In physics, reduced mass is a measure of the effective inertial mass of a system with two or more particles when the particles are interacting with each other. Reduced mass 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
\mu
\mu
Reduced mass is particularly useful in classical mechanics.
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]
\mu=m1\parallelm2=\cfrac{1}{\cfrac{1}{m1}+\cfrac{1}{m2}}=\cfrac{m1m2}{m1+m2},
The reduced mass is always less than or equal to the mass of each body:
\mu\leqm1, \mu\leqm2
1 | |
\mu |
=
1 | |
m1 |
+
1 | |
m2 |
In the special case that
m1=m2
{\mu}=
m1 | |
2 |
=
m2 | |
2 |
If
m1\ggm2
\mu ≈ m2
The equation can be derived as follows.
See also: Newtonian mechanics.
Using Newton's second law, the force exerted by a body (particle 2) on another body (particle 1) is:
F12=m1a1
The force exerted by particle 1 on particle 2 is:
F21=m2a2
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:
F12=-F21
Therefore:
m1a1=-m2a2 ⇒ a2=-{m1\overm2}a1
The relative acceleration arel between the two bodies is given by:
a\rm:=a1-a2=\left(1+
m1 | |
m2 |
\right)a1=
m2+m1 | |
m1m2 |
m1a1=
F12 | |
\mu |
Note that (since the derivative is a linear operator) the relative acceleration
a\rm
x\rm
a\rm=a1-a2=
| |||||||
dt2 |
-
| |||||||
dt2 |
=
d2 | |
dt2 |
(x1-x2)=
| |||||||
dt2 |
This simplifies the description of the system to one force (since
F12=-F21
x\rm
\mu
\mu
See main article: Lagrangian mechanics.
Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of
l{L}={1\over2}m1
2 | |||
| |||
1 |
+{1\over2}m2
2 | |||
| |||
2 |
-V(|r1-r2|)
{r
mi
i
r=r1-r2
m1r1+m2r2=0
r1=
m2r | |
m1+m2 |
, r2=-
m1r | |
m1+m2 |
.
Then substituting above gives a new Lagrangian
l{L}={1\over2}\mu
2 | |||
|
-V(r),
\mu=
m1m2 | |
m1+m2 |
Reduced mass can be used in a multitude of two-body problems, where classical mechanics is applicable.
In a system with two point masses
m1
m2
r1
r2
r1=R
m2 | |
m1+m2 |
r2=R
m1 | |
m1+m2 |
R
R=r1+r2
This holds for a rotation around the center of mass.The moment of inertia around this axis can be then simplified to
In a collision with a coefficient of restitution e, the change in kinetic energy can be written as
\DeltaK=
1 | |
2 |
\mu
2 | |
v | |
\rmrel |
(e2-1)
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
V(|r1-r2|)=-
Gm1m2 | |
|r1-r2| |
,
m1m2=(m1+m2)\mu
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
me →
memp | |
me+mp |
mp → me+mp
This idea is used to set up the Schrödinger equation for the hydrogen atom.