An Introduction to the Generation of Mass from Energy

[neilparker62]

Table of Contents
ToggleIntroductionThe equation for an Inelastic CollisionRatio of Pre-Collision and Post-Collision MassesRelativistic Kinetic Energy of Reduced MassWorked Example...

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[zoki85]

Will there be a discussion about photon-photon collisions?

 

[neilparker62]

I'm sure there could be but such an article would definitely be past my very limited understanding of relativity / quantum theory. Introduction to Relativistic Collisions is probably worth more careful study - it does at least deal with photon/matter interaction in the Compton effect. Quite candidly I did not understand much of what was going on with photon interactions which is why I began by 'cherry picking' Equation 33.

Thanks for taking a look at this article all the same. Would any other "Insights" writer perhaps like to pick up on the challenge posed above ?!

 

[zoki85]

Well, the generation of pure EM energy from mass is known to exists (ie. particle-antiparticle annihilation).
I wondered if the reverse was possible ( ie. high energy photons "collide" and create stable massive particles).
I guess, if possible, probabilities of such interactions are small.
Before reading the article I thought, judging by the title, the article discuss such possibilities.

 

[vanhees71]

Yes, it's possible. It's called pair creation. It's well established for scattering a single photon at a charged heavy particle (like an atomic nucleus). What's very challenging is to measure the pure-QED process $\gamma + \gamma \rightarrow e^+ + e^-$. Also elastic two-photon scattering (a radiative correction in QED) is hard to get, though it has been recently observed in ultraperipheral Pb-Pb collisions at the LHC (ATLAS Collaboration).

 

[SiennaTheGr8]

In exactly the same way a uranium atom flies apart when the atom is split because the positively charged nucleus goes dumbell shaped and the two ends electrostatically repel each other enough to overcome the strong nuclear force holding them together. The energy liberated is the stored electrostatic energy which pushes them apart.

No, it isn't. It's the difference in binding energy per nucleon between the uranium nucleus and the nuclei of the fission products.

Huh. I was under the impression that, during fission, some of the potential energy associated with the repulsive electrostatic forces between nucleons gets converted into the kinetic energy of the fission products. Is that really not the case?

 

[PeterDonis]

I was under the impression that, during fission, some of the potential energy associated with the repulsive electrostatic forces between nucleons gets converted into the kinetic energy of the fission products. Is that really not the case?

It is true that there is an electrostatic interaction between protons in the nucleus, which, at least in a highly oversimplified model, should contribute potential energy to the overall nucleus considered as a bound system in a stationary state.

However, there is so much else going on in a nucleus, particularly a nucleus undergoing fission, that to think of fission as "oh, it's just electrostatic repulsion pushing things apart because we released the thing that was keeping them together", which is the model the OP was describing, is not really useful.

 

[Frodo]

I attach below pages 5, 6 and 7 from The Los Alamos Primer which describe how the Los Alamos Project calculated an estimated yield for the weapons they were inventing based on the release of the electrostatic energy stored in the nucleus when it is assembled from its constituent protons.

Serber, Oppenheimer's protégé in 1943, gave lectures to the Los Alamos scientific and engineering team on the Manhattan Project in 1943 and this book describes those lectures. A copy of The Primer was given to each new arrival. All attended a lecture including Feynman, Bethe etc.

The book was published in 1992 when Serber wrote the Preface and added annotations and comments throughout, including those in these images.

An anecdote: Oppenheimer was in the audience for the first lecture and became concerned when Serber used the word bomb as it might be overheard by non-security classified construction workers (one of whom put a foot through the ceiling). He interrupted Serber and asked him to use the word "gadget" instead of "bomb" and that is why Los Alamos always referred to the device as "the gadget". Serber's original notes have been published here.

I quote (written by Serber in 1992):

"The origin of energy released in fission is exactly the same as the origin of the energy released when two atoms or molecules react chemically. It's the electrostatic energy between two similarly charged particles."

and

"Somehow the popular notion took hold long ago that Einstein's Theory of Relativity, in particular his famous equation E = mc^2, plays some essential role in the theory of fission ... but his theory of relativity is not required in discussing fission. The theory of fission is what physicists call a non-relativistic theory, meaning that relativistic effects are too small to affect the dynamics of the fission process significantly."

