Examples of R in the following topics:
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- $\displaystyle {\bf A}({\bf r},t) = \frac{1}{c} \int d^3 r' \frac{{\bf J}\left ({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c} \right )}{|{\bf r} - {\bf r}'|} \\ \displaystyle \phi({\bf r},t) = \int d^3 r' \frac{\rho\left ({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c} \right )}{|{\bf r} - {\bf r}'|} .$
- $\displaystyle \phi({\bf r},t) = \int d^3 r' \int dt' \frac{\rho \left ({\bf r}',t'\right )}{|{\bf r} - {\bf r}'|} \delta ( t' - t + |{\bf r}-{\bf r}'|/c )$
- $\rho({\bf r},t) = q \delta({\bf r} -{\bf r}_0(t))~~\text{ and }~~{\bf j}(r, t) = q {\bf u} \delta({\bf r} - {\bf r}_0(t))$
- $\displaystyle \phi({\bf r},t) = \int d^3 r' \int dt' \frac{q \delta({\bf r}'-{\bf r}_0(t'))}{|{\bf r} - {\bf r}'|} \delta ( t' - t + |{\bf r}-{\bf r}'|/c ) \\ \displaystyle = q \int dt' \frac{1}{|{\bf r} - {\bf r}_0(t')|} \delta ( t' - t + |{\bf r}-{\bf r}_0(t')|/c )$
- $2 R(t) {\dot R}(t) = -2 {\bf R}(t) \cdot {\bf u}(t) ~~{NB:} ~~{\bf R} = {\bf r}-{\bf r}_0(t)$
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- $\displaystyle \frac{1}{r} \frac{\delta ^2}{\delta r^2} \left [ r \hat G(r,\omega) \right ] + \frac{\omega^2}{c^2} \hat G(r,\omega) = -4\pi \delta(r)$
- $\displaystyle \hat \phi^{\pm}({\bf r}, \omega) = \int d^3 r' \frac{\exp \left ( \pm i\omega/c |{\bf r}-{\bf r}'|/c \right ) }{|{\bf r}-{\bf r}'|} \hat \rho({\bf r}',\omega)$
- $\displaystyle \phi^{\pm}({\bf r}, t) = \int d^3 r' d \omega \frac{\exp \left [ -i\omega \left ( t \mp |{\bf r}-{\bf r}'|/c \right ) \right ]}{|{\bf r}-{\bf r}'|} \hat \rho({\bf r}',\omega) .$
- $\displaystyle \phi^{\pm}({\bf r}, t) = \int d^3 r' \rho \left ({\bf r}, t \mp \frac{|{\bf r}-{\bf r}'|}{c} \right ) \frac{1}{|{\bf r}-{\bf r}'|}.$
- $\displaystyle {\bf A}({\bf r},t) = \frac{1}{c} \int d^3 r' \frac{{\bf J}\left ({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c} \right )}{|{\bf r} - {\bf r}'|} \\ \displaystyle \phi({\bf r},t) = \int d^3 r' \frac{\rho\left ({\bf r}',t-\frac{|{\bf r}-{\bf r}'|}{c} \right )}{|{\bf r} - {\bf r}'|}
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- Essentially the singular value decomposition $\mathbf{R}^{m}$ and $\mathbf{R}^n$ simultaneously.
- Further, there are exactly $r$ nonzero singular values, where $r$ is the rank of $A$ .
- Let $U_r$$V_r$ and $V_r$ be the matrices whose columns are the $r$ model and data eigenvectors associated with the $r$ nonzero eigenvalues and $U_0$$V_0$ and $V_0$$r$ be the matrices whose columns are the eigenvectors associated with the zero eigenvalues, and let $r$ be the diagonal matrix containing the nonzero eigenvalues.
- $ A = \left[U_r , U_0 \right] \left[ \begin{array}{cc} \Lambda _ r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] \left[ \begin{array}{c} V^T _r \\ V^T _0 \end{array} \right] = U_r \Lambda _r V^T _r ,$
- Since $\Lambda_r$ is $r \times r$ and $\Lambda$ is $n \times m$ then the lower left block of zeros must be $n-r \times r$$r \times m-r$ , the upper right must be $r \times m-r$ and the lower right must be $n-r \times m-r$ .
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- For a continuous mass distribution, the position of center of mass is given as $\mathbf R = \frac 1M \int_V\rho(\mathbf{r}) \mathbf{r} dV$ .
