Edits to “Statement”, “Solution”, “Answer”
en/14.3.19.md
+3 −7
| @@ -1,13 +1,11 @@ | |||
| ### Statement | |||
| − | $14.3.19.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| $14.3.19$ Electron velocity in a parallel beam $\beta c$. How will the electron density change | |||
| when moving relative to the beam with the velocity $\beta_1 c$ in the longitudinal | |||
| direction? | |||
| + | ### Solution | ||
| + | |||
| To solve this problem we can use the Lorentz transformation for current density and charge density. | |||
| Also, we need to keep in mind that the current density in the Earth frame is related to the charge density in the Earth frame as follows: | |||
| \begin{equation} | |||
| j = -\rho \beta c | |||
| \end{equation} | |||
| The negative sign is due to the sign of the charge, because it is the current density of the electrons. | |||
| Now, the Lorentz transformations for current density and charge density are: | |||
| \begin{equation} | |||
| j' = \gamma (j - \rho \beta c) \qquad \text{and} \qquad \rho' = \gamma (\rho - \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | |||
| \end{equation} | |||
| Using the equation for the charge density we can calculate it in the moving frame: | |||
| \begin{equation} | |||
| @@ -27,8 +25,6 @@Solution | |||
| \rho' = \gamma (\rho - \frac{j \beta_1}{c}) = \gamma (\rho + \frac{\rho \beta c \beta_1}{c}) =\rho \frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}} | |||
| \end{equation} | |||
| − | So, the charge density becomes $\frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}}$ times the original. | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | So, the charge density becomes $\frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}}$ times the original. | ||
| @@ -1,13 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.19.$ [Insert the problem statement] | |||
| ### Solution | |||
| $14.3.19$ Electron velocity in a parallel beam $\beta c$. How will the electron density change | $14.3.19$ Electron velocity in a parallel beam $\beta c$. How will the electron density change | ||
| when moving relative to the beam with the velocity $\beta_1 c$ in the longitudinal | when moving relative to the beam with the velocity $\beta_1 c$ in the longitudinal | ||
| direction? | direction? | ||
| ### Solution | |||
| To solve this problem we can use the Lorentz transformation for current density and charge density. | To solve this problem we can use the Lorentz transformation for current density and charge density. | ||
| Also, we need to keep in mind that the current density in the Earth frame is related to the charge density in the Earth frame as follows: | Also, we need to keep in mind that the current density in the Earth frame is related to the charge density in the Earth frame as follows: | ||
| \begin{equation} | \begin{equation} | ||
| j = -\rho \beta c | j = -\rho \beta c | ||
| \end{equation} | \end{equation} | ||
| The negative sign is due to the sign of the charge, because it is the current density of the electrons. | The negative sign is due to the sign of the charge, because it is the current density of the electrons. | ||
| Now, the Lorentz transformations for current density and charge density are: | Now, the Lorentz transformations for current density and charge density are: | ||
| \begin{equation} | \begin{equation} | ||
| j' = \gamma (j - \rho \beta c) \qquad \text{and} \qquad \rho' = \gamma (\rho - \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | j' = \gamma (j - \rho \beta c) \qquad \text{and} \qquad \rho' = \gamma (\rho - \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | ||
| \end{equation} | \end{equation} | ||
| Using the equation for the charge density we can calculate it in the moving frame: | Using the equation for the charge density we can calculate it in the moving frame: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -27,8 +25,6 @@Solution | |||
| \rho' = \gamma (\rho - \frac{j \beta_1}{c}) = \gamma (\rho + \frac{\rho \beta c \beta_1}{c}) =\rho \frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}} | \rho' = \gamma (\rho - \frac{j \beta_1}{c}) = \gamma (\rho + \frac{\rho \beta c \beta_1}{c}) =\rho \frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}} | ||
| \end{equation} | \end{equation} | ||
| So, the charge density becomes $\frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}}$ times the original. | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | So, the charge density becomes $\frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}}$ times the original. | ||