The Energy - Momentum Equation vs the Energy - Mass Equation

In summary: I'm not sure what you want me to say. Do you want me to list everyone ever who has ever published a paper in a peer-reviewed journal? There are literally millions of them, maybe tens of millions by now.
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htam9876
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First, introduce the energy – momentum equation E² = p²c² + (m0c²)².

Next, just think it in natural way.

  • If the energy – momentum equation reflects the stationary situation, then, momentum p naturally equals to zero. Then, we got E² = 0 + (m0c²)², namely: E = m0c². It can be denoted exactly as E0 = m0c². This is the energy - mass equation in stationary situation;

  • If the energy – momentum equation reflects the dynamic situation, then, momentum p ≠ 0.
  • Transform the energy – momentum equation E² = p²c² + (m0c²)² into p² – E² / c² = - m0²c²,
  • - m0²c² = m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²),
  • Because m² = m0² / (1 – v² / c²), then, - m0²c² = m²v² – m²c² = p² – E² / c²,
  • Because m²v² = p², then, – m²c² = – E² / c²,
  • Then E² = m²c ^4, namely: E = mc². This is the energy – mass equation in dynamic situation.
Since the energy – momentum equation E² = p²c² + (m0c²)² is generally applicable (to any particle), the stationary situation E0 = m0c² as well as the dynamic situation E = mc² is generally applicable (to any particle) too.Below is the transformation in counter way:

E = mc² is Dynamic mass – energy relationship in nature. It’s a basic natural property. (“Dynamic” means the particle is moving and hints it has momentum);

Next square both sides: E² = m²c ^4, namely: m²c² = E² / c², then, add a negative mark on both side: – m²c² = – E² / c²;

Add two “redundant” items m²v² on both sides: m²v² – m²c² = m²v² – E² / c²;

Because m²v² = p², then, m²v² – m²c² = p² – E² / c²;

Because m = γm0, (pay attention here, it just means that you can consider the magnitude of the moving mass m is γm0, but not means the particle be rest), square both sides: m² = m0² / (1 – v² / c²);

Then, m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²) = p² – E² / c²;

A mathematical calculation: m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²) = - m0²c²;

Then, - m0²c² = p² – E² / c²;

Transform it in math, then, E² = p²c² + (m0c²)².

It’s the so called energy – momentum equation.

If it’s the stationary situation, v = 0, so, no math game can be played. Then, it’s just an energy - mass equation in stationary situation: E0 = m0c².

Liqiang Chen
Sept 19, 2020
 
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Moderator's note: Thread level changed to "I".
 
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htam9876 said:
Because m² = m0² / (1 – v² / c²), then, - m0²c² = m²v² – m²c² = p² – E² / c²
Relativistic mass is a concept that has been abandoned by the scientific community now for several decades. I wouldn’t recommend using it.
 
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  • #5
@ Dale:
Who is " the scientific community "?
 
  • #6
More or less everyone doing science (edit: i.e. publishing in professional journals). You can find people who still use relativistic mass, but they are very rare outside of pop-sci sources.
 
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  • #7
@ Peter:
Piggy doesn't recommend undergraduates to research the issues of this thread...
 
  • #8
@PAllen :
My question is:
There shouldn’t be two kinds of energy – mass relationship in nature:
  • Einstein’s energy – mass equation;
  • Such representation of “for a massless particle, its energy E = pc”
 
  • #9
htam9876 said:
@ Peter:
Piggy doesn't recommend undergraduates to research the issues of this thread...
Whyever not? It's straightforward algebra.
htam9876 said:
@PAllen :
My question is:
There shouldn’t be two kinds of energy – mass relationship in nature:
  • Einstein’s energy – mass equation;
  • Such representation of “for a massless particle, its energy E = pc”
That isn't a question, it's a statement. And both of your cases are just ##m^2c^4=E^2-p^2c^2## in different special circumstances, so I don't see your point here.
 
  • #10
@Ibix:
" More or less everyone doing science. You can find people who still use relativistic mass, "

Yes, so we put aside the issue of what " obsolete notion " in this thread.
 
