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Binding Energy?

  1. Jul 22, 2007 #1
    In the book I'm studying, there is a graph for the potential energy of a system made by a free proton and a free neutron, the graph shows that the potential energy of the system as a function of the distance between the particles is zero for large distances, it has a minimum(aprox. -40MeV) at about 1fm, increases with the distance between 1fm and 3 fm and is very large for distances smaller than 0.5fm. As I understood this happened because of the nuclear force and i think that the large value of the potential energy for distances smaller than 0.5fm wasn't explained(in the book I'm studying), but that for now i don't care. After explaining nuclear force the book explains the binding energy, and it states that a nucleus as a smaller mass than the sum of the masses of it's constituents if they were apart, it also says that because the mass is a measure of energy, the total energy of the nucleus is less than the combined energy of the separated nucleons. And that this difference in energy is called the binding energy. And that I could calculate the binding energy using E=m(c^2).
    What i don't understand is this:
    Does the graph includes solely the change in the potential energy caused by the existence of the nuclear force?, or it includes too the binding energy?, or is the binding energy related with the nuclear force?, if so why does the mass changes with the union of the nucleons?.
    The book also says that the binding energy is the energy that must be added to the nucleus to break it apart, and with this statement it looks like that the binding energy is caused by the nuclear force, because as i understood the nuclear force is what keeps the nucleus together.
    What really confuses me is, as i look at the graph i would expect that if a proton and a neutron were apart and then joined together, the potential energy of the system would be changed from zero to a minimum, the variation of the energy of the system would be released somehow, don't see why should exist any change of the mass, and the energy required to take the particles apart then would be the -40MeV as i referred, that i would expect to be the binding energy.
     
  2. jcsd
  3. Jul 23, 2007 #2
    Suppose that initially your two nucleons with masses m_p and m_n were far apart and not moving. The total mass of the system was m_p + m_n. The total energy was E_1 = (m_p + m_n)c^2. When you let the nucleons to approach each other, their energy (as a function of distance) will decrease by the amount shown in the graph your mentioned. The equilibrium will be reached when the potential energy reaches the minimum. So, in the bound state the total energy is E_2 = E_1 - 40 MeV and the total mass is m_2 = E_2/c^2, which is lower than the original mass. The potential energy in the minimum (40 MeV) is called the "binding energy".

    Of course, there is the energy conservation law. So, in fact, the energy does not disappear when two nucleons bind together. The energy is changed to other forms. You can guess what are these other forms by looking at a picture of an atomic bomb explosion.

    After your nucleons formed the bound state and the excess energy of 40 MeV has been released (in light, heat, kinetic energy of particles, etc.) you will need to spend 40 MeV of energy if you want to separate the nucleons.

    Eugene.
     
  4. Jul 23, 2007 #3
    i didn't understood yet

    What i didn't understood in your explanation is:
    when you sat that E_2=E_1-40MeV, your saying the total energy is E_1 because of the mass of the particles less 40MeV because of their position in space, you also say in the end that the change in the total energy in the system is released somehow and possibly leaves the system, and the amount of energy released is 40MeV. Then you say that the mass of the system(after the particles joined together) is equal to E_2/(c^2), now if i think to calculate the total energy of the system, and if the potential energy of particles because of their position is -40MeV, the total energy is m_2(c^2)-40MeV=E_2-40MeV=E_1-80MeV. If i remembered that as the particles joined together there was 40MeV leaving the system somehow, there is still 40MeV missing in the system with your calculations. So in order to exist conservation of energy what it looks like you are saying is that as the particles join together the potential energy because of their position(caused as i understood only by the nuclear force) becomes zero for the distance were before their potential energy was -40MeV.
    Is that what you are saying?, if so could you explain how that happens?, that part is still confusing me.
     
  5. Jul 23, 2007 #4
    No, this is not correct. You double-counted the binding energy. The correct answer is that in the bound state the total energy is

    E_2 = m_pc^2 + m_nc^2 - 40 MeV

    The total mass is

    m^2 = E_2/c^2

    This is written under condition that no part of the system is moving, or that kinetic energy of this movement can be neglected.

