# Reduced mass

In physics, the reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem. Note, however, that the mass determining the gravitational force is not reduced. In the computation one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses. The reduced mass is frequently denoted by ${\displaystyle \mu }$ (mu), although the standard gravitational parameter is also denoted by ${\displaystyle \mu }$ (as are a number of other physical quantities). It has the dimensions of mass, and SI unit kg.

## Equation

Given two bodies, one with mass m1 and the other with mass m2, the equivalent one-body problem, with the position of one body with respect to the other as the unknown, is that of a single body of mass[1][2]

${\displaystyle \mu ={\cfrac {1}{{\cfrac {1}{m_{1}}}+{\cfrac {1}{m_{2}}}}}={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}},\!\,}$

where the force on this mass is given by the force between the two bodies.

### Properties

The reduced mass is always less than or equal to the mass of each body:

${\displaystyle \mu \leq m_{1},\quad \mu \leq m_{2}\!\,}$

and has the reciprocal additive property:

${\displaystyle {\frac {1}{\mu }}={\frac {1}{m_{1}}}+{\frac {1}{m_{2}}}\,\!}$

which by re-arrangement is equivalent to half of the harmonic mean.

In the special case that ${\displaystyle m_{1}=m_{2}}$:

${\displaystyle {\mu }={\frac {m_{1}}{2}}={\frac {m_{2}}{2}}\,\!}$

If ${\displaystyle m_{1}\gg m_{2}}$, then ${\displaystyle \mu \approx m_{2}}$.

## Derivation

The equation can be derived as follows.

### Newtonian mechanics

Using Newton's second law, the force exerted by a body (particle 2) on another body (particle 1) is:

${\displaystyle \mathbf {F} _{12}=m_{1}\mathbf {a} _{1}}$

The force exerted by particle 1 on particle 2 is:

${\displaystyle \mathbf {F} _{21}=m_{2}\mathbf {a} _{2}}$

According to Newton's third law, the force that particle 2 exerts on particle 1 is equal and opposite to the force that particle 1 exerts on particle 2:

${\displaystyle \mathbf {F} _{12}=-\mathbf {F} _{21}}$

Therefore:

${\displaystyle m_{1}\mathbf {a} _{1}=-m_{2}\mathbf {a} _{2}\;\;\Rightarrow \;\;\mathbf {a} _{2}=-{m_{1} \over m_{2}}\mathbf {a} _{1}}$

The relative acceleration arel between the two bodies is given by:

${\displaystyle \mathbf {a} _{\rm {rel}}:=\mathbf {a} _{1}-\mathbf {a} _{2}=\left(1+{\frac {m_{1}}{m_{2}}}\right)\mathbf {a} _{1}={\frac {m_{2}+m_{1}}{m_{1}m_{2}}}m_{1}\mathbf {a} _{1}={\frac {\mathbf {F} _{12}}{\mu }}}$

Note that (since the derivative is a linear operator), the relative acceleration ${\displaystyle \mathbf {a} _{\rm {rel}}}$ is equal to the acceleration of the separation ${\displaystyle \mathbf {x} _{\rm {rel}}}$ between the two particles.

${\displaystyle \mathbf {a} _{\rm {rel}}=\mathbf {a} _{1}-\mathbf {a} _{2}={\frac {d^{2}\mathbf {x} _{1}}{dt^{2}}}-{\frac {d^{2}\mathbf {x} _{2}}{dt^{2}}}={\frac {d^{2}}{dt^{2}}}(\mathbf {x} _{1}-\mathbf {x} _{2})={\frac {d^{2}\mathbf {x} _{\rm {rel}}}{dt^{2}}}}$

This simplifies the description of the system to one force (since ${\displaystyle \mathbf {F} _{12}=-\mathbf {F} _{21}}$), one coordinate ${\displaystyle \mathbf {x} _{\rm {rel}}}$, and one mass ${\displaystyle \mu }$. Thus we have reduced our problem to a single degree of freedom, and we can conclude that particle 1 moves with respect to the position of particle 2 as a single particle of mass equal to the reduced mass, ${\displaystyle \mu }$.

### Lagrangian mechanics

Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of

${\displaystyle L={1 \over 2}m_{1}\mathbf {\dot {r}} _{1}^{2}+{1 \over 2}m_{2}\mathbf {\dot {r}} _{2}^{2}-V(|\mathbf {r} _{1}-\mathbf {r} _{2}|)\!\,}$

where ${\displaystyle {\mathbf {r} }_{i}}$ is the position vector of mass ${\displaystyle m_{i}}$ (of particle ${\displaystyle i}$). The potential energy V is a function as it is only dependent on the absolute distance between the particles. If we define

${\displaystyle \mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2}}$

and let the centre of mass coincide with our origin in this reference frame, i.e.

${\displaystyle m_{1}\mathbf {r} _{1}+m_{2}\mathbf {r} _{2}=0}$,

then

${\displaystyle \mathbf {r} _{1}={\frac {m_{2}\mathbf {r} }{m_{1}+m_{2}}},\;\mathbf {r} _{2}=-{\frac {m_{1}\mathbf {r} }{m_{1}+m_{2}}}.}$

Then substituting above gives a new Lagrangian

${\displaystyle L={1 \over 2}\mu \mathbf {\dot {r}} ^{2}-V(r),}$

where

${\displaystyle \mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}}$

is the reduced mass. Thus we have reduced the two-body problem to that of one body.

