Examples of delta-v in the following topics:
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- $\displaystyle \gamma m \frac{\Delta {\bf v}}{\Delta t} = \frac{q}{c} {\bf v} \times {\bf B}$
- $\gamma m \frac{2 v}{\gamma \Delta t} = \frac{q}{c} v \sin \alpha B \\ \Delta t = \frac{2 m c}{q B \sin\alpha} = \frac{2}{\gamma \omega_B \sin\alpha}$
- We also need to calculate how long between when the radiation emitted at $t$ and $t+\Delta t$ arrives at the distant observer.
- The difference between the observed times is less than $\Delta t$ by $v \Delta t/c$ so we get
- $\Delta t^A = \frac{2}{\gamma \omega_B \sin\alpha} \left ( 1 - \frac{v}{c} \right )$
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- where V is the volume of the material, and is dV/dT the rate of change of that volume with temperature.
- The original volume will be V = L3,and the new volume, after a temperature increase, will be:
- $\begin{aligned} V+ \Delta V &= (L + \Delta L)^3 \\ &= L^3 + 3L^2\Delta L + 3L(\Delta L )^2 +(\Delta L)^3 \\ &\approx L^3 + 3L^2\Delta L \\ &= V + 3 V \frac {\Delta L}{L} \end{aligned}$.
- The approximation holds for a sufficiently small $\Delta L$ compared to L.
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- $\displaystyle \frac{{\partial \bf V}}{\partial t} + \left ( {\bf V} \cdot \nabla \right ) {\bf V} = \frac{d\bf V}{dt} = -\nabla w + \frac{\langle {\bf F} \rangle}{m}$
- $\displaystyle \frac{1}{2} \nabla v^2 = {\bf v} \times (\nabla \times {\bf v}) + ({\bf v} \cdot \nabla ) {\bf v}$
- $\displaystyle \frac{\partial \Gamma}{\partial t} = \frac{\partial }{\partial t}\oint {\bf v} \cdot d{\bf l} = \frac{\partial }{\partial t} \oint {\bf v} \cdot \delta {\bf r} = \oint \frac{\partial \bf v}{\partial t} \cdot \delta {\bf r} + \oint {\bf v} \cdot \frac{\partial }{\partial t} \delta {\bf r}.$
- Because $\delta {\bf r}$ is the difference between two positions moving with the fluid we have
- $\displaystyle {\bf v} \cdot \frac{\partial }{\partial t} \delta {\bf r} = {\bf v} \cdot \delta \frac{\partial \bf r}{\partial t} = {\bf v} \cdot \delta {\bf v} = \delta v^2$
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- $\displaystyle v_1-v_2=j\left ( V_1 - V_2 \right )~\textrm{so}~\left (v_1-v_2\right)^2 = \left ( p_2 - p_1 \right ) \left (V_1 - V_2\right ) = -\Delta p \Delta V.$
- $\displaystyle p_2 = p_1 \frac{\frac{\gamma}{\gamma-1} V_1 - \frac{1}{2} \left ( V_1 + V_2 \right ) }{\frac{\gamma}{\gamma-1} V_2 - \frac{1}{2} \left ( V_1 + V_2 \right ) }.$
- $\displaystyle v_1^2 = j^2 V_1^2 = -\frac{\Delta p}{\Delta V} V_1^2~\textrm{and}~ v_2^2 = -\frac{\Delta p}{\Delta V} V_2^2.$
- $\displaystyle M_1^2 =\frac{\Delta p}{\Delta V} \Big / {\left .
- \frac{\partial P}{\partial V} \right |_{a}}~\textrm{and}~M_2^2 = { \frac{\Delta p}{\Delta V} } \Big / {\left .
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- $I = s_0 = |\epsilon_1 \cdot {\bf E}|^2 + |\epsilon_2 \cdot {\bf E}|^2 = a_1^2 + a_2^2 \\ Q = s_1 = |\epsilon_1 \cdot {\bf E}|^2 - |\epsilon_2 \cdot {\bf E}|^2 = a_1^2 - a_2^2 \\ U = s_2 = 2 \Re \left [ (\epsilon_1 \cdot {\bf E})^* (\epsilon_2 \cdot {\bf E}) \right ] = 2 a_1 a_2 \cos \left (\delta_2 - \delta_1 \right ) \\ V = s_3 = 2 \Im \left [ (\epsilon_1 \cdot {\bf E})^* (\epsilon_2 \cdot {\bf E}) \right ] = 2 a_1 a_2 \sin \left (\delta_2 - \delta_1 \right )$
- $I = s_0 = |\epsilon_+ \cdot {\bf E}|^2 + |\epsilon_- \cdot {\bf E}|^2 = a_+^2 + a_-^2 \\ Q = s_1 = 2 \Re \left [ (\epsilon_+ \cdot {\bf E})^* (\epsilon_- \cdot {\bf E}) \right ] = 2 a_+ a_- \cos \left (\delta_- - \delta_+ \right ) \\ U = s_2 = 2 \Im \left [ (\epsilon_+ \cdot {\bf E})^* (\epsilon_- \cdot {\bf E}) \right ] = 2 a_+ a_- \sin \left (\delta_- - \delta_+ \right ) \\ V = s_3 = |\epsilon_+ \cdot {\bf E}|^2 - |\epsilon_- \cdot {\bf E}|^2 = a_+^2 - a_-^2$
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- In symbols, this is $\omega = \frac{\Delta \theta}{\Delta t}$, where an angular rotation $\Delta\theta$ takes place in a time $\Delta t$.
