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I Why does a Mass Defect Exist if |PE| Increases?

  1. May 17, 2017 #1
    Hello. Can you shed a different light on why mass defects exist please, so that I might finally grasp it intuitively?!

    I've had a look at these nuclear threads and one about GPE,
    https://www.physicsforums.com/threads/why-is-there-a-mass-defect-in-the-nucleous.374443/
    https://www.physicsforums.com/threads/why-mass-defect.601024/
    https://www.physicsforums.com/threads/gravitaional-field-can-store-mass.371845/

    ...but they don't exactly answer why the mass of the constituents of a nucleus - whether it be the sun and the earth or an atomic nucleus and its nucleons - is greater when separate than combined.

    So far, I've gathered the following (in the case of an atomic nucleus - tell me if I'm wrong on any of these points):
    1) By mass-energy equivalence, the masses decrease as a result of losing energy, usually by photon emission (I'm guessing this isn't a result of losing matter from the fundamental particle - making it a different particle - but just energy?)
    2) The strong nuclear force (SNF) is responsible for this energy change and F = -dV/dr
    3) There is always a minimum energy in a bound state that the system will lean towards.
    4) It is only the change in energy that is relevant, not the quantity (hence we can have negative energy).
    5) Binding energy is another term for nuclear potential energy.
    6) Energy is only released in fusion when the binding energy per nucleon of the product is greater than that of the reactants, and vice versa for fission.

    What my mind keeps playing out that makes me question my sanity here is the following. A neutron and proton are at rest. The system has total energy E = mnc2 + mpc2. An external force then nudges the proton into the neutron for them to bind via the SNF and make a deuteron.

    The energy dissipated is the kinetic energy of the proton when it reaches the repulsive range of the SNF. Since its rest energy from before the external force acted is not affected, where has its mass gone? Am I missing a trick with my 3rd point above here?
     
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  3. May 17, 2017 #2

    mfb

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    It just loses energy, yes.
    The electromagnetic interaction shouldn't be neglected, it contributes as well.
    Right.
    Up to a sign, depending on definition.
    As you start with multiple nuclei in fusion, and end with multiple nuclei in fission, there is no 1:1 comparison. The total binding energy has to be larger (using positive values for the binding energy) after the reaction.

    In the deuterium nucleus, it still behaves like a particle with the proton mass. The naive expectation that you can add masses of all components just doesn't work.
     
  4. May 17, 2017 #3

    Drakkith

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    The most fundamental answer you can get is that because of the mass-energy equivalence principle, if energy is released from an object, that object loses mass. Since energy is released when nucleons bind together, their mass must be less when in a bound state than when separated. We don't know why mass-energy equivalence exists. We only know that it does.

    The mass of a deuteron is less than the sum of the masses of a free proton and free neutron. So you will have an additional amount of energy released equal to this mass deficit.
     
  5. May 18, 2017 #4

    vanhees71

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    The mass defect is an example what's truely behind what's called "mass-energy equivalence", and that's the way Einstein understood it from his first paper on the subject on. Only inaccurate popular-science and bad textbooks perpertuate a misunderstanding of the early relativists from a time before Minkowski clarified the issue. The iconic formula ##E=mc^2## is the most misunderstood formula in the whole universe!

    The point is that for a closed system total energy and total momentum build a four-vector, ##p^{\mu}=(E/c,\vec{p})##, which is conserved, and mass is defined (sic!) as the invariant quantity ##m^2 c^2=p_{\mu} p^{\mu}##.

    Now take as an example an excited nucleus which shows ##\gamma## decay, i.e., it spontaneously emits a photon. Suppose initially the nucleus was at rest. It's total four-momentum is
    $$p_1=(M c,0,0,0).$$
    Now the photon is emitted and you are left with the unexcited nucleus. Let's determine its mass ##m##. Since the system of the unexcited nucleus and the photon is closed we have energy-momentum conservation, i.e.,
    $$p_2+k=p_1,$$
    where ##p_2## is the four-momentum of the unexcited nucleus and ##k## the four-momentum of the photon. The photon is massless, and we can assume without loss of generality that the photon momentum is in ##x## direction, i.e., you have
    $$k=(E_{\gamma}/c,E_{\gamma}/c,0,0)$$
    and thus
    $$p_1=(Mc,0,0,0)=k+p_2=(E_{\gamma}/c,E_{\gamma}/c,0,0)+(E_2/c,\vec{p}_2).$$
    Comparing the spatial components you get
    $$\vec{p}_2=(-E_{\gamma}/c,0,0)$$
    and comparing the temporal component gives
    $$(E_2+E_{\gamma})/c=M c.$$
    Thus our four-vector of the unexcited nucleus becomes
    $$p_2=((M c^2-E_{\gamma})/c,-E_{\gamma}/c,0,0).$$
    Now the mass of the unexcited nucleus is given by
    $$m^2 c^2=p_2^2=(M c^2-E_{\gamma})^2/c^2-E_{\gamma}^2/c^2=M^2 c^2 - 2ME_{\gamma} <M^2 c^2.$$
    The invariant mass of the unexcited nucleus is smaller than the mass of the excited nucleus. The difference is given by the excitation energy divided by ##c^2##.
     
