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en/14.4.15.md
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| + | ### Statement | ||
| + | |||
| + | $14.4.15.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Center-of-mass energy in the symmetric collider | ||
| + | |||
| + | Each electron has total energy $ E = N m_e c^2$. | ||
| + | When two identical particles collide head‑on with opposite momenta, the total energy in the center‑of‑mass (CM) system is simply the sum of their total energies: | ||
| + | |||
| + | $E_{\text{CM}} = 2E = 2N m_e c^2$ | ||
| + | |||
| + | This is the energy available for the interaction. | ||
| + | |||
| + | Center-of-mass energy in the fixed target | ||
| + | |||
| + | Now we have an incident electron with total energy$ E_{\text{lab}}$ and a target electron at rest (energy $m_e c^2)$ | ||
| + | The square of the center‑of‑mass energy for this system is calculated using the relativistic invariant s | ||
| + | |||
| + | $s = (E_{\text{lab}} + m_e c^2)^2 - (p_{\text{lab}} c)^2$ | ||
| + | |||
| + | where$ p_{\text{lab}} $is the momentum of the incident electron. | ||
| + | |||
| + | Using the relation | ||
| + | $E_{\text{lab}}^2 - (p_{\text{lab}} c)^2 = (m_e c^2)^2$ | ||
| + | we expand: | ||
| + | |||
| + | $s = E_{\text{lab}}^2 + 2E_{\text{lab}}m_e c^2 + (m_e c^2)^2 - (p_{\text{lab}} c)^2 | ||
| + | = 2(m_e c^2)^2 + 2E_{\text{lab}}m_e c^2$ | ||
| + | |||
| + | The center‑of‑mass energy is E_{\text{CM}} = \sqrt{s}. | ||
| + | |||
| + | For high energies $(E_{\text{lab}} \gg m_e c^2) $this approximates to: | ||
| + | |||
| + | $E_{\text{CM}} \approx \sqrt{2m_e c^2 E_{\text{lab}}}$ | ||
| + | |||
| + | Equating the center‑of‑mass energies | ||
| + | |||
| + | We want the available energy in the fixed target to be the same as in the symmetric collider: | ||
| + | |||
| + | $\sqrt{2m_e c^2 E_{\text{lab}}} = 2N m_e c^2$ | ||
| + | |||
| + | Squaring both sides: | ||
| + | |||
| + | $2m_e c^2 E_{\text{lab}} = 4 N^2 (m_e c^2)^2 | ||
| + | \quad\Rightarrow\quad | ||
| + | E_{\text{lab}} = 2 N^2 m_e c^2$ | ||
| + | |||
| + | The total energy that the incident electron must have is $E_{\text{lab}} = 2 N^2 m_e c^2$ | ||
| + | Since N = 1000, we get | ||
| + | |||
| + | $E_{\text{lab}} = 2 \times (1000)^2\, m_e c^2 = 2 \times 10^6\, m_e c^2$ | ||
| + | |||
| + | That is, the electron energy must be 2 million times its rest energy. | ||
| + | |||
| + | Comparing with the energy of each electron in the collider $(N m_e c^2 = 1000\, m_e c^2)$, the increase factor is: | ||
| + | |||
| + | $\frac{E_{\text{lab}}}{N m_e c^2} = \frac{2 N^2}{N} = 2N = 2000$ | ||
| + | |||
| + | Therefore, an energy 2000 times greater than that of each electron in the colliding beams is needed. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.4.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| Center-of-mass energy in the symmetric collider | |||
| Each electron has total energy $ E = N m_e c^2$. | |||
| When two identical particles collide head‑on with opposite momenta, the total energy in the center‑of‑mass (CM) system is simply the sum of their total energies: | |||
| $E_{\text{CM}} = 2E = 2N m_e c^2$ | |||
| This is the energy available for the interaction. | |||
| Center-of-mass energy in the fixed target | |||
| Now we have an incident electron with total energy$ E_{\text{lab}}$ and a target electron at rest (energy $m_e c^2)$ | |||
| The square of the center‑of‑mass energy for this system is calculated using the relativistic invariant s | |||
| $s = (E_{\text{lab}} + m_e c^2)^2 - (p_{\text{lab}} c)^2$ | |||
| where$ p_{\text{lab}} $is the momentum of the incident electron. | |||
| Using the relation | |||
| $E_{\text{lab}}^2 - (p_{\text{lab}} c)^2 = (m_e c^2)^2$ | |||
| we expand: | |||
| $s = E_{\text{lab}}^2 + 2E_{\text{lab}}m_e c^2 + (m_e c^2)^2 - (p_{\text{lab}} c)^2 | |||
| = 2(m_e c^2)^2 + 2E_{\text{lab}}m_e c^2$ | |||
| The center‑of‑mass energy is E_{\text{CM}} = \sqrt{s}. | |||
| For high energies $(E_{\text{lab}} \gg m_e c^2) $this approximates to: | |||
| $E_{\text{CM}} \approx \sqrt{2m_e c^2 E_{\text{lab}}}$ | |||
| Equating the center‑of‑mass energies | |||
| We want the available energy in the fixed target to be the same as in the symmetric collider: | |||
| $\sqrt{2m_e c^2 E_{\text{lab}}} = 2N m_e c^2$ | |||
| Squaring both sides: | |||
| $2m_e c^2 E_{\text{lab}} = 4 N^2 (m_e c^2)^2 | |||
| \quad\Rightarrow\quad | |||
| E_{\text{lab}} = 2 N^2 m_e c^2$ | |||
| The total energy that the incident electron must have is $E_{\text{lab}} = 2 N^2 m_e c^2$ | |||
| Since N = 1000, we get | |||
| $E_{\text{lab}} = 2 \times (1000)^2\, m_e c^2 = 2 \times 10^6\, m_e c^2$ | |||
| That is, the electron energy must be 2 million times its rest energy. | |||
| Comparing with the energy of each electron in the collider $(N m_e c^2 = 1000\, m_e c^2)$, the increase factor is: | |||
| $\frac{E_{\text{lab}}}{N m_e c^2} = \frac{2 N^2}{N} = 2N = 2000$ | |||
| Therefore, an energy 2000 times greater than that of each electron in the colliding beams is needed. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||