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

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

Yes, of course

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And this equation can be rewritten in the following beautiful form

E = |P4|c

|P4| - four-vector momentum module

Accordingly, m0c is the projection of the four-vector momentum on the ct axis

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On 2/5/2022 at 2:34 AM, SergUpstart said:

Accordingly, m0c is the projection of the four-vector momentum on the ct axis

...which of course vanishes for photons. Given all that has been said, in what way is it meaningful or consistent to talk about the ‘mass of an EM field’? Such a concept creates far more problems than solutions. It’s simply not helpful.

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4 hours ago, Markus Hanke said:

Given all that has been said, in what way is it meaningful or consistent to talk about the ‘mass of an EM field’? Such a concept creates far more problems than solutions. It’s simply not helpful.

The momentum of a particle is conventionally represented by the letter p. It is the product of two quantities, the particle's mass (represented by the letter m) and its velocity (v):[1]

p=mv.

According to this definition of momentum, a particle with zero mass can have a momentum other than zero only when moving at an infinite greater speed. Naturally, here we are talking about the total mass M, and not about the rest mass m0.

Edited by SergUpstart
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58 minutes ago, SergUpstart said:

The momentum of a particle is conventionally represented by the letter p. It is the product of two quantities, the particle's mass (represented by the letter m) and its velocity (v):[1]

p=mv.

According to this definition of momentum, a particle with zero mass can have a momentum other than zero only when moving at an infinite greater speed. Naturally, here we are talking about the total mass M, and not about the rest mass m0.

How about not ignoring relativity? We know that massless particles move at c and have momentum E/c

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21 minutes ago, swansont said:

How about not ignoring relativity? We know that massless particles move at c and have momentum E/c

Where am I ignoring special relativity?

For massless particles, the rest mass is 0 and the total mass M = E/c^2, respectively, the momentum p =E/c = Mc. Everything is in accordance with both SRT and the definition of momentum.(quoted from wikipedia)

Einstein even distinguished between longitudinal and transverse masses

Edited by SergUpstart
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Quote

If we briefly call this force as "the force acting upon the electron," and maintain the equation :—

,

and if we further fix that the accelerations are measured in the stationary system K, then from the above equations, we obtain :

[1]

Naturally, when other definitions are given of the force and the acceleration, other numbers are obtained for the mass ; hence we see that we must proceed very carefully in comparing the different theories of the motion of the electron.

And, in the 115+ years since then, the physics has proceeded very carefully and has found a clear and efficient way to deal with these concepts. The "mass" is generally a term for the invariant mass, an inherent property of a particle (or, a specific coefficient in a QFT Lagrangian), i.e. "the rest mass". The "other numbers [] obtained for the mass" are not in use, nor are the "other definitions [] of the force and the acceleration".

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Exactly as @Genady says. Taylor-Wheeler's Spacetime Physics was instrumental in making the whole thing more transparent. For all I know Einstein formulated these ideas very much like groping attempts towards the right concept. Not even Einstein himself stuck to these longitudinal and transversal masses. It's all taken care of by the relativistic definition of momentum.

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And, in the 115+ years since then, the physics has proceeded very carefully and has found a clear and efficient way to deal with these concepts. The "mass" is generally a term for the invariant mass, an inherent property of a particle (or, a specific coefficient in a QFT Lagrangian), i.e. "the rest mass". The "other numbers [] obtained for the mass" are not in use, nor are the "other definitions [] of the force and the acceleration".

Then it is necessary to change the definition of momentum, since it contradicts the fact that mass is a rest mass. So what is momentum if not p=mv?

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12 minutes ago, SergUpstart said:

Then it is necessary to change the definition of momentum, since it contradicts the fact that mass is a rest mass. So what is momentum if not p=mv?

There are more than one way to arrive to a definition of momentum, e.g. as a conserved quantity associated with space translation invariance as per Noether's theorem. But as a conserved quantity in the Lorentz covariant 4-momentum for a massive particle, it is

p=mγv

"p(v) = mγ v = mv[1 + v2/2c2 +···]." (Weinberg, Steven. Foundations of Modern Physics (p. 109).)

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18 minutes ago, SergUpstart said:

Then it is necessary to change the definition of momentum, since it contradicts the fact that mass is a rest mass. So what is momentum if not p=mv?