  • #11
htam9876 said:
@Ibix:
" More or less everyone doing science. You can find people who still use relativistic mass, "

Yes, so we put aside the issue of what " obsolete notion " in this thread.
Do you mean that you intend to carry on using relativistic mass? Bad idea. It was dropped (decades ago) for a reason - almost everybody finds it just confusing to have multiple different masses in play (and it gets worse if you want to consider more than one spatial dimension, because then you end up needing longitudinal and transverse relativistic masses as well). If you don't think it's confusing, fine. But everybody you are trying to communicate with does think it's confusing - so you're going to end up with less confusion if you adopt the consensus position.
 
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  • #12
@htam9876 why are you here?
If you are here to learn, you might want to take the advice of the people you want to learn from.

If you are here to teach, it helps to learn first. Then see above.
 
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  • #13
htam9876 said:
Piggy doesn't recommend undergraduates to research the issues of this thread...

If you think the subject matter of this thread is graduate level, you are mistaken.

I actually learned this stuff in high school (a special summer program I was lucky enough to get into). It is certainly well within the scope of an undergraduate physics course on relativity.
 
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htam9876 said:
@Ibix:
" More or less everyone doing science. You can find people who still use relativistic mass, "

Yes, so we put aside the issue of what " obsolete notion " in this thread.
Just FYI, within a couple of years of introducing relativistic mass, Einstein considered it a mistake that should never have been introduced. So, per Einstein, the notion was obsolete as of 1908.
 
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  • #16
htam9876 said:
@ Dale:
Who is " the scientific community "?
The people who do science professionally, including in particular publishing in the professional scientific literature.

Probably the best known for elucidating the community's broader opinion on this topic is LB Okun. Here are a couple of free papers by him:

https://arxiv.org/abs/1010.5400
https://arxiv.org/abs/hep-ph/0602037

htam9876 said:
There shouldn’t be two kinds of energy – mass relationship in nature:
  • Einstein’s energy – mass equation;
  • Such representation of “for a massless particle, its energy E = pc”
No. Only the invariant mass is needed. The concept of relativistic mass has been discarded by the scientific community for decades
 
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Ibix said:
(and it gets worse if you want to consider more than one spatial dimension, because then you end up needing longitudinal and transverse relativistic masses as well)
That's not the case, if you define the relativistic mass as defined by Gilbert N. Lewis and Richard C. Tolman in 1909:
##m_r= \gamma * m_0##, which preserved ##F= dp/ dt = \frac{d}{ dt}(m_r * v)##.

It's the case, if you define the longitudinal and transverse relativistic masses as defined by A. Einstein in 1905 (with a slight mistake for transverse mass), which was intended to preserve ##F= m * a##.

Wikipedia said:
Planck (1906a) defined the relativistic momentum and gave the correct values for the longitudinal and transverse mass by correcting a slight mistake of the expression given by Einstein in 1905. Planck's expressions were in principle equivalent to those used by Lorentz in 1899.[79] Based on the work of Planck, the concept of relativistic mass was developed by Gilbert Newton Lewis and Richard C. Tolman (1908, 1909) by defining mass as the ratio of momentum to velocity. So the older definition of longitudinal and transverse mass, in which mass was defined as the ratio of force to acceleration, became superfluous. Finally, Tolman (1912) interpreted relativistic mass simply as the mass of the body.[80] However, many modern textbooks on relativity do not use the concept of relativistic mass anymore, and mass in special relativity is considered as an invariant quantity.
Source:
https://en.wikipedia.org/wiki/History_of_special_relativity#Relativistic_momentum_and_mass
 
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  • #18
htam9876 said:
@PAllen :
My question is:
There shouldn’t be two kinds of energy – mass relationship in nature:
  • Einstein’s energy – mass equation;
  • Such representation of “for a massless particle, its energy E = pc”
Einstein abandoned quite quickly the idea of relativistic masses. If you want to introduce this very confusing idea that the mass depends on the velocity of a particle, then you must introduce not only one such quantity but at least two, i.e., "transverse and longitudinal mass".