The approximate estimate of the yield of the bomb, 20,000 tons of TNT for a 1 kg of Uranium completely fissioned, follows directly from the calculation of the electrostatic energy - it takes just a few lines.

Wiki Nuclear_fission says:

"For uranium-235 (total mean fission energy 202.79 MeV), typically ~169 MeV appears as the kinetic energy of the daughter nuclei, which fly apart at about 3% of the speed of light, due to Coulomb repulsion"

and

"... a nuclear fission explosion or criticality accident emits about 3.5% of its energy as gamma rays, less than 2.5% of its energy as fast neutrons (total of both types of radiation ~ 6%), and the rest as kinetic energy of fission fragments (this appears almost immediately when the fragments impact surrounding matter, as simple heat)."

In Chapter 1 - Electromagnetism in Volume II of his lectures from the 60s Feynman says the "nuclear energy" released is really "electrical energy" (Feynman's quotes), albeit he ignores the rearrangement into the daughter nuclei:

"There is another question: “What holds the nucleus together”? In a nucleus there are several protons, all of which are positive. Why don’t they push themselves apart? It turns out that in nuclei there are, in addition to electrical forces, nonelectrical forces, called nuclear forces, which are greater than the electrical forces and which are able to hold the protons together in spite of the electrical repulsion. The nuclear forces, however, have a short range—their force falls off much more rapidly than 1/r^2. And this has an important consequence. If a nucleus has too many protons in it, it gets too big, and it will not stay together. An example is uranium, with 92 protons. The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all of the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just “tapped” lightly (as can be done by sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called “nuclear” energy, but it is really “electrical” energy released when electrical forces have overcome the attractive nuclear forces." (my emphasis).

 

[PeterDonis]

I attach pages 5, 6 and 7 from The Los Alamos Primer, a book by Robert Serber.

Very interesting, thanks for providing this reference!

Please note that this reference was produced before any bomb had been detonated, and the only fission reaction that had been produced at that point, IIRC, was Fermi's experiment in Chicago, from which not much data had been obtained. So the number given for energy released per fission was not based on measurements; it was simply a rough theoretical calculation that was the best that could be done at the time.

Serber's estimate for the energy released per fission was 170 MeV. (Note that he is careful to say "of the order of", indicating that he realizes this is a rough estimate, not an exact number.) AFAIK the actual value from modern measurements is higher--about 215 MeV per fission, if I'm remembering correctly from the nuclear physics courses I took way back in college.

The point of my post was that it is very helpful and insightful to say that "energy has mass and if the energy goes away, the mass goes down".

This is correct as far as it goes, but electrostatic energy is not the only kind of energy involved. (Note, btw, that Serber, as you note, comments that, while $E = m c^2$ comes from relativity, his model of fission is actually non-relativistic.)

Can you please describe what else you think is going on which affects the energy release.

The strong interaction. You mention it in the first of your two items on what you think is going on--the two fragments "rearrange themselves" into new bound states in which the strong interaction holds them together. But this does not just involve a change in electrostatic energy, as you say, but also a change in strong interaction energy. This affects how much energy is left over to go into the kinetic energy of the fission fragments (the second of your two items). I don't see where Serber took that into account anywhere.

I'm actually not sure why Serber (apparently) did not take into account the strong interaction, since its effects are included in the semi-empirical mass formula [1], which had already been proposed prior to 1943 (although the best known values of the coefficients in the formula probably were different back then than they are today).

[1] https://en.wikipedia.org/wiki/Semi-empirical_mass_formula

 

[PeterDonis]

we see that ~84% of the energy comes from electrostatic repulsion.

No, that's not what we see.

If you actually run some numbers using the semi-empirical mass formula (note that, while I linked to Wikipedia, the article I linked to has several references which are peer-reviewed papers--or textbooks, which would also be acceptable--in fact, one of the sources also has a more modern form of the formula with updated coefficients), you will see that the Coulomb term in that formula, or more precisely the difference between the Coulomb term for the uranium nucleus and the sum of the Coulomb terms for the two fission fragment nuclei--is quite a bit larger than 170 MeV (in the mid 300's, the exact value depends on which version of the formula and which set of coefficients you use).

