... | ... | @@ -30,10 +30,11 @@ to the lag, latexmath:[\varphi] and set the result to zero: |
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[latexmath]
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++++
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\begin{gathered}
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\mathrm{d}{\Delta E(\varphi,r)}{\varphi} = -\int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \sin(\omega t(z,\varphi) + \varphi) E_z(z,r)\\
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= -\cos(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \sin(\omega t(z,\varphi)) E_z(z,r) dz \\
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-\sin(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \cos(\omega t(z,\varphi)) E_z(z,r) dz \stackrel{!}{=} 0.\end{gathered}
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\begin{aligned}
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\mathrm{d}{\Delta E(\varphi,r)}{\varphi} &= -\int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \sin(\omega t(z,\varphi) + \varphi) E_z(z,r)\\
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&= -\cos(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \sin(\omega t(z,\varphi)) E_z(z,r) dz \\
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&-\sin(\varphi) \int_{z_\text{begin}}^{z_\text{end}} (1 + \omega \frac{\partial t(z,\varphi)}{\partial \varphi}) \cos(\omega t(z,\varphi)) E_z(z,r) dz \stackrel{!}{=} 0.
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\end{aligned}
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++++
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Thus to get the maximum energy the lag has to fulfill
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... | ... | @@ -84,7 +85,8 @@ where |
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\Gamma_{11,i} &= \int_0^1 \sin(\omega(t_{i-1} + \tau \Delta t_{i})) d\tau = - \frac{\cos(\omega t_{i}) - \cos(\omega t_{i-1})}{\omega \Delta t_{i}}\\
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\Gamma_{12,i} &= \int_0^1 \sin(\omega(t_{i-1} + \tau \Delta t_{i})) \tau d\tau = \frac{-\omega \Delta t_{i} \cos(\omega t_{i}) + \sin(\omega t_{i}) - \sin(\omega t_{i-1})}{\omega^2 (\Delta t_{i})^2}\\
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\Gamma_{21,i} &= \int_0^1 \cos(\omega(t_{i-1} + \tau \Delta t_{i})) d\tau = \frac{\sin(\omega t_{i}) - \sin(\omega t_{i-1})}{\omega \Delta t_{i}}\\
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\Gamma_{22,i} &= \int_0^1 \cos(\omega(t_{i-1} + \tau \Delta t_{i})) \tau d\tau = \frac{\omega \Delta t_{i} \sin(\omega t_{i}) + \cos(\omega t_{i}) - \cos(\omega t_{i-1})}{\omega^2 (\Delta t_{i})^2}\end{aligned}
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\Gamma_{22,i} &= \int_0^1 \cos(\omega(t_{i-1} + \tau \Delta t_{i})) \tau d\tau = \frac{\omega \Delta t_{i} \sin(\omega t_{i}) + \cos(\omega t_{i}) - \cos(\omega t_{i-1})}{\omega^2 (\Delta t_{i})^2}
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\end{aligned}
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++++
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... | ... | @@ -126,11 +128,11 @@ Figure 1. |
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[latexmath]
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++++
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\begin{gathered}
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\Delta E(\varphi,r) = q\, V_{0}\,\int_{z_\text{begin}}^{z_\text{beginCore}} \cos(\omega t(z,\varphi) + \varphi) E_z(z, r) dz \\
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+ q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c1} + \varphi) E_z(z, r) dz \\
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+ q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c2} + \varphi) E_z(z + s, r) dz \\
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+ q\, V_{0}\,\int_{z_\text{endCore}}^{z_\text{end}} \cos(\omega t(z,\varphi) + \varphi_\text{ef} + \varphi) E_z(z, r) dz,\end{gathered}
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\begin{aligned}
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\Delta E(\varphi,r) &= q\, V_{0}\,\int_{z_\text{begin}}^{z_\text{beginCore}} \cos(\omega t(z,\varphi) + \varphi) E_z(z, r) dz \\
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&+ q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c1} + \varphi) E_z(z, r) dz \\
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&+ q\, V_\text{core}\,\int_{z_\text{beginCore}}^{z_\text{endCore}} \cos(\omega t(z,\varphi) + \varphi_\text{c2} + \varphi) E_z(z + s, r) dz \\
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&+ q\, V_{0}\,\int_{z_\text{endCore}}^{z_\text{end}} \cos(\omega t(z,\varphi) + \varphi_\text{ef} + \varphi) E_z(z, r) dz,\end{aligned}
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++++
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where latexmath:[s] is the cell length. Instead of one sum as in
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... | ... | @@ -155,7 +157,8 @@ For this example we find |
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V_\text{core} &= \frac{V_{0}}{\sin(2.0/3.0 \pi)} = \frac{2 V_{0}}{\sqrt{3.0}}\\
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\varphi_\text{c1} &= \frac{\pi}{6}\\
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\varphi_\text{c2} &= \frac{\pi}{2}\\
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\varphi_\text{ef} &= - 2\pi \cdot(\mathbf{NUMCELLS} - 1) \cdot \mathbf{MODE} = 26\pi\end{aligned}
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\varphi_\text{ef} &= - 2\pi \cdot(\mathbf{NUMCELLS} - 1) \cdot \mathbf{MODE} = 26\pi
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\end{aligned}
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++++
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... | ... | @@ -171,11 +174,12 @@ example from above the energy gain is approximately |
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[latexmath]
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++++
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\begin{gathered}
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\Delta E(\varphi,r) = q\;V_0 \int_{0}^{1.5\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz\\
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+ \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z(z\;\;\quad,r) dz\\
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+ \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z(z+s,r) dz \\
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+ q\;V_{0} \int_{40.5\cdot s}^{42\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz.\end{gathered}
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\begin{aligned}
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\Delta E(\varphi,r) &= q\;V_0 \int_{0}^{1.5\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz\\
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&+ \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z(z\;\;\quad,r) dz\\
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&+ \frac{2 q\;V_{0}}{\sqrt{3}} \int_{1.5\cdot s}^{40.5\cdot s}\cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z(z+s,r) dz \\
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&+ q\;V_{0} \int_{40.5\cdot s}^{42\cdot s} \cos\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z(z,r)\, dz.