  6. May 18, 2017 #5
    Thank you for your replies. However, they don't answer why if you put energy (i.e. the KE of the proton) into the system you get more out - and based on the fact we know the exact amount we get out using the mass defect, surely we know what causes this effect? I know you're going to say 'binding energy increases', but this doesn't mean much - surely the kinetic energy of the proton goes into increasing the binding energy, so the mass of the system should increase, not decrease. The only energy out is the photon, which I can understand is due to quantisation but its energy would surely not exceed the kinetic energy of the proton.

    In the case of a proton and neutron the electromagnetic interaction shouldn't be ignored because of the non-zero charge of the quarks that constitute them not always summing to zero in a neutron depending on its position?

    Why is it naive?

    Energy is released equal to a part of the proton's initial kinetic energy (following my example above), not some arbitrary amount of the nucleons' masses which we can't explain but can quantify, surely?

    Great reply vanhees71; I followed the maths despite not having done four vectors/tensors for 2 years! Although I hate to say that I understand why the example you give loses mass-energy already, because the loss of energy is obviously attributed to the nucleus' excited state, not some random decrease in mass due to an interaction which should conserve mass-energy, as with my fusion example.


    I feel like I'm missing a fundamental concept here - which does normally turn out to be the case, so please treat me like I'm 5. You're putting work into a system which then gives out energy. Does the energy come from the SNF / EM field of the particles, and if so why does it increase again when they're separated? It can't be because you're doing work on the system, as that only goes to decrease the potential energy and increase the kinetic energy, cancelling each other out.
     
  7. May 18, 2017 #6

    vanhees71

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    My point was that you should work with the concept of four-tensors (including scalars and vectors as 0th and 1st-rank tensors). Then it's clear that mass is a scalar and energy the time-component of the energy-momentum four-vector. That's why the kinetic energy doesn't attribute to mass as falsely often stated. The correct relation is
    $$p_{\mu} p^{\mu}=m^2 c^2$$
    or in components
    $$E=c \sqrt{m^2 c^2+\vec{p}^2}.$$
    Only in the restframe of the particle (if ##m>0##!) you do have ##E=E_0=m c^2##.

    Mass, which is unambiguously in all modern settings the invariant mass, for itself is not conserved. This was shown with my example and is a slightly modernized version of what Einstein considered in his original note on "energy-mass equivalence" of 1905.
     
  8. May 18, 2017 #7

    Drakkith

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    It does not. Binding energy is defined as the energy required to disassemble a system into its separate parts. If you take a deuteron and disassemble it into a proton and neutron, the amount of energy needed is equal to the binding energy. When you take a proton and neutron and let them bind together, the energy released is equal to the binding energy plus any other energy, such as kinetic energy, the two particles may have had prior to binding.

    Notice that in the definition of binding energy there is nothing about kinetic energy. There is, in fact, no way to increase the binding energy of a system except by changing what the system itself is made up of.

    Consider this example:

    You release a bowling ball down into a depression such that it impacts another bowling ball stationary in the center. The 1st ball is initially stationary and you do nothing except let gravity accelerate it during its descent. After colliding both return to being stationary after jostling about for a bit. Clearly energy has been released. In this case the energy was gravitational potential energy that was released as heat and sound during and after the collision.

    Now repeat the scenario but give the 1st ball a push such that it collides with the 2nd at a higher velocity than before. Did the GPE of the 1st ball change in any way just because it was moving? No. The GPE of the 1st ball is still exactly the same as before. The energy released in the collision is now equal to the GPE of the 1st ball plus an additional amount equal to the kinetic energy you gave the 1st ball upon release.

    Now repeat both scenarios except drop the 2nd ball from the opposite side of the depression. This is analogous to the binding between a proton and neutron. The binding energy in this case is the energy needed to roll the balls back up the sides of the depression to their starting points (that's not the technical, 100% accurate definition, but for this thought experiment it's close enough). The extra kinetic energy you give them doesn't factor into it at all, even though it does add to the total energy released once both balls have come to a standstill.

    Note that while the balls obviously have their GPE converted into KE in this example, it isn't a requirement. We could make the sides of the depression very muddy such that the balls have nearly zero kinetic energy the whole way down and the energy released for the whole process would be equal to the GPE.

    Putting work into pulling apart the nucleons (increasing their potential energy) does require that you give them some initial KE to move them apart, but that KE is lost as soon as you bring them to a halt. So the initial and final KE are both zero, but final PE is NOT zero. This change in PE is the binding energy.
     
  9. May 18, 2017 #8

    mfb

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    To fuse a proton and a neutron, you don't need kinetic energy. The particles have to meet each other of course, but that can happen as slow as you want. You get energy out because the particles attract each other. You get some sound and heat if you let a magnet attach itself to another magnet - a very similar concept, energy is released.
    It does exceed the initial energy of the fusing particles. That is the point.
    In this special case it can be ignored.
    The idea comes from our everyday life. It doesn't have to apply everywhere.
     
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