“Derive” the Lorentz force law, as follows: Let charge q be at rest in S¯, so F¯ = qE¯ , and let S¯ move with velocity v = υ xˆ with respect to S. Use the transformation rules (Eqs. 12.67 and 12.109) to rewrite F¯ in terms of F, and E¯ in terms of E and B. From these, deduce the formula for F

Question

“Derive” the Lorentz force law, as follows: Let charge q be at rest in [latex]\overline{ S }, ext { so } \overline{ F }=q \overline{ E }, ext { and let } \overline{ S }[/latex] move with velocity [latex]v =v \hat{ x }[/latex] with respect to S. Use the transformation rules (Eqs. 12.67 and 12.109) to rewrite [latex] \overline{ F } [/latex] in terms of F, and [latex]\overline{ E }[/latex] in terms of E and B. From these, deduce the formula for F in terms of E and B.

[latex]\overline{ F }_{\perp}=\frac{1}{\gamma} F _{\perp}, \quad \bar{F}_{\|}=F_{\|}[/latex] (12.67)

[latex]\begin{aligned}&\bar{E}_{x}=E_{x}, \quad \bar{E}_{y}=\gamma\left(E_{y}-v B_{z} ight), \quad \bar{E}_{z}=\gamma\left(E_{z}+v B_{y} ight) \&\bar{B}_{x}=B_{x}, \quad \bar{B}_{y}=\gamma\left(B_{y}+\frac{v}{c^{2}} E_{z} ight), \quad \bar{B}_{z}=\gamma\left(B_{z}-\frac{v}{c^{2}} E_{y} ight)\end{aligned}[/latex] (12.109)

Accepted Answer

Equation 12.67 assumes the particle is (instantaneously) at rest in S. Here the particle is at rest in [latex]\overline{ S }[/latex].

So [latex]F _{\perp}=\frac{1}{\gamma} \overline{ F }_{\perp}, F_{\|}=\bar{F}_{\|} \cdot \operatorname{Using} \overline{ F }=q \overline{ E }[/latex] , then,

[latex]F_{x}=\bar{F}_{x}=q \bar{E}_{x}, \quad F_{y}=\frac{1}{\gamma} \bar{F}_{y}=\frac{1}{\gamma} q \bar{E}_{y}, \quad F_{z}=\frac{1}{\gamma} \bar{F}_{z}=\frac{1}{\gamma} q \bar{E}_{z}[/latex].

Invoking Eq. 12.109:

[latex]F_{x}=q E_{x}, \quad F_{y}=\frac{1}{\gamma} q \gamma\left(E_{y}-v B_{z} ight)=q\left(E_{y}-v B_{z} ight) \quad F_{z}=\frac{1}{\gamma} q \gamma\left(E_{z}+v B_{y} ight)=q\left(E_{z}+v B_{y} ight)[/latex].

[latex] ext { But } \quad v imes B =-v B_{z} \hat{ x }+v B_{y} \hat{ z }, \quad ext { so } F =q( E + v imes B ) . \quad ext { qed }[/latex]

Question 12.68: “Derive” the Lorentz force law, as follows: Let charge q be ...

Related Answered Questions