Show that the (ordinary) acceleration of a particle of mass m and charge q, moving at velocity u under the influence of electromagnetic fields E and B, is given by

$\mathbf{a}\mathbf{=}\frac{\mathbf{q}}{\mathbf{m}}\sqrt{\mathbf{1}\mathbf{‐}{\mathbf{u}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}\left[E+u\times B-\frac{1}{c^2}u\left(u‐E\right)\right]$

[Hint: Use Eq. 12.74.]

Write the expression for the force acting on a particle under the influence of an electromagnetic fields.

$\mathbf{F}\mathbf{=}\mathbf{}\frac{\mathbf{m}}{\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}\left[a+\frac{u\left(u\times a\right)}{c^2‐u^2}\right]$

.............(1)

Here, a is the ordinary acceleration.

It is known that:

$\mathrm{F}=\mathrm{q}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right]$

Substitute $\mathrm{q}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right]$for $\mathrm{F}$in equation (1).

$$\begin{gathered} \mathrm{q}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right]=\frac{\mathrm{m}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}·\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2}\right] \\ \left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}·\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2}\right]=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right] \end{gathered}$$ …… (2)

Take the dot product ofuon L.H.S of equation (2).

$$\begin{gathered} \mathrm{u}·\left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}·\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2}\right]=\mathrm{u}.\mathrm{a}+\frac{{\mathrm{u}}^2\left(\mathrm{u}·\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2} \\ =\frac{\left(\mathrm{u}.\mathrm{a}\right){\mathrm{c}}^2-\left(\mathrm{u}.\mathrm{a}\right){\mathrm{u}}^2+{\mathrm{u}}^2\left(\mathrm{u}.\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2} \\ =\frac{\left(\mathrm{u}.\mathrm{a}\right){\mathrm{c}}^2}{{\mathrm{c}}^2-{\mathrm{u}}^2} \\ =\frac{\mathrm{u}.\mathrm{a}}{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}} \end{gathered}$$

Similarly, take the dot product ofuon R.H.S of equation (2).

$$\begin{gathered} \frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right]=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{u}.\mathrm{E}+\mathrm{u}.\left(\mathrm{u}\times \mathrm{B}\right)\right] \\ =\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left(\mathrm{u}.E\right) \end{gathered}$$

From equation (2).

$$\begin{gathered} \frac{\mathrm{u}.\mathrm{a}}{1-{\displaystyle \frac{{\mathrm{u}}^2}{c^2}}}=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left(\mathrm{u}.E\right) \\ \frac{\mathrm{u}\left(\mathrm{u}.\mathrm{a}\right)}{c^2-u^2}=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\frac{\mathrm{u}\left(\mathrm{u}.E\right)}{c^2} \end{gathered}$$

Substitute $\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\frac{\mathrm{u}\left(\mathrm{u}.E\right)}{c^2}$for $\frac{\mathrm{u}\left(\mathrm{u}.\mathrm{a}\right)}{c^2-u^2}$in equation (2).

role="math" localid="1654672556955" $\mathrm{a}+\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\frac{\mathrm{u}\left(\mathrm{u}.\mathrm{E}\right)}{{\mathrm{c}}^2}=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{E}+\mathrm{u}\times \mathrm{B}\right]$

$\mathrm{a}=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{E}=\mathrm{u}\times \mathrm{B}-\frac{\mathrm{u}\left(\mathrm{u}.\mathrm{E}\right)}{{\mathrm{c}}^2}\right]$

Therefore, the ordinary acceleration of a particle of mass m and charge q is

$\mathrm{a}=\frac{\mathrm{q}}{\mathrm{m}}\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[\mathrm{E}=\mathrm{u}\times \mathrm{B}-\frac{\mathrm{u}\left(\mathrm{u}.\mathrm{E}\right)}{{\mathrm{c}}^2}\right]$