# Does a Static EM Field Acquire Mass Due to Stored Energy?

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3 hours ago, studiot said:

If you want a classical answer you could consider the decrease in mass due to changing a silicon atom for a boron one in a silicon lattice, since this would decrease the charge by 1,thus reducing the field slightly. Since we are then talking about a solid lattice, momentum would not be involved the simple e = mc2 would suffice.

Not sure that's a great example, as elemental silicon has a giant covalent structure, in which the bonding involves electrons in motion in orbitals shared between atoms, but no doubt one could consider changes to a purely ionic structure that would alter the energy of the lattice and thus its mass. So I do take your point.

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58 minutes ago, exchemist said:

Not sure that's a great example, as elemental silicon has a giant covalent structure, in which the bonding involves electrons in motion in orbitals shared between atoms, but no doubt one could consider changes to a purely ionic structure that would alter the energy of the lattice and thus its mass. So I do take your point.

Maybe not a great example, but I meant it as a ballpark example along the following lines.

Classically, (without QM) bonds are just links of electrostatic origin. That is the bond energy is contained in an electrostatic field of some sort between the atoms.
Si is 4 valent and B is 3 valent.
So one can measure or look up the bond energies.
And also the mass difference between a silicon atom and a boron atom.
So one can get the energy difference in substituting 1 in X silicons by a boron and reducing the bond energy by four Si-Si bonds and adding back three Si-B bonds.
However this substitution will also introduce strain energy into the lattice which will tend to zero as X tends to infinity.

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9 minutes ago, studiot said:

Maybe not a great example, but I meant it as a ballpark example along the following lines.

Classically, (without QM) bonds are just links of electrostatic origin. That is the bond energy is contained in an electrostatic field of some sort between the atoms.
Si is 4 valent and B is 3 valent.
So one can measure or look up the bond energies.
And also the mass difference between a silicon atom and a boron atom.
So one can get the energy difference in substituting 1 in X silicons by a boron and reducing the bond energy by four Si-Si bonds and adding back three Si-B bonds.
However this substitution will also introduce strain energy into the lattice which will tend to zero as X tends to infinity.

Oh I see what you mean. But as there isn't really a classical picture of covalent bonding in chemistry, it's a tiny bit artificial.

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It may be interesting to notice that the EM contribution to mass is negligible in most macroscopic cases. If you plug in the values of $$\varepsilon_{0}$$, $$c^{2}$$ and assume 'typical' values for the field $$\boldsymbol{E}$$ the order of Volts/metre, volumes the order of cm3, you get for this charged macroscopic object a correction to mass of its uncharged state the order of one proton mass or thereabouts.

This is, of course, due to the high value of the speed of light.

Edited by joigus
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8 hours ago, joigus said:

It may be interesting to notice that the EM contribution to mass is negligible in most macroscopic cases. If you plug in the values of ε0 , c2 and assume 'typical' values for the field E the order of Volts/metre, volumes the order of cm3, you get for this charged macroscopic object a correction to mass of its uncharged state the order of one proton mass or thereabouts.

This is, of course, due to the high value of the speed of light.

Sure. My interest in the issue was merely that I often,  when explaining E=mc² to lay people, use the example of charging and discharging a battery, to show that the formula  says mass and energy go hand in hand, rather one being converted into the other, which is what the uninitiated  frequently seem to think, probably due to the association of Einstein's formula with  the mass defect in nuclear fission. And then it struck me suddenly that, while it may seem comprehensible that an object with mass, like a battery, may  in theory gain or lose a tiny amount of mass, it is less obvious what happens to something nebulous and apparently massless, like the magnetic field of a solenoid when it is energised. So I wanted to make sure my way of explaining it covered that case as well.

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They do "go hand in hand" ( @exchemist) but they are different nevertheless. Energy is additive, mass is not. Energy with no mass, such as light, can be added to something and that something will acquire extra mass.

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They do "go hand in hand" ( @exchemist) but they are different nevertheless. Energy is additive, mass is not. Energy with no mass, such as light, can be added to something and that something will acquire extra mass.

Sort of, ish. Light is not energy of course: it has energy (E=pc = νλ), which can be added to that of an entity that absorbs it, which will then gain mass according to E=mc².

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26 minutes ago, exchemist said:

Sort of, ish. Light is not energy of course: it has energy (E=pc = νλ), which can be added to that of an entity that absorbs it, which will then gain mass according to E=mc².

Of course. I should've said "such as that of light" or something similar. Otherwise, yes, this is what I mean.

