# An Introduction to the Generation of Mass from Energy

**Estimated Read Time:**6 minute(s)

**Common Topics:**energy, mass, collision, particles, colliding

Table of Contents

#### Introduction

This article is essentially an addition to the previous one on (mainly) inelastic collisions to include the particular case of inelastic relativistic collisions. Reasons for writing a separate article are first that this author is not particularly well qualified to write on the topic and so may well need to request the scrutiny of specialists in the field. And secondly, that equations which govern the generation of mass from energy during collisions may be of sufficient interest to readers to warrant a separate treatment.

#### Equation for an Inelastic Collision

A detailed theoretical treatment of “Relativistic Collisions” in general is provided in the following paper available online:

Introduction to Relativistic Collisions

Our starting point is Equation 33 on page 9 of the article. At this point the article’s author (Frank W.K. Kirk) is dealing specifically with inelastic collisions:

$$ {{E_3}^0}^{ \;2} = {({E_1}^0 + {E_2}^0)}^2 + 2T_1{E_2}^0 \text{ with } {E_3}^0 > {E_1}^0 + {E_2}^0 $$

A brief explanation of notation may be in order. The superscript zeros indicate the “rest energy” of particles involved in the collision while the superscript ‘twos’ indicate the square of quantities as usual. ##T_1## indicates the relativistic kinetic energy of particle 1 given by the expression ##T_1=(\gamma-1){E_1}^0##. The Lorentz factor ##\gamma## is given by the equation:$$\gamma=\frac{1}{\sqrt{(1-\frac{v^2}{c^2})}} $$ … where v is the * relative* velocity of the colliding particles and should (where applicable) be calculated using the Einstein velocity addition formula. The subscripted numbers indicate the particles involved in the collision – here ##E_1## and ##E_2## refer to the energy of pre-collision particles which (in a perfectly inelastic collision) collide and coalesce to form the (nominally) composite particle having energy ##E_3##. In Newtonian mechanics, such a collision results in dissipative energy losses quantified as ½μΔv² where μ is the reduced mass of the colliding particles and Δv their relative velocity. But in the corresponding relativistic collision, the equation shows that instead of being “lost” , collision energy manifests as an increase in mass or creation of new particles as indicated by ## {E_3}^0 > {E_1}^0 + {E_2}^0 ## in the above equation.

#### Ratio of Pre-Collision and Post-Collision Masses

We begin by dividing through by ##{({E_1}^0 + {E_2}^0)}^2## obtaining:

$$ \frac{{{E_3}^0}^{ \;2}}{{({E_1}^0 + {E_2}^0)}^2} = 1 + \frac{2T_1{E_2}^0}{{({E_1}^0 + {E_2}^0)}^2}= 1 + \frac{2(\gamma-1){E_1}^0{E_2}^0}{{({E_1}^0 + {E_2}^0)}^2}$$

As we are now dealing with a ratio equation in which all the quantities correspond to rest mass energies, we can simply replace energy with mass since the factor ##c^4## cancels out everywhere:

$$ \frac{{m_3}^{ \;2}}{{(m_1+m_2)}^2} = 1 + \frac{2(\gamma-1)m_1m_2}{{(m_1+m_2)}^2}$$

Denoting ##m_3## as ##m_f## – the final post collision mass – and the sum ##(m_1+m_2)## as the pre-collision mass, we obtain:

$$ {\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2}{{m_i}^2}$$

In the case where the colliding particles are the same – ie have equal mass, the expression simplifies still further:

$$ {\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{\gamma-1}{2}=\frac{\gamma+1}{2}$$

More generally where the rest masses of colliding particles are m and km respectively:

$$ {\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2k}{{(k+1)}^2}(\gamma-1)$$

#### Relativistic Kinetic Energy of Reduced Mass

The equation: ## {\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2}{{m_i}^2}## may be re-written in the following form: $${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$

Whether or not there is any particular advantage or insight to be gained from writing the equation in this form is debatable but it re-introduces the ubiquitous ‘reduced mass’ parameter ##\mu=\frac{m_1m_2}{m_1+m_2}## which we have previously found so useful in solving classical collision problems.

