Help in steps on derivation of potential involve moving point charge.

[yungman]

I mainly have problem is in the final step of the derivation, but I am going to list the steps leading to that:

The question is: Find the potentials of a point charge moving with constant velocity.

Relevant equation:

$$\vec w(t) = \;\hbox { is position of q at time t.}$$

$$| \vec r - \vec w (t_r) |=c(t-t_r) \;\hbox { and }\; \eta =\vec r - \vec w(t_r)$$

$$V_{(\vec r, t)} = \frac 1 {4\pi \epsilon _0} \frac {qc}{(\eta c -\vec {\eta} \cdot \vec v)} \;\hbox { and }\; \vec A_{(\vec r,t)} = \frac {\vec v}{c^2} V_{(\vec r, t)}$$



Below is the solution from the book:

Let the particle passes through the origin at t=0 so that $$\vec w(t) = \vec v t$$

$$\vec w(t_r)=\vec vt_r \Rightarrow \; |\vec r -\vec v_t_r|=c(t-t_r) \Rightarrow \; t_r=t\;^+_-\frac r c$$

I skip a few step above to arrive to the solution of the retarded time tr.




$$\eta = |\vec r - \vec v t_r| \;\hbox { and } \hat { \eta} =\frac {\vec r -\vec v t_r}{c(t-t_r)}$$

$$\Rightarrow \; \eta \left (1-\hat {\eta} \cdot \frac{\vec v(t_r)}{c}\right ) = \frac 1 c [(c^2t -\vec r \cdot \vec v)-(c^2-v^2)t_r]$$ (1)

$$\Rightarrow \; \eta \left (1-\hat {\eta} \cdot \frac{\vec v(t_r)}{c}\right ) = \frac 1 c \sqrt {(c^2t-\vec r \cdot \vec v)^2 + (c^2-v^2)(r^2 – c^2 t^2)}$$ (2)


I don’t get how to get from (1) to (2).

I tried squaring (1) and take the square root. But I don't know how to deal with the tr=t-(+/-)(v/c).

 

[NateD]

Can someone please explain how to get from (1) to (2)? The expression in (2) is obtained by using the Pythagorean theorem. The first term in the parentheses is the hypotenuse and the second term is the sum of squares of the other two sides. Note that we have t_r = t +/- (r/c). Therefore, we can expand the expression in (1) as follows: \eta \left (1-\hat {\eta} \cdot \frac{\vec v(t_r)}{c}\right ) = \frac 1 c [(c^2t -\vec r \cdot \vec v)-(c^2-v^2)(t + r/c)]= \frac 1 c [(c^2t -\vec r \cdot \vec v)-(c^2-v^2)t + (c^2-v^2)r/c]Now, we can use the Pythagorean theorem to obtain the expression in (2):\eta \left (1-\hat {\eta} \cdot \frac{\vec v(t_r)}{c}\right ) = \frac 1 c \sqrt {(c^2t-\vec r \cdot \vec v)^2 + (c^2-v^2)^2[t^2 + (r/c)^2]}