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Please explain Taylor expansion in radiation.

  1. Jun 27, 2011 #1
    For retarded scalar potential of arbigtrary source around origin:

    [tex]V(\vec r, t) = \frac 1 {4\pi\epsilon_0}\int \frac { \rho(\vec r\;',t-\frac {\eta}{c}) }{\eta} d\;\tau' \;\hbox { where }\;\eta =\sqrt{r^2 + r'^2 - 2 \vec r \cdot \vec r\;' }[/tex]

    Where [itex]\;\vec r \;[/itex] point to the field point where V is measured. And [itex]\;\vec r\;' \;[/itex] points to the source point.

    For [itex]\;\vec r\;' \;[/itex] << [itex]\;\vec r \;[/itex]:

    [tex] \eta \approx \; r- \hat r \cdot \vec r\;' \Rightarrow \rho(\vec r\;',\;t-\frac {\eta}{c}) \approx \rho (\vec r\;',\;t-\frac {r}{c} + \frac {\vec r \cdot \vec r\;'}{c}) [/tex]

    This next step is where I don't understand how the book do the Taylor expansion. I am going to type the exact word from the book:

    Expanding [itex]\rho \;[/itex] as a Taylor series in t about the retarded time at the origin,

    [tex]t_0=t-\frac r c [/tex]

    We have

    [tex]\rho(\vec r\;',\;t-\frac {\eta}{c}) \approx \rho (\vec r\;',\; t_0) + \dot{\rho} (\vec r\;',\; t_0)\left ( \frac {\vec r \cdot \vec r\;'}{c}\right ) + \frac 1 {2!} \ddot{\rho} \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^2 + \frac 1 {3!} \rho^{...}_{ } \left ( \frac {\vec r \cdot \vec r\;'}{c}\right )^3 ........[/tex]

    Why are they use [itex]\left ( \frac {\vec r \cdot \vec r\;'}{c}\right )\;[/itex] as x for the expansion. I just don't follow this. Please help.


  2. jcsd
  3. Jun 27, 2011 #2


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    What step don't you understand?

    You have some function f(x) that you wish to do the Taylor's expansion about the point x_0, hence you wish to approximate f(x+x_0) for small values of x << 1.

    So, if we choose r' << r, then:

    [tex] x+x_0 \approx t-\frac{r}{c}+\frac{\mathbf{r}'\cdot\mathbf{r}}{c} [/tex]
    We choose x_0 to be t-r/c and x to be r' dot r/c and do our Taylor's expansion about our x.
  4. Jun 27, 2011 #3
    I don't understand why it is in power of [itex] \frac {\vec r-\vec r\;'}{c}[/itex]. How does this fit into x-x0?
  5. Jun 27, 2011 #4


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    The Taylor's expansion of f(x) about zero is the Maclaurin Series.

    [tex] f(x) \approx f(0) + f'(0)x + 0.5f''(0)x^2 + \dots [/tex]

    If we want to expand about some point x_0 then

    [tex] f(x_0+x) \approx f(x_0) + f'(x_0)x + \dots[/tex]

    Or something like that. So it seems that the author desires to do a Taylor's expansion of the charge distribution about the retarded time at the origin, t_0. That is, he is expanding the charge about the charge picture as time progressed at the origin where the charges are clustered. So you can think of \rho(r', t_0) as the charge distribution that occurred if the charges about the origin instantly communicated their information to the origin.

    So if t-r/c is our x_0, then our x is r' \cdot r/c. So we expand out in powers of r' \cdot r/c.
  6. Jun 27, 2011 #5
    Again thanks for your time. I thought Taylor expansion is:

    [tex]f(x)\approx f(x_0)+f'(a)(x-a) +\frac { f''(x_0)(x-x_0)^2 }{2!} +......\frac { fn(x_0)(x-x_0)^n }{n!}........[/tex]

    In this case, it should be x=t and [itex]x_0=\frac r c -\frac {\vec r-\vec r\;'}{c}[/itex] so the Taylor expansion should be:

    [tex]f(t)\approx f(\frac r c -\frac {\vec r-\vec r\;' } c )+f' (\frac r c -\frac {\vec r-\vec r\;'} c)( t-\frac r c -\frac {\vec r-\vec r\;'} c) +

    \frac { f''(\frac r c -\frac {\vec r-\vec r\;'} c)( t-\frac r c + \frac {\vec r-\vec r\;'} c )^2 }{2!} +......\frac { f^n(\frac r c -\frac {\vec r-\vec r\;'} c)( t-\frac r c +\frac {\vec r-\vec r\;'} c )^n }{n!}........[/tex]

    But the book use [itex] t_0=t-\frac r c [/itex] this mean the independent variable is [itex]\frac {\vec r-\vec r\;'} c[/itex]

    [tex]\Rightarrow f'(t)\;=\; \frac {d (f(\frac {\vec r-\vec r\;'} c)}{d(\frac {\vec r-\vec r\;'} c)} \;\hbox { also}\; \frac {\vec r-\vec r\;'} c \;\hbox { is a constant at the given source and field point location!!!}[/tex]

    I understand how the book come up with this, it is just:

    [tex] x_0=t-\frac r c \;\;\hbox { and };\; x=\frac {\vec r-\vec r\;' } c [/tex]

    The question is why!!
    Last edited: Jun 27, 2011
  7. Jun 28, 2011 #6
    Anyone please?
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