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Does the sum of the mass of two bodies vary with the distance between them ?

  1. Sep 14, 2007 #1
    Say we have 2 bodies of m1 , m2 .
    The distance between them is
    a) r1 Ma=m1+m2
    b) r2>r1 Mb=m1 +m2
    In each case the mass was measured by an inertial observer . Is it true that Ma<Mb
    I saw this kind of thinking in measuring the energy neccesar to separate a nucleon from a nucelous . It said that a nucleus mass is less than the mass of the component nucleons added .
    I only read special relativity . Is this a general relativity case ?
  2. jcsd
  3. Sep 14, 2007 #2
    Suppose that two bodies [itex]m_1[/itex] and [itex]m_2[/itex] are at rest and there is interaction between them with potential energy [itex]V(r)[/itex] . Then the total energy of the system is

    [tex] E(r) = m_1c^2 + m_2c^2 + V(r) [/tex]

    and the total mass

    [tex] M(r) = E(r)/c^2 = m_1 + m_2 + V(r)/c^2 [/tex]

    If the interaction is attractive [itex]V(r) < 0[/itex], then the total mass of the system is less than the sum of masses of its constituents. These rules are valid for all kinds of interactions: electromagnetic, nuclear, and (I believe) gravitational as well.

  4. Sep 14, 2007 #3


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    Basically, the answer is yes, though it turns out that GR defines more than one type of mass.

    See for instance This wiki article or this one

    for some of the details on defintions of mass in GR. (Note also that I wrote the second article above).

    If you use the appropriate GR defintion of mass (say the ADM mass), then if you separate m1 and m2, you can measure their individual masses in isolation (as long as they are in asymptotically flat spacetimes so that the ADM mass concept applies).

    Then one can allow m1 to approach m2, and keep the assumption that spacetime is asymptotically flat so that the ADM concept applies. The mass of the system, m_total, will then be lower than m1 + m2, and for weak fields m_total - m1 - m2 will be a negative number equal to the Newtonian binding energy of the system.
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