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\end{aligned}
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++++
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Here latexmath:[\beta c = 2.9886774\cdot10^8\;\text{m s}^{-2}],
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... | ... | @@ -189,11 +193,12 @@ cell), latexmath:[E_z^{(2)}]. The core fringe field is periodic with a |
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period of latexmath:[3\,s]. We thus find
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[latexmath]
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++++
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\begin{gathered}
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0 \stackrel{!}{=} \int_{0}^{1.5\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz \\
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+ \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z^{(1)}(z \text{ mod}(3\,s)\;\;\qquad,r)\,dz \\
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+ \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z^{(1)}((z + s) \text{ mod} (3\,s),r)\, dz \\
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+ \int_{1.5\cdot s}^{3\cdot s} \sin\left(\omega\left(\frac{z + 39\,s}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz\end{gathered}
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\begin{aligned}
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0 \stackrel{!}{=}& \int_{0}^{1.5\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz \\
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&+ \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi \right) E_z^{(1)}(z \text{ mod}(3\,s),r)\,dz \\
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&+ \frac{2}{\sqrt{3}} \int_{0}^{39\cdot s}\sin\left(\omega \left(\frac{z + 1.5\,s}{\beta c} + t_{0}\right) + \frac{\pi}{2} + \varphi \right) E_z^{(1)}((z + s) \text{ mod} (3\,s),r)\, dz \\
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&+ \int_{1.5\cdot s}^{3\cdot s} \sin\left(\omega\left(\frac{z + 39\,s}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z,r)\, dz
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\end{aligned}
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++++
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This equation is much simplified if we take into account that
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... | ... | @@ -201,20 +206,20 @@ latexmath:[\omega / \beta c \approx 10\pi]. We then get |
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[latexmath]
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++++
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\begin{gathered}
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0 \stackrel{!}{=} \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z)\, dz \\
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+ \frac{26}{\sqrt{3}} \int_{0}^{3\cdot s}\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{7\pi}{6} + \varphi \right)
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\begin{aligned}
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0 \stackrel{!}{=}& \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) E_z^{(2)}(z)\, dz \\
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+ \frac{26}{\sqrt{3}} & \int_{0}^{3\cdot s}\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{7\pi}{6} + \varphi \right)
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+ \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{5\pi}{6} + \varphi \right)\right) E_z^{(1)}(z)\, dz \\
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= \int_{0}^{3\cdot s}\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \left(E_z^{(2)} - 26\cdot E_z^{(1)}\right)(z)\,dz\end{gathered}
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= & \int_{0}^{3\cdot s}\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \left(E_z^{(2)} - 26\cdot E_z^{(1)}\right)(z)\,dz\end{aligned}
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++++
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where we used
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[latexmath]
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++++
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\begin{gathered}
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\int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{3\pi}{2} + \varphi\right) E_z^{(1)}((z + s) \text{ mod}(3\,s),r) dz \\
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\stackrel{z' = z + s}{\longrightarrow} \int_{s}^{4\cdot s} \sin\left(\omega \left(\frac{z'-s}{\beta c} + t_{0} \right) + \frac{3\pi}{2} + \varphi\right)E_z^{(1)}(z' \text{ mod}(3\,s),r)dz' \\
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= \int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{5\pi}{6} + \varphi\right) E_z^{(1)}(z,r)\,dz.\end{gathered}
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\begin{aligned}
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&\int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{3\pi}{2} + \varphi\right) E_z^{(1)}((z + s) \text{ mod}(3\,s),r) dz \\
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\stackrel{z' = z + s}{\longrightarrow}& \int_{s}^{4\cdot s} \sin\left(\omega \left(\frac{z'-s}{\beta c} + t_{0} \right) + \frac{3\pi}{2} + \varphi\right)E_z^{(1)}(z' \text{ mod}(3\,s),r)dz' \\
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= &\int_{0}^{3\cdot s} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{5\pi}{6} + \varphi\right) E_z^{(1)}(z,r)\,dz.\end{aligned}
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++++
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In the last equal sign we used the fact that both functions,
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... | ... | @@ -266,12 +271,13 @@ Here we also used some trigonometric identities: |
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[latexmath]
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++++
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\begin{gathered}
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\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \pi + \frac{\pi}{6} + \varphi \right) +
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\begin{aligned}
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&\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \pi + \frac{\pi}{6} + \varphi \right) +
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\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \pi - \frac{\pi}{6} + \varphi \right) \\
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= -\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi\right) +
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&= -\left(\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \frac{\pi}{6} + \varphi\right) +
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\sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) - \frac{\pi}{6} + \varphi\right)\right) \\
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= -2\cdot \cos\left(\frac{\pi}{6}\right) \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \\
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= -\sqrt{3} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right)\end{gathered}
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&= -2\cdot \cos\left(\frac{\pi}{6}\right) \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right) \\
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&= -\sqrt{3} \sin\left(\omega \left(\frac{z}{\beta c} + t_{0}\right) + \varphi\right)
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\end{aligned}
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++++
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