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