    Eugene.
     
  6. Jul 23, 2007 #5
    When i look at E_2 = m_p(c^2) + m_n(c^2) - 40 MeV, i'm interpreting it as: the total energy in situation 2 (particles together) is the energy because of the particles masses(individual mass m_p and m_n) less 40Mev because of their position (they have -40MeV when they are together because of the nuclear force), is that correct?
    If that is correct, then mass in situation 2 would be m_p+m_n, but that shouldn't be correct because you wrote m_2=E_2/(c^2)<=>E_2=m_2(c^2), which is equivalent to say: all of the total energy in situation 2 is because of the mass of the joined particles (were is the potential energy caused by the nuclear force?). Or am i confusing everything?
     
  7. Jul 23, 2007 #6
    I think you are still confused. The fundamental rules are these:

    1. the total energy of a compound system is the sum of energies of constituents plus potential energy of their interaction

    2. If system is at rest, then its total mass m is equal to E/c^2, where E is system's total energy.

    Apllying these rules to the dissociated pair (proton + neutron) we obtain:

    from rule 2: energy of isolated proton E_p = m_pc^2; energy of isolated neutron E_n = m_nc^2;
    from rule 1: total energy of the system E_1 = E_p + E_n (the potential energy of interaction is zero, because particles are separated)
    from rule 2: total mass of the system M_1 = E_1/c^2.

    Applying these rules to the deuteron (the bound state of a proton and a neutron) we obtain:

    from rule 1: total energy of the system E_2 = m_pc^2 + m_nc^2 - 40MeV (the potential energy of interaction is -40 MeV; the negative sign corresponds to attraction)
    from rule 2: total mass of the system M_2 = E_2/c^2

    The mass of the deuteron is 40 MeV/c^2 lower than the mass of separated proton and neutron.

    Eugene.
     
  8. Jul 24, 2007 #7
    So, joining rule 1 and 2, you are saying that the total mass of a system isn't just the sum of the individual masses, the total mass of the system is also affected by the potential energy of it's constituents. Is that what i was missing?
    Could you explain why that happens?, or possibly, if that is too complex, could you just give me a clue. I think, now i understand the mistake i was doing, thank you.
     
  9. Jul 24, 2007 #8
    Exactly. The mass of a compound system is not equal to the sum of masses of constituents


    You may find useful the following general formula to calculate the total mass of any 2-particle system

    [tex] M = \frac{1}{c^2}\sqrt{E^2 - \mathbf{P}^2c^2} [/tex]

    Here [itex] \mathbf{P} = \mathbf{p}_1 + \mathbf{p}_2 [/itex] is the total momentum of the system (the sum of one-particle momenta)

    [tex] E = E_0 + V [/tex]

    is the total energy of the system, where

    [tex] E_0 = \sqrt{m_1^2c^4 + \mathbf{p}^2_1 c^2} + \sqrt{m_2^2c^4 + \mathbf{p}^2_2 c^2} [/tex]

    is the sum of one-particle energies, and [itex] V [/itex] is the potential energy of interaction, which is a function that can depend on particle positions and momenta. The rule I gave you earlier is a simplified version of this rule for the case of particles at rest [itex] \mathbf{p}_1 = \mathbf{p}_2 = 0 [/itex]

    It is not that easy to explain why the above rule should be valid. My favorite explanation involves quantum mechanics and unitary representations of the Poincare group. Are you prepared to listen to this explanation?

    Eugene.
     
  10. Jul 24, 2007 #9
    Thank you for your explanation and effort. I don't think i'm prepared to understand your favorite explanation, because i am using one of those big physics books that covers a lot of subjects and the chapter about nuclear structure is just an introduction to the subject and so is very simplified and superficial.
     
  11. Jul 24, 2007 #10
    You are welcome. I understand your frustration. In my opinion, it is impossible to understand nuclear physics without quantum mechanics (Hilbert spaces) and relativity (the Poincare group). Most "big physics books" do not cover these concepts well, and try to substitute them with "handwavings". For a serious reader this could be very frustrating.

    Eugene.
     
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