It was changed in time immemorial --to me at least--. It is:

$p^{\mu}=m\frac{1}{\sqrt{1-v^{2}/c^{2}}}\left(c,\boldsymbol{v}\right)$

And you've been here for long enough that you should know about it already.

But I don't think it contradicts a fact; it contradicts a definition.

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8 hours ago, SergUpstart said:

Where am I ignoring special relativity?

particle with zero mass can have a momentum other than zero only when moving at an infinite greater speed”

Infinite speed is Newtonian and not possible in SR. The latter limits speeds to c, which is the speed of massless particles

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Forgot to mention that the definition that both @Genady and I mentioned is not valid for a massless particle. For such massless particle the definition must be changed to,

$p^{\mu}=\left(\frac{E}{c},\boldsymbol{k}\right)$

with the "length" of $$\boldsymbol{k}$$ being,

$\left\Vert \boldsymbol{k}\right\Vert =E/c$

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1 hour ago, joigus said:

k=E/c

To be even more precise, we should add that the last condition is required only for the on-shell i.e. non-virtual massless particles

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To be even more precise, we should add that the last condition is required only for the on-shell i.e. non-virtual massless particles

If we're talking QFT, yes. I think we're talking classical here though. If we're talking QFT, even massive-particle momentum has to be generalised, and off-shell it can take all values when you don't look at them. The moment you want to measure the actual momentum of a virtual particle, you bring it back on shell.

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11 hours ago, SergUpstart said:

So what is momentum if not p=mv?

It’s the conserved current associated with spatial translation invariance under Noether’s theorem - as Grenady has pointed out.

I say it again - it is not helpful to speak of the ‘mass of an EM field’.

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9 hours ago, Markus Hanke said:

It’s the conserved current associated with spatial translation invariance under Noether’s theorem - as Grenady has pointed out.

I say it again - it is not helpful to speak of the ‘mass of an EM field’.

Surely that may depend on what one is talking about. When we charge a battery we say it very slightly gains mass. That mass gain is in the form of chemical potential energy, due to the electrostatic potential of the electrons and nuclei that comprise the atoms or ions involved. Isn't it?  So isn't that mass gain in the form of the electric fields they experience.

Edited by exchemist
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17 minutes ago, exchemist said:

Surely that may depend on what one is talking about. When we charge a battery we say it very slightly gains mass. That mass gain is in the form of chemical potential energy, due to the electrostatic potential of the electrons and nuclei that comprise the atoms or ions involved. Isn't it?  So isn't that mass gain in the form of the electric fields they experience.

These potential energy gains are due to interactions between constituent particles. It's a problem trying to ignore them and assign the mass to the field. The mass belongs to the system as a whole.

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14 minutes ago, exchemist said:

Surely that may depend on what one is talking about. When we charge a battery we say it very slightly gains mass. That mass gain is in the form of chemical potential energy, due to the electrostatic potential of the electrons and nuclei that comprise the atoms or ions involved. Isn't it?  So isn't that mass gain in the form of the electric fields they experience.

Mass is not additive. A mass of the whole is not equal sum of masses of its components. In the particular example, the mass of a charged battery is not equal the sum of mass of the uncharged battery and mass of the electric fields. The mass in this example is gained "in the form of energy" of the electric fields, but not in the form of their mass.

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Mass is not additive. A mass of the whole is not equal sum of masses of its components. In the particular example, the mass of a charged battery is not equal the sum of mass of the uncharged battery and mass of the electric fields. The mass in this example is gained "in the form of energy" of the electric fields, but not in the form of their mass.

Eh? The gain in stored energy is accompanied by a gain in rest mass of the battery, yes or no?

20 minutes ago, swansont said:

These potential energy gains are due to interactions between constituent particles. It's a problem trying to ignore them and assign the mass to the field. The mass belongs to the system as a whole.

Yes, that makes sense, I suppose.

Edited by exchemist
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1 minute ago, exchemist said:

Eh? The gain in stored energy is accompanied by a gain in rest mass of the battery, yes or no?

Yes, the rest mass of the battery, but not the rest mass of the electric fields.

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Yes, the rest mass of the battery, but not the rest mass of the electric fields.

OK, as @swansont says, maybe I can't separate the fields from the particles experiencing them.

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