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diff --git a/content/know/concept/reduced-mass/index.pdc b/content/know/concept/reduced-mass/index.pdc new file mode 100644 index 0000000..feaddd6 --- /dev/null +++ b/content/know/concept/reduced-mass/index.pdc @@ -0,0 +1,140 @@ +--- +title: "Reduced mass" +firstLetter: "R" +publishDate: 2021-07-05 +categories: +- Physics + +date: 2021-07-03T11:39:03+02:00 +draft: false +markup: pandoc +--- + +# Reduced mass + +Problems with two interacting objects can be simplified +by combining them into a pseudo-object with **reduced mass** $\mu$, +whose position equals the relative position of the objects. +For bodies 1 and 2 with respective masses $m_1$ and $m_2$: + +$$\begin{aligned} + \boxed{ + \mu \equiv \frac{m_1 m_2}{m_1 + m_2} + } +\end{aligned}$$ + +If $\vec{x}_1$ and $\vec{x}_2$ are the objects' respective positions, +then we define +the relative position $\vec{x}_r$, +the relative velocity $\vec{v}_r$, +and the relative acceleration $\vec{a}_r$: + +$$\begin{aligned} + \vec{x}_r + \equiv \vec{x}_1 - \vec{x}_2 + \qquad + \vec{v}_r + \equiv \vec{v}_1 - \vec{v}_2 + = \dv{\vec{x}_r}{t} + \qquad \quad + \vec{a}_r + \equiv \vec{a}_1 - \vec{a}_2 + = \dv[2]{\vec{x}_r}{t} +\end{aligned}$$ + +We now choose the coordinate system's origin +to be the center of mass of both objects: + +$$\begin{aligned} + m_1 \vec{x}_1 + m_2 \vec{x}_2 = \vec{0} +\end{aligned}$$ + +Rearranging and differentiating then yields the following useful equations: + +$$\begin{aligned} + \vec{x}_2 = - \frac{m_1}{m_2} \vec{x}_1 + \qquad \quad + \vec{v}_2 = - \frac{m_1}{m_2} \vec{v}_1 + \qquad \quad + \vec{a}_2 = - \frac{m_1}{m_2} \vec{a}_1 +\end{aligned}$$ + +Using these relations, we can rewrite the relative quantities we defined earlier: + +$$\begin{aligned} + \vec{x}_r + = \Big( 1 + \frac{m_1}{m_2} \Big) \vec{x}_1 + = \frac{m_1 + m_2}{m_2} \vec{x}_1 + \qquad + \vec{v}_r + = \frac{m_1 + m_2}{m_2} \vec{v}_1 + \qquad + \vec{a}_r + = \frac{m_1 + m_2}{m_2} \vec{a}_1 +\end{aligned}$$ + +Meanwhile, Newton's third law states that +if object 1 experiences a force $\vec{F}_1 = m_1 \vec{a}_1$ caused by object 2, +then object 2 experiences an opposite and equal force $\vec{F}_2 = - \vec{F}_1$. +In fact, our earlier relation between $\vec{a}_1$ and $\vec{a}_1$ +boils down to Newton's third law: + +$$\begin{aligned} + \vec{F}_2 = m_2 \vec{a}_2 = - m_1 \vec{a}_1 = - \vec{F}_1 + \quad \implies \quad + \vec{a}_2 = - \frac{m_1}{m_2} \vec{a}_1 +\end{aligned}$$ + +With all that in mind, let us take a closer look at the relative acceleration $\vec{a}_r$: + +$$\begin{aligned} + \vec{a}_r + = \frac{m_1 + m_2}{m_2} \Big( \frac{m_1}{m_1} \Big) \vec{a}_1 + = \frac{m_1 + m_2}{m_1 m_2} \big( m_1 \vec{a}_1 \big) + = \frac{\vec{F}_1}{\mu} + = - \frac{\vec{F}_2}{\mu} +\end{aligned}$$ + +Where $\mu$ is the reduced mass, as defined above. +In other words, the relative acceleration $\vec{a}_r$ +is just $\vec{a}_1 = \vec{F}_1 / m_1$ multiplied by $m_1 / \mu$. +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 $\vec{x}_1$ and $\vec{x}_2$ +from $\vec{x}_r$, which you can easily verify for yourself: + +$$\begin{aligned} + \vec{x}_1 + = \frac{\mu}{m_1} \vec{x}_r + = \frac{m_2}{m_1 + m_2} \vec{x}_r + \qquad \quad + \vec{x}_2 + = \frac{\mu}{m_2} \vec{x}_r + = - \frac{m_1}{m_1 + m_2} \vec{x}_r +\end{aligned}$$ + +With this, we can rewrite the total kinetic energy $T$ in an elegant way: + +$$\begin{aligned} + T + &= \frac{1}{2} m_1 \vec{v}_1^2 + \frac{1}{2} m_2 \vec{v}_2^2 + = \frac{1}{2} m_1 \Big( \frac{\mu}{m_1} \vec{v}_r \Big)^2 + \frac{1}{2} m_2 \Big( \frac{\mu}{m_2} \vec{v}_r \Big)^2 + \\ + &= \frac{1}{2} \frac{\mu^2}{m_1} \vec{v}_r^2 + \frac{1}{2} \frac{\mu^2}{m_2} \vec{v}_r^2 + = \frac{1}{2} \Big( \frac{m_2 \mu^2}{m_1 m_2} + \frac{m_1 \mu^2}{m_1 m_2} \Big) \vec{v}_r^2 + \\ + &= \frac{1}{2} \frac{(m_1 + m_2) \mu^2}{m_1 m_2} \vec{v}_r^2 + = \frac{1}{2} \frac{\mu^2}{\mu} \vec{v}_r^2 + = \frac{1}{2} \mu \vec{v}_r^2 +\end{aligned}$$ + +Then, assuming that the system's potential energy $V$ +only depends on the distance between the two objects, +i.e. $V = V(|\vec{x}_1 - \vec{x}_2|) = V(|\vec{x}_r|)$, +we just showed that we can rewrite both $T$ and $V$ +to contain only $\mu$ and relative quantities. +This is relevant for both [Lagrangian mechanics](/know/concept/lagrangian-mechanics/) +and [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/), +where $L = T - V$ and $H = T + V$ respectively. |