I have a picture in mind where the light remains light, i.e. is not absorbed in a sense of being converted into something else. It will be there with no mass, but the entity will gain mass still.

minor grammar correction
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Of course. I should've said "such as that of light" or something similar. Otherwise, yes, this is what I mean.

I have a picture in mind where the light remains light, i.e. is not absorbed in a sense of being converted into something else. It will be there with no mass, but the entity will gain mass still.

Hmm, do you mean that a glass prism gains mass if you shine a light through it? I struggle to see how that would work, I must admit.

Unless..........you mean that the coupling of the radiation to the medium "lends" some of its energy to it as it passes through, which I guess it does if its refractive index deviates from unity.

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Just now, exchemist said:

Hmm, do you mean that a glass prism gains mass if you shine a light through it? I struggle to see how that would work, I must admit.

Unless..........you mean that the coupling of the radiation to the medium "lends" some of its energy to it as it passes through, which I guess it does if its refractive index deviates from unity.

No, I just mean mirrors. I take a box with internal walls being mirrors, let light in, say, through a little opening. Light just bounces inside from mirror to mirror (we can close the opening to make sure it doesn't escape). The mass of the box increased.

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• 3 weeks later...
On 1/16/2022 at 9:51 AM, exchemist said:

Hmm, do you mean that a glass prism gains mass if you shine a light through it?

On 1/16/2022 at 9:54 AM, Genady said:

No, I just mean mirrors. I take a box with internal walls being mirrors, let light in, say, through a little opening. Light just bounces inside from mirror to mirror (we can close the opening to make sure it doesn't escape). The mass of the box increased.

I don't think the light has to be confined, and a system consisting of a glass prism with light shining through it should have more mass than just the prism.

As a thought experiment, consider a massive particle at rest, and a photon moving with +x velocity. The photon has no mass, but considering the two particles as one system, this is not its rest frame. In the system's rest frame, the massive particle has some -x momentum balancing the photon's momentum. As a system, the particles' kinetic energy contributes to the system's rest mass.

Likewise the prism+light's rest frame would be different from the prism's rest frame. I don't know if this is meaningful in general, since spacetime curvature depends on how the mass is distributed, and I can't imagine how to describe the effects of unconfined photons. However there are cases where it is meaningful, such as with a "kugelblitz", a black hole created by a dense concentration of light energy for example.

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24 minutes ago, md65536 said:

such as with a "kugelblitz", a black hole created by a dense concentration of light energy for example.

Nice little gizmos to think about relativity.

I totally agree that photons don't have to be confined to be able to assign them a mass. A bunch of photons radiating away from a point would have a mass. If you tried to accelerate the bunch of photons, you would find an inertia. Doing the experiment is another matter, of course.

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If I may ...

"In theoretical general relativity, a geon is a nonsingular electromagnetic or gravitational wave which is held together in a confined region by the gravitational attraction of its own field energy. They were first investigated theoretically in 1955 by J. A. Wheeler, who coined the term as a contraction of "gravitational electromagnetic entity"

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The electrostatic field must have mass. If an electron moves in a field created by another charged particle, then it experiences acceleration due to the action of this field, therefore its momentum changes. In accordance with the Law of Conservation of Momentum, the momentum of the particle creating the field must also change. But the momentum of this particle will not begin to change immediately, but only after a time r / c (r is the distance between the particles). Therefore, it turns out that the first particle should take the momentum from the field and then the field should transmit the momentum to the second particle. But for a field to have momentum, it must have mass.

For these reasons, even 30 years before the creation of the SRT, physicists Heinrich Schramm and Umov obtained the formula E=kmc^2

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55 minutes ago, SergUpstart said:

But for a field to have momentum, it must have mass.

I think it is not particularly helpful to think about the situation in this way, because if you look at the Lagrangian that describes the electromagnetic field, you will find that it does not contain any mass terms - neither in classical field theory nor in QED. Saying that “the field has mass” is thus misleading at best.

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19 minutes ago, Markus Hanke said:

I think it is not particularly helpful to think about the situation in this way, because if you look at the Lagrangian that describes the electromagnetic field, you will find that it does not contain any mass terms - neither in classical field theory nor in QED. Saying that “the field has mass” is thus misleading at best.

+1

1 hour ago, SergUpstart said:

The electrostatic field must have mass. If an electron moves in a field created by another charged particle, then it experiences acceleration due to the action of this field, therefore its momentum changes. In accordance with the Law of Conservation of Momentum, the momentum of the particle creating the field must also change. But the momentum of this particle will not begin to change immediately, but only after a time r / c (r is the distance between the particles). Therefore, it turns out that the first particle should take the momentum from the field and then the field should transmit the momentum to the second particle. But for a field to have momentum, it must have mass.