- In the case of a system of particles $P_i, i = 1, \cdots, n$, each with a mass, $m_i$, which are located in space with coordinates $r_i, i = 1, \cdots, n$, the coordinates $\mathbf{R}$ of the center of mass satisfy the following condition:
- If the mass distribution is continuous with respect to the density, $\rho (r)$, within a volume, $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass, $\mathbf{R}$, is zero, that is:
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- $\displaystyle{\nabla ^2 \psi(r, \theta, z) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2\psi}{\partial \theta^2} + \frac{\partial^2\psi}{\partial z^2}=0. }$
- $\displaystyle{\frac{r}{R} \frac{d}{dr} \left( r \frac{dR}{dr}\right) + \frac{r^2}{Z} \frac{d^2Z}{dz^2} = -\frac{1}{Q} \frac{d^2Q}{d\theta^2} = m^2. }$
- $\displaystyle{r \frac{d}{dr} \left( r \frac{dR}{dr} \right) + \left(r^2 k^2 -m^2\right)R = 0 }$
- $\displaystyle{r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} - (k^2 r^2 + m^2)R = 0. }$
- $\displaystyle{V(r, \theta) = -E_0 r \cos \theta + E_0 a^3 \frac{\cos \theta}{r^2} = -E_0 \left( 1 - \left(\frac{a}{r}\right)^3 \right) r \cos \theta . }$
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- Given a set $D \subseteq R^n$ and an integrable function $f$ over $D$, the average value of $f$ over its domain is given by:
- The gravitational potential associated with a mass distribution given by a mass measure $dm$ on three-dimensional Euclidean space $R^3$ is:
- In the following example, the electric field produced by a distribution of charges given by the volume charge density $\rho (\vec r)$ is obtained by a triple integral of a vector function:
- $\displaystyle{\vec E = \frac {1}{4 \pi \epsilon_0} \iiint \frac {\vec r - \vec r'}{\| \vec r - \vec r' \|^3} \rho (\vec r')\, {d}^3 r'}$
- Points $\mathbf{x}$ and $\mathbf{r}$, with $\mathbf{r}$ contained in the distributed mass (gray) and differential mass $dm(\mathbf{r})$ located at the point $\mathbf{r}$.
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- $\displaystyle l = \Omega r^2 = \left ( GM r \right )^{1/2}.$
- $\displaystyle \tau = f_\phi \left ( 2 \pi r \right ) \left ( 2 h \right ) ( r ) = \dot L^+ - \dot L^- = \dot M \left [ \left ( GM r \right )^{1/2} - \beta \left ( GM r_I \right )^{1/2} \right ]$
- $\displaystyle f_\phi = -\eta \frac{d \Omega}{d \ln r} = - \eta r \frac{d}{d r} \left ( \sqrt{GM} r^{-3/2} \right )= \frac{3}{2} \eta \Omega.$
- $\displaystyle \eta = \frac{\dot M}{6 \pi r^2 h \Omega } \left [ \left ( GM r \right )^{1/2} - \beta \left ( GM r_I \right )^{1/2} \right ].$
- $\displaystyle 2 h Q \approx 2 h \frac{\left(f_\phi\right)^2}{\eta} = \frac{3 \dot M}{4\pi r^2}\frac{GM}{r} \left [1 - \left (\frac{r_I}{r} \right )^{1/2} \right ]
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- $\displaystyle \frac{1}{r^2} \frac{\partial }{\partial r} \left ( r^2 \rho v \right ) = 0$
- $\displaystyle v\frac{\partial v}{\partial r} + \frac{1}{\rho} \frac{\partial p}{\partial r} + \frac{GM}{r^2} = 0$
- $\displaystyle v\frac{\partial v}{\partial r} - \frac{c_s^2}{vr^2} \frac{\partial }{\partial r} (vr^2) + \frac{GM}{r^2} = 0.$
- It then accelerates for $r<r_c$ as well.
- $\displaystyle \frac{c_s^2(r_c)}{2} + \frac{c_s^2(r_c) - c_s^2(\infty)}{\gamma-1} - 2 c_s^2(r_c) = 0 .$
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- $\displaystyle \left [ -\frac{\hbar^2}{2 \mu_{AB}} \nabla^2_{\bf R} -\frac{\hbar^2}{2 \mu_e} \nabla^2_{\bf r} - \frac{e^2}{r_A} - \frac{e^2}{r_B} + \frac{e^2}{R} - E \right ] \psi({\bf r},{\bf R}) = 0 $
- $\displaystyle \left [ -\frac{\hbar^2}{2 \mu_e} \nabla^2_{\bf r} - \frac{e^2}{r_A} - \frac{e^2}{r_B} + \frac{e^2}{R} - E_j({\bf R}) \right ] \chi_j({\bf r};{\bf R}) = 0 $
- We try various values of ${\bf R}$ and solve for $\chi_j({\bf r};{\bf R})$ each time.
- $\displaystyle \left [ -\frac{\hbar^2}{2 \mu_{AB}} \nabla^2_{\bf R} + E_j({\bf R}) - E \right ] F_j({\bf R}) = 0 $
- $\displaystyle E_{gu} (R) = E_{1s} + \frac{e^2}{R} \frac{(1 + R/a_0)e^{-2R/a_0}\pm[1-(2/3)(R/a_0)^2] e^{-R/a_0}}{1\pm[1+R/a_0+(R/a_0)^2/3]e^{-R/a_0}}$
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- $\displaystyle \frac{1}{r^2} \frac{\partial }{\partial r} \left ( \frac{r^2}{\rho} \frac{\partial p}{\partial r} \right ) = -4\pi G \rho.$
- $\displaystyle \nabla w = -\nabla \phi + \Omega^2({\bf r}) {\bf R}$
- where ${\bf R}$ is a vector pointing from the rotation axis to the point in the fluid.
- $\displaystyle 0 = 0 + \nabla \times \left ( \Omega^2({\bf r}) {\bf R} \right )$