From 0d3574cf5cdb0c7aebe596b1035a2ea64b5327b6 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 7 Jul 2021 18:52:47 +0200
Subject: Expand knowledge base

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+---
+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.
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