This is utmost confusing and thus not done anymore by physicists using the special or general theory of relativity in contemporary research. The reason is that relativistic physics is most simply expressed in terms of tensor (and in high-energy physics also spinor) notation, where you only work with invariant quantities, i.e., tensors and spinors (or their covariant components).

Einstein's "energy-mass equation" in modern form reads ##p_{\mu} p^{\mu}=m^2 c^2=(E/c)^2-\vec{p}^2##. For massless "particles" (there are no such things in nature though) ##m=0##, and that's it. Here, ##m## is the invariant mass and a scalar, and that's the only notion of mass which makes sense in a manifestly covariant formalism, and that's how it's done in modern formulations of relativistic theories.
 
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  • #19
vanhees71 said:
If you want to introduce this very confusing idea that the mass depends on the velocity of a particle, then you must introduce not only one such quantity but at least two, i.e., "transverse and longitudinal mass".
No, see my above posting #17.
 
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  • #20
Ok, in this way you save at least this part of the confusion, but still I don't know, why some people insist on confusing oldfashioned notions abandoned for decades (!).
 
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  • #21
Sagittarius A-Star said:
That's not the case, if you define the relativistic mass as defined by Gilbert N. Lewis and Richard C. Tolman in 1909:
##m_r= \gamma * m_0##, which preserved ##F= dp/ dt = \frac{d}{ dt}(m_r * v)##.

It's the case, if you define the longitudinal and transverse relativistic masses as defined A. Einstein in 1905 (with a slight mistake for transverse mass), which was intended to preserve ##F= m * a##.
Fair enough, but what you are saying here is that "relativistic mass" can mean two different things, only one of which requires longitudinal and transverse masses. Neither option is a Lorentz scalar, and there is no term for the modulus of the four momentum (important because it differentiates between null and timelike four momenta) unless you also introduce invariant mass. I'm not sure that reduces the confusion.
 
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  • #22
PeterDonis said:
It is certainly well within the scope of an undergraduate physics course on relativity.
See, for example, chapter 7 of Taylor and Wheeler.
 
  • #23
PeterDonis said:
If you think the subject matter of this thread is graduate level, you are mistaken.

I actually learned this stuff in high school (a special summer program I was lucky enough to get into). It is certainly well within the scope of an undergraduate physics course on relativity.
Special relativity is taught in the 1st semester of the theoretical physics course, and I'm not aware of any of my colleagues who'd introduce the outdated idea of relativistic mass in this lecture. So it's clearly taught successfully for decades in the modern way to undergraduates!
 
  • #24
Ibix said:
Neither option is a Lorentz scalar, and there is no term for the modulus of the four momentum (important because it differentiates between null and timelike four momenta) unless you also introduce invariant mass. I'm not sure that reduces the confusion.
Yes. Wolfgang Rindler did it like this in equation (40):
Wolfgang Rindler said:
## \mathbf {P} = m_0 \mathbf {U} = m_0\gamma(u) (\mathbf {u},c) =: (\mathbf {p}, m_uc)##

where, in the last equation, we have introduced the symbols
##m_u = \gamma(u) m_0##,
##\mathbf {p} = m_u \mathbf {u}##.
Source:
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Relativistic_Mechanics

This approach is critizised in the following paper:
This relativistic shell game would not be too objectionable if the improper 4-velocity were not to be considered. However, the geometric formulation has been usurped by the introduction of an improper velocity that does not transform under a Lorentz transformation.
Source:
https://arxiv.org/pdf/physics/0504110.pdf
 
  • #25
In the above paper the proponents of "relativistic mass" claim Einstein "did it". Yes, he did in a very early stage introducing not only one relativistic mass but at least two (a transverse and a longitudinal mass). A little later he knew better and voted against the use of this utmost superfluous and confusing quantity, which indeed has not a simple transformation behavior under Lorentz transformations.

https://physicstoday.scitation.org/doi/10.1063/1.881171
https://arxiv.org/abs/hep-ph/0602037
 
