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| Average accelerating electric field:$ E = 10^5\ \text{V/cm}$ | | Average accelerating electric field:$ E = 10^5\ \text{V/cm}$ |
| Final energy of the$ \pi^+ meson: \mathcal{E} = 10^{10}\ \text{eV}$ | | Final energy of the$ \pi^+ meson: \mathcal{E} = 10^{10}\ \text{eV}$ |
| Rest energy of the meson: $m_\pi c^2 = 10^8\ \text{eV}$ | | Rest energy of the meson: $m_\pi c^2 = 10^8\ \text{eV}$ |
| Charge of the meson: e (same elementary charge). | | Charge of the meson: e (same elementary charge). |
| The meson starts from rest. | | The meson starts from rest. |
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| In a uniform electric field directed along the motion, the force on the meson is constant: $ F = eE$ | | In a uniform electric field directed along the motion, the force on the meson is constant: $ F = eE$ |
| The work done by the field over the distance l is $W = eEl.$ | | The work done by the field over the distance l is $W = eEl.$ |
| This work is converted into relativistic kinetic energy. Since the particle starts from rest, the final kinetic energy is: | | This work is converted into relativistic kinetic energy. Since the particle starts from rest, the final kinetic energy is: |
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| $\mathcal{E}_{\text{kin}} = \mathcal{E} - m_\pi c^2$ | | $\mathcal{E}_{\text{kin}} = \mathcal{E} - m_\pi c^2$ |
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| But$ \mathcal{E} = 10^{10}\ \text{eV} \gg m_\pi c^2 = 10^8\ \text{eV}$ so the final energy is practically all kinetic. Thus: | | But$ \mathcal{E} = 10^{10}\ \text{eV} \gg m_\pi c^2 = 10^8\ \text{eV}$ so the final energy is practically all kinetic. Thus: |
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| $eEl \approx \mathcal{E} \quad\Rightarrow\quad l = \frac{\mathcal{E}}{eE}$. | | $eEl \approx \mathcal{E} \quad\Rightarrow\quad l = \frac{\mathcal{E}}{eE}$. |
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| Substituting the values gives: | | Substituting the values gives: |
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| $\boxed{l = 1\ \text{km}}$ | | $\boxed{l = 1\ \text{km}}$ |
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| Acceleration time | | Acceleration time |
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| Newton's second law in relativistic form is: | | Newton's second law in relativistic form is: |
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| $\frac{dp}{dt} = eE$ | | $\frac{dp}{dt} = eE$ |
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| where$ p = \gamma m_\pi v$ is the linear momentum and $\gamma = 1/\sqrt{1 - v^2/c^2}$ | | where$ p = \gamma m_\pi v$ is the linear momentum and $\gamma = 1/\sqrt{1 - v^2/c^2}$ |
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| Integrating from rest $(p(0)=0) $ up to the final momentum $p_f $corresponding to the energy \mathcal{E} | | Integrating from rest $(p(0)=0) $ up to the final momentum $p_f $corresponding to the energy \mathcal{E} |
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| $\tau = \frac{p_f}{eE}$ | | $\tau = \frac{p_f}{eE}$ |
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| The final momentum is related to the total energy by | | The final momentum is related to the total energy by |
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| $p_f c = \sqrt{\mathcal{E}^2 - (m_\pi c^2)^2}$. | | $p_f c = \sqrt{\mathcal{E}^2 - (m_\pi c^2)^2}$. |
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| Since$ \mathcal{E} \gg m_\pi c^2 f | | Since$ \mathcal{E} \gg m_\pi c^2 f |
| one can approximate $p_f \approx \mathcal{E}/c $ However, for greater precision, the exact expression can be used. | | one can approximate $p_f \approx \mathcal{E}/c $ However, for greater precision, the exact expression can be used. |
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| An alternative way to find \tau is to integrate the velocity as a function of time. From$ dp/dt = eE $we get$ p = eEt$. The velocity is: | | An alternative way to find \tau is to integrate the velocity as a function of time. From$ dp/dt = eE $we get$ p = eEt$. The velocity is: |
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| $v = \frac{p}{m_\pi \sqrt{1 + \dfrac{p^2}{m_\pi^2 c^2}}}$ | | $v = \frac{p}{m_\pi \sqrt{1 + \dfrac{p^2}{m_\pi^2 c^2}}}$ |
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| The total energy as a function of time is: | | The total energy as a function of time is: |
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| $\mathcal{E}(t) = \sqrt{(m_\pi c^2)^2 + (eEtc)^2}$ | | $\mathcal{E}(t) = \sqrt{(m_\pi c^2)^2 + (eEtc)^2}$ |
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| The time to reach the final energy $\mathcal{E}$ is found by inverting this relation or by directly integrating $dt = \frac{dp}{eE}$ in the expression for $dx = v\,dt$ | | The time to reach the final energy $\mathcal{E}$ is found by inverting this relation or by directly integrating $dt = \frac{dp}{eE}$ in the expression for $dx = v\,dt$ |
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| The resulting integral is | | The resulting integral is |
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| $\tau = \frac{1}{eE} \int_0^{\mathcal{E}} \left(2 + \frac{\mathcal{E}}{m_\pi v^2}\right) m_\pi v\,dv$ | | $\tau = \frac{1}{eE} \int_0^{\mathcal{E}} \left(2 + \frac{\mathcal{E}}{m_\pi v^2}\right) m_\pi v\,dv$ |
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| Carrying out the numerical calculations gives: | | Carrying out the numerical calculations gives: |
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| $\boxed{\tau = 0.34\ \text{ms}}$ | | $\boxed{\tau = 0.34\ \text{ms}}$ |
