This is why I was regretting trying to use analogies without the ability to using actual math,
Momentum a conserved quantity of every physical system that is transitionally invariant.
At slow speeds K ~ mv^2/2
At relativistic speeds the following is more correct.
K = mc^2(1/sqrt(1-v^2)/c^2)-1)
Momentum is (force)(time), and kenetic energy is related to (force)(distance), but a matter of frame of reference, either time or distance. But we chose those frames of reference based on convenience and they are not invariant.
It is important to not confuse our chosen frame from the fundamental backing realities.
Heat is a described as a scalar because it is useful and easier to conceptualize in that scalar form, but heat is movement, and it is not an invalid concept to think of it as relative velocity even if it is difficult to describe in those terms.
As an example, consider the best current methods for cooling material to produce states of matter like the Bose-Einstein Condensate. To produce BEC rubidium atoms were cooled to 170 nanokelvins, and they did so with laser confinement, where photons were used to reduce the lateral speed of those atoms, by slowing their movement.
But the main error is to think that classical physics is descriptive of the best currently accepted theories, they are better through as approximations and tools to simplify solving problems.
It may be helpful to think that four-momentum in SR is commonly used for Lorentz transformations, but the only numberthat are invariant in that collection of vectors is the scalar.
I am regretting trying to move past just talking about energy states but I want to try and clarify my position here as much as possible within the limitations of this message board.
Please feel free to replace “potential” with “mechanical” in my previous posts:
since feeling is first
who pays any attention
to the syntax of things
will never wholly kiss you
Or feel free to provide a more suitable analogy to describe the interaction of the Coulomb force, strong force and tunneling.
There are tons of pitfalls when trying to describe concepts that are best communicated via math, and I don’t know if trying to argue dozens of connotative and multiple denotative variations of a few words is a productive way of helping the OP.
I personally found the stone rolling down a hill analogy as useful, but I will avoid trying to use it as a tool here in the future.
Perhaps I should error to answering everything on stack exchange and providing links to avoid confusion in the future.
As a moderator, I would assume you would be able to provide a post that provided information on how my explanation was wrong vs just saying no.
I agree that my analogy was flawed, but what are you saying is wrong?
That the mass of two free hydrogen atoms and an oxygen atom is not more than that of a water molecule?
Or that the relation of the binding energy of two fused hydrogen atoms is not less than the independent hydrogen atoms?
Or that fusion above FE is endothermic because of the distance that the nuclear force works, and that below that threshold it releases energy due to confinement?
This is General Questions, and I want to learn where I am wrong too, but I am about really close to giving up on contributing to this site in general when moderators just resort to "nu uh " posts.
Yes I am breaking the TOC with this post, but I am going to take a break and move to sites that aren’t as math phobic and will enable simple tools to help fight ignorance.
I am sure we are just talking past each other here but I wish the dope the best, it is useless to try and “fight” ignorance, even if it is my own when the site has regressed to simple “nu uh” exchanges in GQ.
That said I still want to know why I am in error so please respond, but I am done with trying to help add content.
The way your explanation is wrong is that you keep on referring to P as energy, when it’s not. Even if you can (in some systems of units) measure E and P in the same units, they’re still different things.
Actually, that’s perfectly okay. It’s a very standard textbook model for energy potentials, quantum tunneling, the binding curve of nuclear energy, etc.
You do have a tendency to use complex and confusing language. My advice would be to study Isaac Asimov’s science essays, and to strive for greater clarity.
No, I was simplifying to try and explain the EXACT formula used for approximating yields, using terms that may be ambiguous based on domain, due to the lack of mathtex support.
While trying to respond to dismissals that lacked any form of cite.
But please state the exact domain, or field of science to explain my mistakes, but unless you are limiting your definition of “Energy” to “Kinetic Energy”
Are you trying to conflate Newtonian differences between energy and momentum to fission?
Please provide **cites ** from the scientific domain of the OP, because my understanding is the classic ordinary mechanical quantities do not apply in this domain.
And that one must resort to the relativistic energy-momentum relationship of pc.
Momentum and energy are different things in every domain of science. Neither momentum nor energy can be defined in the same way as in Newtonian physics, when one is working in relativity, but they are both still defined, and still different from each other.
In extremely relativistic situations, with energy much greater than the mass, the kinetic energy is approximately equal to the magnitude of p*c (this is exact, for photons and other massless particles which travel at c). But we’re not talking about an extremely relativistic situation, here; we’re talking about fusion, where the kinetic energy is less than 1% of the mass. And even in extremely relativistic systems, momentum and energy are still different things (for starters, momentum has three components, while energy only has one).
Heck I will lower the bar, provide any fusion reaction that isn’t dependent on energy-momentum 4-vectors.
The OP is related to the yield, and not some non-calculable solution. And even if you ignore the Coulomb barrier through some form of tunneling you will still have to deal with the proton or neutron particle that is produced in a relativistic fashion.
Momentum is a conserved quantity of every physical system that is translationally invariant. The (momentum or energy) of a photon in such a reaction just does come close to what the proton, neutron. or electron take away. The problem is mainly related to the momentum side, and confining your rebuttal to the photon ignores that point.