Also, if you just do a simple calculation of the Coulomb potential energy between the two fission fragment nuclei separated by approximately the size of a uranium nucleus (which is approximately $r_0 A^{1/3}$, where a good value of $r_0$ seems to be about $1.2 \times 10^{-15}$ meters), you get a value which is more than twice 170 MeV (I get 39`1 MeV for $A = 235$).

These numbers tell me that the final kinetic energy of the fission fragments, about 170 MeV, cannot be due to a simple process of their Coulomb potential energy being converted to kinetic energy--if it were, their final kinetic energy would be roughly twice as large as is observed. There must be significant other effects involved.

 

[neilparker62]

My ha'penny's worth:

When writing this article I was quite taken with the idea of an inelastic (collide and coalesce) collision in which there were no dissipative energy losses. I had come across this concept in one of the problems I dealt with in a previous article - it comes from an MIT problem set on collisions:

Because the small block reaches a point where the two masses are at rest with respect to each other, the conditions for an inelastic collision are momentarily met. So part a) of the problem can easily be solved by setting the standard expression for energy "loss" in an inelastic collision equal to the PE gained with the point being that in this case the collision energy is not lost - it is simply converted to PE.

Similarly in the problem dealt with in this article we again have an "inelastic" collision in so far as all the particles are at rest after the collision. But once again the collision energy is not lost - it simply ends up in the mass of the kaons. What would be fascinating is if one could somehow have an energy storage system in which those kaons (instead of decaying) are maintained as "mass" until the stored energy is needed. What more "compact" means of storing energy can there be than as mass ?!

 

[PAllen]

Maybe the point was made earlier in this thread, but it seems trivial to me that there is much more to fission and binding energy than Coulomb force. Admitting that energy has mass as even @Frodo does, consider the process of bringing all the nucleons of a uranium nucleus together. You would have to do work against Coulomb repulsion and thus expect that the nucleus would be more massive than the separate nucleons, in the same way a compressed spring is more massive than an uncompressed spring. It is the strong nuclear force that changes this picture from mass excess to mass defect. It is the difference in the degree to which strong force overcompensates Coulomb repulsion between the parent and daughter nuclei that produces the fission yield. That an analysis only looking at coulomb repulsion yielded a ballpark correct value for the yield only reflects that the strong force effect is of the same order of magnitude as the coulomb. I would view it as something of a lucky coincidence.

I think one might say that relativity is not directly involved except for the convenience of relating mass observations to energy yields, and that the more complete explanation involves the strong force, specifically as it contributes to nuclear binding energy models. However, I don’t believe there is any such thing as a non-relativistic treatment of QCD, so it seems to me that fission is, in fact, intrinsically a relativistic process.

Similar to what @PeterDonis said, without doubt, I am nothing as a physicist compared to Serber, except that I have had the opportunity to read what all the great physicists have written since 1943.

[edit: but looking at the history, I think Serber was making an erroneous oversimplification even for his time. Already, in the late 1930s, it was realized that nuclei had a mass defect due to a new force believed mediated by mesons. Already, at that time, it was realized that a fission yield must be based on the differing degree to which this force over compensates Coulomb repulsion between parent and daughter nuclei, that produces fission yield. Thus, even in historical context, I am amazed at what Serber wrote. I would think, for example, that Oppenheimer, Bethe, and Feynman would have all disagreed with this analysis.]

 

[PAllen]

To make my last post more concrete, let’s suppose nuclear binding force was barely sufficient to compensate for Coulomb repulsion. Suppose, as a result, that for uranium, the mass defect was only 1% of the coulomb potential energy. And suppose, for the daughter nuclei, it happened to be 2%. Then the fission yield would only be about 1% of what Serber calculated. Thus, ignoring nuclear force and mass defect, Serber’s calculation could be wrong by orders of magnitude. That it wasn’t is luck that does not repair its conceptual fallacy.

