- #1
- 1,125
- 642
Frodo said: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.
PeterDonis said:No, it isn't. It's the difference in binding energy per nucleon between the uranium nucleus and the nuclei of the fission products.
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?
Frodo said:I attach pages 5, 6 and 7 from The Los Alamos Primer, a book by Robert Serber.
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".
Frodo said:Can you please describe what else you think is going on which affects the energy release.
Frodo said:we see that ~84% of the energy comes from electrostatic repulsion.
PAllen said:looking at the history, I think Serber was making an erroneous oversimplification even for his time
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.
I think this sort of reaction violates conservation laws. I could be wrong though.zoki85 said: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.
Vitani1 said:I think this sort of reaction violates conservation laws.
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).neilparker62 said:
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 said:What am I missing?
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.Ibix said: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 said: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.
Example:weirdoguy said:Rest energy is equal to mass in units where ##c=1##. So if one changes, the other one also.
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.lightarrow said:So the system's invariant mass is conserved.
Don't know what "extra conserved" means.vanhees71 said: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.
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) .Ibix said: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.
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.Ibix said:...
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.
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:Don't know what "extra conserved" means.
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.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.
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.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
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.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
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:neilparker62 said: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.
The start point for this article is the following diagram from the given referencelightarrow said: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.
--
lightarrow