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  • #26
vanhees71 said:
In the above paper the proponents of "relativistic mass" claim Einstein "did it". Yes, he did in a very early stage introducing not only one relativistic mass but at least two (a transverse and a longitudinal mass). A little later he knew better and voted against the use of this utmost superfluous and confusing quantity, which indeed has not a simple transformation behavior under Lorentz transformations.

https://physicstoday.scitation.org/doi/10.1063/1.881171
https://arxiv.org/abs/hep-ph/0602037
It agree. Einstein never called the term ##\gamma*m_0## the "relativistic mass". He call for example the term
##m(\gamma-1)## the "kinetic energy", see equation (4) - in units with ##c:=1##:
https://www.ams.org/journals/bull/2000-37-01/S0273-0979-99-00805-8/S0273-0979-99-00805-8.pdf
 
  • #27
Yes, and that's all right. Everything can be put in a nice covariant framework. There's the mass of a particle, which is a scalar quantity and there are energy and momentum, building a four-vector, ##(E/c,\vec{p})##. The relation between them is the covariant on-shell condition, ##p \cdot p=(E/c)^2-\vec{p}^2=m^2 c^2##.
 
  • #28
vanhees71 said:
Yes, and that's all right. Everything can be put in a nice covariant framework. There's the mass of a particle, which is a scalar quantity and there are energy and momentum, building a four-vector, ##(E/c,\vec{p})##. The relation between them is the covariant on-shell condition, ##p \cdot p=(E/c)^2-\vec{p}^2=m^2 c^2##.
In 1935, Einstein proved ##E_0 = mc^2## from Relativity:
https://projecteuclid.org/download/pdf_1/euclid.bams/1183498131

Therefore, one can also write the formula, you provided, as:

##p \cdot p=(E/c)^2-\vec{p}^2=(E_0/c)^2##

So, using the term "mass" in Relativity is only a convention, but it is not needed. It is redundant to "rest energy".
 
  • #29
It's more subtle from the point of view of relativistic QFT, because there you deal with the unitary representations of the proper orthochronous Poincare group, and ##m^2=p\cdot p/c^2## is a Casimir operator. Thus one of the parameters determining an irrep. is ##m^2##. For physical purposes ##m^2 \geq 0##.

So you cannot simply eliminate the notion of invariant mass and substitute it by "rest energy". Mass is an intrinsic property of an elementary particle (elementary meaning it's desribed by a quantum field built from an irrep. of the Poincare group), i.e., it's one of the parameters which defines a particle, e.g., an electron is dinstinguished from a muon only by the masses of these elementary particles (leptons).
 
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  • #30
vanhees71 said:
##m^2=p\cdot p/c^2## is a Casimir operator.
Wouldn't be ##(\frac{E_0}{c^2})^2=p\cdot p/c^2## also a Casimir operator?

Wikipedia said:
The invariant mass of an electron is approximately 9.109×10−31 kilograms,[67] or 5.489×10−4 atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV.
Source:
https://en.wikipedia.org/wiki/Electron#Fundamental_properties
vanhees71 said:
Mass is an intrinsic property of an elementary particle

My question is then, if this is only a convention, or if ##E_0## could be regarded as the intrinsic property of an elementary particle. Reason: If you create for example an electron-positron pair, you need to store a certain energy in the electron and the positron. To my understanding, this energy threshold is defined by the interaction with the Higgs field. When the electron and the positron meet later again, you will get the energy back.
 
  • #31
Sagittarius A-Star said:
My question is then, if this is only a convention, or if ##E_0## could be regarded as the intrinsic property of an elementary particle.

Nomeclature used in physics is what is needed to communicate effectively. The invariant quantity of the energy-momentum four-vector is such an important concept and quantity that giving it a special symbol ##m## must be worthwhile. Likewise, we have the invariant mass ##M## of a system of particles.

The problem with relativistic mass is that it's not a Lorentz Invariant, so it adds very little. Or, at least, this is the modern view. It does, of course, allow you to write ##m_R## instead of ##\gamma m##, but that seems (to most of us) insufficient advantage to complicate things and introduce more Lorentz Invariance quantities than we need.
 
