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The reduced mass in the two body problem is $\mu= \cfrac{m_1 m_2}{m_1 + m_2}$. Is there any analog to this with interacting charged particles (or at least that is of use somewhere in physics)? I have not seen anything like this anywhere, but was curious if someone else has. I would imagine that this is not possible for the repulsive case.

Edit: To clarify, I understand that the same equations hold for the attractive force since they are both $\frac{1}{r^2}$ force laws. I was curious if anyone had used this for charges before.

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What do you mean by two body problem? And in the title of the question you are asking if the formula is applicable to a system of two charged particles? –  NowIGetToLearnWhatAHeadIs Dec 16 '13 at 15:49
    
For an attractive force it should be the same formula since its just a 1/r^2 force law. I was curious if it shows up anywhere. –  Anode Dec 16 '13 at 15:51
    
Does solving the hydrogen atom count? Also what about two masses attached by a spring? –  NowIGetToLearnWhatAHeadIs Dec 16 '13 at 15:52
    
Where does this show up? Its basically the same thing as writing an effective potential correct? –  Anode Dec 16 '13 at 15:55
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The reduced mass appears in the context of transforming to the CoM frame (i.e. isolating the motion of a system from the motion in the system). It has nothing to do with the force between the bodies, and always applies to the mass. We do transform to the CoM frame when solving the hydrogen atom, but that only affects the mass. –  dmckee Dec 16 '13 at 16:32
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The answer is in the comment by dmckee: the reduced mass appears when you consider a two body problem, and transform it into two independent one-body problems (motion of the center of mass, and motion of body 2 with respect to the position of body 1), by way of combining the equations provided by the fundamental principle of dynamics (F=ma), which hold whatever the interaction is.

In other terms, the reduced mass is an inertial mass, whereas in the realm of the equations of dynamics, charge plays a role analogous to gravitational mass, when determining some force magnitude. Gravitational mass and inertial mass, even though equal numerical quantities, as far as anyone can tell thus far, are conceptually two different concepts.

There is therefore no reason why any equivalent reduced quantity should be useful with charges.

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