Правка раздела «Answer»
en/14.2.2.md
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| − | ### Statement | ||
| − | |||
| − | $14.2.2.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| ### Problem | |||
| $14.2.2.$ Lifetime of a stationary particle \( \tau \). At what speed must this particle move in order to travel a distance \( l \) before decaying? | |||
| ### Solution | |||
| In the laboratory frame, the lifetime of a moving particle is increased due to the relativistic time dilation effect: | |||
| \[ | |||
| \tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}} | |||
| \] | |||
| The particle must cover the distance \( l \) within the time \( \tau' \): | |||
| \[ | |||
| l < v \tau' | |||
| \] | |||
| Substituting the expression for \( \tau' \): | |||
| \[ | |||
| l < \frac{v \tau}{\sqrt{1 - \frac{v^2}{c^2}}} | |||
| \] | |||
| Let's solve this inequality for \( v \). First, divide both sides by \( \tau \) and square both sides: | |||
| \[ | |||
| \frac{v^2}{1 - \frac{v^2}{c^2}} > \left( \frac{l}{\tau} \right)^2 | |||
| \] | |||
| Multiply both sides by \( 1 - \frac{v^2}{c^2} \): | |||
| \[ | |||
| v^2 > \left( \frac{l}{\tau} \right)^2 \left( 1 - \frac{v^2}{c^2} \right) | |||
| \] | |||
| Distribute and bring all terms to one side: | |||
| \[ | |||
| v^2 + \left( \frac{l}{\tau c} \right)^2 v^2 > \left( \frac{l}{\tau} \right)^2 | |||
| \] | |||
| Factor out \( v^2 \): | |||
| \[ | |||
| v^2 \left( 1 + \left( \frac{l}{\tau c} \right)^2 \right) > \left( \frac{l}{\tau} \right)^2 | |||
| \] | |||
| Now solve for \( v \): | |||
| \[ | |||
| v^2 > \frac{\left( \frac{l}{\tau} \right)^2}{1 + \left( \frac{l}{\tau c} \right)^2} | |||
| \] | |||
| \[ | |||
| v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | |||
| \] | |||
| ### Answer | |||
| @@ -69,7 +63,3 @@Answer | |||
| \[ | |||
| v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | |||
| \] | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,9 +1,3 @@ | |||
| ### Statement | |||
| $14.2.2.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | ### Problem | ||
| $14.2.2.$ Lifetime of a stationary particle \( \tau \). At what speed must this particle move in order to travel a distance \( l \) before decaying? | $14.2.2.$ Lifetime of a stationary particle \( \tau \). At what speed must this particle move in order to travel a distance \( l \) before decaying? | ||
| ### Solution | ### Solution | ||
| In the laboratory frame, the lifetime of a moving particle is increased due to the relativistic time dilation effect: | In the laboratory frame, the lifetime of a moving particle is increased due to the relativistic time dilation effect: | ||
| \[ | \[ | ||
| \tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}} | \tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}} | ||
| \] | \] | ||
| The particle must cover the distance \( l \) within the time \( \tau' \): | The particle must cover the distance \( l \) within the time \( \tau' \): | ||
| \[ | \[ | ||
| l < v \tau' | l < v \tau' | ||
| \] | \] | ||
| Substituting the expression for \( \tau' \): | Substituting the expression for \( \tau' \): | ||
| \[ | \[ | ||
| l < \frac{v \tau}{\sqrt{1 - \frac{v^2}{c^2}}} | l < \frac{v \tau}{\sqrt{1 - \frac{v^2}{c^2}}} | ||
| \] | \] | ||
| Let's solve this inequality for \( v \). First, divide both sides by \( \tau \) and square both sides: | Let's solve this inequality for \( v \). First, divide both sides by \( \tau \) and square both sides: | ||
| \[ | \[ | ||
| \frac{v^2}{1 - \frac{v^2}{c^2}} > \left( \frac{l}{\tau} \right)^2 | \frac{v^2}{1 - \frac{v^2}{c^2}} > \left( \frac{l}{\tau} \right)^2 | ||
| \] | \] | ||
| Multiply both sides by \( 1 - \frac{v^2}{c^2} \): | Multiply both sides by \( 1 - \frac{v^2}{c^2} \): | ||
| \[ | \[ | ||
| v^2 > \left( \frac{l}{\tau} \right)^2 \left( 1 - \frac{v^2}{c^2} \right) | v^2 > \left( \frac{l}{\tau} \right)^2 \left( 1 - \frac{v^2}{c^2} \right) | ||
| \] | \] | ||
| Distribute and bring all terms to one side: | Distribute and bring all terms to one side: | ||
| \[ | \[ | ||
| v^2 + \left( \frac{l}{\tau c} \right)^2 v^2 > \left( \frac{l}{\tau} \right)^2 | v^2 + \left( \frac{l}{\tau c} \right)^2 v^2 > \left( \frac{l}{\tau} \right)^2 | ||
| \] | \] | ||
| Factor out \( v^2 \): | Factor out \( v^2 \): | ||
| \[ | \[ | ||
| v^2 \left( 1 + \left( \frac{l}{\tau c} \right)^2 \right) > \left( \frac{l}{\tau} \right)^2 | v^2 \left( 1 + \left( \frac{l}{\tau c} \right)^2 \right) > \left( \frac{l}{\tau} \right)^2 | ||
| \] | \] | ||
| Now solve for \( v \): | Now solve for \( v \): | ||
| \[ | \[ | ||
| v^2 > \frac{\left( \frac{l}{\tau} \right)^2}{1 + \left( \frac{l}{\tau c} \right)^2} | v^2 > \frac{\left( \frac{l}{\tau} \right)^2}{1 + \left( \frac{l}{\tau c} \right)^2} | ||
| \] | \] | ||
| \[ | \[ | ||
| v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | ||
| \] | \] | ||
| ### Answer | ### Answer | ||
| @@ -69,7 +63,3 @@Answer | |||
| \[ | \[ | ||
| v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}} | ||
| \] | \] | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||