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So this article got me thinking: https://www.scientificamerican.com/article/do-electric-charges-and-m/

It states that according to GR, the energy in an electric field should curve spacetime and therefore produce a gravitational effect. And so I started wondering if a charged capacitor would weigh more than an uncharged capacitor, meaning the electric field would be adding gravitational pull to the capacitor.

It seems that it must since it is curving spacetime itself. However, to my understanding electric fields would follow the null geodesic, like light, and be bent by curved spacetime, not accelerated by it.

So to be clear, I'm not talking about the gravitational effect FROM the capacitor. The article above makes it clear the charged capacitor would exert more gravitational pull on objects around it. I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth.

Qmechanic
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HardlyCurious
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    I'm confused about what you mean by "get pulled harder". For example, a hammer has more weight than a feather, but if you drop them from the same height, they'll hit the ground at the same time – Níckolas Alves Nov 22 '21 at 06:49
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    Sure they accelerate at the same speed. But if you put them on a scale they have different weights, and that is the property I'm asking about. – HardlyCurious Nov 22 '21 at 13:59
  • Yes. e=mc^2. The electricity has mass, so it has weight, so a charged capacitor weighs more. (I know there are real answers, and this "answer" is a cop out, but I love how many questions that formula can answer, quickly and simply, when you attack a problem critically from that perspective, so I do think it's worth adding a note) – TCooper Nov 22 '21 at 22:55
  • @TCooper I've heard that professional relativists, since about 1970, no longer use "mass" and "energy" to mean what Einstein did, instead they mean "rest mass" or "invariant mass" to use the favored term, and "total energy including rest mass", respectively. – Matthew Christopher Bartsh Jul 13 '22 at 17:45
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    @MatthewChristopherBartsh First, thank you, as a layman I'm interested but ignorant (generally speaking). Does your comment imply that with a sufficiently accurate scale a charged capacitor would not have a different weight than the same capacitor without a charge? – TCooper Jul 13 '22 at 19:19
  • @TCooper The question asks about the field, not about a comparison between a charged and an uncharged capacitor. A charged capacitor would probably have different weight quite apart from the field, because the charge carriers have mass and therefore weight. The field is a lot less tangible than the charges that create it, and for all I know is inseparable from them, so you question is very good one. AFAIK know, there's no way to instantly remove or destroy the charge and watch what happens to the field, and the weight in the time that follows. How about quickly crushing +ve and -ve together? – Matthew Christopher Bartsh Jul 13 '22 at 20:34

5 Answers5

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I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth.

General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) and thus the energy stored by the capacitor is $\frac{1}{2}\epsilon_0 E_0^2$ and by $E=mc^2$ (assuming the capacitor to be at rest) the mass of the capacitor will increase to $m_{new}=m_{initial}+\epsilon_0\frac{E_0^2}{2c^2}$ meaning by Newtonian gravity it will have a greater weight.

However, to my understanding electric fields would follow the null geodesic, like light, and be bent by curved spacetime, not accelerated by it.

First of all in the realm of General Relativity, any object solely under the influence of gravity follows geodesics and is unaccelerated irrespective of whether the geodesic is null or timelike. Light follows null geodesics in the geometric optics approximation. In case of electric/magnetic fields no path as such is followed we are just concerned with evolution of the field values at any given point.