 

[PeterDonis]

looking at the history, I think Serber was making an erroneous oversimplification even for his time

Yes, that's why I mentioned earlier the semi-empirical mass formula, an early version of which already existed in 1943 (it was originally proposed in the mid-1930s, I believe), and wondered why Serber did not use it.

 

[PeterDonis]

consider the process of bringing all the nucleons of a uranium nucleus together. You would have to do work against Coulomb repulsion and thus expect that the nucleus would be more massive than the separate nucleons, in the same way a compressed spring is more massive than an uncompressed spring.

Not only that, but such a system would not be stable by itself. To hold a spring compressed, something else has to be present to hold it. To hold charged particles together against Coulomb repulsion, something else has to be present to hold them. And whatever is holding them together, if the system as a whole is going to be stable, has to provide enough mass defect to compensate for the mass excess.

In short, any system with an overall mass excess would not be stable. And, as you point out, in the case of an actual uranium nucleus, there is a mass defect, because the strong interaction is also present.

 

[Vitani1]

Well, the generation of pure EM energy from mass is known to exists (ie. particle-antiparticle annihilation).
I wondered if the reverse was possible ( ie. high energy photons "collide" and create stable massive particles).
I guess, if possible, probabilities of such interactions are small.
Before reading the article I thought, judging by the title, the article discuss such possibilities.

I think this sort of reaction violates conservation laws. I could be wrong though.

 

[PeterDonis]

I think this sort of reaction violates conservation laws.

No, it doesn't. A single photon creating a particle-antiparticle pair, or a particle-antiparticle pair annihilating each other to create a single photon, would violate conservation laws (it's impossible to conserve both energy and momentum). But the article is talking about pairs of photons creating particle-antiparticle pairs, or particle-antiparticle pairs annihilating to create pairs of photons. Those reactions can obey conservation laws just fine.

 

[lightarrow]

Continue reading...

Sorry, but can't understand what you mean with "generation of mass from energy". I thought mass is conserved in SR (yes I'm serious and I'm speaking of invariant mass).
Let's take the example of vanhees: creation of an electron-positron pair by head-on collision between an high-energy photon and a heavy nucleus: the system's mass M before the process is:
$$M=\frac{1}{c^2}\sqrt{E^2-c^2p^2}$$
Where E, p are, respectively, the system's energy and momentum.
But energy and momentum are conserved, whatever process will take place, so the system's mass is the same at the end.
What am I missing?

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[Ibix]

What am I missing?

The mass of a closed system is conserved, yes, but the sum of the invariant masses of its components is not. So looking at the component masses you can have mass-from-energy or vice versa.

 

[lightarrow]

The mass of a closed system is conserved, yes, but the sum of the invariant masses of its components is not. So looking at the component masses you can have mass-from-energy or vice versa.

You can have kinetic or potential energy from rest energy, so convertion of a kind of energy into another kind, not "convertion of mass into energy" as your statement implicitly suggest. Anyway, I imagine is just a matter of words. I don't like it however, I think it's source of many misunderstandings in the concepts of mass and energy from people.

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[vanhees71]

Invariant mass is not conserved. The most famous example are nuclei. There the mass of the nucleus is smaller by the amount $|E_{\text{binding}}/c^2|$, where $E_{\text{binding}}$ is the binding energy.

The same is true for, e.g., a gas, whose invariant mass depends on temperature. Changing the temperature (meausured in the rest frame of the gas) changes the heat energy by $\Delta Q$. The invariant mass of the gas changes by $\Delta Q/c^2$.

It's a quite deep difference between mass in Newtonian and relativistic physics, based on the different space-time symmetries (Galilei vs. Poincare invariance, respectively) and their realization in quantum (field) theory.

 

[weirdoguy]

You can have kinetic or potential energy from rest energy, so convertion of a kind of energy into another kind, not "convertion of mass into energy" as your statement implicitly suggest.

Rest energy is equal to mass in units where $c=1$. So if one changes, the other one also.

 

[lightarrow]

Rest energy is equal to mass in units where $c=1$. So if one changes, the other one also.

