4
$\begingroup$

As we know the mass-energy equivalence relation $E=mc^2$ originally came from special relativity. And the binding energy is $\Delta mc^2$. How do we know that the extra mass coming from theoretical calculation is associated with the binding energy? What is special relativity inside the nucleus?

$\endgroup$
4
  • $\begingroup$ Maybe first we measured the mass of the whole nucleus, then only of the protons. So, then we could conclude that the missing part must be binding energy. $\endgroup$
    – Yashbhatt
    Jul 31 '14 at 15:53
  • $\begingroup$ possible duplicate of About mass defect $\endgroup$
    – Void
    Jul 31 '14 at 17:05
  • $\begingroup$ I think you would help your question by elaborating more on the "special relativity" part about the nucleus. The binding energy has several things lumped into it. It's not only from some distance-potential relationship. Kinetic energy within the nucleus is also a part of it to some degree. There are some finer accounting details you can get into. However, there's not any singular confusion that obviously needs to be answered as far as I can tell here. $\endgroup$ Jul 31 '14 at 17:28
  • $\begingroup$ Having answered both questions I disagree that they are duplicates. This question is about the relationship between binding energy and mass, while the other is about the relationship between binding energy and arbitrary zero definition associated with any sort of potential energy. $\endgroup$
    – rob
    Jul 31 '14 at 19:06
5
$\begingroup$

The relationship between nuclear masses and mass differences and binding energies has been confirmed by many decades of careful nuclear spectroscopy.

It's possible to measure an atom's mass by purely mechanical means: you ionize the atoms, accelerate them to a known energy, and use a magnetic field to measure their momentum. This lets you come up with an independent measurement of the mass of any particular nucleus. There are other methods, too.

It's also possible to absolutely measure the energy released in a nuclear reaction, by capturing all of the decay products and measuring their energy.

Now you have two datasets: a set of masses, and a set of transition energies. There's a lot of complicated feedback that goes into connecting the one to the other, but the long and short of it is that the equivalence between mass difference and binding energy difference is an experimental result, consistent with the prediction of relativity.

$\endgroup$
6
  • $\begingroup$ the mass of a N-particle non-interacting system should be equal to $M^2=\left( \sum\limits_{i=1}^N E_i \right)^2-\left( \sum \limits_{i=1}^N {\bf p } _i \right)^2$. Which part of this formula is changing inside the Uranium nucleus to get the mass defect? I would say that it is the sum of the energies? $\endgroup$ Nov 30 at 12:25
  • $\begingroup$ You’ve almost said it yourself: inside of a nucleus, the non-interacting approximation is terrible. See this answer for a way to think about the energy stored in the fields which mediate the interactions. $\endgroup$
    – rob
    Nov 30 at 16:25
  • $\begingroup$ Can we think about this using the following analogy: Suppose that we had a gas in a balloon. Since the motion is random, $\sum\limits_{i=1}^N{\bf p}_i=0$. If we would start to heat up the gas, however, $\sum\limits_{i=1}^NE_i$ and consequently $M$ will increase. Cooling the gas does the opposite. It should be something similar in a heavy nucleus, where nucleons are like energetic gas particles. Once split into smaller pieces, the energy of the nucleons decreases due to stronger interaction holding them together and therefor the mass is smaller (things get cooler). Does this make sense? $\endgroup$ Dec 1 at 21:08
  • $\begingroup$ It does not. You seem to still be thinking only about the particles and neglecting the (possibly negative) energy stored in the field. If the answer I linked in my other comment isn't helpful for you, please ask a follow-up question and I'll try to answer there. $\endgroup$
    – rob
    Dec 1 at 21:12
  • $\begingroup$ Well this energy should reduce the term $\sum\limits_{i=1}^N E_i$. $\endgroup$ Dec 1 at 22:35
5
$\begingroup$

Rob's explanation of how we know is bang on, but I wanted to address a part of your question that might point to a basic misunderstanding.

What is special relativity inside the nucleus?

Everything is always relativity. Everything. Always.

All those Newtonian equations like $T = \frac12 m v^2$ for the kinetic energy can be properly understood as approximations (to the correct relativistic expression) that apply in the limit of low relative velocity. Including the notion that binding two objects of masses $m_1$ and $m_2$ gives you a object of mass $m = m_1 + m_2$.

There is no magic point at which relativity suddenly begins applying: it always applies. It's just that the Newtonian math is often easier, so people talk about the point at which they can no longer tolerate the imprecision that comes with using Newtonian physics.

The reason we can't tolerate the Newtonian view in nuclear physics is that binding energies are a non-trivial fraction of the mass-energy of the constituents. It's not a big fraction, but we can get access to it and that factor of $c^2$ is huge.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.