The Energy - Momentum Equation vs the Energy - Mass Equation

[htam9876]

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

 

[PAllen]

What is your question? You are doing simple, well understood manipulations showing equivalence of different relativistic formulas. One thing you introduce is $m$ versus $m_0$, which is generally considered an obsolete notion. See:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[PeterDonis]

Moderator's note: Thread level changed to "I".

 

[Dale]

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.

 

[htam9876]

@ Dale:
Who is " the scientific community "?

 

[Ibix]

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.

 

[htam9876]

@ Peter:
Piggy doesn't recommend undergraduates to research the issues of this thread...

 

[htam9876]

@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”

 

[Ibix]

@ Peter:
Piggy doesn't recommend undergraduates to research the issues of this thread...

Whyever not? It's straightforward algebra.

@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.

 

[htam9876]

@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.

 

[Ibix]

@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.

 

[Vanadium 50]

@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.

 

[PeterDonis]

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.

 

[PeterDonis]

@htam9876 do you have a question or not? If you don't, this thread is pointless and will be closed.

 

[PAllen]

@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.

 

[Dale]

@ 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

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

 

[Sagittarius A-Star]

(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 $F= m * a$.

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

 

[vanhees71]

@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.

 

[Sagittarius A-Star]

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.

 

[vanhees71]

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 (!).

 

[Ibix]

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.

 

[Ibix]

It is certainly well within the scope of an undergraduate physics course on relativity.

See, for example, chapter 7 of Taylor and Wheeler.

 

[vanhees71]

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!

 

[Sagittarius A-Star]

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):

$\mathbf {P} = m_0 \mathbf {U} = m_0\gamma(u) (\mathbf {u},c) =: (\mathbf {p}, m_uc)$

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

 

[vanhees71]

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

 

[Sagittarius A-Star]

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

 

[vanhees71]

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, $p \cdot p=(E/c)^2-\vec{p}^2=m^2 c^2$.

 

[Sagittarius A-Star]

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".

 

[vanhees71]

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).

 

[Sagittarius A-Star]

$m^2=p\cdot p/c^2$ is a Casimir operator.

Wouldn't be $(\frac{E_0}{c^2})^2=p\cdot p/c^2$ also a Casimir operator?

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

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.