopy technique somewhat similar to the one I've suggested. $\endgroup$ Apr 20, 2011 at 3:11
  • 1
    $\begingroup$ Why would a neutron beam deflection tell you about the electron's gravitational field? $\endgroup$
    – rob
    Jul 27, 2014 at 13:01
  • $\begingroup$ @rob: the electronic distribution is not symmetric in a large condenser and any possible deviation of the neutrons path (by reversing the polarity) could be attributed to gravity, and slow to maximize the exposure to the field. $\endgroup$ Jul 28, 2014 at 12:52
  • $\begingroup$ the program APOLLO, using improved Lunar Laser Ranging is working in this subject. The results are model dependent. $\endgroup$ Jul 28, 2014 at 13:04

According to General Relativity, energy is equivalent to inertial mass, and all inertial mass generates gravity. Since electrons have measureable inertial mass, they should have a small influence 0.1% contribution to the gravitational force in neutral matter. That being said however, the question cannot be answered experimentally because the electrical coulomb force is on the order of $10^{30}$ times greater than the gravitational force for particles, hence comparing these in the lab is not feasible since we have no experiments that could reduce electrical noise to this degree. For particles, this type of question has only been partly answered for slow neutrons, which are electrically neutral. It is a very interesting question regardless, because some theories speculate charged particles are some kind of quantized topological alterations in space-time, hence their motion within curved spacetime (i.e. a gravitational field) could theoretically vary. These questions hold for antimatter as well, and could effect other questions about the nature of dark energy and dark matter. Some references for measurement of lepton (i.e. electron) gravity are below.

C.S. Unnikrishnan. G.T. Gillies "Do leptons generate gravity? First laboratory constraints obtained from some G experiments and possibility of a new decisive constraint" Physics Letters A 288:161-166, 2001.

Homer G. Ellis Leptons might not generate gravity http://lanl.arxiv.org/abs/gr-qc/0308082


The question you're asking is tantamount to asking what the gravitational field of a single particle is.

There are two directions of interaction for a force or field, including gravity. The "forward reaction" is how the body responds to the given force or field. For gravity and inertia, together, the combined effect - for a given space-time metric - is given by the "geodesic equation" associated with that metric.

The Equivalence Principle, in the form originally stated by Galileo, asserts that the "forward reaction" for gravity is independent of the composition of the body: that the response to gravity is proportional to the body's response in the law of inertia. The principle, in the form re-instated by Einstein, goes one step further to assert that the law of gravity, itself, is but a warped form of the law of inertia. That statement is made in the form of the single equation - the geodesic equation - to account for them both, as one. So, it asserts the equivalence of mass and passive gravitational charge.

The other direction of interaction is "back-reaction"; which pertains to how the body's presence affects, molds or modifies the force or the field. On account of this, technically, the equivalence principle isn't quite true. A pin dropped onto the ground will fall a little bit differently than a space station as large as the moon dropped onto the ground would, because the space station is going to be pulling the ground up a lot harder. That's back-reaction at play.

The back-reaction of the body defines its active gravitational charge.

For gravity, back-reaction is determined by Einstein's field equations, which substantially modifies the pre-relativistic back-reaction law, which was Newton's inverse square force law.

If you play fast and loose with the math, you can sorta derive the geodesic law from Einstein's field equations ... namely by taking the (singular) stress tensor for a source concentrated on a world line and applying the continuity equation to the stress tensor; the continuity equation being both a pre-condition for the formulation of Einstein's field equations and derivable from it as a corollary. So, in that sense the equivalence of the active and passive gravitational charges is mandated by the theory... but not necessarily by nature. It still has to be tested, and those tests are done regularly, too.

The back-reaction described by Einstein's field equations pertains to matter such as would be described from the physics predating Quantum Theory - "classical physics". It is a back-reaction law only for classical matter. There is no known back-reaction law for matter such as would be described by contemporary theories of matter, which are all grounded in Quantum Theory. That's a gap.

