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Two interacting bodies in an external field

  1. Aug 30, 2010 #1
    I wonder about the general solvability of the problem of two point masses, interacting via a central force, exposed to an external field.

    Some special cases seem easy:

    - interaction by a rigid rod
    - any interaction, but the field and interaction forces "decoupled" (example: two spring-connected bodies of equal mass in a constant field: oscillation and acceleration by the external field take place independently)

    A non trivial case:

    - two spring-connected masses in a gravitational field

    What I find especially hard to model in this case is the rigidness of the spring perpendicular to the connecting line (which I feel necessary to presuppose somehow).

    Can anyone give me some hints under which general conditions on the interaction and the external forces the problem has a closed solution - and where I might find an elaborate treatment of the case of two spring-connected masses in a gravitational field.

    Many thanks in advance
    Hans
     
  2. jcsd
  3. Aug 30, 2010 #2
    If the external field is constant and homogeneous (i.e. mgh) then the center of mass moves trivially. It remains to find the relative motion of the two masses with respect to the center of mass. In the case the two masses form a rigid body, conservation of angular momentum says they rotate at all times the way they rotate at the beginning. In the case of a central potential v(x,y,z) = f(r) then you can use the method of reduced mass: you can consider the motion of a single mass m=(m1m2)/(m1+m2) that moves in the same potential. This way you have reduced the 2-body problem to a 1-body problem, easier to solve. In the case of an harmonic potential, you can decompose the motion into 3 independent harmonic oscillations. In the case of 1/r potential, type "Euler motion" in Wikipedia.
     
  4. Aug 30, 2010 #3
    Hello Petr,

    thanks, but I am not quite sure how to understand your answer. You distinguish four cases:

    1. In the case the two masses form a rigid body, conservation of angular momentum says they rotate at all times the way they rotate at the beginning.
    2. In the case of a central potential v(x,y,z) = f(r) then you can use the method of reduced mass: you can consider the motion of a single mass m=(m1m2)/(m1+m2) that moves in the same potential. This way you have reduced the 2-body problem to a 1-body problem, easier to solve.
    3. In the case of an harmonic potential, you can decompose the motion into 3 independent harmonic oscillations.
    4. In the case of 1/r potential, type "Euler motion" in Wikipedia.

    Case 1 is clear to me.
    In case 2, do you mean a central potential "between" the two bodies or as the external potential? (I have to admit, I remember to have learned to know "reduced mass" only in the context of the two-body system without external field.)
    Are cases 3 and 4 intented to be special cases of case 2?

    Thanks for a clarification!
    Hans
     
  5. Aug 30, 2010 #4
    Case 1 is a general hypothesis I make on all other 3 cases, and it's necessary in order to decompose the motion of the center of mass from the motion of the relative coordinates. For example, if you put a system like cases 1,2,3 in a gravitational potential 1/r (not mgh) this decomposition fails, because not all the points in the system feel the same force.
    So yes, when I say "central potential", I mean the potential between the masses, and the external potential is in all cases assumed to be mgh.

    In formulas,

    [tex]E=\frac{1}{2}m_1\mathbf{v}_1^2\,+\,m_1gz_1\,+\,\frac{1}{2}m_2\mathbf{v}_2^2\,+\,m_2gz_2\,+\,V(|\mathbf{x}_1-\mathbf{x}_2|)[/tex]

    If you define the total and reduced mass

    [tex]M=m_1+m_2[/tex]

    [tex]\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}[/tex]

    and introduce the coordinates of the center of mass and the relative coordinates

    [tex]\mathbf{X}=\frac{m_1\mathbf{x}_1+m_2\mathbf{x}_2}{M}[/tex]

    [tex]\mathbf{x}=\mathbf{x}_2-\mathbf{x}_1[/tex]

    then

    [tex]E=\frac{1}{2}M\mathbf{V}^2+MgZ+\frac{1}{2}m\mathbf{v}^2+V(|\mathbf{x}|)[/tex]

    i.e. you have decomposed the coupled motion of m1 and m2 into the uncoupled motion of M and m.
     
  6. Aug 30, 2010 #5
    Sorry for any confusion, but I am dedicatedly interested in the non-mgh case (where separation cannot be performed): I wonder under which circumstances there can be a closed solution nevertheless.
     
  7. Aug 31, 2010 #6
    Uhm, for example with a 1/r external potential we actually obtain a 3-body system (the third one, fixed at the origin, being of infinite mass, but it's still a body)... difficult to solve, generally, but if you put here a particular example we'll have it a look. Generally the concept of "center of force" could be useful, in addition to the center of mass, and the two points don't coincide for an arbitrary external potential.
     
  8. Aug 31, 2010 #7
    You are right, mgh is not constant, it's the force you derive from it being constant. The two bodies connected rigidly won't start rotating if mgh is turned on, regardless if they are at the same height or not, because if you calculate the momentum of the forces the two bodies feel (with respect to the center of mass), you'll find it is zero, so the angular momentum is constant and equal to the initial value (zero or whatever it is).

    So, you want to solve the system of two bodies connected by an ideal spring in an external mgh potential, right?
     
  9. Aug 31, 2010 #8
    Please consider a homogenous, but not constant field, for example given by mgh2.

    Two bodies connected by a rigid rod falling free will start rotating when the field mgh2 is suddenly switched on, except they are at the same height at this moment.

    I guess this case can be calculated easily. But what about two bodies connected by a spring (rigid perpendicular to the connection line)? They will start rotating and oscillating, won't they? And how to calculate this?

    Thanks for your interest!

    Hans
     
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