The natural relativistic generalization of the Abraham-Lorentz formula (Eq. 11.80) would seem to be

${\mathit{K}}_{\mathbf{r}\mathbf{a}\mathbf{}\mathbf{d}}^{\mathbf{\mu }}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{6}\mathbf{\Pi }\mathbf{c}}\frac{\mathbf{d}{\mathbf{\alpha }}^{\mathbf{\mu }}}{\mathbf{d}\mathbf{b}}$

This is certainly a 4-vector, and it reduces to the Abraham-Lorentz formula in the non-relativistic limit$\mathit{v}\mathbf{\ll }\mathit{c}$ .

(a) Show, nevertheless, that this is not a possible Minkowski force.

(b) Find a correction term that, when added to the right side, removes the objection you raised in (a), without affecting the 4-vector character of the formula or its non-relativistic limit.

Write the expression for the natural relativistic generalizations of the Abraham-Lorentz formula.

${\mathit{K}}_{\mathbf{r}\mathbf{a}\mathbf{d}}^{\mathbf{\mu }}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{6}\mathbf{\Pi }\mathbf{c}}\frac{\mathbf{d}{\mathbf{\alpha }}^{\mathbf{v}}}{\mathbf{d}\mathbf{b}}$ …… (1)

Here, $K_{rad}^{\mu }$is the generalization constant, q is the charge on the particle, c is the speed of light and$\frac{d{\alpha }^{\mu }}{d\mathcal{b}}$ is the change in proper acceleration with respect to time.

(a)

Substitute $K_{rad}^{\mu }=\left(\frac{{\mu }_0{q}^2}{6\pi c}\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right)$in the expression ${\eta }_{\mu }{K}_{rad}^{\mu }=0$.

${\eta }_{\mu }=\left(\frac{{\mu }_0{q}^2}{6\pi c}\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right)=0$ …… (2)

Here,${\eta }_{\mu }$ is the proper velocity of the entity.

Clearly, the proper acceleration of the entity is changing with respect to time, so there must be some magnitude of force exerting on the entity. Hence, equation (2) is expressed as,

${\eta }_{\mu }\left(\frac{{\mu }_0{q}^2}{6\pi c}\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right)\ne 0$ …… (3)

This condition is not possible with the Minkowski forcebecause with Minkowski force produces a constant magnitude of effort (external radial force), which is against the result obtained in equation (3).

Therefore, the value of $K_{rad}^{\mu }$cannot be obtained from the Minkowski force.

(b)

Let the 4-vector be $b^{\mu }$, which does not affect the character of the formula. So the property is,

$\left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}+b^{\mu }\right){\eta }_{\mu }=0$ …… (4)

Here, $b^{\mu }$is the 4-vector.

But as per the definition, the expression for the 4-vector is,

$b^{\mu }=K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_{\nu }\right){\eta }^{\mu }$ …… (5)

Here, K is the vector constant, ${\eta }_v$is the vertical component of velocity and ${\eta }^{\mu }$is the proper velocity.

Substitute $b^{\mu }=K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_{\nu }\right){\eta }^{\mu }$in equation (4).

$$\begin{gathered} \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right)\right){\eta }_{\mu }=0 \\ \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right){\eta }_{\mu }+K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right){\eta }^{\mu }{\eta }_{\mu }=0 \end{gathered}$$

Substitute ${\eta }^{\mu }{\eta }_{\mu }=-c^2$in the above expression.

$$\begin{gathered} \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right){\eta }_{\mu }+K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right)\left(-c^2\right)=0 \\ \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right){\eta }_{\mu }-c^{2}K\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right)=0 \end{gathered}$$

Substitute$K=\left(\frac{1}{c^2}\right)$ in the above expression.

$$\begin{gathered} \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right){\eta }_{\mu }-c^2\left(\frac{1}{c^2}\right)\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right)=0 \\ \left(\frac{d{\alpha }^{\mu }}{d\mathcal{b}}\right){\eta }_{\mu }-\left(\frac{d\alpha }{d\mathcal{b}}{\eta }_v\right)=0 \end{gathered}$$

The components of$b^{\mu }$ vanish in the non-relativistic limit (v<<<c). Thus the vector constant still reduces to the Abraham-Lorentz formula,

${\alpha }^v{\eta }_v=0$

Then,

$$\begin{gathered} \frac{d}{d\mathcal{b}}\left({\alpha }^v{\eta }_v\right)=0 \\ \frac{d{\alpha }^v}{d\mathcal{b}}{\eta }_v+{\alpha }^v\frac{d{\eta }_v}{d\mathcal{b}}=0 \end{gathered}$$

Therefore, the correction term added to the right side of the 4-vector character of the formula is $\frac{d{\alpha }^v}{d\mathcal{b}}{\eta }_v+{\alpha }^v\frac{d{\eta }_v}{d\mathcal{b}}=0$.