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Let $q$ be the charge of a particle whose motion is $\mathbf y(t)$; let $\boldsymbol \beta = \dot {\mathbf {y}}/c $. Let also $\mathbf x$ be a point in space, and $r=|\mathbf x|$, $\mathbf n = \mathbf x /|\mathbf x|$.

Consider the Liénard-Wiechert field at first order wrt $\mathbf y / r $: $$\mathbf E(t,\mathbf x)=\frac {q} {4\pi r c} \frac {\mathbf n \times [\left (\mathbf n - \boldsymbol \beta \right) \times \dot {\boldsymbol \beta}]} {(1-\mathbf n\cdot \boldsymbol \beta)^3}\,,$$ where $\boldsymbol \beta$ and $\dot {\boldsymbol \beta}$ are evaluated at the retarded time $t'$ that satisfies the implicit equation $$ct-ct'=|\mathbf x -\mathbf y(t')|\,.$$ Actually, we should expand $t'$ at first order wrt $\mathbf y/r$, too, so that yields $$t'=t-\frac r c + \mathbf n \cdot \frac {\mathbf y(t')} c + O\left(\frac {y^2} r \right)\,.$$ Now to derive Larmor's formula, if I understand correctly, we are taking the limit as $\boldsymbol \beta \to 0$, so the Liénard-Wiechert field simplifies to $$\mathbf E(t,\mathbf x)=\frac {q} {4\pi r c} {\mathbf n \times (\mathbf n \times \dot {\boldsymbol \beta})} \,.$$ The claim I've read in multiple books is that, if $\boldsymbol \beta \to 0$, we can evaluate $\dot {\boldsymbol \beta}$ at $$T=t-\frac r c\,,$$ neglecting the $\mathbf n \cdot \frac {\mathbf y(t')} c $ term. Why is it so?


Edit

The book I'm using is Lechner, which just says "We'll prove later that this term, called microscopic retardation, can be neglected in more generality", and proceeds explaining Larmor's formula.

The "later" proof he exposes is too hand-waving for me, since I'm searching for a rigorous Taylor expansion / limit derivation. He proves that the microscopic retardation can be neglected if it's negligible wrt $$t_0 = \frac l {\beta c}$$ ($l$ is defined as $|\mathbf y|\le l$) which is the "characteristic time of change" of the generic 4-current. So $$\left| \mathbf n \cdot \frac {\mathbf y} c \right| \le \frac l c = \beta t_0 \ll t_0 \iff \beta \ll 1\,.$$

This reference at page 10-11 states the same claim I'm stuck with, but I can't understand how $\beta$ enters the proof.

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    $\begingroup$ Can you refer to the book you are using? $\endgroup$ – eranreches Dec 12 '18 at 11:54
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The limit $\mathbf \beta \to 0$ is used to simplify the expressions, to make the derivation easier. It does not imply that retarded time can be accurately expressed as $t- |\mathbf x|/c$, because even if particle is slow, it may be far from the origin of the coordinate system and in such case, $|\mathbf x|$ does not give accurate value of distance between the spatial point of the generation event to the spatial point of the observation event.

One can neglect $\mathbf y$ in the retarded time if one assumes $|\mathbf y|/|\mathbf x| \ll 1$. In other words, the coordinate system has to be chosen in such a way that the particle was near its origin when the field was generated, and the observation point is far from this origin.

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  • $\begingroup$ So are you saying the Larmor formula does not derive from any Taylor expansion, but it comes from the simple limit of the Liénard-Wiechert field as $\mathbf y \to 0$ and $\boldsymbol \beta \to 0$? $\endgroup$ – renyhp Dec 12 '18 at 15:21
  • $\begingroup$ No. I'm saying that the approximation you asked about isn't justified by assuming velocity is low. It is justified by assuming the observation point is much farther from the origin than the particle at generation event is from it. In simpler words, we are assuming that we are evaluating the field of the particle far from the region where it executes its motion, so we may replace the actual retarded distance by distance to the origin of the coordinate system. In this respect, both derivations you mention seem to involve unnecessary use of the concept of Taylor series. $\endgroup$ – Ján Lalinský Dec 12 '18 at 22:46
  • $\begingroup$ The velocity being low is assumed in the derivation for a different reason: so that other expressions may be simplified, as in Jarrell's document formula 33. There the argument with non-relativistic simplification does make sense. But as an argument for replacing retarded distance by $r$ it does not - in Jarrel's document in the formula 35, even if beta is very small, one can still make the second term arbitrarily large by using $\mathbf x$ that has big enough magnitude. $\endgroup$ – Ján Lalinský Dec 12 '18 at 22:55
  • $\begingroup$ His explanation of the derivation is not very good, for example he uses $\mathbf x$ both for the observation point and for the position of the source particle, which is conducive to confusion. It is best imo to denote the source particle position vector as $\mathbf r$ (radius vector) and the observation point $\mathbf x$ (any point of space). He also introduces a size of system $a$ but does not use it later. If he stated that (in his notation) $|\mathbf x(t)|<a\ll r$, that would make the derivation make sense, but he got confused into thinking that assuming low beta is enough. $\endgroup$ – Ján Lalinský Dec 12 '18 at 23:00
  • $\begingroup$ So can you start from the exact Liénard-Wiechert field and derive the "Larmor field" with a rigorous approach, by clearly stating all the limits and Taylor expansions involved, and what are the actual steps? I apologize for being stubborn and demanding, but I'm not so much into wordy explanations when it comes to calculations $\endgroup$ – renyhp Dec 13 '18 at 7:41

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