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| ### Statement |
| ### Statement |
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| $14.4.9.$ [Insert the problem statement] |
| $14.4.9$ |
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| ### Solution |
| ### Solution |
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| The force on the proton (charge e, rest mass m) is |
| The force on the proton (charge e, rest mass m) is |
| $ \mathbf{F} = e\mathbf{E}$ |
| $ \mathbf{F} = e\mathbf{E}$ |
| The acceleration \mathbf{a} in special relativity is obtained from |
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| the expresion for aceleration $\mathcal{a}$ |
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| is obtained by |
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| $\frac{d\mathbf{p}}{dt} = \mathbf{F}, \qquad \mathbf{p} = \gamma m \mathbf{v}, \quad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ |
| $\frac{d\mathbf{p}}{dt} = \mathbf{F}, \qquad \mathbf{p} = \gamma m \mathbf{v}, \quad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ |
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| Then we differentiate and arrive at the expression |
| Then we differentiate and arrive at the expression |
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| $\mathbf{a} = \frac{\mathbf{F} - \mathbf{v}\,(\mathbf{F}\cdot\mathbf{v})/c^2}{\gamma m}$ | | $\mathbf{a} = \frac{\mathbf{F} - \mathbf{v}\,(\mathbf{F}\cdot\mathbf{v})/c^2}{\gamma m}$ |
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| Motion perpendicular to the field | | Motion perpendicular to the field |
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| If | | If |
| $ \mathbf{v} \perp \mathbf{E}$ | | $ \mathbf{v} \perp \mathbf{E}$ |
| the dot product $\mathbf{F}\cdot\mathbf{v} = 0$. | | the dot product $\mathbf{F}\cdot\mathbf{v} = 0$. |
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| Then: | | Then: |
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| $\mathbf{a}_\perp = \frac{\mathbf{F}}{\gamma m} = \frac{e\mathbf{E}}{\gamma m}$ | | $\mathbf{a}_\perp = \frac{\mathbf{F}}{\gamma m} = \frac{e\mathbf{E}}{\gamma m}$ |
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| The magnitude is | | The magnitude is |
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| $\mathbf{a}_\parallel = \frac{e\mathbf{E} - \mathbf{v}(eEv)/c^2}{\gamma m} = \frac{e\mathbf{E}(1 - v^2/c^2)}{\gamma m} = \frac{e\mathbf{E}}{\gamma m}(1-\beta^2)$ | | $\mathbf{a}_\parallel = \frac{e\mathbf{E} - \mathbf{v}(eEv)/c^2}{\gamma m} = \frac{e\mathbf{E}(1 - v^2/c^2)}{\gamma m} = \frac{e\mathbf{E}}{\gamma m}(1-\beta^2)$ |
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| Since | | Since |
| $1-\beta^2 = 1/\gamma^2$ | | $1-\beta^2 = 1/\gamma^2$ |
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| $\boxed{a_\parallel = \frac{eE}{\gamma^3 m}}$ | | $\boxed{a_\parallel = \frac{eE}{\gamma^3 m}}$ |
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| Now apply a Comparison: | | Now apply a Comparison: |
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| $\frac{a_\perp}{a_\parallel} = \frac{eE/(\gamma m)}{eE/(\gamma^3 m)} = \gamma^2 = \frac{1}{1-\beta^2}$ | | $\frac{a_\perp}{a_\parallel} = \frac{eE/(\gamma m)}{eE/(\gamma^3 m)} = \gamma^2 = \frac{1}{1-\beta^2}$ |
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| $\boxed{\frac{a_\perp}{a_\parallel} = \frac{1}{1-\beta^2}}$ | | $\boxed{\frac{a_\perp}{a_\parallel} = \frac{1}{1-\beta^2}}$ |
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| Motion at an angle$ \alpha (\mathbf{v} forms \alpha with \mathbf{E})$ | | Motion at an angle$ \alpha (\mathbf{v} forms \alpha with \mathbf{E})$ |
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| We choose | | We choose |
| $\mathbf{E} = E\,\hat{\mathbf{x}} $ | | $\mathbf{E} = E\,\hat{\mathbf{x}} $ |
| and $\mathbf{v} = v(\cos\alpha\,\hat{\mathbf{x}} + \sin\alpha\,\hat{\mathbf{y}}). | | and $\mathbf{v} = v(\cos\alpha\,\hat{\mathbf{x}} + \sin\alpha\,\hat{\mathbf{y}}). |
| \mathbf{F}\cdot\mathbf{v} = eE v \cos\alpha.$ | | \mathbf{F}\cdot\mathbf{v} = eE v \cos\alpha.$ |
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| Components of the acceleration | | Components of the acceleration |
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| $a_x = \frac{eE}{\gamma m}(1 - \beta^2\cos^2\alpha), \quad | | $a_x = \frac{eE}{\gamma m}(1 - \beta^2\cos^2\alpha), \quad |
| a_y = -\frac{eE}{\gamma m}\,\beta^2\sin\alpha\cos\alpha$ | | a_y = -\frac{eE}{\gamma m}\,\beta^2\sin\alpha\cos\alpha$ |
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| Magnitude | | Magnitude |
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| $a_\alpha = \frac{eE}{\gamma m}\sqrt{(1-\beta^2\cos^2\alpha)^2 + \beta^4\sin^2\alpha\cos^2\alpha}$ | | $a_\alpha = \frac{eE}{\gamma m}\sqrt{(1-\beta^2\cos^2\alpha)^2 + \beta^4\sin^2\alpha\cos^2\alpha}$ |
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| Simplifying the radicand: | | Simplifying the radicand: |
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| $(1-\beta^2 x)^2 + \beta^4(1-x)x = 1 - 2\beta^2 x + \beta^4 x, \quad x = \cos^2\alpha$ | | $(1-\beta^2 x)^2 + \beta^4(1-x)x = 1 - 2\beta^2 x + \beta^4 x, \quad x = \cos^2\alpha$ |
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| It can be verified that this expression is equivalent to | | It can be verified that this expression is equivalent to |
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| $a_\alpha = \frac{eE}{\gamma m} \cdot \frac{1}{\sqrt{\sin^2\alpha + \dfrac{\cos^2\alpha}{(1-\beta^2)^2}}}$ | | $a_\alpha = \frac{eE}{\gamma m} \cdot \frac{1}{\sqrt{\sin^2\alpha + \dfrac{\cos^2\alpha}{(1-\beta^2)^2}}}$ |
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| This is verified by evaluating the limiting cases$ \alpha = 0$ and $\alpha = 90^\circ $ and the equality can be proven algebraically. | | This is verified by evaluating the limiting cases$ \alpha = 0$ and $\alpha = 90^\circ $ and the equality can be proven algebraically. |
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| Therefore, the comparison with the perpendicular case is | | Therefore, the comparison with the perpendicular case is |
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| $\boxed{\frac{a_\perp}{a_\alpha} = \sqrt{\sin^2\alpha + \frac{\cos^2\alpha}{(1-\beta^2)^2}}}$ | | $\boxed{\frac{a_\perp}{a_\alpha} = \sqrt{\sin^2\alpha + \frac{\cos^2\alpha}{(1-\beta^2)^2}}}$ |