For these reasons, even 30 years before the creation of the SRT, physicists Heinrich Schramm and Umov obtained the formula E=kmc^2

What happens in this model as r tends to infinity since c is still finite ?

Edited by studiot
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2 hours ago, SergUpstart said:

But for a field to have momentum, it must have mass.

EM radiation is massless. One must conclude the associated fields are massless. But the radiation has momentum.

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15 minutes ago, swansont said:

EM radiation is massless. One must conclude the associated fields are massless. But the radiation has momentum.

Absophotonlutely!

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3 hours ago, studiot said:

What happens in this model as r tends to infinity since c is still finite ?

With r tending to infinity, the force of the static field that the second particle creates tends to zero. So the particle interaction just disappears

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5 hours ago, SergUpstart said:

If an electron moves in a field created by another charged particle, then it experiences acceleration due to the action of this field, therefore its momentum changes. In accordance with the Law of Conservation of Momentum, the momentum of the particle creating the field must also change.

11 minutes ago, SergUpstart said:

With r tending to infinity, the force of the static field that the second particle creates tends to zero. So the particle interaction just disappears

Please clarify which is the first particle and which is the second.

Please also clarify why you did not mention force in the post I queried but now introduce it ?

You actually said

5 hours ago, SergUpstart said:

but only after a time r / c (r is the distance between the particles).

Since c is finite but nonzero there is no division by zero and the ratio tends to infinity of time.

Edited by studiot
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6 minutes ago, studiot said:

Since c is finite but nonzero there is no division by zero and the ratio tends to infinity of time.

I was referring to Coulomb's law, according to which the electric field strength decreases inversely proportional to the square of the distance r

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5 hours ago, SergUpstart said:

For these reasons, even 30 years before the creation of the SRT, physicists Heinrich Schramm and Umov obtained the formula E=kmc^2

I did not know of these physicists, can you provide a reference?
(A quick search gave me nothing of value.)

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49 minutes ago, SergUpstart said:

I was referring to Coulomb's law, according to which the electric field strength decreases inversely proportional to the square of the distance r

Coulomb's law does not include c or mention time, though I grant you that r/c has the dimensions of a time.

You have not answered my more important question as to which particle is which ie is the electron you mentioned the first or second particle.

Without this vital information your two statements are just quotations from gobbledegook.

Edited by studiot
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1 hour ago, Ghideon said:
7 hours ago, SergUpstart said:

For these reasons, even 30 years before the creation of the SRT, physicists Heinrich Schramm and Umov obtained the formula E=kmc^2

I did not know of these physicists, can you provide a reference?
(A quick search gave me nothing of value.)

Unfortunately I can only give a link to the Russian Wikipedia https://ru.wikipedia.org/wiki/Умов,_Николай_Алексеевич#cite_note-7

here is a quote from there in translation

Long before A. Einstein, he discussed in his works[6] the formula {\displaystyle E=kmc^{2}}E=kmc^{2}, derived earlier by Heinrich Schramm[7], which, according to his assumption, connected the mass density and energy of a hypothetical luminiferous ether. Subsequently, this dependence was strictly deduced, without any coefficient k and for all types of matter, by Einstein in the special theory of relativity (SRT). One of the first Russian scientists estimated the value of SRT.

Addition. In the USSR, the Poynting vector was called the Umov-Poynting vector

1 hour ago, studiot said:

Coulomb's law does not include c or mention time, though I grant you that r/c has the dimensions of a time.

You have not answered my more important question as to which particle is which ie is the electron you mentioned the first or second particle.

Without this vital information your two statements are just quotations from gobbledegook.

The electron is the first particle, and the particle in which it accelerates in the electrostatic field is the second particle.

Edited by SergUpstart
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16 minutes ago, SergUpstart said:

here is a quote from there in translation

Long before A. Einstein, he discussed in his works[6] the formula {\displaystyle E=kmc^{2}}E=kmc^{2}, derived earlier by Heinrich Schramm[7], which, according to his assumption, connected the mass density and energy of a hypothetical luminiferous ether. Subsequently, this dependence was strictly deduced, without any coefficient k and for all types of matter, by Einstein in the special theory of relativity (SRT). One of the first Russian scientists estimated the value of SRT.

But in SR the correct equation is not E=mc2 but rather E2-p2c2=m2c4, isn't it? Not even mentioning its different meaning.

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