anna v
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user7896
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  • are you mixing up E for Electric field and E for Energy? also the famous relativistic mas needs a velocity to appear https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass – anna v Nov 22 '21 at 10:03
  • Yup I messed up the notation and used the same letter for both Energy and Electric field. Fixed it now. – user7896 Nov 22 '21 at 10:06
  • While I understand that technically what occurs in GR is not acceleration due to gravity, at the end of the day massive objects gain energy when falling into gravity as an increase in the magnitude of their velocity, but light gains energy through a shorter wavelength. If an electric field was permitted to fall toward earth, what property of it would reflect the increase in energy as it descended? – HardlyCurious Nov 22 '21 at 12:50
  • Energy is not a Lorentz invariant it depends on the frame of reference. And as I mentioned, fields don't fall, so there is no such thing as a falling electric field. However the energy density of the field at a certain point can be computed. For this one can construct the stress-energy tensor of the field. The $T_{00}$ component of the tensor will give you the energy density as measured in your frame of reference. The form of the stress energy tensor can be found here – user7896 Nov 22 '21 at 17:07
  • Nitpick: the $h_{00}$ component of the metric responds to the quantity $(T_{00} + \frac{1}{2} T_\sigma{}^\sigma)$, not just $T_{00}$. For conventional matter the difference is negligible, but for an electric field the $T_i {}^i$ components are comparable in magnitude to the energy density, and so they need to be accounted for. – Michael Seifert Nov 22 '21 at 19:59
  • @MichaelSeifert I am just talking about the energy density of a field as measured by an observer with four velocity $u^\mu$. This is given as $T_{\mu \nu}u^\mu u^\nu$. Hence in the observers frame of reference $u^\mu=(1,0,0,0)$ and the energy density is $T_{00}$ – user7896 Nov 23 '21 at 03:19
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Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance.

Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be

$$ U = \frac12 C V^2 = \frac12\times10^6 \rm\,J $$

This is an awful lot of energy for a capacitor, but the gravitational field it creates will correspond to a mass

$$ m_\text{effective} = U/c^2 \approx 10^6\,\mathrm{J}/c^2 \approx 10^{-11}\rm\,kg \approx 10\,ng $$

Supercapacitors are pretty amazing these days, but a farad-scale capacitor starts out at a mass of a few grams, not a few nanograms. And a small farad-scale capacitor that can actually hold a kilovolt? That’s a nontrivial challenge. The nanogram correction to the mass of our hypothetical supercapacitor, due to its electric field, is a part-per-trillion correction at the most. (This is an order-of-magnitude estimate; see the comments below for one factor-of-two correction.)

For what it’s worth, the electric-field correction to the mass of the hydrogen atom, a binding energy of $-13.6\rm\,eV$ on a GeV-scale mass, is a part-per-billion correction.

Why have I written that the electric fields in a capacitor add to its weight, when in the case of the hydrogen atom the binding energy is negative? The relativistic mass of the charged capacitor is actually less than the relativistic mass of the two charged plates: you would have to apply work to the system to pull the plates apart. But the relativistic mass of the charged capacitor is more than the combined mass of the two neutral plates. You can reduce the system’s effective mass by shorting the capacitor and allowing the charges to recombine, confining the strong electric fields to the atomic scale instead of the capacitor-gap length scale.

rob
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  • So, what mechanism is there to convey the 'weight' of the electric field to the charged plates? In the case of two charged plates oriented so the gap between them was parallel to the pull of gravity, do the field lines between them sag slightly so that the fields have a slight downward direction at the point of contact with the plates? – HardlyCurious Nov 22 '21 at 12:57
  • The “mechanism” is that the stress-energy tensor dictates the curvature of spacetime. Rather than sagging like a chain, I would expect that the field lines anti-sag, because that’s the direction of the geodesics followed by light rays. There might be literature about the shape of the electric field in the Kerr-Newman metric. – rob Nov 22 '21 at 13:55
  • I was thinking about that 'anti-sag' shape for the same reason. But since the electric field lines exert a pull on the plates, wouldn't this produce force in the opposite direction of gravity? – HardlyCurious Nov 22 '21 at 14:07
  • If the capacitor is stationary (e.g. sitting on a table) then there are already forces on it opposing its gravitational fall. And if the capacitor is undergoing accelerated motion, there is a magnetic field where the fringes of the electric field are changing. The whole picture gets complicated quickly. – rob Nov 22 '21 at 14:50
  • So I think we only need to be concerned with the stationary case. And I'm not concerned with why the capacitor isn't falling through the table, I know that is because of the normal force from the table. – HardlyCurious Nov 22 '21 at 15:25
  • If the field lines symmetrically sagged or anti-sagged, the angle between the field lines and the plates would produce an upward force on one plate and a downward force on the other. That would in principle produce a torque on the capacitor. It's not immediately clear to me whether this torque can be nonzero without violating charge-conjugation symmetry ($C$ symmetry). But of course in a capacitor whose (matter) plates are identical when uncharged, the negative plate is slightly more massive than the positive plate because of its electron excess, so there is a torque in that situation, too. – rob Nov 22 '21 at 16:51
  • Now I am reminded of this unanswered question of mine and I wonder how "uniform" the electric field in the gap could be, even after you've accounted for known experimental issues like the patch effect. – rob Nov 22 '21 at 16:55
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    Nitpick: unlike the stress-energy tensor for non-relativistic matter, the stress-energy tensor for an electric field has significant "pressure" components. If I've done my sums correctly, the net effect is that the gravitational potential produced by the electric field in the capacitor is twice as large as one would expect from its energy density alone. But 20 ng is still negligible relative to the mass of the capacitor itself. – Michael Seifert Nov 22 '21 at 19:52
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This was tested experimentally by Kreuzer, Phys. Rev. 169 (1968) 1007. The "capacitors" in this experiment were actually atomic nuclei. The electric field of a heavy nucleus makes a fairly significant contribution to its own weight, which is easy to measure. Kreuzer used a Cavendish balance to test whether this was also true for the active gravitational mass, as required by the equivalence principle.

