SergUpstart Posted February 4, 2022 Posted February 4, 2022 On 2/4/2022 at 4:09 PM, Genady said: But in SR the correct equation is not E=mc2 but rather E2-p2c2=m2c4, isn't it? Not even mentioning its different meaning. Expand Yes, of course
SergUpstart Posted February 4, 2022 Posted February 4, 2022 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
Markus Hanke Posted February 6, 2022 Posted February 6, 2022 On 2/4/2022 at 7:34 PM, SergUpstart said: Accordingly, m0c is the projection of the four-vector momentum on the ct axis Expand ...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.
SergUpstart Posted February 6, 2022 Posted February 6, 2022 (edited) On 2/6/2022 at 1:24 AM, 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. Expand 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. https://en.wikipedia.org/wiki/Momentum 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 February 6, 2022 by SergUpstart
swansont Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 5:45 AM, 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. https://en.wikipedia.org/wiki/Momentum 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. Expand How about not ignoring relativity? We know that massless particles move at c and have momentum E/c
SergUpstart Posted February 6, 2022 Posted February 6, 2022 (edited) On 2/6/2022 at 6:46 AM, swansont said: How about not ignoring relativity? We know that massless particles move at c and have momentum E/c Expand 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 https://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies_(1920_edition) Edited February 6, 2022 by SergUpstart
Genady Posted February 6, 2022 Posted February 6, 2022 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. Expand 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". 1
joigus Posted February 6, 2022 Posted February 6, 2022 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.
SergUpstart Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 1:10 PM, Genady said: 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". Expand 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?
Genady Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 1:58 PM, 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? Expand 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).)
joigus Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 1:58 PM, 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? Expand 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.
swansont Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 7:03 AM, SergUpstart said: Where am I ignoring special relativity? Expand “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
joigus Posted February 6, 2022 Posted February 6, 2022 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
Genady Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 4:12 PM, joigus said: ∥k∥=E/c Expand 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
joigus Posted February 6, 2022 Posted February 6, 2022 On 2/6/2022 at 5:21 PM, Genady said: 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 Expand 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.
Markus Hanke Posted February 7, 2022 Posted February 7, 2022 On 2/6/2022 at 1:58 PM, SergUpstart said: So what is momentum if not p=mv? Expand 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’.
exchemist Posted February 7, 2022 Author Posted February 7, 2022 (edited) On 2/7/2022 at 1:50 AM, 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’. Expand 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 February 7, 2022 by exchemist
swansont Posted February 7, 2022 Posted February 7, 2022 On 2/7/2022 at 11:24 AM, 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. Expand 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.
Genady Posted February 7, 2022 Posted February 7, 2022 On 2/7/2022 at 11:24 AM, 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. Expand 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.
exchemist Posted February 7, 2022 Author Posted February 7, 2022 (edited) On 2/7/2022 at 11:52 AM, Genady said: 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. Expand Eh? The gain in stored energy is accompanied by a gain in rest mass of the battery, yes or no? On 2/7/2022 at 11:44 AM, 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. Expand Yes, that makes sense, I suppose. Edited February 7, 2022 by exchemist
Genady Posted February 7, 2022 Posted February 7, 2022 On 2/7/2022 at 12:03 PM, exchemist said: Eh? The gain in stored energy is accompanied by a gain in rest mass of the battery, yes or no? Expand Yes, the rest mass of the battery, but not the rest mass of the electric fields.
exchemist Posted February 7, 2022 Author Posted February 7, 2022 On 2/7/2022 at 12:06 PM, Genady said: Yes, the rest mass of the battery, but not the rest mass of the electric fields. Expand OK, as @swansont says, maybe I can't separate the fields from the particles experiencing them.
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