Новое решение

Alexphysics правка от 23 июн. 2026 г., 16:47

правка #19114 позже →

en/14.4.15.md +65 −0

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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]