user320000
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    I think I found the study you referenced. https://journals.aps.org/pr/abstract/10.1103/PhysRev.169.1007 but the abstract doesn't sound related to electric fields as you describe... Unfortunately I can't access the full paper. – HardlyCurious Nov 22 '21 at 18:33
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Energy and mass are equivalent. If the electric field stores energy, then it stores mass. Weight is proportional to mass, other things being equal, so the electric field will contribute to the weight of the capacitor.

Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal.

It's really very simple, as explained in "Relativity Visualized" by Lewis Carroll Epstein.

N.B. It seems that professional relativists have been using a new convention regarding the terms "mass" and "energy". The physical laws are the same, so Epstein was not wrong, but it seems that his use of "mass" and "energy" (which matches Albert Einstein's use which is also out of date, it seems) is now frowned on as dated by professional relativists, or at least the majority of English speaking ones.

I will let my answer above stand, because it's not wrong, but allegedly uses "mass" and "energy" in a dated way. However I will add another version that uses terms that cannot be faulted by anyone.

In the old convention (Einstein's) mass and energy were the same thing. But in the new convention used by professional relativists since about 1970, "mass" means "rest mass" and "energy" means "total energy including rest mass".

If the electric field stores energy, or if there is some extra energy in or attached to the capacitor or its field or charge in any way, that extra energy means extra weight. The weight is given my W = mg and in this case the part called "m" is replaced by "E" for "energy", where E is the total energy including all forms of energy including KE and mechanical energy and including the rest mass. So we have W = Eg where E is the total energy in kilograms.

Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal.

A simple and crystal clear explanation of this can be found it Lewis Epstein's wonderful book, "Relativity Visualized", but be warned that he uses the terms "mass" and "energy" in the way relativists including Einstein used them up until about 1970, it seems. It's not a problem, an in fact it might even be a strength as the new conventions are highly confusing, to me for one. One reason it is so confusing is that instead of using terms of art that are clear to all physicists and intelligent nonphysicists, such as "rest mass" (professional relativists call it just, "mass") and "total energy including rest mass" (they just say, "energy") they use in effect a private language, AKA "jargon" that not the majority of fellow physics graduates understand. In fact, the majority of them misunderstand it.

HardlyCurious raised a great point in his comment, which is that it is far from clear how the weight of the energy of the electric field pushes down on the capacitor. I don't know how, but I've read that it's figurative to say the energy of a capacitor is in the field.