A lot of things could (and do) lie in that gap which defy contemporary understanding by anyone in this world today. That's why Congress entered the survey blurb

Advanced Space Propulsion Based on Vacuum (Spacetime Metric) Engineering

into its records at the start of the subcommittee hearing on UAP's in July of 2023.

There isn't even so much as a back-reaction law for single particle, such as the photon or electron, nor any consensus (at least in the case of the photon) what such a thing would even mean, since the photon's electromagnetic field is a q-number field. The same goes for the other fundamental bosons, including the gauge bosons and Higgs boson.

In the case of the fundamental fermions, however, this much we can say: a solution to Einstein's field equations that has the same mass, same total charge (meaning to combination of all charges, for all the gauge forces, not just electro-magnetism, so you need to be talking about the Einstein-Yang-Mills-Higgs equation, not Einstein-Maxwell equation) and same angular momentum as given by their intrinsic (spin) angular momentum is a

... (wait, wait) ...

Kerr-Newman Solution

Moreover, it's not just any kind of singularity, but is a ...

Naked Singularity

specifically: a naked Kerr-Newman ring singularity; way beyond the threshold, not even close; and so much so that they had to call in the science police for its indecency.

That's semi-serious, too. It is a total violation of the...

Chronology Protection Conjecture

So, it's not a black hole at all, since by definition, these are sheathed by event horizons, but more like a mini-stargate.

That's even the case for the left neutrinos and right anti-neutrinos. They may be electrically neutral, but in case anyone forgot, they have weak nuclear force charges.

It may or may not be the case for the right neutrinos or left anti-neutrinos - if they exist; but their existence is not established since they have no electric charge, weak nuclear charge or strong nuclear charge and therefore almost nothing to see them with - except via the Higgs field and gravity, and possibly except through their "baryon minus lepton" (B-L) number, if this proves to be a charge for a heretofore unverified gauge force. If there is a B-L force, and a corresponding charge, then the right neutrino and left anti-neutrino would also fall into the naked singularity category with the other fundamental fermions. Otherwise, a Kerr-Newman solution matching the stats of a right neutrino or left anti-neutrino might not be naked, but just a regular rotating, uncharged black hole (a Kerr solution).

Being a naked singularity means there is no event horizon, no ergosphere, and the singularity is directly exposed to and seen by everyone. That also means no Hawking evaporation, since that's predicated on the sheathing by an event horizon. It also means there is causality violation near the singularity and indeterminacy - something that (in retrospect) looks suspiciously like it could serve as a classical underpinning to quantum indeterminacy - an Einstein's Revenge Scenario.

In that vein, though I don't know enough about


in my mind, it raises the question: could such singularities also serve as an underpinning to this?

All of this, of course, raises the question: is that what the fermion actually is? Here are the two cases:

(1) The actual fermions we observe, measure, track in bubble chambers, irradiate CRT screens with, are totally different than these solutions. But if so, that raises the questions: (a) what else would it be if not this, and (b) shouldn't we already be seeing the discrepancies between the two models soon (something we could measure or experimentally test for), since we're talking about the Compton scale?

... or ...

(2) Fundamental fermions are indeed Kerr-Newman ring singularities. In that case, there should already exist some kind of deep correspondence between the Dirac equation and its solutions versus the Kerr-Newman solution.

So, does such a correspondence exist?

The answer is yes - very much so!

And by "very much so", I mean not just at the level of the semi-classical theory of first quantization, but even broaching the level of second quantization, also being able to account for some of the results therein.

The Dirac - Kerr-Newman Electron

It's one of those things that's been sitting in the attic that nobody's really been paying attention to, on account of the attention having been drawn away from this (the question of single-particle back-reaction / gravity fields) by the distraction of other things, such as Loop Quantum Gravity or String Theory, that may (and probably will) prove to be red herrings.


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