Mass Generation

An Introduction to the Generation of Mass from Energy


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 firstly that this author is not particularly well qualified to write on the topic and so may well need to request the careful 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 colllisions:

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

A brief explanation of notation may be in order. The superscript zeros  indicate  “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

We are now ready to apply one of the formulae above (the simplest one!) to solve a problem which 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+):

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


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 expressions which 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).


In this article we have employed a ratio equation to determine the fractional increase in mass following a 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.


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 kinetic energy of colliding protons which – along with the (plentiful) responses from PF staff members – provided the background material for this article.


[1]Frank W.K. Kirk.
Introduction to Relativistic Collisions.
(Accessed on 02/19/2020).
bib ]
[2]Wikipedia Contributors.
Lorentz factor – Wikipedia.
(Accessed on 02/19/2020).
bib ]
Einstein Velocity Addition.
(Accessed on 02/19/2020).
bib ]
[4]PF User Kharrid and PF Staff.
Kinetic Energy of Colliding Protons | physics forums.
(Accessed on 02/19/2020).
bib ]

Comment Thread


16 replies
  1. PeterDonis says:
    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.

  2. PeterDonis says:
    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.

  3. PAllen says:
    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.
  4. PAllen says:
    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.]

  5. PeterDonis says:
    I had never realised that time made science wrong in the way you think.

    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 what the Manhattan Project was and whom Serber was addressing? They developed the bomb!

    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 you have 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 never come across anyone who cannot understand that the two electrically charged fragments in a fission repel each other with Coulomb repulsion

    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.

  6. neilparker62 says:
    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:

    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 ?!

  7. PeterDonis says:
    It’s plain simple electrostatics. Binding energies and lost masses are a later description which I think unfortunately hide what is really happening.

    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.

  8. PeterDonis says:
    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.

  9. PeterDonis says:
    the 170 MeV energy change

    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.

  10. PeterDonis says:
    I made no comment about the rearrangement energy because it is pretty trivial compared with the 170 MeV energy change

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

    The points I have been trying to make are

    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.

  11. PeterDonis says:
    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).


  12. PeterDonis says:
    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.

  13. PeterDonis says:
    When you get them very close together they are strongly repelling each other and would fly apart if you did not hold them.

    When you measure their combined mass

    …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".

    In exactly the same way a uranium atom flies apart when the atom is split

    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.

    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.

  14. vanhees71 says:
    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).
  15. neilparker62 says:
    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 ?!

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