# Nuclear binding energy

1. Jan 28, 2013

### springwave

Hey,

I'm having a hard time trying to understand what nuclear binding energy really means.
In most of the introductory texts I have, they say some of the mass of the nucleus appears as binding energy (mass defect).

According to the definition, shouldn't it just be the work done by strong interaction minus work done by electrostatic forces while assembling the nucleus. If we had a mathematical equation for the strong interaction, could we calculate the binding energy by integrating it over displacement from infinity.

I don't understand how exactly it is related to the mass defect. Unfortunately I still don't fully understand the theory of relativity. Any idea how I should proceed? I'd like to getbatleast the basic idea of what it really is.

2. Jan 28, 2013

### mathman

3. Jan 29, 2013

### springwave

Okay let's consider this simple situation.

If I have two isolated protons initially at an infinitely large distance apart.(of course there is a uniform gravitational field, so that we can measure weight) (let's say it is possible to weigh them using a machine whose mechanism is purely mechanical or to be specific based on gravity like the common weighing machines we find at home, I know it's not possible but for now if we consider it to be possible)
If I weight these isolated protons, say I obtain a weight of w° of each.

Now if I bring these two close to each other, I have to do positive work, ie inject energy into the system. After this, if I weigh the system using the same weighing machine, according to what I understand, the new weight (w) should be more than than the sum of the initials (2w°)

If this is true which of the following would be a correct explanation?

1. The mass of each of the protons has increased, and hence they are more strongly attracted by the gravitational field, leading to a larger weight when measured.

2. The mass of the protons is still the same, but the extra positive energy the system has (electrostatic potential energy), behaves exactly like mass ie is also "attracted" by the gravitational field, and hence leads to the extra weight of the system when measured.

3. I am totally confused, and I should go back to the basics again.

4. Jan 29, 2013

### Khashishi

I don't think there's any real difference between 1 and 2, since they give the same measurement. You can pick one that works better for your calculation.