#### Worked Example 1

We are now ready to apply one of the formulae above (the simplest one!) to solve a problem that appeared in the Physics Homework Help forum recently. The thread title was Kinetic Energy of Colliding Protons.

*In high-energy physics, new particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon (K−) and a positive kaon (K+):*

*p+p→p+p+K−+K+*

*Part A:*

*Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV.*

*Part C: Suppose that instead, the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur.*

*The answer given for part A is 2494 MeV*

#### Solution

##### Part A

Use formula ## {\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{\gamma-1}{2} ## to determine a value for ##\gamma-1##. Then multiply this value by the given rest energy of a proton.

$${\left(\frac{938.3+493.7}{938.3}\right)}^2=1 + \frac{\gamma-1}{2} $$

$$⇒\gamma-1=\left({\left(\frac{938.3+493.7}{938.3}\right)}^2-1\right)\times 2=2.658$$

$$KE_{min}=2.658\times 938.3 \approx 2494 MeV $$

##### Part C

In this case (centre of mass frame) all the kinetic energy of the colliding protons is converted to mass (of kaons). So the combined kinetic energy of protons is just ##2(\gamma_{com}-1) \times 938.3 = 2 \times 493.7 = 987.4 \; MeV ##. From which we can determine ##\gamma_{com}## as ##1+\frac{493.7}{938.3} = 1.526 ##.

As an aside to Part C, it may be noted that since we previously had an expression for ##\gamma_0## in the frame where one proton is at rest, we can determine an expression that relates ##\gamma_0## and ##\gamma_{com}##: $$\gamma_0=2{\gamma_{com}}^2-1$$ $$\gamma_0-1=2{\gamma_{com}}^2-2$$ $$\frac{\gamma_0-1}{\gamma_{com}-1}=2(\gamma{com}+1)=5.052$$ Threshold energy of a proton colliding with a stationary proton will be more than 5 times the threshold energy of protons colliding in a zero momentum frame (equal and opposite velocities).

#### Worked Example 2

From Feynmann Lectures:

A particle of mass m, moving at speed v = 4c/5, collides inelastically with a similar particle at rest.

(a) What is the speed ##v_c## of the composite particle?

(b) What is its mass ##m_c##?

We will begin with part (b). Let the mass of the identical particles be m so that pre-collision mass is 2m. Post collision mass ##m_f=m_c##. Then:

$$ {\left(\frac{m_f}{2m}\right)}^2 = \frac{\gamma+1}{2},$$ where $$\gamma=\frac{1}{\sqrt{1-{\frac{4}{5}}^2}}=\frac{5}{3}.$$ Hence: $${\left(\frac{m_f}{m}\right)}^2=\frac{16}{3}\implies m_c=\frac{4m}{\sqrt{3}}$$. For part (a) we apply conservation of momentum. Let ##\frac{v_i}{c}=\sin\beta_i=\frac{4}{5}##. Then $$p_i=mc\tan\beta_i=mc\times\frac{4}{3}$$ and $$p_f=m_fc\tan\beta_f$$. $$p_f=p_i\implies\tan\beta_f=\frac{m}{m_f}\times\frac{4}{3}=\frac{\sqrt{3}}{4}\times\frac{4}{3}=\frac{1}{\sqrt{3}}.$$ Hence: $$\sin\beta_f=\frac{v_c}{c}=\frac{1}{2}\implies v_c=\frac{c}{2}$$

#### Summary

In this article, we have employed a ratio equation to determine the increase in mass following an inelastic collision of particles travelling at relativistic speed(s). We have also derived a relationship between the Lorentz factor ##\gamma## as determined in a frame of zero momentum (centre of mass frame) and a frame in which one of the colliding particles is at rest. The equations govern a ‘minimum energy’ situation in which collision energy is entirely converted to mass so that there is no post-collision relative velocity between generated and generator particles.

#### Acknowledgements

Sincere thanks to PF Staff members Perok and mfb for reviewing this article as well as providing this “Insights” author with a lot of extra insights which will need some time to digest! As mentioned in the introduction, this article has been a somewhat ambitious project for an ‘amateur’ in the field of relativity. Thanks also to PF user Kharrid for his homework question on the kinetic energy of colliding protons which – along with the (plentiful) responses from PF staff members – provided the background material for this article.