Matthew Christopher Bartsh
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  • Sure, I get the basic concept that mass and energy are the same thing... However I still don't really understand a couple points here. 1. How does an electric field follow the null geodesic if it has mass? And 2. What mechanism for the mass of the electric field to push down on the capacitor? I made a follow up question about the field lines in a capacitor to try to get to the bottom of that. https://physics.stackexchange.com/questions/678640/how-does-a-static-electric-field-behave-in-curved-spacetime – HardlyCurious Nov 22 '21 at 20:47
  • @HardlyCurious It seems "mass" and "energy" have new meanings since 1970 for relativists. See the edit to my answer for details, by in short a relativist wouldn't say "mass and energy are the same thing", now. No idea about null geodesics. In the new terminology, it is the (total) "energy" that has weight (without having "mass" in the new sense). Nevertheless it is associated with / attached to the capacitor. How does it push down on the capacitor and thus add to its weight? Great question. I don't have the answer, but I think I read somewhere that the energy stored in the capacitor may only – Matthew Christopher Bartsh Jul 13 '22 at 17:35
  • @HardlyCurious be figuratively stored in the field, while actually being stored in the positions of charge carriers relative to the capacitor. That still doesn't explain how the capacitor gets pushed down, but it's worth bearing in mind. – Matthew Christopher Bartsh Jul 13 '22 at 17:37
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IMO No. The electromagnetic energy from the source is used to coherently rearrange the existing charges in the circuit. It does not add more matter or energy in this open system. The electric potential energy of the source is transformed in kinetic energy of the moving charges and all of this energy is expelled out of the system as heat. After the capacitor is fully charged there is no more work done in the system. To prove conclusively that the capacitor gained mass after it was fully charged you must prove that there is an excess energy given by the source minus the expelled heat from the electrons motion that is stored in the capacitor. Notice the stored electromagnetic potential energy is not given by the source but sources from the electrons which where there in the capacitor before the source was applied. No extra energy was stored in the capacitor.

Part of the capacitor's own matter energy was made more coherent (less entropy). You use the energy given by the source to make the capacitor with less entropy on its matter field (polarization) and this energy given by the source is then all transformed into heat and expelled out of the system.

The confined energy of the capacitor remains the same no extra energy is stored in the capacitor given from the source. The extra energy is all expelled out as heat.

Markoul11
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    It's not an opinion question. – hobbs Nov 22 '21 at 15:57
  • @hobbs Please reference an experimental result in the literature that verifies the gain in weight of a capacitor after it is charged? The question was practical and not theoretical. If experiment does not verify the theoretical analysis, the analysis is incomplete. – Markoul11 Nov 22 '21 at 16:51
  • That's like saying that because nobody has directly, specifically measured a change in the mass of the Earth when we build some of the Earth's matter into a rocket and launch it into space, no such change occurs. – user2357112 Nov 22 '21 at 18:07
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    Nanogram mass changes, to an object with a base mass in the grams, should be measurable with current technology. But the real problem with this answer is that the assertion that "no extra energy is stored in the capacitor" is just wrong. – zwol Nov 22 '21 at 20:30
  • @zwol The capacitor expels electromagnetic energy from it own matter field during its discharge going from a low entropy state to a high entropy state of its charges. Its matter field energy is conserved all times and replenished by its charges (free electrons) and not by the voltage source in the circuit. The voltage source is only used to make work by setting the already existing charges in the capacitor from a high entropy to a low entropy state (polarization) and is then transformed and expelled out of the system as heat. The electrons conserve their rest mass all the time. – Markoul11 Nov 23 '21 at 07:34
  • @Markoul11 I think you may be insisting on an overly narrow definition of "the capacitor". Yes, the energy stored by a capacitor is stored in the electric field near the plates, rather than in excited states of the atoms of the plates or anything like that. But the energy in the field will still contribute to the mass you observe if you put the charged capacitor on a sufficiently sensitive balance. We know this because field energy in a confined space is what gives protons some 2/3 of their observed mass. – zwol Nov 23 '21 at 13:40
  • @Markoul11 If that's not what you meant, then you're going to have to explain yourself better, because the only other possible interpretation of what you're saying that I can see is "capacitors don't store energy" which is just not true. – zwol Nov 23 '21 at 13:43