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  • #32
PeroK said:
The problem with relativistic mass is that it's not a Lorentz Invariant, so it adds very little

I would say it confuses energy, which is one component of a four-vector, with mass, the four-vector's length. (Or norm if you prefer) If we did this with Newtonian mechanics and three-vectors people would call us crazy.
 
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  • #33
Sagittarius A-Star said:
Wouldn't be ##(\frac{E_0}{c^2})^2=p\cdot p/c^2## also a Casimir operator?Source:
https://en.wikipedia.org/wiki/Electron#Fundamental_propertiesMy question is then, if this is only a convention, or if ##E_0## could be regarded as the intrinsic property of an elementary particle. Reason: If you create for example an electron-positron pair, you need to store a certain energy in the electron and the positron. To my understanding, this energy threshold is defined by the interaction with the Higgs field. When the electron and the positron meet later again, you will get the energy back.
The point is that I want a clear distinction between quantities of different meaning. There's no need to call ##m## all of a sudden ##E_0/c^2##. In natural units ##c=1## anyway.

Mass is an important quantity to be distinguished from energy. The important difference between Newtonian and relativistic mechanics is that in relativistic physics the inertia of a system is due to its (internal) energy and not mass as in Newtonian physics. It's also important to note that mass is not conserved for composite systems in relativistic physics. In Newtonian physics mass is separately conserved due to the somewhat more complicated structure of the Galilei group in comparison to the Poincare group. In non-relatistic physics mass is a central charge of the Galilei Lie algebra and thus gives rise to a superselection rule for mass, while the Poincare group has no non-trivial central extensions and mass is just a Casimir operator of the Poincare Lie algebra and superpositions of different mass states make sense.
 
  • #34
PeroK said:
Nomeclature used in physics is what is needed to communicate effectively.
Yes. I know, that it is often very difficult to estabish standards, and it is good, that the physicists community agreed on a common concept of mass. Also, I find it important, to understand the advantages and disadvantages of a concept.

vanhees71 said:
The important difference between Newtonian and relativistic mechanics is that in relativistic physics the inertia of a system is due to its (internal) energy and not mass as in Newtonian physics.
That statement could be expressed best by writing the 4-momentum as ##\mathbf {P} = \frac{E_0}{c^2} \mathbf {U}## (instead of ##\mathbf {P} = m_0 \mathbf {U}##).

vanhees71 said:
It's also important to note that mass is not conserved for composite systems in relativistic physics.
Yes and no:
Wikipedia said:
Again, in special relativity, the rest mass of a system is not required to be equal to the sum of the rest masses of the parts (a situation which would be analogous to gross mass-conservation in chemistry). For example, a massive particle can decay into photons which individually have no mass, but which (as a system) preserve the invariant mass of the particle which produced them. Also a box of moving non-interacting particles (e.g., photons, or an ideal gas) will have a larger invariant mass than the sum of the rest masses of the particles which compose it. This is because the total energy of all particles and fields in a system must be summed, and this quantity, as seen in the center of momentum frame, and divided by c², is the system's invariant mass.
Source:
https://en.wikipedia.org/wiki/Mass_...individual_rest_masses_of_parts_of_the_system

vanhees71 said:
The point is that I want a clear distinction between quantities of different meaning.
  • In Newton's theory, "mass" and "energy" are names for different physical quantities.
  • In SR, the concept of "relativistic mass" visualizes best, via ##E=mc^2## in all inertial frames, that "mass" and "energy" can be regarded as two names for exactly the same physical quantity. That is a major achievement of SR over Newton's theory. According to Occam's razor, either "mass" or "energy" is needless from a physical viewpoint.
But I also see reasons to keep the name "mass" for rest mass. For example to use similar words as in the forum "Classical Physics", and to stay according to how most people intuitively understand the words "mass" and "energy".
 
  • #35
In SR mass and energy are different physical quantities. Energy is the temporal component of a four-vector, mass is a scalar. Neither mass nor energy cannot be eliminated from relativistic physics. It's dictated by the structure of the spacetime continuum to have both for a complete description of a system.
 

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