Example:
In an inertial frame where it's stationary, a neutron disintegrates in the particles we know (proton, $e^-$, electron antineutrino); the energy is conserved.
The original neutron's rest energy is transformed in:
1) the other particles' rest energy
2) the other particles' kinetic energy
3) the proton-electron's potential energy.
The system's momentum is conserved, it was 0 before the process, it's 0 after the process. So the system's invariant mass is conserved.

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[Ibix]

So the system's invariant mass is conserved.

I'd have put the emphasis on "system's". I don't disagree with what you say, but it's also possible to think about the total mass of the components, which is not conserved. The energy those components have comes from the mass of the original component in your example.

I think it's just that "mass" is a word that doesn't have a single meaning in relativity, even if you (quite sensibly) disregard relativistic mass. So it's important to be clear whether you are talking about the system's total mass, which is constant, or the components' total mass, which is not.

 

[vanhees71]

The point is that in relativity invariant mass is not an extra conserved quantity as it is in Newtonian physics. What's described in #23 is energy-momentum conservation in the center-of-mass frame. What's not conserved is the sum of the invariant masses before and after the reaction.

Take the decay of a neutral pion (mass about 140 MeV) to two photons (mass 0). Of course, what's conserved is the total energy and momentum.

 

[lightarrow]

The point is that in relativity invariant mass is not an extra conserved quantity as it is in Newtonian physics. What's described in #23 is energy-momentum conservation in the center-of-mass frame. What's not conserved is the sum of the invariant masses before and after the reaction.

Take the decay of a neutral pion (mass about 140 MeV) to two photons (mass 0). Of course, what's conserved is the total energy and momentum.

Don't know what "extra conserved" means.
Anyway, I don't think we have to invoke a "non conservation of the sum of invariant masses before and after a reaction", we can simply say: "invariant mass is not additive".

About the pion, it's an emblematic example of this fact: the sum of the masses of the two photons is zero, but their system's mass is not!
And since I'm completely confident that you already knew it, what are we discussing about? :-)

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[lightarrow]

I'd have put the emphasis on "system's". I don't disagree with what you say, but it's also possible to think about the total mass of the components, which is not conserved. The energy those components have comes from the mass of the original component in your example.

I think it's just that "mass" is a word that doesn't have a single meaning in relativity, even if you (quite sensibly) disregard relativistic mass. So it's important to be clear whether you are talking about the system's total mass, which is constant, or the components' total mass, which is not.

Ok, with the terminology "components' total mass" I can agree with you. But I believe we should be aware that not focusing the attention on the system or on the single particle, a lot of people, those reading popular books, e.g., can have misunderstandings on the "transformation of mass into energy" and on the concept of mass, energy, matter, "pure energy " which would be light according to them, and so on (as in my experience) .

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[lightarrow]

...
So it's important to be clear whether you are talking about the system's total mass, which is constant, or the components' total mass, which is not.

Sorry I forgot to answer this. Clearly I was talking exactly of system's total mass, which is, ok I don't want to say " the only" but "the most significant"... meaning of "total mass", for me, in order not to complicate things too much.
Regards and thanks for the discussion.

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[Ibix]

Don't know what "extra conserved" means.

Extra conserved doesn't mean anything. @vanhees71 is using "an extra conserved quantity" in the sense of "another conserved quantity". Mass is a separate quantity with its own independent conservation law in Newtonian physics, but in relativity its conservation (or otherwise) follows from its relationship to the four momentum, as we've been discussing.

Sorry I forgot to answer this. Clearly I was talking exactly of system's total mass, which is, ok I don't want to say " the only" but "the most significant"... meaning of "total mass", for me, in order not to complicate things too much.

I'm not sure I agree. One important application of this subject is nuclear power, where the energy theoretically available for electricity generation is the difference between the invariant mass of the fuel and the waste. In this case, the conversion of mass to energy is an extremely useful concept.

I agree that there's potential for confusion (and definitely "converted to pure energy" should be suppressed, IMO), but I think the concept of energy/mass conversion is useful in some contexts, unlike relativistic mass which is more trouble than it's worth.

