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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu } (mu), although the standard gravitational parameter is also denoted by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu } (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.

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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},} 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu \leq m_1, \quad \mu \leq m_2} and has the reciprocal additive property: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1 = m_2} : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu = \frac{m_1}{2} = \frac{m_2}{2}}

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1 \gg m_2} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu \approx m_2} .

Derivation

The equation can be derived as follows.

Newtonian mechanics

See also: Newtonian mechanics

Using Newton's second law, the force exerted by a body (particle 2) on another body (particle 1) is: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{F}_{12} = m_1 \mathbf{a}_1}

The force exerted by particle 1 on particle 2 is: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{F}_{12} = - \mathbf{F}_{21}}

Therefore: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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_2}\mathbf{a}_1 = \frac{\mathbf{F}_{12}}{\mu} = \frac{m_1\mathbf{a}_1}{\mu}}

Note that (since the derivative is a linear operator) the relative acceleration Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{a}_\text{rel}} is equal to the acceleration of the separation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{x}_\text{rel}} between the two particles. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

This simplifies the description of the system to one force (since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{F}_{12} = - \mathbf{F}_{21}} ), one coordinate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{x}_\text{rel}} , and one mass Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu} .

Lagrangian mechanics

Main article: Lagrangian mechanics

Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathbf{r}}_{i}} is the position vector of mass Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_{i}} (of particle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} ). The potential energy V is a function as it is only dependent on the absolute distance between the particles. If we define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 = 0, } then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L} = \frac{1}{2} \mu \mathbf{\dot{r}}^2 - V(r), } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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

File:Rigid rotor point masses.svg

Two point masses rotating around the center of mass.

In a system with two point masses Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_2} such that they are co-linear, the two distances Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_2} to the rotation axis may be found with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_1 = R \frac{m_2}{m_1+m_2}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_2 = R \frac{m_1}{m_1+m_2}} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} is the sum of both distances Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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.}

Collisions of particles

In a collision with a coefficient of restitution e, the change in kinetic energy can be written as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta K = \frac{1}{2}\mu v^2_\text{rel} \left(e^2 - 1\right),} 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.

Motion of two massive bodies under their gravitational attraction

In the case of the gravitational potential energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 (M) equal to the one particular sum equal to the sum of these two masses, because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1 m_2 = \left(m_1+m_2\right) \mu;} but all those other pairs whose sum is M would have the wrong product of their masses.

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_\text{e} \rightarrow \frac{m_\text{e} m_\text{p}}{m_\text{e} + m_\text{p}} }

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

See also

• 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

External links

• Reduced Mass on HyperPhysics