Categories: Physics.

Reduced mass

Problems with two interacting objects can be simplified by combining them into a pseudo-object with reduced mass μ\mu, whose position equals the relative position of the objects. For bodies 1 and 2 with respective masses m1m_1 and m2m_2:

μ≡m1m2m1+m2\begin{aligned} \boxed{ \mu \equiv \frac{m_1 m_2}{m_1 + m_2} } \end{aligned}

If x⃗1\va{x}_1 and x⃗2\va{x}_2 are the objects’ respective positions, then we define the relative position x⃗r\va{x}_r, the relative velocity v⃗r\va{v}_r, and the relative acceleration a⃗r\va{a}_r:

x⃗r≡x⃗1−x⃗2v⃗r≡v⃗1−v⃗2=dx⃗rdta⃗r≡a⃗1−a⃗2=d2x⃗rdt2\begin{aligned} \va{x}_r \equiv \va{x}_1 - \va{x}_2 \qquad \va{v}_r \equiv \va{v}_1 - \va{v}_2 = \dv{\va{x}_r}{t} \qquad \quad \va{a}_r \equiv \va{a}_1 - \va{a}_2 = \dvn{2}{\va{x}_r}{t} \end{aligned}

We now choose the coordinate system’s origin to be the center of mass of both objects:

m1x⃗1+m2x⃗2=0\begin{aligned} m_1 \va{x}_1 + m_2 \va{x}_2 = 0 \end{aligned}

Rearranging and differentiating then yields the following useful equations:

x⃗2=−m1m2x⃗1v⃗2=−m1m2v⃗1a⃗2=−m1m2a⃗1\begin{aligned} \va{x}_2 = - \frac{m_1}{m_2} \va{x}_1 \qquad \quad \va{v}_2 = - \frac{m_1}{m_2} \va{v}_1 \qquad \quad \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1 \end{aligned}

Using these relations, we can rewrite the relative quantities we defined earlier:

x⃗r=(1+m1m2)x⃗1=m1+m2m2x⃗1v⃗r=m1+m2m2v⃗1a⃗r=m1+m2m2a⃗1\begin{aligned} \va{x}_r = \Big( 1 + \frac{m_1}{m_2} \Big) \va{x}_1 = \frac{m_1 + m_2}{m_2} \va{x}_1 \qquad \va{v}_r = \frac{m_1 + m_2}{m_2} \va{v}_1 \qquad \va{a}_r = \frac{m_1 + m_2}{m_2} \va{a}_1 \end{aligned}

Meanwhile, Newton’s third law states that if object 1 experiences a force F⃗1=m1a⃗1\va{F}_1 = m_1 \va{a}_1 caused by object 2, then object 2 experiences an opposite and equal force F⃗2=−F⃗1\va{F}_2 = - \va{F}_1. In fact, our earlier relation between a⃗1\va{a}_1 and a⃗1\va{a}_1 boils down to Newton’s third law:

F⃗2=m2a⃗2=−m1a⃗1=−F⃗1  ⟹  a⃗2=−m1m2a⃗1\begin{aligned} \va{F}_2 = m_2 \va{a}_2 = - m_1 \va{a}_1 = - \va{F}_1 \quad \implies \quad \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1 \end{aligned}

With all that in mind, let us take a closer look at the relative acceleration a⃗r\va{a}_r:

a⃗r=m1+m2m2(m1m1)a⃗1=m1+m2m1m2(m1a⃗1)=F⃗1μ=−F⃗2μ\begin{aligned} \va{a}_r = \frac{m_1 + m_2}{m_2} \Big( \frac{m_1}{m_1} \Big) \va{a}_1 = \frac{m_1 + m_2}{m_1 m_2} \big( m_1 \va{a}_1 \big) = \frac{\va{F}_1}{\mu} = - \frac{\va{F}_2}{\mu} \end{aligned}

Where μ\mu is the reduced mass, as defined above. In other words, the relative acceleration a⃗r\va{a}_r is just a⃗1=F⃗1/m1\va{a}_1 = \va{F}_1 / m_1 multiplied by m1/μm_1 / \mu. This can be regarded as focusing on the dynamics of body 1, while correcting for the effects of body 2.

This also suggests the following way to recover the original positions x⃗1\va{x}_1 and x⃗2\va{x}_2 from x⃗r\va{x}_r, which you can easily verify for yourself:

x⃗1=μm1x⃗r=m2m1+m2x⃗rx⃗2=μm2x⃗r=−m1m1+m2x⃗r\begin{aligned} \va{x}_1 = \frac{\mu}{m_1} \va{x}_r = \frac{m_2}{m_1 + m_2} \va{x}_r \qquad \quad \va{x}_2 = \frac{\mu}{m_2} \va{x}_r = - \frac{m_1}{m_1 + m_2} \va{x}_r \end{aligned}

With this, we can rewrite the total kinetic energy TT in an elegant way:

T=12m1v⃗12+12m2v⃗22=12m1(μm1v⃗r)2+12m2(μm2v⃗r)2=12μ2m1v⃗r2+12μ2m2v⃗r2=12(m2μ2m1m2+m1μ2m1m2)v⃗r2=12(m1+m2)μ2m1m2v⃗r2=12μ2μv⃗r2=12μv⃗r2\begin{aligned} T &= \frac{1}{2} m_1 \va{v}_1^2 + \frac{1}{2} m_2 \va{v}_2^2 = \frac{1}{2} m_1 \Big( \frac{\mu}{m_1} \va{v}_r \Big)^2 + \frac{1}{2} m_2 \Big( \frac{\mu}{m_2} \va{v}_r \Big)^2 \\ &= \frac{1}{2} \frac{\mu^2}{m_1} \va{v}_r^2 + \frac{1}{2} \frac{\mu^2}{m_2} \va{v}_r^2 = \frac{1}{2} \Big( \frac{m_2 \mu^2}{m_1 m_2} + \frac{m_1 \mu^2}{m_1 m_2} \Big) \va{v}_r^2 \\ &= \frac{1}{2} \frac{(m_1 + m_2) \mu^2}{m_1 m_2} \va{v}_r^2 = \frac{1}{2} \frac{\mu^2}{\mu} \va{v}_r^2 = \frac{1}{2} \mu \va{v}_r^2 \end{aligned}

Then, assuming that the system’s potential energy VV only depends on the distance between the two objects, i.e. V=V(∣x⃗1−x⃗2∣)=V(∣x⃗r∣)V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|), we just showed that we can rewrite both TT and VV to contain only μ\mu and relative quantities. This is relevant for both Lagrangian mechanics and Hamiltonian mechanics, where L=T−VL = T - V and H=T+VH = T + V respectively.