# Active gravitational mass of the electron

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?

from here activ/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.

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""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. –  Georg Mar 9 '11 at 18:06
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. –  Luboš Motl Mar 9 '11 at 18:11
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. –  Luboš Motl Mar 9 '11 at 18:12
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. –  user566 Mar 9 '11 at 20:52
@all - I hope that the question is more clear now. –  Helder Velez Mar 10 '11 at 1:06

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.

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I'm sorry for the inconvenience of the not so clear question and thanks for your answer. Your 1st paragraph is about what I asked. –  Helder Velez Mar 10 '11 at 1:22

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

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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.

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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. –  Helder Velez Apr 20 '11 at 3:11
Why would a neutron beam deflection tell you about the electron's gravitational field? –  rob Jul 27 at 13:01
@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. –  Helder Velez Jul 28 at 12:52
the program APOLLO, using improved Lunar Laser Ranging is working in this subject. The results are model dependent. –  Helder Velez Jul 28 at 13:04

## protected by Qmechanic♦Jul 27 at 11:33

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