# Why does fusion create so much energy? **URL:** \ **Category:** Factual Questions\ **Created:** [August 12, 2017, 1:31am UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398 "2017-08-12T01:31:30Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 14, 2017, 10:03pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/21 "2017-08-14T22:03:30Z")
> [@Itself](#): > > Momentum is not energy. (Classical 3-) momentum is a vector; energy is a scalar. Photons carry momentum (they have to, since they’re massless); heat is a scalar. > > I don’t know what you’re trying to say here. 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. [![](https://img.youtube.com/vi/hFkiMWrA2Bc/maxresdefault.jpg "How does laser cooling work?") ](https://www.youtube.com/watch?v=hFkiMWrA2Bc&t=80) 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 _number_that 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. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 14, 2017, 10:18pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/22 "2017-08-14T22:18:36Z")
> [@John\_Mace](#): > > Each component of that equation has the units of (energy squared). If “p” is energy, the units don’t balance. Your link has nothing to do with the equation we are discussing in which “p” is momentum. Physicists routinely use “p” to mean momentum, and if they meant “potential energy”, they’d use E[sub]P[/sub]. But the “p” in the equation is momentum no matter what symbols are used. Saying that it’s “potential energy” is like saying “m” must be “momentum” because momentum begins with m. 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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 14, 2017, 10:58pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/23 "2017-08-14T22:58:44Z")
P isn’t mechanical energy, either. The problem isn’t that we can’t tell what you’re saying; the problem is that what you are saying is simply wrong. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 14, 2017, 11:13pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/24 "2017-08-14T23:13:32Z")
> [@Chronos](#): > > P isn’t mechanical energy, either. The problem isn’t that we can’t tell what you’re saying; the problem is that what you are saying is simply wrong. 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. I wish everyone the best. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 14, 2017, 11:35pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/25 "2017-08-14T23:35:22Z")
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. ---
**Author:** ![Trinopus](https://avatars.discourse-cdn.com/v4/letter/t/2bfe46/32.png) [@Trinopus](https://boards.straightdope.com/u/Trinopus)\ **Post date:** [August 14, 2017, 11:44pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/26 "2017-08-14T23:44:16Z")
> [@rat\_avatar](#): > > . . . . 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. . . . 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. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 12:03am UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/27 "2017-08-15T00:03:36Z")
> [@Chronos](#): > > 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. 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. [http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html) > [@](#): > > The relativistic energy of a particle can also be expressed in terms of its momentum in the expression > [@](#): > > The relativistic energy expression is the tool used to calculate binding energies of nuclei and the energy yields of nuclear fission and fusion. [http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4) 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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 15, 2017, 1:32am UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/28 "2017-08-15T01:32:32Z")
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. > [@](#): > > And that one must resort to the relativistic energy-momentum relationship of pc. 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). ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 2:55am UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/29 "2017-08-15T02:55:55Z")
> [@Chronos](#): > > 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). Cite and call out what you include in “Energy”, and the math to produce it. This is General Questions and not IMHO, I can provide a cite to proton-proton fusion, which requires relativistic calculations. [http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/solarpp.html#c1](http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/solarpp.html#c1) Once again, you are not providing cites, but only opinions with zero cites. 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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 15, 2017, 1:38pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/30 "2017-08-15T13:38:23Z")
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? > [@](#): > > 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. 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. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 4:34pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/31 "2017-08-15T16:34:17Z")
> [@Chronos](#): > > 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. Or are you equating inertia with momentum? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 15, 2017, 4:41pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/32 "2017-08-15T16:41:37Z")
What mass loss? The mass of any closed system never changes. And yes, momentum and energy are closely related. They’re still not the same thing. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 5:29pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/33 "2017-08-15T17:29:48Z")
> [@Chronos](#): > > What mass loss? The mass of any closed system never changes. > > And yes, momentum and energy are closely related. They’re still not the same thing. Momentum is a part of Total Energy, 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, [![](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/imgrel/relmecon.gif) ](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/imgrel/relmecon.gif) 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. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 5:41pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/34 "2017-08-15T17:41:08Z")
