Lienard-Wiechert fields can be derived by directly differentiating the Lienard-Wiechert potentials. But for convenience many textbook authors choose to differentiate under the integration sign of the integrations that lead to the Lienard-Wiechert potentials. To do this, there is an integration by parts involved. For example, in Jackson (revision 1) text book on page 466 it is said, quote, $${\mathbf B}({\mathbf x},t)=e\int({\mathbf n}\times{\mathbf \beta})\left[-\frac{\delta(t^\prime+\frac{R}{c}-t)}{R^2}+\frac{1}{cR}\delta^\prime(t^\prime+\frac{R}{c}-t)\right]dt^\prime$$ The primes on the delta functions mean differentiation with respect to their arguments. If the variable of integration is changed to $f(t^\prime)=t^\prime + [R(t^\prime)/c]$, we can integrate by parts on the derivative of the delta function. Then we find readily $${\mathbf B}({\mathbf x},t)=e\left[\frac{{\mathbf \beta}\times{\mathbf n}}{\kappa R^2}+\frac{1}{c\kappa}\frac{d}{dt^\prime}\left(\frac{{\mathbf \beta}\times{\mathbf n}}{\kappa R}\right)\right]_\mbox{ret}.$$ End quote. Though Jackson did not show how this integration by parts worked, Mark Jarrell showed more details in one of his lecture notes, http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap14/chap14.pdf (for some unknown reason, he wrote Sir Joseph Larmor in place of the author----the same style for his other notes), in his eqn (11), (12), (13),and (14), and noting that $d{\delta(t^\prime+\frac{R}{c}-t)}/{dt^\prime}=\kappa$, $${\mathbf B}({\mathbf x},t)=e\int({\mathbf n}\times{\mathbf \beta})\left[-\frac{\delta(t^\prime+\frac{R}{c}-t)}{R^2}+\frac{1}{cR}\delta^\prime(t^\prime+\frac{R}{c}-t)\right]dt^\prime$$ $$=e\left\{\left[\frac{({\mathbf \beta}\times{\mathbf n})}{\kappa R^2}\right]_\mbox{ret}+\int dt^\prime\frac{\kappa}{cR}\left(\frac {\delta^\prime(t^\prime+\frac{R}{c}-t)}{\kappa}\right)({\mathbf n}\times{\mathbf \beta})\right\}$$ $$=e\left\{\left[\frac{({\mathbf \beta}\times{\mathbf n})}{\kappa R^2}\right]_\mbox{ret}+\int d(t^\prime+\frac{R}{c}-t)\left(\frac{{\mathbf n}\times{\mathbf \beta}}{c\kappa R}\right) \delta^\prime(t^\prime+\frac{R}{c}-t)\right\}$$ $$\stackrel{\mbox{(by parts)}}{=}e\left\{\left[\frac{({\mathbf \beta}\times{\mathbf n})}{\kappa R^2}\right]_\mbox{ret}-\int d(t^\prime+\frac{R}{c}-t)\frac{\partial [({\mathbf n}\times{\mathbf \beta})/c\kappa R]}{\partial (t^\prime+{R}/{c}-t) } \delta(t^\prime+\frac{R}{c}-t)\right\}$$ $$=\cdots=e\left[\frac{{\mathbf \beta}\times{\mathbf n}}{\kappa R^2}+\frac{1}{c\kappa}\frac{d}{dt^\prime}\left(\frac{{\mathbf \beta}\times{\mathbf n}}{\kappa R}\right)\right]_\mbox{ret}.$$ Here comes my question. In the integration by parts part, apparently the authors thought the "whole" part equaled to zero so they omitted it, $$\left(\frac{{\mathbf n}\times{\mathbf \beta}}{c\kappa R}\right) \delta(t^\prime+\frac{R}{c}-t)\stackrel{\mbox{?}}{=}0,$$ but we know that although the delta function is zero elsewhere, it is infinite at the position where its argument is zero, here at $t^\prime=t-{R}/{c}$. How could they assume it is zero everywhere and omit it?

  • $\begingroup$ What are the limits of integration ?? I think they are $\boldsymbol{-\infty},\boldsymbol{+\infty}$ so $$\left[\left(\frac{{\mathbf n}\times{\mathbf \beta}}{c\kappa R}\right) \delta(t^\prime+\frac{R}{c}-t)\right]_{-\infty}^{+\infty}\boldsymbol{=}0$$ $\endgroup$
    – Frobenius
    Nov 23, 2019 at 20:49
  • $\begingroup$ @Frobenius Thank you, I think tthis is the answer. How could I miss that? I don't know. $\endgroup$
    – verdelite
    Nov 23, 2019 at 21:21


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.