Problem 14.4.9^∗

Solution

The force on the proton (charge e, rest mass m) is
$\mathbf{F} = e\mathbf{E}$

the expresion for aceleration $\mathcal{a}$ is obtained by

$\frac{d\mathbf{p}}{dt} = \mathbf{F}, \qquad \mathbf{p} = \gamma m \mathbf{v}, \quad \gamma = \frac{1}{\sqrt{1-\beta^2}}$

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

$\boxed{a_\perp = \frac{eE}{\gamma m}}$

Now the 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

$\boxed{\frac{a_\perp}{a_\alpha} = \sqrt{\sin^2\alpha + \frac{\cos^2\alpha}{(1-\beta^2)^2}}}$