As an example, the de Broglie wavelength of an emitted electron λ = h/p = h/m\*v Note the p, or mass\*velocity, or momentum? This can also be written as: p = E/c = h / λ Does that make the relation to _total energy_ more clear? The de Broglie relations \*directly \*relate the wavelength, momentum, and frequency to the total energy of a particle with mass. This is why I am confused about the claim of there being no relation. This message board’s Unicode support limits my ability to take this relation further. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 15, 2017, 6:57pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/35 "2017-08-15T18:57:40Z")
> [@](#): > > Note the p, or mass\*velocity, or momentum? This can also be written as: > > p = E/c = h / λ Not for an electron, it can’t. That’s only for a photon. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 8:56pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/36 "2017-08-15T20:56:22Z")
> [@Chronos](#): > > Not for an electron, it can’t. That’s only for a photon. 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. [http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html](http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html) [http://www.sjsu.edu/faculty/watkins/debroglie.htm](http://www.sjsu.edu/faculty/watkins/debroglie.htm) > **[Deriving the de Broglie Wavelength](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/Deriving_the_de_Broglie_Wavelength)** > > In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure. By using a series of substitution de Broglie hypothesizes particles to hold … [http://scienceworld.wolfram.com/physics/deBroglieWavelength.html](http://scienceworld.wolfram.com/physics/deBroglieWavelength.html) But ya, it is quite clear that cites do little, but how about the KE equation? E = (1/2) mv^2 2E = mv^2 2Em = m^2v^2 2Em = (mv)^2 As mv = p 2Em = p^2 p = sqrt(2Em) As λ = h/p λ = h/sqrt(2Em) Add in the 2nd form ν = E/h And with substitution you can solve for the phase velocity of the wave for E for a massive particle. E = (1/2) (mv)^2/m _Can be written as_ (1/2) p^2/m E = (1/2) m^2v^2/m E = (1/2) mv^2 Note how it is changed to the Kinetic form with the last step there? This is just another form where P relates to E, and disproves your claim that it is not related to energy. Are you maybe thinking about classical wave _p_, which is the group velocity? Because I can’t find anything this side of the mid 1800’s that matches your claims. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [August 15, 2017, 9:17pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/37 "2017-08-15T21:17:44Z")
So what? You wrote > [@](#): > > p = E/c = h / λ 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. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 9:30pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/38 "2017-08-15T21:30:56Z")
And all of this goes back to: E = mc^2 = sqrt(p^2c^2+(mc^2)^2) Which is not some form of pseudoscience: [http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4) > **[Energy–momentum relation](https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation)** > > In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating total energy (which is also called relativistic energy) to invariant mass (which is also called rest mass) and momentum. It is the extension of mass–energy equivalence for bodies or systems with non-zero momentum. It can be written as the following equation: > > > > > > E > > 2 > > > = > ( > p > > ... And I am not sure why I have bothered to keep responding when those cites have been posted before and handwaved away. but note the second link particularly, where it states > [@](#): > > energy is typically given in units of electron volts (eV), momentum in units of eVc^−1, and mass in units of eVc^−2 And remember that the norm is to use natural units where c = 1; c^-1 and c^-2 all == 1. 1 = 1 1^-1 = 1 1^-2 = 1 it the energy–momentum equation reduces to: E^2 = p^2 + m^2 Which results in nice methods for invariance. But anyway, I need to move on, as it appears I will not discover the nature of the claim of my misunderstanding in this thread. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 9:33pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/39 "2017-08-15T21:33:15Z")
> [@Francis\_Vaughan](#): > > So what? > > You wrote > > 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. The second version was the Planck–Einstein relation, the first was the de Broglie hypothesis. I am not going to re-prove the de Broglie hypothesis, especially without any claim any cites or claims outside of dismissal. ---
**Author:** ![rat\_avatar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/rat_avatar/32/255_2.png) [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)\ **Post date:** [August 15, 2017, 9:44pm UTC](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398/40 "2017-08-15T21:44:49Z")
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 I will provide them yet again. [http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4](http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4) > **[Energy–momentum relation](https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation)** > > In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating total energy (which is also called relativistic energy) to invariant mass (which is also called rest mass) and momentum. It is the extension of mass–energy equivalence for bodies or systems with non-zero momentum. It can be written as the following equation: > > > > > > E > > 2 > > > = > ( > p > > ... 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. [Previous page](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398.md?page=1) [Next page](https://boards.straightdope.com/t/why-does-fusion-create-so-much-energy/793398.md?page=3)