Правка разделов «Solution», «Answer»
en/14.5.14.md
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| ### Statement | |||
| $14.5.14^∗ $. The mass and momentum of a state obtained when a state with mass $M$ and zero momentum moves with velocity $v$ are $\gamma M$ and $\gamma Mv_0$, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$. Prove this statement for a state in which two non-interacting particles are moving. | |||
| ### Solution | |||
| + | As in many other problems in this section, the author considers mass to be a measure of total energy. Then, let the center-of-mass system of two particles | ||
| + | $$ | ||
| + | E_1+E_2=Mc^2 \qquad \vec{p_1}+\vec{p_2}=0 \quad | ||
| + | $$ | ||
| + | $Note \ 1$: It follows from this that the sum of the projections of the momenta on any axis is equal to 0 | ||
| + | Now, using the Lorentz transformations, we move to a system moving with velocity $v$ along the $x$ axis. For each particle, we can write: | ||
| + | $$\begin{pmatrix} | ||
| + | E'/c \\\\ | ||
| + | p'_x \\\\ | ||
| + | p'_y \\\\ | ||
| + | p'_z | ||
| + | \end{pmatrix}= | ||
| + | \begin{pmatrix} | ||
| + | E/c\\\\ | ||
| + | p_x \\\\ | ||
| + | p_y \\\\ | ||
| + | p_z | ||
| + | \end{pmatrix} | ||
| + | \cdot | ||
| + | \begin{pmatrix} | ||
| + | \gamma& \beta\gamma & 0 & 0 \\\\ | ||
| + | \beta\gamma &\gamma& 0 & 0 \\\\ | ||
| + | 0 & 0 & 1 & 0 \\\\ | ||
| + | 0 & 0 & 0 & 1 | ||
| + | \end{pmatrix}= | ||
| + | \begin{pmatrix} | ||
| + | \frac{\gamma}{c}(E+vp_x)\\\\ | ||
| + | \gamma(p_x+\beta\frac{E}{c}) \\\\ | ||
| + | p_y \\\\ | ||
| + | p_z | ||
| + | \end{pmatrix}\tag{2} | ||
| + | $$ | ||
| + | Summing, we obtain: | ||
| + | $$ | ||
| + | E'=E'_1-E'_2=\gamma(E_1+vp_{x1}+E_2+vp_{x1})\tag{3} | ||
| + | $$ | ||
| + | $$ | ||
| + | \vec{P'}=\hat{x}(p'_{x1}+p'_{x2})+\hat{y}(p'_{y1}+p'_{y2})+\hat{z}(p'_{z1}+p'_{z2}) | ||
| + | $$ | ||
| + | $$ | ||
| + | \vec{P'}=\gamma\hat{x}\left(p_{x1}+\beta\frac{E_1}{c}+p_{x2}+\beta\frac{E_2}{c}\right)+\hat{y}(p_{y1}+p_{y2})+\hat{z}(p_{z1}+p_{z2})\tag{4} | ||
| + | $$ | ||
| + | Concidering $Note \ 1$, | ||
| + | $$ | ||
| + | E'=M'c^2=\gamma(E_1+E_2+vp_{x1})=\gamma Mc^2 \quad so\quad M'=\gamma M \tag{5} | ||
| + | $$ | ||
| + | $$ | ||
| + | \vec{P'}=\gamma\hat{x}\left(\beta\frac{E_1+E_2}{c}\right)=\gamma \vec{v} \frac{Mc^2}{c^2}=\gamma M\vec{v} \tag{6} | ||
| + | $$ | ||
| + | |||
| + | |||
| + | Which is what we needed. | ||
| + | |||
| + | #### Answer | ||
| + | $$ | ||
| + | \boxed{M'=\gamma M\qquad \vec{P'}=\gamma M\vec{v} } | ||
| + | $$ | ||
| @@ -1,7 +1,64 @@ | |||
| ### Statement | ### Statement | ||
| $14.5.14^∗ $. The mass and momentum of a state obtained when a state with mass $M$ and zero momentum moves with velocity $v$ are $\gamma M$ and $\gamma Mv_0$, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$. Prove this statement for a state in which two non-interacting particles are moving. | $14.5.14^∗ $. The mass and momentum of a state obtained when a state with mass $M$ and zero momentum moves with velocity $v$ are $\gamma M$ and $\gamma Mv_0$, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$. Prove this statement for a state in which two non-interacting particles are moving. | ||
| ### Solution | ### Solution | ||
| As in many other problems in this section, the author considers mass to be a measure of total energy. Then, let the center-of-mass system of two particles | |||
| $$ | |||
| E_1+E_2=Mc^2 \qquad \vec{p_1}+\vec{p_2}=0 \quad | |||
| $$ | |||
| $Note \ 1$: It follows from this that the sum of the projections of the momenta on any axis is equal to 0 | |||
| Now, using the Lorentz transformations, we move to a system moving with velocity $v$ along the $x$ axis. For each particle, we can write: | |||
| $$\begin{pmatrix} | |||
| E'/c \\\\ | |||
| p'_x \\\\ | |||
| p'_y \\\\ | |||
| p'_z | |||
| \end{pmatrix}= | |||
| \begin{pmatrix} | |||
| E/c\\\\ | |||
| p_x \\\\ | |||
| p_y \\\\ | |||
| p_z | |||
| \end{pmatrix} | |||
| \cdot | |||
| \begin{pmatrix} | |||
| \gamma& \beta\gamma & 0 & 0 \\\\ | |||
| \beta\gamma &\gamma& 0 & 0 \\\\ | |||
| 0 & 0 & 1 & 0 \\\\ | |||
| 0 & 0 & 0 & 1 | |||
| \end{pmatrix}= | |||
| \begin{pmatrix} | |||
| \frac{\gamma}{c}(E+vp_x)\\\\ | |||
| \gamma(p_x+\beta\frac{E}{c}) \\\\ | |||
| p_y \\\\ | |||
| p_z | |||
| \end{pmatrix}\tag{2} | |||
| $$ | |||
| Summing, we obtain: | |||
| $$ | |||
| E'=E'_1-E'_2=\gamma(E_1+vp_{x1}+E_2+vp_{x1})\tag{3} | |||
| $$ | |||
| $$ | |||
| \vec{P'}=\hat{x}(p'_{x1}+p'_{x2})+\hat{y}(p'_{y1}+p'_{y2})+\hat{z}(p'_{z1}+p'_{z2}) | |||
| $$ | |||
| $$ | |||
| \vec{P'}=\gamma\hat{x}\left(p_{x1}+\beta\frac{E_1}{c}+p_{x2}+\beta\frac{E_2}{c}\right)+\hat{y}(p_{y1}+p_{y2})+\hat{z}(p_{z1}+p_{z2})\tag{4} | |||
| $$ | |||
| Concidering $Note \ 1$, | |||
| $$ | |||
| E'=M'c^2=\gamma(E_1+E_2+vp_{x1})=\gamma Mc^2 \quad so\quad M'=\gamma M \tag{5} | |||
| $$ | |||
| $$ | |||
| \vec{P'}=\gamma\hat{x}\left(\beta\frac{E_1+E_2}{c}\right)=\gamma \vec{v} \frac{Mc^2}{c^2}=\gamma M\vec{v} \tag{6} | |||
| $$ | |||
| Which is what we needed. | |||
| #### Answer | |||
| $$ | |||
| \boxed{M'=\gamma M\qquad \vec{P'}=\gamma M\vec{v} } | |||
| $$ | |||