 

[vanhees71]

Don't know what "extra conserved" means.
Anyway, I don't think we have to invoke a "non conservation of the sum of invariant masses before and after a reaction", we can simply say: "invariant mass is not additive".

About the pion, it's an emblematic example of this fact: the sum of the masses of the two photons is zero, but their system's mass is not!
And since I'm completely confident that you already knew it, what are we discussing about? :-)

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In Newtonian physics you have in addition to the 10 conservation laws arising from Noether's theorem due to Galilei invariance (energy and momentum from translation symmetry of time and space, angular momentum from isotropy of space and center-of-mass velocity from invariance under Galileo boosts) another independent conservation law for mass, which however is of a quite special kind and arises from quantum theory.

Analyzing Galileo invariance in quantum mechanics you realize that the irreducible unitary ray representations of the Galileo group allow for a non-trivial central charge and instead of the rotation subgroup SO(3) you can use its covering group SU(2). The latter leads to the possibility of half-integer spin. The former leads to an extension of the Galileo algebra with the mass as a central charge. As it turns out, indeed the irreducible unitary representations of the original ("classical") Galileo group doesn't lead to a physically meaningfull quantum dynamics. Thus one has to use the extended group (or rather its Lie algebra) with mass as a central charge and investigate its unitary representations. That analysis leads to the usual non-relativistic quantum theory we know from our QM 1 lectures. The fact that in non-relativistic QM mass enters as a central charge implies a socalled superselection rule, i.e., there are no transitions between representations of different mass, and that's why mass is conserved, i.e., you get an 11th independent conservation law but not from Noether symmetry but due to a superselection rule for a non-trivial central charge of the central extension of the Galileo group.

This is nicely covered in

L. E. Ballentine, Quantum Mechanics, World Scientific,
Singapore, New Jersey, London, Hong Kong (1998).

In contradistinction to that the (proper orthochronous) Poincare group, i.e., the space-time symmetry group of Minkowski space, has no non-trivial central charges and thus you directly consider the unitary representations of the covering group of the Poincare group (which boils down to use the covering group $\text{SL}(2,\mathbb{C})$ of the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$. So you have half-integer and integer spin but no additional conservation law for mass. Mass in relativistic physics is just a Casimir operator given by $p_{\mu} p^{\mu}=m^2 c^2$. So there are just the "Noether conservation laws" for energy, momentum, angular momentum, and center-of-energy (!) velocity and no extra conservation law for mass. Of course I use mass in the only meaningful definition, i.e., as invariant mass. This is all covered in in great detail in

S. Weinberg, The Quantum Theory of Fields, vol. 1,
Cambridge University Press (1995).

 

[lightarrow]

Thank you, vanheese71.
Unfortunately I am able to understand just a part of your very interesting post. I think the "centre of energy" is the key concept here, anyway I can't understand exactly where is the mistake in the computation of the system's (invariant) mass in my two examples.
Regards

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[vanhees71]

There is no error. What you did was using the usual "kinematics" of collisions/decays, i.e., the energy-momentum conservation for the transition from the asymptotic free initial to the asymptotic free final state. What you then called "mass" is in fact the total energy of the system as measured in the center-of-momentum frame. For collisions that's "Mandelstam s", defined by $s=(p_1+p_2)^2$ (where $p_1$ and $p_2$ are then four-momenta of the two incoming particles). For a decay you have only the one unstable particle in the initial state. So there indeed $s=M^2 c^2$, where $M$ is the invariant mass of the unstable particle.

What I wanted to stress is that there's no additional conservation law for invariant mass as in Newtonian physics, and how this can be understood from very fundamental symmetry considerations.

 

[neilparker62]

Sorry, but can't understand what you mean with "generation of mass from energy". I thought mass is conserved in SR (yes I'm serious and I'm speaking of invariant mass).

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The specific example you bought up in your post has been discussed in some detail by others much better qualified than myself so I won't go into that. But I'll just try and explain what I had in mind.

