Правка разделов «Statement», «Solution», «Answer»
en/14.1.24.md
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| ### Statement | |||
| − | $14.1.24.$ | ||
| + | $14.1.24.$ | ||
| + | $\pi^0 $-Mesons having the same velocity βc decay into $\gamma$-quants: | ||
| + | $$ | ||
| + | \pi^0\to \gamma + \gamma | ||
| + | $$ | ||
| + | What part of the γ-quanta moves at angles to the velocity $\beta c$ less than $\frac{\pi}{2}$ ? | ||
| ### Solution | |||
| + | First, let us go to the rest frame of the $\pi^0$ mesons. In this frame, denote the angles to the velocity $\beta c$ as $\theta$. Obviously, for the conservation of momentum, the photons of each pair fly apart strictly in opposite directions, but overall isotropically – all $\theta$ for the line along which the photons fly apart are equally probable. | ||
| − | π 0 -Mesons having the same velocity βc decay into vhquants: π 0 → γ +γ. What | ||
| − | part of the γ-quanta moves at angles to the velocity βc less than π2 ? | ||
| + | Now let us switch to the laboratory frame. We need to find how the angles change. This can be done using the ready‑made formula for light aberration, but I will show the derivation using the Lorentz transformations for the 4‑momentum. Taking into account that for a photon $E=pc$, it is not necessary to write the full transformations; it is sufficient to write | ||
| + | $$ | ||
| + | \begin{pmatrix} | ||
| + | p' \\\\ | ||
| + | p_x' | ||
| + | \end{pmatrix}= | ||
| + | \begin{pmatrix} | ||
| + | p \\\\ | ||
| + | p_x | ||
| + | \end{pmatrix}\cdot | ||
| + | \begin{pmatrix} | ||
| + | \gamma & \beta\gamma \\\\ | ||
| + | \beta\gamma & \gamma | ||
| + | \end{pmatrix} | ||
| + | $$ | ||
| + | $$ | ||
| + | p' \begin{pmatrix} | ||
| + | 1\\\\ | ||
| + | \cos\theta' | ||
| + | \end{pmatrix}=\gamma p | ||
| + | \begin{pmatrix} | ||
| + | 1 \\\\ | ||
| + | \cos\theta | ||
| + | \end{pmatrix}\cdot | ||
| + | \begin{pmatrix} | ||
| + | 1 & \beta \\\\ | ||
| + | \beta& 1 | ||
| + | \end{pmatrix}=\gamma p | ||
| + | \begin{pmatrix} | ||
| + | 1+\beta\cos\theta \\\\ | ||
| + | \beta+\cos\theta | ||
| + | \end{pmatrix} | ||
| + | $$ | ||
| + | $$ | ||
| + | \cos\theta'=\frac{p_x'}{p'}=\frac{\beta+\cos\theta}{1+\beta\cos\theta } | ||
| + | $$ | ||
| + | |||
| + | Working with the full matrix gives the same result. | ||
| + | |||
| + | The condition $\theta'<\frac{\pi}{2}$ is equivalent to $\cos\theta'>0$. | ||
| + | $$ | ||
| + | 0<\beta<1;\ \cos\theta\ge -1\quad \to \quad 1+\beta\cos\theta>0 \text{ – always.} | ||
| + | $$ | ||
| + | Then the condition can be rewritten as | ||
| + | $$ | ||
| + | \cos\theta>-\beta. | ||
| + | $$ | ||
| + | The probability, taking into account the isotropy of $\theta$ , is then found as follows ($k$ is a normalisation factor chosen so that the total probability equals 1): | ||
| + | $$ | ||
| + | N=\frac{\int_{-\beta}^1 k\, d(\cos\theta)}{\int_{-1}^1 k\, d(\cos\theta)}=\frac{1+\beta}{2}. | ||
| + | $$ | ||
| + | |||
| + | |||
| + | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$ | ||
| + | \boxed{\frac{1+\beta}{2}} | ||
| + | $$ | ||
| @@ -1,12 +1,72 @@ | |||
| ### Statement | ### Statement | ||
| $14.1.24.$ |
$14.1.24.$ | ||
| $\pi^0 $-Mesons having the same velocity βc decay into $\gamma$-quants: | |||
| $$ | |||
| \pi^0\to \gamma + \gamma | |||
| $$ | |||
| What part of the γ-quanta moves at angles to the velocity $\beta c$ less than $\frac{\pi}{2}$ ? | |||
| ### Solution | ### Solution | ||
| First, let us go to the rest frame of the $\pi^0$ mesons. In this frame, denote the angles to the velocity $\beta c$ as $\theta$. Obviously, for the conservation of momentum, the photons of each pair fly apart strictly in opposite directions, but overall isotropically – all $\theta$ for the line along which the photons fly apart are equally probable. | |||
| π 0 -Mesons having the same velocity βc decay into vhquants: π 0 → γ +γ. What | Now let us switch to the laboratory frame. We need to find how the angles change. This can be done using the ready‑made formula for light aberration, but I will show the derivation using the Lorentz transformations for the 4‑momentum. Taking into account that for a photon $E=pc$, it is not necessary to write the full transformations; it is sufficient to write | ||
| part of the γ-quanta moves at angles to the velocity βc less than π2 ? | $$ | ||
| \begin{pmatrix} | |||
| p' \\\\ | |||
| p_x' | |||
| \end{pmatrix}= | |||
| \begin{pmatrix} | |||
| p \\\\ | |||
| p_x | |||
| \end{pmatrix}\cdot | |||
| \begin{pmatrix} | |||
| \gamma & \beta\gamma \\\\ | |||
| \beta\gamma & \gamma | |||
| \end{pmatrix} | |||
| $$ | |||
| $$ | |||
| p' \begin{pmatrix} | |||
| 1\\\\ | |||
| \cos\theta' | |||
| \end{pmatrix}=\gamma p | |||
| \begin{pmatrix} | |||
| 1 \\\\ | |||
| \cos\theta | |||
| \end{pmatrix}\cdot | |||
| \begin{pmatrix} | |||
| 1 & \beta \\\\ | |||
| \beta& 1 | |||
| \end{pmatrix}=\gamma p | |||
| \begin{pmatrix} | |||
| 1+\beta\cos\theta \\\\ | |||
| \beta+\cos\theta | |||
| \end{pmatrix} | |||
| $$ | |||
| $$ | |||
| \cos\theta'=\frac{p_x'}{p'}=\frac{\beta+\cos\theta}{1+\beta\cos\theta } | |||
| $$ | |||
| Working with the full matrix gives the same result. | |||
| The condition $\theta'<\frac{\pi}{2}$ is equivalent to $\cos\theta'>0$. | |||
| $$ | |||
| 0<\beta<1;\ \cos\theta\ge -1\quad \to \quad 1+\beta\cos\theta>0 \text{ – always.} | |||
| $$ | |||
| Then the condition can be rewritten as | |||
| $$ | |||
| \cos\theta>-\beta. | |||
| $$ | |||
| The probability, taking into account the isotropy of $\theta$ , is then found as follows ($k$ is a normalisation factor chosen so that the total probability equals 1): | |||
| $$ | |||
| N=\frac{\int_{-\beta}^1 k\, d(\cos\theta)}{\int_{-1}^1 k\, d(\cos\theta)}=\frac{1+\beta}{2}. | |||
| $$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$ | ||
| \boxed{\frac{1+\beta}{2}} | |||
| $$ | |||