#### References

[1] | Frank W.K. Kirk. Introduction to Relativistic Collisions. https://arxiv.org/ftp/arxiv/papers/1011/1011.1943.pdf. (Accessed on 02/19/2020). [ bib ] |

[2] | Wikipedia Contributors. Lorentz factor – Wikipedia. https://en.wikipedia.org/wiki/Lorentz_factor. (Accessed on 02/19/2020). [ bib ] |

[3] | Hyperphysics. Einstein Velocity Addition. http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html. (Accessed on 02/19/2020). [ bib ] |

[4] | PF User Kharrid and PF Staff. Kinetic Energy of Colliding Protons | physics forums.https://www.physicsforums.com/threads/kinetic-energy-of-colliding-protons.980267/. (Accessed on 02/19/2020). [ bib ] |

- BSc (Elec Eng) University of Cape Town, HDE University of South Africa
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Not explicitly no. But I did indicate somewhere in the article that in the Lorentz factor ##\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}## the 'v' was to be taken as a relative velocity in which case ##T_2=0##.

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.

simplistically we had the following equation in the article:Very$${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$ 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.

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.

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).

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.

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.

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.

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'stotal mass, which is constant, or thecomponents'total mass, which is not.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.

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.

No, it doesn't. A

singlephoton 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 aboutpairsof photons creating particle-antiparticle pairs, or particle-antiparticle pairs annihilating to create pairs of photons. Those reactions can obey conservation laws just fine.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.

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.

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,

exceptthat 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.]

I have said no such thing. But science does progress. Quoting science from 1943 as though it must still be perfectly valid and nothing we have learned since can change any of it, is not a good practice.

I take it you do know that Serber's primer was written in 1943, before the bomb was developed, and at which time there was very little data on fission, and practically no data on precise measurements of the masses of various nuclides (which is the kind of precise data on which the current coefficients in the semi-empirical mass formula are based)?

It is certainly not Serber's fault that his rough analysis–which, btw, was not meant as a final definitive statement on how fission works, but only as a rough estimate on which to base the initial work on the bomb–did not take into account all the things we have learned since 1943 about fission and about what factors contribute to the masses of nuclides. But

youhave no excuse for not taking all those things into account.In any case, arguments from authority carry no weight. The only valid response to the substantive points I have raised would be to point out other substantive points. Simply saying that Serber was on the Manhattan Project is irrelevant.

I have said no such thing. You apparently do not even understand the point I am making, so it's a good thing you don't plan to post any more in this thread.

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:

View attachment 261783

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 ?!

Based on the numbers I have given in my previous post just now, I think this view is mistaken.

If you want to convince me otherwise, you're going to have to do better than just quoting numbers from Wikipedia or giving a reference from 1943, which, however interesting historically, cannot be taken as reflecting our best current understanding, given how much additional data we have accumulated since then and how much additional work has been done with that data.

No, that's

notwhat 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

largerthan 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,

cannotbe 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.Wikipedia is not a valid source for the claim you are making. You need to find an actual peer-reviewed paper.

Note that, as I have already mentioned, this estimate by Serber is not the same (AFAIK) as our best current value for the energy released per fission, based on all available data now.

Please give a reference with some actual numbers to support this claim.

I have no issue with these as general points.

But your original claim regarding fission went beyond these general points. You made a specific analogy that relied upon the only significant stored energy in a uranium nucleus that affects the energy released in fission being electrostatic energy; you viewed the strong force as only analogous to something that was holding the two fission fragments together but then gets released when fission occurs, not as a significant contribution to the actual energy released. I do not think that specific claim is correct.

Very interesting, thanks for providing this reference!

Please note that this reference was produced

beforeany 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.

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.)

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

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.

…you are not just measuring the combined mass of the two protons, because, as you yourself note, there has to be something else present to hold them together. So it is impossible to measure just "the combined mass of two protons close together". The best you can do is to measure "the mass of two protons close together plus whatever is holding them together".

No, this is not "exactly the same" as your other scenario above. A uranium atom is a bound system; you don't need to have something else holding it together.

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

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 ?!