Very simplistically we had the following equation in the article:

$${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$ Hence:
$$\frac{m_f}{m_i} = \sqrt{1+\frac{2T_{\mu}}{m_ic^2}}$$ For a very rough approximation we can take a one tem binomial expansion of the square root expression: $$\frac{m_f}{m_i} \approx 1+\frac{T_{\mu}}{m_ic^2}\implies m_f \approx m_i+\frac{T_{\mu}}{c^2}$$ The term $\frac{T_{\mu}}{c^2}$ is essentially the 'system's' kinetic energy expressed in units of mass. In a typical 'collide and coalesce' collision in the macro world, this energy is dissipated as heat/sound/deformation etc but here it is not. It results (for example) in the production of kaons (ie new 'massive' particles) as per worked example.

 

[lightarrow]

The specific example you bought up in your post has been discussed in some detail by others much better qualified than myself so I won't go into that. But I'll just try and explain what I had in mind.

Very simplistically we had the following equation in the article:

$${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$ Hence:
$$\frac{m_f}{m_i} = \sqrt{1+\frac{2T_{\mu}}{m_ic^2}}$$ For a very rough approximation we can take a one tem binomial expansion of the square root expression: $$\frac{m_f}{m_i} \approx 1+\frac{T_{\mu}}{m_ic^2}\implies m_f \approx m_i+\frac{T_{\mu}}{c^2}$$ The term $\frac{T_{\mu}}{c^2}$ is essentially the 'system's' kinetic energy expressed in units of mass. In a typical 'collide and coalesce' collision in the macro world, this energy is dissipated as heat/sound/deformation etc but here it is not. It results (for example) in the production of kaons (ie new 'massive' particles) as per worked example.

In your article (and the previous one) you begin with a particle 1 moving at $v_1$ against a particle 2 initially at rest and all the computations are (implicitly) made in the rest frame of particle 2. Later you write:
"Denoting $m_3$ as $m_f$ – the final post collision mass – and the sum $(m_1+m_2)$ as the pre-collision mass [which you then call it "$m_i$"], we obtain"...
and you write an equation involving the relative velocity v of the two particles.
1) Which is the reference frame here? Is the same as the previous article, that is the particle 2 rest frame? The phrase "relative velocity" could suggest it's not.
2) You call $(m_1+m_2)$ "pre-collision mass" but I can't see which is the physical system which have this mass.

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[neilparker62]

In your article (and the previous one) you begin with a particle 1 moving at $v_1$ against a particle 2 initially at rest and all the computations are (implicitly) made in the rest frame of particle 2. Later you write:
"Denoting $m_3$ as $m_f$ – the final post collision mass – and the sum $(m_1+m_2)$ as the pre-collision mass [which you then call it "$m_i$"], we obtain"...
and you write an equation involving the relative velocity v of the two particles.
1) Which is the reference frame here? Is the same as the previous article, that is the particle 2 rest frame? The phrase "relative velocity" could suggest it's not.
2) You call $(m_1+m_2)$ "pre-collision mass" but I can't see which is the physical system which have this mass.

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The start point for this article is the following diagram from the given reference

Equation 33 which I have 're-worked' follows from the above so yes the rest frame is where particle 2 is at rest. "Relative" velocity in this case is just the velocity of the other particle. But the technique I have used in my classical articles does not - in principle - require either particle to be stationary since it is based on their velocity relative to each other. Collision impulse is calculated according to the formula $\Delta P=\mu \Delta v$ for completely inelastic collisions and $\Delta P=2\mu \Delta v$ for elastic collisions. $\mu$ is the reduced mass of the colliding objects and $\Delta v$ is their relative velocity.

In this article I sought to find a similar approach for a relativistic collision - hence the formula referred to in my previous post (which brings in the ubiquitous 'reduced mass' parameter). I think there was one example I found somewhere and solved where the two colliding particles were both moving in some reference frame. Perhaps I need to dig it up and include it as part of this article.

For point 2 I would refer you to the worked example where the pre-collision mass is that of the two colliding protons and the post-collision mass is that of the two protons plus the 'generated' kaons.

 

[neilparker62]

Have added another worked example which (I hope) puts the formula through its paces!