Правка разделов «Statement», «Solution», «Answer»
en/14.3.25.md
+11 −11
| @@ -1,20 +1,22 @@ | |||
| ### Statement | |||
| − | $14.3.25.$ | ||
| + | $14.3.25.$ | ||
| + | Solve problem 14.3.24 for a round long solenoid | ||
| + | |||
| ### Solution | |||
| − | Analyse the rest frame S' | ||
| + | first Analyse the rest frame S' | ||
| We have a long cylindrical solenoid, with axis along z'. | |||
| rest magnetic moment | |||
| $\mathbf{M}' = M\,\hat{\mathbf{z}}'$ | |||
| There is no electric dipole moment: $\mathbf{p}' = 0$ | |||
| In the laboratory system S, the solenoid moves with velocity | |||
| − | $\mathbf{v} = v\,\hat{\mathbf{x}} | ||
| + | $\mathbf{v} = v\,\hat{\mathbf{x}}$ | ||
| − | This is the same situation as in 14.3.24 but with the round solenoid | ||
| + | This is the same situation as in the problem 14.3.24 but with the round solenoid | ||
| Under a boost with velocity $\mathbf{v}$ | |||
| the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as | |||
| $\begin{aligned} | |||
| \mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\ | |||
| \mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right). | |||
| @@ -25,6 +27,7 @@Solution | |||
| \end{aligned}$ | |||
| With of course | |||
| + | |||
| $\mathbf{p}' = 0$ $\mathbf{v}\cdot\mathbf{m}' = 0 $ | |||
| so this simplifies to | |||
| $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad | |||
| \mathbf{m} = \gamma\,\mathbf{M}.$ | |||
| Expression in SI | |||
| And finaly In the International System, the equivalent transformation is | |||
| @@ -39,13 +42,10 @@Solution | |||
| $\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad | |||
| \mathbf{m} = \gamma\,\mathbf{M}$ | |||
| − | Conclusion | ||
| − | The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. In both cases, a moving magnetic dipole acquires an electric dipole moment given by: | ||
| − | |||
| − | \boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad | ||
| − | \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}. | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. | ||
| + | |||
| + | $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad | ||
| + | \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$ | ||
| @@ -1,20 +1,22 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.25.$ |
$14.3.25.$ | ||
| Solve problem 14.3.24 for a round long solenoid | |||
| ### Solution | ### Solution | ||
| Analyse the rest frame S' | first Analyse the rest frame S' | ||
| We have a long cylindrical solenoid, with axis along z'. | We have a long cylindrical solenoid, with axis along z'. | ||
| rest magnetic moment | rest magnetic moment | ||
| $\mathbf{M}' = M\,\hat{\mathbf{z}}'$ | $\mathbf{M}' = M\,\hat{\mathbf{z}}'$ | ||
| There is no electric dipole moment: $\mathbf{p}' = 0$ | There is no electric dipole moment: $\mathbf{p}' = 0$ | ||
| In the laboratory system S, the solenoid moves with velocity | In the laboratory system S, the solenoid moves with velocity | ||
| $\mathbf{v} = v\,\hat{\mathbf{x}} |
$\mathbf{v} = v\,\hat{\mathbf{x}}$ | ||
| This is the same situation as in 14.3.24 but with the round solenoid | This is the same situation as in the problem 14.3.24 but with the round solenoid | ||
| Under a boost with velocity $\mathbf{v}$ | Under a boost with velocity $\mathbf{v}$ | ||
| the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as | the electric and magnetic dipole moments$ \mathbf{p} $and $\mathbf{m} $(in CGS) transform as | ||
| $\begin{aligned} | $\begin{aligned} | ||
| \mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\ | \mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\\ | ||
| \mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right). | \mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1}\,\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right). | ||
| @@ -25,6 +27,7 @@Solution | |||
| \end{aligned}$ | \end{aligned}$ | ||
| With of course | With of course | ||
| $\mathbf{p}' = 0$ $\mathbf{v}\cdot\mathbf{m}' = 0 $ | $\mathbf{p}' = 0$ $\mathbf{v}\cdot\mathbf{m}' = 0 $ | ||
| so this simplifies to | so this simplifies to | ||
| $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad | $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M}}, \qquad | ||
| \mathbf{m} = \gamma\,\mathbf{M}.$ | \mathbf{m} = \gamma\,\mathbf{M}.$ | ||
| Expression in SI | Expression in SI | ||
| And finaly In the International System, the equivalent transformation is | And finaly In the International System, the equivalent transformation is | ||
| @@ -39,13 +42,10 @@Solution | |||
| $\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad | $\boxed{\mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M}}, \qquad | ||
| \mathbf{m} = \gamma\,\mathbf{M}$ | \mathbf{m} = \gamma\,\mathbf{M}$ | ||
| Conclusion | |||
| The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. In both cases, a moving magnetic dipole acquires an electric dipole moment given by: | |||
| \boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad | |||
| \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}. | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | The geometry (flat or round) does not affect the vector relation between the magnetic moment at rest and the electric moment induced by motion. | ||
| $\boxed{\mathbf{p} = \frac{\gamma}{c}\,\mathbf{v} \times \mathbf{M} \;\;(\text{CGS}), \qquad | |||
| \mathbf{p} = \frac{\gamma}{c^2}\,\mathbf{v} \times \mathbf{M} \;\;(\text{SI})}$ | |||