Help in derivation on a trig function in dipole radiation.

[yungman]

This is part of derivation of electric dipole potential where:

V(\vec r,t)=\frac 1 {4\pi \epsilon_0} \left [ \frac {q_0 cos [\omega(t- \frac {\eta_+} c )]}{\eta_+}- \frac {q_0 cos [\omega(t- \frac {\eta_- } c)]}{\eta_-} \right ]

\vec {\eta_+} = \vec r - \vec r^+ (t_r) \;,\; \eta =\sqrt{r^2+(r^+) ^2}

Where r points to the field point where V is measured and \vec r^+ (t_r)\; is position of the positive point charge. \vec {\eta}_+ (t_r)\; is the vector from positive charge to field point. There is of cause a \vec {\eta}_- (t_r)\; for the negative charge also.

For d<< r, The book claimed without proof that:

cos [\omega(t-\frac {\eta_+} c)] = cos \left [\omega(t-\frac r c )+\frac{\omega d}{2c} cos \theta \right ]

Please help me on how to derive this. I know this is more a trig problem, but it has to be in the context of dipole radiation, this why I post it here instead.

Thanks

Alan

 

[yungman]

I tried and this is the only one I come up with and it is quite lame:

Assuming the vector from the center of the dipole to the field point is \vec{\eta} and vector from the +ve charge to field point is \vec{\eta}_+. Using Law of Cosine:

(\eta_+)^2=\eta^2 + (\frac d 2)^2 - 2\frac d 2 \eta cos \theta \;\Rightarrow\; cos\theta = \frac { \eta^2 +(\frac d 2)^2-(\eta_+)^2}{ d\eta} \cong \frac {\eta^2 -(\eta_+)^2}{ d\eta}

cos \left [\omega(t-\frac {\eta} c)+\frac{\omega d}{2c}cos\theta \right]= cos\left [\omega(t-\frac {\eta} c)+\frac{\omega d}{2c}\frac {\eta^2 -(\eta_+)^2}{ d\eta}\right ]= cos \left [ \omega \left ( t-\frac {\eta} c + \frac d {2c} \frac {\eta^2 -(\eta_+)^2}{ d\eta}\right ) \right ] = cos \left [ \omega \left ( t- \frac {2\eta^2 - \eta^2 + (\eta_+)^2}{2c\eta}\right ) \right ]

For d<<\eta \;\Rightarrow \; \eta_+\cong \eta.

cos \left [\omega(t-\frac {\eta} c)+\frac{\omega d}{2c}cos\theta \right] = cos \left [ \omega \left ( t- \frac { \eta^2 + (\eta_+)^2}{2c\eta}\right ) \right ] \cong cos \left [ \omega \left ( t- \frac { \eta + \eta_+}{2c}\right ) \right ] \cong cos \left [ \omega \left ( t- \frac { \eta_+}{c}\right ) \right ]

But this is very lame and stretching it!

 

[Bill_K]

Piece of cake.

η₊ = √(r² + d² -2dr cos θ) [Law of cosines]

≈ r - d cos θ [Binomial theorem]

so cos [ω(t - η₊/c)] ≈ cos[ωt - ωr/c + ωd/c cos θ]

(missing factor of 2 is apparently because I defined d as the distance of each charge from the origin, whereas you defined it as the distance between them.)

 

[yungman]

Piece of cake.

η₊ = √(r² + d² -2dr cos θ) [Law of cosines]

≈ r - d cos θ [Binomial theorem]

so cos [ω(t - η₊/c)] ≈ cos[ωt - ωr/c + ωd/c cos θ]

(missing factor of 2 is apparently because I defined d as the distance of each charge from the origin, whereas you defined it as the distance between them.)

Thanks for your reply, can you explain a little more on Binomial theorem?

Alan