What is the radiation terms and the non-radiation terms in the field tensor F ??

Question

What is the radiation terms and the non-radiation terms in the field
tensor F?? of equation B for a moving charged particle ?

(Express in tensor form)

Sect. 14.1 Liénard-Wiechert Potentials and Fields for a Point Charge 663 where n is a unit vector in the direction of x-r(n) and ? (14.6) can thus be written ?(?)/c. The potentials ?(x, t) A(x, t) (14.8) (1 - B n)R re re The subscript "ret" means that the quantity in the square brackets is to be eval- uated at the retarded time To, given by, /o(To) -Xo R. It is evident that for nonrelativistic motion the potentials reduce to the well-known results The electromagnetic fields FaP(x) can be calculated directly from (14.6) or (14.8), but it is simpler to return to the integral over dT, (14.3). In computing FaP the differentiation with respect to the observation point x will act on the theta and delta functions. Differentiation of the theta function will give ?[Xo-O(n)] and so constrain the delta function to be S(-R2). There will be no contribution from this differentiation except at R-0. Excluding R- 0 from consideration, the derivative ???? is (14.9) he partial derivative can be written where f- [x r(T)]2. The indicated differentiation gives (x - r)" d ?[f] When this is inserted into (14.9) and an integration by parts performed, the result is dt V -(x -r) In the integration by parts the differentiation of the theta function gives no con tribution, as already indicated. The form of (14.10) is the same as (14.3), witlh V(T) replaced by the derivative term. The result can thus be read off by substi tution from (14.6). The field strength tensor is Here rde and Va are functions of T. After differentiation the whole expression is to be evaluated at the retarded proper time To- The field-strength tensor Fa explicit. It is sometimes useful to have the fields E and B exhibited as explicit functions of the charge's velocity and acceleration. Some of the ingredients needed to carry out the differentiation in (14.11) are (14.11) is manifestly covariant, but not overly (14.12) d Vo

Explanation / Answer

The term which is under the differentiation is radiation term because we know that the radiation is the radiation or transmission of energy. Which is strongly dependent on time. That's why the time-dependent portion is radiation portion. And the 1st term which is not under the differentiation is the non-radiation term.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.