I've asked before, but I'm still confused as to what the mass energy equivalence implies. I've taken an introductory course in relativity, so I only covered special relativity. From what I gather, all things have a rest mass. As you increase your energy, your mass increases as well, and the rest mass behaves as some potential. But then when we talk about fusion and fission, we are splitting the atoms to generate energy-breaking apart the potential stored in forces between atomic components. This kind of makes sense to me. But then are all energies and masses equivalent? An electron has a rest mass, is this considered energy as well? Is the mass stored in objects I hold due to the gravitational field simply too small for me to notice? Furthermore, if mass is energy, then why do nuclei weigh less? Is some of their mass going into the bonds? The fission and fusion also confuse me, both are building towards a more stable atom (32?) but somehow both paths generate energy.


As you increase your energy, your mass increases as well, and the rest mass behaves as some potential.

I assume you mean "as you increase your velocity"; correct me if I'm wrong. Rest mass does not depend on velocity. Be careful to distinguish between rest mass and relativistic mass. Rest mass, denoted as $m_0$, is the one that is equivalent to energy via $E_0 = m_0 c^2$. Relativistic mass is something like the resistance of the object to acceleration, but as a concept it's not used today. To quote einstein,

It is not good to introduce the concept of the mass $M = m_0/\sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.


But then are all energies and masses equivalent? An electron has a rest mass, is this considered energy as well?

Yes. In order to measure that energy, simply annihilate that electron with a positron. Without doing so, you can't "feel" the energy.


Here is how energy works in nuclear fission. Let's take the reaction

$$ n + _{92}^{235}\textrm{U} \to _{56}^{141}\textrm{Ba} + _{36}^{92}\textrm{Kr} + 3n $$

The total mass on the left hand side is

$$ (92 m_p + 143 m_n + E_U/c^2) + (m_n) $$

where $E_U$ is the binding energy of Uranium, or about -1783870 keV. The total mass on the right hand side is

$$ (56 m_p + 85 m_n + E_{Ba}/c^2) + (36 m_p + 56 m_n + E_{Kr}/c^2) + 3 m_n $$

comparing, the $m_n$ and $m_p$ cancel, as they should. The mass difference is $\frac{1}{c^2} (E_{Kr} + E_{Ba} - E_{U})$ which you can look up if you wish. This is the energy liberated by fission.

  • $\begingroup$ My apologies if I was unclear. The rest mass of some object is a potential energy. So then when you have some atom it can either undergo fusion or fission. With fusion, the atoms are small enough that they want to come together. But where is the energy? Isn't the difference in mass going into the bonds? Conversely, when fission happens, the atoms are too large, so they fall apart- but isn't energy return to the mass of the constituent atoms? Why would energy be released? $\endgroup$ – user24082 Sep 19 '13 at 3:51
  • $\begingroup$ The rest mass of a nucleus isn't simply the sum of the rest mass of the constituent protons, neutrons and electrons, it also includes the binding energy, which is the energy that comes from the strong-force interaction between the nucleons. The binding energy is what changes in fission or fusion processes. $\endgroup$ – xuanji Sep 19 '13 at 3:57
  • $\begingroup$ But why is the binding energy released, as opposed to turning back into the mass of particles? For instance, nuclei have less mass than the nucleons-why wouldn't the binding energy just return to the nucleons? $\endgroup$ – user24082 Sep 19 '13 at 4:12
  • $\begingroup$ The rest mass of a proton cannot change. It is a fundamental constant of nature and identical for all protons. Likewise for nucleons. $\endgroup$ – xuanji Sep 19 '13 at 4:19
  • $\begingroup$ So the rest mass does not change, but there is still mass lost when the nucleons are bound. But when they undergo fission, wouldn't the lost potential just "return" to the proton? I'm missing something that should be being conserved. $\endgroup$ – user24082 Sep 19 '13 at 4:21

It all depends on the scale at which you examine things. Just like thermal energy being resolved as the randomized kinetic energy of the parts if examined closely enough, so mass can be resolved into binding and kinetic energies plus some intrinsic masses of the bits if examined at the right scale.

  • $\begingroup$ But how do we know at what level we are probing? Mass is called mass for my body... If we look at the molecules in me will they weight less then my weight? And when atoms come together does their weight evaporate into the bond? Why is that bond energy released, then, in fission, rather then turning back into the mass of the constituent particles? $\endgroup$ – user24082 Sep 19 '13 at 3:32
  • $\begingroup$ Oddly, the weight of the system doesn't change. Nor does it's inertia. These things are functions of energy not of mass. If you can see the molecules in motion that you have resolved part of the energy (which you would previously have measured purely as mass) into internal kinetic modes. I am not following you fission arguments, but I imagine that you are not figuring out exactly where all the energy came from in making up the fissile nucleus in the first place. You might want to read What is so special about iron? and related questions. $\endgroup$ – dmckee Sep 19 '13 at 3:54
  • $\begingroup$ I will read that, but I feel like I just have some really basic misunderstanding. With fissile nuclei, the energy being let out comes from the broken bonds right? But the broken bonds took their potential from the rest mass? So why wouldn't it just return to the particle? $\endgroup$ – user24082 Sep 19 '13 at 4:23
  • $\begingroup$ In fissile materials the whole nuclear configuration is a local minimum, but is not a global minimum (think about a ball in a cup-shaped depression on the top of a hill). Some external energy was necessary to put it into that configuration in the first place. $\endgroup$ – dmckee Sep 19 '13 at 4:45

Well I'm not up on current practice, but when I studied such things, some particle physicists used a system of units where c = hbar = 1

So in that system of units you simply have E = m and also E = (nu)

When accelerated particles approach the speed of light (c) , they don't add much in the way of speed, but their mass increases.

And in any radioactive decays, or Nuclear fission or fusion, the mass balance has to involve the energies of the various components, treating mass and energy as identical.


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