Please explain Taylor expansion in radiation.

[yungman]

For retarded scalar potential of arbigtrary source around origin:

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\;' }

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



For \;\vec r\;&#039; \; << \;\vec r \;:

\eta \approx \; r- \hat r \cdot \vec r\;&#039; \Rightarrow \rho(\vec r\;&#039;,\;t-\frac {\eta}{c}) \approx \rho (\vec r\;&#039;,\;t-\frac {r}{c} + \frac {\vec r \cdot \vec r\;&#039;}{c})

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 \rho \; as a Taylor series in t about the retarded time at the origin,

t_0=t-\frac r c

We have

\rho(\vec r\;&#039;,\;t-\frac {\eta}{c}) \approx \rho (\vec r\;&#039;,\; t_0) + \dot{\rho} (\vec r\;&#039;,\; t_0)\left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right ) + \frac 1 {2!} \ddot{\rho} \left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right )^2 + \frac 1 {3!} \rho^{...}_{ } \left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right )^3 ...

Why are they use \left ( \frac {\vec r \cdot \vec r\;&#039;}{c}\right )\; as x for the expansion. I just don't follow this. Please help.

thanks

Alan

 

[Born2bwire]

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:

x+x_0 \approx t-\frac{r}{c}+\frac{\mathbf{r}&#039;\cdot\mathbf{r}}{c}
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.

 

[yungman]

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:

x+x_0 \approx t-\frac{r}{c}+\frac{\mathbf{r}&#039;\cdot\mathbf{r}}{c}
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.

I don't understand why it is in power of \frac {\vec r-\vec r\;&#039;}{c}. How does this fit into x-x0?

 

[Born2bwire]

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

f(x) \approx f(0) + f&#039;(0)x + 0.5f&#039;&#039;(0)x^2 + \dots

If we want to expand about some point x_0 then

f(x_0+x) \approx f(x_0) + f&#039;(x_0)x + \dots

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.

 

[yungman]

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

f(x) \approx f(0) + f&#039;(0)x + 0.5f&#039;&#039;(0)x^2 + \dots

If we want to expand about some point x_0 then

f(x_0+x) \approx f(x_0) + f&#039;(x_0)x + \dots

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.

Again thanks for your time. I thought Taylor expansion is:

f(x)\approx f(x_0)+f&#039;(a)(x-a) +\frac { f&#039;&#039;(x_0)(x-x_0)^2 }{2!} +...\frac { fn(x_0)(x-x_0)^n }{n!}...

In this case, it should be x=t and x_0=\frac r c -\frac {\vec r-\vec r\;&#039;}{c} so the Taylor expansion should be:

f(t)\approx f(\frac r c -\frac {\vec r-\vec r\;&#039; } c )+f&#039; (\frac r c -\frac {\vec r-\vec r\;&#039;} c)( t-\frac r c -\frac {\vec r-\vec r\;&#039;} c) +<br /> <br /> \frac { f&#039;&#039;(\frac r c -\frac {\vec r-\vec r\;&#039;} c)( t-\frac r c + \frac {\vec r-\vec r\;&#039;} c )^2 }{2!} +...\frac { f^n(\frac r c -\frac {\vec r-\vec r\;&#039;} c)( t-\frac r c +\frac {\vec r-\vec r\;&#039;} c )^n }{n!}...But the book use t_0=t-\frac r c this mean the independent variable is \frac {\vec r-\vec r\;&#039;} c

\Rightarrow f&#039;(t)\;=\; \frac {d (f(\frac {\vec r-\vec r\;&#039;} c)}{d(\frac {\vec r-\vec r\;&#039;} c)} \;\hbox { also}\; \frac {\vec r-\vec r\;&#039;} c \;\hbox { is a constant at the given source and field point location!}

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

x_0=t-\frac r c \;\;\hbox { and };\; x=\frac {\vec r-\vec r\;&#039; } c

The question is why!

 

[yungman]

Anyone please?