Difficult function of a function problem

[contempquant]

Homework Statement

I'm trying to solve a double derivative of a function of a function problem, but can't seem to be able to differentiate once. I have the formula given by the chain rule,

find $$\frac{d}{dz}{E}(\bar{r})$$ when $${E}(\bar{r})={E_0}(\bar{r})e^{ikz}$$ and $$\bar{r}$$ is a function of z.

Homework Equations

I have used $$\frac{df}{dz}=\frac{dg}{df}\frac{df}{dz}h(x)+\frac{dh}{df}\frac{df}{dz}g(x)$$

where $$g(x)={E_0}(\bar{r})$$ and $$h(x)=e^{ikz}$$

and $$f=E(\bar{r})=E_{0}(\bar{r})$$

The Attempt at a Solution

I have then substituted in for these giving,

$$\frac{d{E_0}(\bar{r})exp^{ikz}}{dz}$$=$$\frac{\partial{E_0}(\bar{r})}{\partial{E}(\bar{r})}\frac{\partial{E}(\bar{r})}{\partial{z}}*e^{ikz}+\frac{\partial{e^{ikz}}}{\partial{E}(\bar{r})}\frac{\partial{E}(\bar{r})}{\partial{z}}*E_{0}(\bar{r})$$

The problem is i don't know how to differentiate any of the individual partial differentials because $$\bar{r}=r(z)$$, r is dependent on z. Any ideas? i really don't know how to start this one, and when this is done i have to differentiate it again. I'm sure if i can find out how to do this once i can do the second derivative but can't seem to do this one.

apparently the final solution is,

$$(\frac{\partial^{2}E_{0}}{\partial{z}^{2}}+2ik\frac{\partial{E_0}}{\partial{z}}-k^{2}E_0)e^{ikz}$$

 

[contempquant]

anyone?

 

[hunt_mat]

So in $$\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{i}$$ and $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$ then you use the chain rule in the following way:

$$\frac{d}{dz}E_{0}(r)=\frac{dE_{0}}{dr}\frac{dr}{dz}$$

 

[contempquant]

sorry, that third line under relevant equations: $$f=E(\bar{r})=E_{0}(\bar{r})$$
should read

$$f=E(\bar{r})=E_{0}(\bar{r})e^{ikz}$$


I think that might change the answer you gave hunt_mat?

 

[hunt_mat]

define $$\bar{r}$$ for me.

 

[contempquant]

well that's the thing because $$\bar{r}$$ isn't defined in the book (Milonni, Lasers) but only the scalar version of r is defined, which is as you wrote it above,
$$r=\sqrt{x^{2}+y^{2}+R^{2}}$$

Where $$z=R$$, R is the distance along the z axis

 

[hunt_mat]

So this r is a vector? I think that either you can take E_0 as a function of x, y, and z or that it only has radial dependence.

 

[contempquant]

so the differential of $$\frac{\partial{E_0}(\bar{r})}{\partial{Z}$$ is just one?

how do i do the differential,
$$\frac{\partial{E_0}(\bar{r})}{\partial{E}(\bar{r}) }$$? as it's taken w.r.t the starting function?

 

[contempquant]

* $$\frac{\partial{E_0}(\bar{r})}{\partial{Z}}$$

 

[hunt_mat]

I think that may be the case, having had a little look at the book on amazon, it seems like it's poor notation. I think the notation means that E_0 is a function of x,y and z and you just write down: the derivative as you have, that is how I would view the problem.

 

[contempquant]

ok i'll try that, thanks