Reduced mass

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 ${\displaystyle \mu \approx m_{2}}$.

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 a_rel 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, ${\displaystyle \mu }$.

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

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

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

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

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.

Two point masses rotating around the center of mass.

In a system with two point masses ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ such that they are co-linear, the two distances ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$ to the rotation axis may be found with

$${\displaystyle r_{1}=R{\frac {m_{2}}{m_{1}+m_{2}}}}$$

$${\displaystyle r_{2}=R{\frac {m_{1}}{m_{1}+m_{2}}}}$$

where ${\displaystyle R}$ is the sum of both distances ${\displaystyle R=r_{1}+r_{2}}$.

This holds for a rotation around the center of mass. The moment of inertia around this axis can be then simplified to

$${\displaystyle I=m_{1}r_{1}^{2}+m_{2}r_{2}^{2}=R^{2}{\frac {m_{1}m_{2}^{2}}{(m_{1}+m_{2})^{2}}}+R^{2}{\frac {m_{1}^{2}m_{2}}{(m_{1}+m_{2})^{2}}}=\mu R^{2}.}$$

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 v_rel 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.

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

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

Consider the electron (mass m_e) and proton (mass m_p) 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.

"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.