# Why is binding energy $\Delta mc^2$?

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?

• 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. – Yashbhatt Jul 31 '14 at 15:53
• possible duplicate of About mass defect – Void Jul 31 '14 at 17:05
• 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. – Alan Rominger Jul 31 '14 at 17:28
• 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. – rob Jul 31 '14 at 19:06

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.

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.