Think of D+T fusion, 80% of the energy yield is in the energy of the neutron, and lets assume the following for easy math:
D = mass 2
T = mass 3
D + T = mass 4.98 + 17.6 MeV “Energy” (in neutrino)
But of course D+D is more practical for Earth based fusion as the proton, through interaction converts most of it’s kinetic energy into thermal energy due to interactions with the electromagnetic force thermal energy.
Once again in this case it is the momenta of that proton that results in the yield, and not some mythical, yet undefined “Energy”.
This is all trivial to show (with math) but impossible to debate when “Energy” is some undefined, magical property of a system.
So I will ask again for cites, which you have yet to provide, that describe this mythical “Energy” that cannot be described or converted into something similar to “kinematic momentum”.
If this is so utterly basic and trivial it should also be trivial to provide a link to non-relativistic, non-momenta centrist methods of accurately describing and calculating the yield of fusion, which is what the OP was about.
Please help me understand what I am missing here, that goes beyond Mass-energy equivalence, to some special form of energy that doesn’t work for p, or the total energy E.
Note you can’t invoke classic physics when talking about energy, which is part of the reason I am dismissing your claims,* E = pc* is simply invalid in E = mc = (1/2)mu^2 (rest mass here).
But if you are going to claim the frame-independent intrinsic mass that will cause issues with energies and momenta.
Are you referencing the de Broglie relations for relativistic momentum and relativistic mass energy? Because I am pretty sure those are limited to free non-relativistic particles.
In a nuclear reaction, the total (relativistic) energy is conserved, and in every simple calculation that I can find is calculated in momenta, or in Lagrangian and Hamiltonian mechanics as generalized momentum, but as your arguments seem focused on some hard tie to “kinetic or potential” energy, which don’t actually apply to the OP’s problem directly I still want cites instead of random hand waving arguments.
Simply provide me with what “type” and domain of “Energy” you are using that can’t possibly be described as p in relation to nuclear reaction yield…this doesn’t seem to be that hard of a request.
Without cites I will just chock these arguments up to pedantry and move on.
You want a cite that says… that momentum is not the same thing as energy? I’m not sure how to find that. I can find cites that say what momentum is, and cites that say what energy is, but few bother to say that they’re not the same thing, because why would one need to say that in the first place?
The domain is fusion, and in the domain of fusion, there is no type of energy that is the same as momentum. The answer to “what kind of energy isn’t momentum” is “all of them”. In extremely relativistic regimes (again, not fusion, which is relativistic but not extremely relativistic), energy is approximately equal to the magnitude of momentum, but that “magnitude” is in there for a reason.
Momentum is a function of an object’s mass and velocity.
If you are not measuring the magnitude of momentum, you are measuring the inertia, or mass.
Momentum (Mass x Velocity), the collective momenta are a critical component of all the formula I can fine, and critical for total energy to be conserved (first law of thermodynamics).
The math simply doesn’t work out if you don’t account for the conservation of momentum in the context of total energy.
Besides the difference that it is conserved, Momentum (Mass x Velocity) and Kinetic (Mass x Velocity^2) are very very similar terms.
In every cite above that is how momenta was use, as a way to ensure that Total energy was conserved.
Without a cite, your argument does nothing but focus on minutiae.
But once again, feel free to provide formula that ignore momenta in the Total energy of a the system in this case…but be sure you can account for the ‘mass loss’ problem.
But if you are still thinking of an ideal box, and energy content of that ideal box that does explain a lot, I am talking past that basic concept.
To avoid the math, let me link to this image that was in my above cites,
Remember that with practical fusion reactions, and not idealized perfect enclosed boxes there is a reduction in “mass” during fusion.
A small portion of this energy may be released via gamma radiation, where the momentum directly relates to the frequency and thus the energy of the release particle.
But most of the released energy in exothermic fusion is typically in a Proton, Neutron or helium nucleus (Alpha particle). These particles are often traveling at a significant portion of the speed of light depending on the reaction type (10%+) and in general baryons are heavy enough to be relativistic in these interactions anyway.
I think that momentum is hidden inside another term like the de Broglie wavelength and you are maybe overlooking the importance of it to the equation.
But momentum being conserved is critical to the core concept of the reduced mass of the main product of fusion.
You are changing the goal post but Louis De Broglie got his Nobel Prize for the idea of matter having a wavelength. Is there a reason you glossed over the first series related to massive particles? This was confirmed in the Davisson–Germer experiment
But I will provide yet more cites, despite zero being returned my direction.
Nowhere in the above do you justify p = E/c
As chronos notes, that is only valid for a photon, and is not valid for an electron. Heck it is on one of the pages you cite earlier. Explicitly listed for a photon. Indeed a number of your cites actually contradict what you claim if you read them with any care.
Otherwise you just started with E = (1/2) mv^2 and ended up with it.
And to be clear, the main reduction is E = pc, is the main reduction for massless simple calcs for the energy momentum relation, and the cites have been handed several times.
But the above claim was that “momentum isn’t energy”, when it is a critical portion of total energy as described with E = pc even if it is for a massless particle.