## Applications

Reduced mass can be used in a multitude of two-body problems, where classical mechanics is applicable.

### Moment of inertia of two point masses in a line

In a system with two point masses m1 and m2 such that they are co-linear, the center of mass may be found via

${\displaystyle x_{cm}={\frac {m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}}}$.

The moment of inertia of this system in the center of mass frame of reference is a two-body problem with two masses, each with its own distance to the center of mass. By converting to the equivalent reduced mass frame of reference, the system may be treated as a single reduced mass, μ, with a distance, d, equal to the sum of x1 and x2. This is because the rotational axis of the reduced mass frame of reference is through the combined mass, M (the sum of m1 and m2)

${\displaystyle M=m_{1}+m_{2}}$

${\displaystyle d=x_{1}+x_{2}}$

${\displaystyle \mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}={\frac {m_{1}m_{2}}{M}}}$

The combined mass M is eliminated in the determination of the moment of inertia of the system since it is a point particle (making M the same size as μ, but with greater mass. Images not drawn to scale) along the axis of rotation (via the parallel axis theorem, it's contribution of MR2 would be zero since R=0 because M and μ are also point masses). Determining the moment of inertia of the reduced mass (μ) system is equivalent to determining the moment of inertia of the center of mass system.

${\displaystyle {\displaystyle I={\frac {m_{1}m_{2}}{m_{1}\!+\!m_{2}}}d^{2}=\mu d^{2}}}$

### Collisions of particles

In a collision with a coefficient of restitution e, the change in kinetic energy can be written as

${\displaystyle \Delta K={\frac {1}{2}}\mu v_{\rm {rel}}^{2}(e^{2}-1)}$,

where vrel is the relative velocity of the bodies before collision.

For typical applications in nuclear physics, where one particle's mass is much larger than the other the reduced mass can be approximated as the smaller mass of the system. The limit of the reduced mass formula as one mass goes to infinity is the smaller mass, thus this approximation is used to ease calculations, especially when the larger particle's exact mass is not known.

### Motion of two massive bodies under their gravitational attraction

In the case of the gravitational potential energy

${\displaystyle V(|\mathbf {r} _{1}-\mathbf {r} _{2}|)=-{\frac {Gm_{1}m_{2}}{|\mathbf {r} _{1}-\mathbf {r} _{2}|}}\,,}$

we find that the position of the first body with respect to the second is governed by the same differential equation as the position of a body with the reduced mass orbiting a body with a mass equal to the sum of the two masses, because

${\displaystyle m_{1}m_{2}=(m_{1}+m_{2})\mu \!\,}$

### Non-relativistic quantum mechanics

Consider the electron (mass me) and proton (mass mp) in the hydrogen atom.[3] They orbit each other about a common centre of mass, a two body problem. To analyze the motion of the electron, a one-body problem, the reduced mass replaces the electron mass

${\displaystyle m_{e}\rightarrow {\frac {m_{e}m_{p}}{m_{e}+m_{p}}}}$

and the proton mass becomes the sum of the two masses

${\displaystyle m_{p}\rightarrow m_{e}+m_{p}}$

This idea is used to set up the Schrödinger equation for the hydrogen atom.

### Other uses

"Reduced mass" may also refer more generally to an algebraic term of the form

${\displaystyle x^{*}={1 \over {1 \over x_{1}}+{1 \over x_{2}}}={x_{1}x_{2} \over x_{1}+x_{2}}\!\,}$

that simplifies an equation of the form

${\displaystyle \ {1 \over x^{*}}=\sum _{i=1}^{n}{1 \over x_{i}}={1 \over x_{1}}+{1 \over x_{2}}+\cdots +{1 \over x_{n}}.\!\,}$

The reduced mass is typically used as a relationship between two system elements in parallel, such as resistors; whether these be in the electrical, thermal, hydraulic, or mechanical domains. A similar expression appears in the transversal vibrations of beams for the elastic moduli.[4] This relationship is determined by the physical properties of the elements as well as the continuity equation linking them.

## References

1. Encyclopaedia of Physics (2nd Edition), R.G. Lerner, G.L. Trigg, VHC publishers, 1991, (Verlagsgesellschaft) 3-527-26954-1, (VHC Inc.) 0-89573-752-3
2. Dynamics and Relativity, J.R. Forshaw, A.G. Smith, Wiley, 2009, ISBN 978-0-470-01460-8
3. Molecular Quantum Mechanics Parts I and II: An Introduction to Quantum Chemistry (Volume 1), P.W. Atkins, Oxford University Press, 1977, ISBN 0-19-855129-0
4. Experimental study of the Timoshenko beam theory predictions, A.Díaz-de-Anda J.Flores, L.Gutiérrez, R.A.Méndez-Sánchez, G.Monsivais, and A.Morales.Journal of Sound and Vibration Volume 331, Issue 26, 17 December 2012, Pages 5732-5744 https://doi.org/10.1016/j.jsv.2012.07.041
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