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When we have two bodies and a central force acting towards the center of each other, we could treat the whole problem as a one body problem by introducing the relative coordinate. My question is, when you were to calculate the relative coordinate, you are working out the acceleration of a fictitious particle, which does not exist.

However, if you work out the coupled differential equations, you would get exact acceleration of both particles.

Therefore I think reduced mass is not a reliable method , but I was told that the reduced mass method can get the exact solution as the coupled differential equation. Therefore I think I am wrong, could anyone shed the light on this please?

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Reducing a two body problem into a effective one body problem is just an mathematical manipulation to simplify the coupled differential equations, that you have to solve, and convert them to an easier problem (two uncoupled differential equations). There is no approximation involved. Actually, it is just a transformation of variables and after you have solved the problem in these variables you can transform back and you have the exact solution of your original problem. The concept of a reduced mass is only introduced, because after the transformation the differential equation of the relative coordinate looks like one of a fictitious particle with this particular mass.

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  • $\begingroup$ Hi, pressure. Do you mean that we can get the exact solution of each mass (i.e. acceleration) from the acceleration of fictitious particle? And is the acceleration of fictitious particle useful? $\endgroup$
    – el psy Congroo
    Commented Jan 21, 2014 at 19:24
  • $\begingroup$ Yes, if you have solved the equation of the fictitious particle and the second one of the center of mass, which performs a uniform motion, you can get the exact solution for each mass (for example their acceleration). The concept of this fictitious particle with the reduced mass is extremely useful, because as have shown on mathematical level you can think about this problem (when neglecting the uniform motion of the center of mass) as the motion of only one particle in an external potential. $\endgroup$
    – pressure
    Commented Jan 22, 2014 at 7:34
  • $\begingroup$ Sorry, I don't quite get the term external potential, do you mean only one particle was under a force? $\endgroup$
    – el psy Congroo
    Commented Jan 22, 2014 at 10:05

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