- From the relation of $s$ and ($\Delta s = r\Delta \theta$), we see:
- $\displaystyle v = \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} = r \omega$
- $\displaystyle a_c = \frac{dv}{dt} = \omega \frac{dr}{dt} = \omega v = r \omega^2 = \frac{v^2}{r}$
- The arc length $\Delta s$ is described on the circumference.
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- Angular velocity ω is the rate of change of an angle, mathematically defined as ω = $\Delta \theta$$/\Delta t$ .
- From $\Delta \theta = (\Delta s)/r$ we see that $\Delta s = r\cdot \Delta \theta$.
- Substituting this into the expression for v gives $v = (r\cdot\Delta \theta)/(\Delta t) = r(\Delta \theta/\Delta t) = r\omega$.
- We can write this relationship in two different ways: v=rω or ω=v/r.
- The faster the car moves, the faster the tire spins—large v means a large ω, because v=rω.
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- $\frac{\Delta X^2 +\Delta Y^2 + \Delta Z^2}{\Delta T^2}=c^2 \rightarrow 0=-c^2 \Delta T^2 + \Delta X^2 + \Delta Y ^2 + \Delta Z^2$
- $s^2 = -c^2 \Delta t^2 + \Delta x^2 +\Delta y ^2 + \Delta z^2$
- Finally, let's discuss an important result of special relativity -- that the energy $E$ of an object moving with speed $v$ is:
- As $v \rightarrow c$ , $m \rightarrow \infty$, and it takes an infinite amount of energy to accelerate the object further.
- Two coordinate systems in which the primed frame moves with velocity v with respect to the unprimed frame
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- $s_1 = Q = \Pi I \cos2\psi \cos 2\chi \\ s_2 = U = \Pi I \sin2\psi \cos 2\chi \\ s_3 = V = \Pi I \sin 2\chi$
- $I = s_0 = |\epsilon_1 \cdot {\bf E}|^2 + |\epsilon_2 \cdot {\bf E}|^2 \\ \displaystyle = \frac{1}{\Delta t} \int_0^{\Delta t} \biggr [ \left | \epsilon_1 \cdot {\bf E}_a e^{-i\omega_a t} + \epsilon_1 \cdot {\bf E}_b e^{-i\omega_b t} \right |^2 + \nonumber \\ \displaystyle ~~~~~~ \left | \epsilon_2 \cdot {\bf E}_a e^{-i\omega_a t} + \epsilon_2 \cdot {\bf E}_b e^{-i\omega_b t} \right |^2 \biggr ] d t \\ \displaystyle = \frac{1}{\Delta t} \int_0^{\Delta t} \biggr \{ \left | \epsilon_1 \cdot {\bf E}_a \right |^2 + \left | \epsilon_1 \cdot {\bf E}_b \right |^2 + \left | \epsilon_1 \cdot {\bf E}_a \right |^2 + \left | \epsilon_1 \cdot {\bf E}_b \right |^2 + \nonumber \\ \displaystyle ~~~ 2 \left [ \left ( \epsilon_1 \cdot {\bf E}_a \right ) \left ( \epsilon_1 \cdot {\bf E}_b \right ) + \left ( \epsilon_2 \cdot {\bf E}_a \right ) \left ( \epsilon_2 \cdot {\bf E}_b \right ) \right ] \cos \left [ \left (\omega_a-\omega_b\right ) t \right ] \biggr \} \\ \displaystyle = s_{0,a} + s_{0,b} + \left \{ \begin{array}{cl} \mathcal{O} \left [ \sqrt{s_{0,a} s_{0,b}} \left ( \Delta \omega \Delta t \right)^{-1} \right ] \Delta\omega \Delta t \gg 1 \\\displaystyle \mathcal{O} \left [ \sqrt{s_{0,a} s_{0,b}} \left ( \Delta \omega \Delta t \right ) \right ] \Delta\omega \Delta t \ll 1\end{array} \right.$
- where $\Delta \omega = \omega_a-\omega_b$.
- For example, if we look at a star over a wide range of frequencies (the definition of wide is $\Delta \omega \Delta t \gg\ 1$), the phase of waves at one end of the frequency range will not correlate with waves at the other end.
- $s_2 = 2 \left < a_1 a_2 \cos \left (\delta_2 - \delta_1 \right ) \right >.$
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- $\displaystyle {\bf F} = q \left ( {\bf E} + \frac{\bf v}{c} \times {\bf B} \right ).$
- ${\bf v}\cdot {\bf F} = q {\bf v} \cdot {\bf E}.$
- One can imagine an ensemble of charged particles of charge density $\rho$ and define the current to be ${\bf J}=\rho {\bf v}$.
- $\displaystyle 0 = \frac{4\pi}{c} {\bf \nabla} \cdot {\bf J} + \frac{4\pi}{c} \frac{\delta \rho}{\delta t}$
- $\displaystyle {\bf J} \cdot {\bf E} = \frac{1}{4\pi} \left [ - {\bf H} \cdot \frac{\delta \bf B}{\delta t} - {\bf E} \cdot \frac{\delta \bf D}{\delta t} - c {\bf \nabla} \cdot \left ({\bf E} \times {\bf H} \right ) \right ]$