Reduced mass
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 <math> \mu </math> (mu), although the standard gravitational parameter is also denoted by <math> \mu </math> (as are a number of other physical quantities). It has the dimensions of mass, and SI unit kg.
Reduced mass is particularly useful in classical mechanics.
EquationEdit
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<ref>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</ref><ref>Dynamics and Relativity, J.R. Forshaw, A.G. Smith, Wiley, 2009, Template:ISBN</ref> <math display="block">\mu = m_1 \parallel m_2 = \cfrac{1}{\cfrac{1}{m_1} + \cfrac{1}{m_2}} = \cfrac{m_1 m_2}{m_1 + m_2},</math> where the force on this mass is given by the force between the two bodies.
PropertiesEdit
The reduced mass is always less than or equal to the mass of each body: <math display="block">\mu \leq m_1, \quad \mu \leq m_2</math> and has the reciprocal additive property: <math display="block">\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}</math> which by re-arrangement is equivalent to half of the harmonic mean.
In the special case that <math>m_1 = m_2</math>: <math display="block">\mu = \frac{m_1}{2} = \frac{m_2}{2}</math>
If <math>m_1 \gg m_2</math>, then <math>\mu \approx m_2</math>.
DerivationEdit
The equation can be derived as follows.
Newtonian mechanicsEdit
Using Newton's second law, the force exerted by a body (particle 2) on another body (particle 1) is: <math display="block">\mathbf{F}_{12} = m_1 \mathbf{a}_1</math>
The force exerted by particle 1 on particle 2 is: <math display="block">\mathbf{F}_{21} = m_2 \mathbf{a}_2</math>
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: <math display="block">\mathbf{F}_{12} = - \mathbf{F}_{21}</math>
Therefore: <math display="block">m_1 \mathbf{a}_1 = - m_2 \mathbf{a}_2 \;\; \Rightarrow \;\; \mathbf{a}_2=-{m_1 \over m_2} \mathbf{a}_1</math>
The relative acceleration arel between the two bodies is given by: <math display="block">\mathbf{a}_\text{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}</math>
Note that (since the derivative is a linear operator) the relative acceleration <math>\mathbf{a}_\text{rel}</math> is equal to the acceleration of the separation <math>\mathbf{x}_\text{rel}</math> between the two particles. <math display="block">\mathbf{a}_\text{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}\left(\mathbf{x}_1 - \mathbf{x}_2\right) = \frac{d^2\mathbf{x}_\text{rel}}{dt^2}</math>
This simplifies the description of the system to one force (since <math>\mathbf{F}_{12} = - \mathbf{F}_{21}</math>), one coordinate <math>\mathbf{x}_\text{rel}</math>, and one mass <math>\mu</math>. 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, <math>\mu</math>.
Lagrangian mechanicsEdit
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Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of <math display="block"> \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 | ) </math> where <math>{\mathbf{r}}_{i}</math> is the position vector of mass <math>m_{i}</math> (of particle <math>i</math>). The potential energy V is a function as it is only dependent on the absolute distance between the particles. If we define <math display="block">\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2 </math> and let the centre of mass coincide with our origin in this reference frame, i.e. <math display="block"> m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 = 0, </math> then <math display="block"> \mathbf{r}_1 = \frac{m_2 \mathbf{r}}{m_1 + m_2} , \; \mathbf{r}_2 = -\frac{m_1 \mathbf{r}}{m_1 + m_2}.</math>
Then substituting above gives a new Lagrangian <math display="block"> \mathcal{L} = \frac{1}{2} \mu \mathbf{\dot{r}}^2 - V(r), </math> where <math display="block">\mu = \frac{m_1 m_2}{m_1 + m_2} </math> is the reduced mass. Thus we have reduced the two-body problem to that of one body.
ApplicationsEdit
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 lineEdit
In a system with two point masses <math>m_1</math> and <math>m_2</math> such that they are co-linear, the two distances <math>r_1</math> and <math>r_2</math> to the rotation axis may be found with <math display="block">r_1 = R \frac{m_2}{m_1+m_2}</math> <math display="block">r_2 = R \frac{m_1}{m_1+m_2}</math> where <math> R</math> is the sum of both distances <math>R = r_1 + r_2 </math>.
This holds for a rotation around the center of mass. The moment of inertia around this axis can be then simplified to <math display="block"> 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.</math>
Collisions of particlesEdit
In a collision with a coefficient of restitution e, the change in kinetic energy can be written as <math display="block">\Delta K = \frac{1}{2}\mu v^2_\text{rel} \left(e^2 - 1\right),</math> 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 attractionEdit
In the case of the gravitational potential energy <math display="block">V(| \mathbf{r}_1 - \mathbf{r}_2 | ) = - \frac{G m_1 m_2}{| \mathbf{r}_1 - \mathbf{r}_2 |} \, ,</math> 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 (M) equal to the one particular sum equal to the sum of these two masses , because <math display="block">m_1 m_2 = \left(m_1+m_2\right) \mu;</math> but all those other pairs whose sum is M would have the wrong product of their masses.
Non-relativistic quantum mechanicsEdit
Consider the electron (mass me) and proton (mass mp) in the hydrogen atom.<ref>Molecular Quantum Mechanics Parts I and II: An Introduction to Quantum Chemistry (Volume 1), P.W. Atkins, Oxford University Press, 1977, Template:ISBN</ref> 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 <math display="block">m_\text{e} \rightarrow \frac{m_\text{e} m_\text{p}}{m_\text{e} + m_\text{p}} </math>
This idea is used to set up the Schrödinger equation for the hydrogen atom.
See alsoEdit
- Parallel (operator) - the general operation, of which reduced mass is just one case
- Center-of-momentum frame
- Momentum conservation
- Harmonic oscillator
- Chirp mass, a relativistic equivalent used in the post-Newtonian expansion