General Virial Theorem:
In a $3$-dimensional space of $N$ point masses with masses $\{m_i\}_{i \in N}$, positions $\{\vec{q}_i\}_{i \in N}$ and velocities $\{\vec{v}_i\}_{i \in N}$, if $\left < \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right >_{T} = 0$, then
$$2 \left < K(t) \right >_{T} + \left < \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right >_{T} =0$$
where $N \in \mathbb{N}$, $G(t) = \displaystyle \sum_{i=1}^{N} \left ( m_i \vec{q}_i(t) \cdot \vec{v}_i(t) \right )$, $\displaystyle K(t) = \dfrac{1}{2} \sum_{i=1}^{N} \left ( m_i \|\vec{v}_i(t)\|^2 \right )$, $\displaystyle \left < f(t) \right >_{T} = \frac{1}{T} \int_0^T f(t) \ \mathrm{d} t$
Proof:
Based on the definition of virial $G(t)$:
\[
G(t) = \sum_{i=1}^{N} \left ( m_i \vec{q}_i(t) \cdot \vec{v}_i(t) \right )
\]
Taking the time derivative on both sides, we have:
$$
\begin{align}
\dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} &= \sum_{i=1}^{N} \left ( \dfrac{\mathrm{d} \left ( m_i \vec{q}_i(t) \cdot \vec{v}_i(t) \right )}{\mathrm{d} t} \right ) \\
&= \sum_{i=1}^{N} \left ( m_i \left ( \dfrac{\mathrm{d} \left ( \vec{q}_i(t)\right )}{\mathrm{d} t} \cdot \vec{v}_i(t) + \vec{q}_i(t) \cdot \dfrac{\mathrm{d} \left ( \vec{v}_i(t)\right )}{\mathrm{d} t} \right ) \right ) \\
&= \sum_{i=1}^{N} \left ( m_i \left ( \vec{v}_i(t) \cdot \vec{v}_i(t) + \vec{q}_i(t) \cdot \vec{a}_i(t) \right ) \right ) \\
&= \sum_{i=1}^{N} \left ( m_i \left \| \vec{v}_i(t) \right \|^2 + m_i \vec{a}_i(t) \cdot \vec{q}_i(t) \right ) \\
&= \sum_{i=1}^{N} \left ( m_i \left \| \vec{v}_i(t) \right \|^2 \right ) + \sum_{i=1}^{N} \left ( m_i \vec{a}_i(t) \cdot \vec{q}_i(t) \right ) \\
&= 2K(t) + \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right )
\end{align}
$$
$$
\begin{align}
\implies \frac{1}{T} \int_0^T \left ( \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right ) \ \mathrm{d} t &= \frac{1}{T} \int_0^T \left ( 2K(t) + \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right ) \ \mathrm{d} t\\
\frac{1}{T} \int_0^T \left ( \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right ) \ \mathrm{d} t &= 2 \left ( \frac{1}{T} \int_0^T \left ( K(t) \right ) \ \mathrm{d} t \right )+ \frac{1}{T} \int_0^T \left ( \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right ) \ \mathrm{d} t\\
\left < \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right >_T &= 2 \left < K(t) \right >_T + \left < \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right >_T
\end{align}
$$
If $\left < \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right >_{T} = 0$, then
$$
2 \left < K(t) \right >_T + \left < \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right >_T=0
$$
$\Box$
Virial Theorem with conservative system and homogeneous potential:
In a $3$-dimensional space, for a conservative system of $N$ point masses with masses $\{m_i\}_{i \in N}$, positions $\{\vec{q}_i\}_{i \in N}$ and velocities $\{\vec{v}_i\}_{i \in N}$, if $\left < \dfrac{\mathrm{d}\left ( G(t) \right )}{\mathrm{d} t} \right >_{T} = 0$ and $\forall i, \quad U_i(\vec{q}_i(t))$ are homogeneous of degree $\alpha$ (which $ \forall \lambda \in \mathbb{R}, \quad U_i(\lambda \vec{q}_i(t)) = \lambda^{\alpha} U_i(\vec{q}_i(t))$), then
$$
2 \left< K(t) \right>_T -\alpha \left< U(t) \right>_T = 0
$$
where $N \in \mathbb{N}$, $\alpha \in \mathbb{R}$, $G(t) = \displaystyle \sum_{i=1}^{N} \left ( m_i \vec{q}_i(t) \cdot \vec{v}_i(t) \right )$, $\displaystyle K(t) = \dfrac{1}{2} \sum_{i=1}^{N} \left ( m_i \|\vec{v}_i(t)\|^2 \right )$, $\displaystyle U(t) = \sum_{i=1}^{N} \left ( U_i(\vec{q}_i(t)) \right )$, $\displaystyle \left < f(t) \right >_{T} = \frac{1}{T} \int_0^T f(t) \ \mathrm{d} t$
Proof:
Applying the General Virial Theorem and the definition of conservative system which $\vec{F}_i(t) = -\nabla U_i(\vec{q}_i(t))$, we have
$$
\begin{align}
2 \left < K(t) \right >_T + \left < \sum_{i=1}^{N} \left ( \vec{F}_i(t) \cdot \vec{q}_i(t) \right ) \right >_T &= 0 \\ \\
2 \left< K(t) \right>_T + \left< \sum_{i=1}^{N} \Bigg( \Big( -\nabla U_i(\vec{q}_i(t)) \Big) \cdot \vec{q}_i(t) \Bigg) \right>_T &= 0 \\
\end{align}
$$
Applying the Euler’s Homogeneous Function Theorem as $\forall i, \quad U_i(\vec{q}_i)$ are homogeneous of degree $\alpha$, which results in $\nabla U_i(\vec{q}_i(t)) \cdot \vec{q}_i(t) = \alpha U_i(\vec{q}_i(t))$, then we have
$$
\begin{align}
2 \left< K(t) \right>_T + \left< \sum_{i=1}^{N} \Bigg( \Big( -\nabla U_i(\vec{q}_i(t)) \Big) \cdot \vec{q}_i(t) \Bigg) \right>_T &= 0 \\ \\
2 \left< K(t) \right>_T + \left< \sum_{i=1}^{N} \Bigg( -\Big( \nabla U_i(\vec{q}_i(t)) \cdot \vec{q}_i(t) \Big) \Bigg) \right>_T &= 0 \\ \\
2 \left< K(t) \right>_T + \left< \sum_{i=1}^{N} \Bigg( -\alpha U_i(\vec{q}_i(t)) \Bigg) \right>_T &= 0 \\ \\
2 \left< K(t) \right>_T -\alpha \left< \sum_{i=1}^{N} \Bigg( U_i(\vec{q}_i(t)) \Bigg) \right>_T &= 0 \\ \\
2 \left< K(t) \right>_T -\alpha \left< U(t) \right>_T &= 0
\end{align}
$$
$\Box$
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