Reduced mass

In physics, the Reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. This 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 by the reduced mass, if this is compensated by replacing the other mass by the sum of both masses. The reduced mass is frequently denoted by $\scriptstyle \mu$ (Greek lower case mu); note however that the standard gravitational parameter is also denoted by $\scriptstyle \mu$. 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

$m_\text{red} = \mu = \cfrac{1}{\cfrac{1}{m_1}+\cfrac{1}{m_2}} = \cfrac{m_1 m_2}{m_1 + m_2},\!\,$

where the force on this mass is given by the force between the two bodies. The (mathematical) replacement of masses, stated above is as follows: for the particle of interest, its mass becomes the reduced mass

$m_1 \rightarrow m_\text{red}$

and for the other particle, its mass becomes

$m_2 \rightarrow m_1 + m_2$

Properties

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

$m_\text{red} \leq m_1, \quad m_\text{red} \leq m_2 \!\,$

and is half of the harmonic mean of the two masses:

$m_\text{red} = \frac{1}{2}\left(\frac{2}{\frac{1}{m_1} + \frac{1}{m_2}}\right) \,\!$

Derivation

The equation can be derived as follows.

Newtonian mechanics

Main article: Newtonian mechanics

Using Newton's second law, the force exerted by body 2 on body 1 is

$\bold{F}_{12} = m_1 \bold{a}_1. \!\,$

The force exerted by body 1 on body 2 is

$\bold{F}_{21} = m_2 \bold{a}_2. \!\,$

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

$\bold{F}_{12} = - \bold{F}_{21}.\!\,$

Therefore,

$m_1 \bold{a}_1 = - m_2 \bold{a}_2. \!\,$

and

$\bold{a}_2=-{m_1 \over m_2} \bold{a}_1. \!\,$

The relative acceleration a_rel between the two bodies is given by

$\bold{a}_{\rm rel}= \bold{a}_1-\bold{a}_2 = \left(1+\frac{m_1}{m_2}\right) \bold{a}_1 = \frac{m_2+m_1}{m_1 m_2} m_1 \bold{a}_1 = \frac{\bold{F}_{12}}{m_{\rm red}}.$

So we conclude that body 1 moves with respect to the position of body 2 as a body of mass equal to the reduced mass.

Lagrangian mechanics

Main article: Lagrangian mechanics

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

$L = {1 \over 2} m_1 \mathbf{\dot{r}}_1^2 + {1 \over 2} m_2 \mathbf{\dot{r}}_2^2 - V(| \mathbf{r}_1 - \mathbf{r}_2 | ) \!\,$

where r is the position vector of mass m_i (of particle i). The potential energy V is a function as it is only dependent on the absolute distance between the particles. If we define

$\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$

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

$m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 = 0$,

then

$\mathbf{r}_1 = \frac{m_2 \mathbf{r}}{m_1 + m_2} , \mathbf{r}_2 = \frac{-m_1 \mathbf{r}}{m_1 + m_2}.$

Then substituting above gives a new Lagrangian

$L = {1 \over 2}m_\text{red} \mathbf{\dot{r}}^2 - V(r),$

where

$m_\text{red} = \frac{m_1 m_2}{m_1 + m_2}$

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

Applications

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

Collisions of particles

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

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

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

Motions of masses in gravitational fields

In the case of the gravitational potential energy

$V(| \mathbf{r}_1 - \mathbf{r}_2 | ) = - \frac{G m_1 m_2}{| \mathbf{r}_1 - \mathbf{r}_2 |} \, ,$

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

$m_1 m_2 = (m_1+m_2) m_\text{red}\!\,$

Non-relativistic quantum mechanics

Consider the electron (mass m_e) and proton (mass m_p) in the hydrogen atom. 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

$m_e \rightarrow \frac{m_em_p}{m_e+m_p}$

and the proton mass becomes the sum of the two masses

$m_e \rightarrow m_e + m_p$

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^{[citation needed]}

$x_\text{red} = {1 \over {1 \over x_1} + {1 \over x_2}} = {x_1 x_2 \over x_1 + x_2}\!\,$

that simplifies an equation of the form

$\ {1\over x_\text{eq}} = \sum_{i=1}^n {1\over x_i} = {1\over x_1} + {1\over x_2} + \cdots+ {1\over x_n}.\!\,$

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. This relationship is determined by the physical properties of the elements as well as the continuity equation linking them.

See also

• Center-of-momentum frame
• Momentum conservation
• Defining equation (physics)

External links

• Reduced Mass on HyperPhysics