An Introduction to the Generation of Mass from Energy

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

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 ]

28 replies
1. neilparker62 says:

lightarrow said

Sorry, I don't have seen your example yet, anyway, concerning the transformation from a ref. frame where both particle 1 and particle 2 have non zero velocity, I find, instead of equation (33) of "Introduction to Relativistic Collisions":

##{E_3^0}^2={E_1^0}^2+{E_2^0}^2+2E_1^0 E_2^0+2T_1E_2^0+2T_1T_2+2T_2E_1^0-2c^2\vec{p_1}·\vec{p_2}## (*)

Then we change frame on K', co-moving with particle 2; ##\vec{v_1}'## is now the relativistic relative velocity between particle 1 and particle 2 and:
##T_2'=\vec{p_2}'=0##
so (*) becomes equal to (33).
I apologize if you have already described this in your last example.

lightarrow

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##.

2. neilparker62 says:
Energy is neither created nor destroyed – simply changed from one form to another. I think that would apply to just about any form of energy 'generation'. In a nuclear power plant we generate energy from mass, here we generate mass from energy.
3. Neil Parker says:
Have added another worked example which (I hope) puts the formula through its paces!
4. neilparker62 says:

lightarrow said

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).

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.

5. vanhees71 says:
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.

6. vanhees71 says:

lightarrow said

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? :-)

lightarrow

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).

7. Ibix says:

lightarrow said

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.

lightarrow said

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.

8. vanhees71 says:
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.

9. Ibix says:

lightarrow said

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.

10. vanhees71 says:
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.

11. Ibix says:

lightarrow said

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.

12. PeterDonis says:

Vitani1 said

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.

13. PeterDonis says:

PAllen said

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.

14. PeterDonis says:

PAllen said

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.

15. 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.
16. 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.]

17. PeterDonis says:

Frodo said

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.

Frodo said

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.

Frodo said

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.

18. 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 ?!

19. PeterDonis says:

Frodo said

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.

20. PeterDonis says:

Frodo said

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.

21. PeterDonis says:

Frodo said

Certainly.

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

22. PeterDonis says:

Frodo said

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.

23. PeterDonis says:

Frodo said

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.

Frodo said

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.

24. PeterDonis says:

Frodo said

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.

Frodo said

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.)

Frodo said

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).

25. PeterDonis says:

SiennaTheGr8 said

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.

26. PeterDonis says:

Frodo said

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".

Frodo said

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.

Frodo said

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.

27. 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).
28. 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 ?!