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Electromagnetic Waves. Radiation

Problem 4.218

A charged particle moves uniformly with velocity vv along a circle of radius RR in the plane xyx y (Fig. 4.40 ). An observer is located on the xx axis at a point PP which is removed from the centre of the circle by a distance much exceeding RR. Find: (a) the relationship between the observed values of the yy projection of the particle's acceleration and the yy coordinate of the particle; (b) the ratio of electromagnetic radiation flow densities S1/S2S_{1} / S_{2} at the point PP at the moments of time when the particle moves, from the standpoint of the observer PP, toward him and away from him, as shown in the figure.

Reveal Answer
4.218. (a) Suppose that tt is the moment of time when the particle is at a definite point x,yx, y of the circle, and t′t^{\prime} is the moment when the information about that reaches the point P.P . Denoting the observed values of the yy coordinate at the point PP by y′y^{\prime} (see Fig. 4.40), we shall write t′=t+l−x(t)c,y′(t′)=y(t) t^{\prime}=t+\frac{l-x(t)}{c}, \quad y^{\prime}\left(t^{\prime}\right)=y(t) The sought acceleration is found by means of the double differentiation of y′y^{\prime} with respect to t′:t^{\prime}: dy′dt′=dydt′=dydtdtdt′,d2ydt′2=dtdt′ddt(dy′dt′)=v2Rv/c−y/R(1−vy/cR)3 \frac{d y^{\prime}}{d t^{\prime}}=\frac{d y}{d t^{\prime}}=\frac{d y}{d t} \frac{d t}{d t^{\prime}}, \quad \frac{d^{2} y}{d t^{\prime 2}}=\frac{d t}{d t^{\prime}} \frac{d}{d t}\left(\frac{d y^{\prime}}{d t^{\prime}}\right)=\frac{v^{2}}{R} \frac{v / c-y / R}{(1-v y / c R)^{3}} where the following relations are taken into account: x=Rsin⁡ωtx=R \sin \omega t y=Rcos⁡ωt, and ω=v/R y=R \cos \omega t, \text { and } \omega=v / R (b) Energy flow density of electromagnetic radiation SS is proportional to the square of the yy projection of the observed acceleration of the particle. Consequently, S1/S2=(1+v/c)4/(1−v/c)4S_{1} / S_{2}=(1+v / c)^{4} /(1-v / c)^{4}.
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