Does this problem warrant a taylor expansion? (solid state physics)

In summary: It seems like this may be what is wanted. It would make sense to use a taylor expansion on the limiting cases where \(\theta\) is not a factor.
  • #1
skate_nerd
176
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The whole problem I'm doing here is not even really relevant, so I won't go too much into it...I'm told to find an atomic form factor given some certain conditions, and I do a big gross integral and got this:
$$f=(\frac{4}{4+(a_oG)^2})^2$$
where \(a_o\) is the Bohr radius and \(G\) is the magnitude of an arbitrary reciprocal lattice vector.
After finding this, I am asked to find the atomic form factor as
\(a_o>>\lambda\) and as
\(a_o<<\lambda\).
This seems pretty simple because there is a relation between \(\lambda\) and \(G\):
$$G=\frac{4\pi}{\lambda}\sin\theta$$
Plugging that into my formula for the form factor, I quickly assumed that \(a_o>>\lambda\) will just make \(\frac{a_o}{\lambda}\) really large, so the denominator of the form factor gets really large and then is approximately zero. And as \(a_o<<\lambda\), \(\frac{a_o}{\lambda}\) is approximately zero, so the form factor ends up being 1.

The problem, however, also asks for both of the limiting cases if the form factor ends up depending on \(\theta\). I'm not sure why this would be asked for each case if one of them didn't end up with a \(\theta\) dependence...
So I'm asking for some advice, would it make sense to use a taylor expansion on either or both of these limiting cases? Why or why not?

Thanks for your attention!
 
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  • #2
skatenerd said:
The whole problem I'm doing here is not even really relevant, so I won't go too much into it...I'm told to find an atomic form factor given some certain conditions, and I do a big gross integral and got this:
$$f=(\frac{4}{4+(a_oG)^2})^2$$
where \(a_o\) is the Bohr radius and \(G\) is the magnitude of an arbitrary reciprocal lattice vector.
After finding this, I am asked to find the atomic form factor as
\(a_o>>\lambda\) and as
\(a_o<<\lambda\).
This seems pretty simple because there is a relation between \(\lambda\) and \(G\):
$$G=\frac{4\pi}{\lambda}\sin\theta$$
Plugging that into my formula for the form factor, I quickly assumed that \(a_o>>\lambda\) will just make \(\frac{a_o}{\lambda}\) really large, so the denominator of the form factor gets really large and then is approximately zero. And as \(a_o<<\lambda\), \(\frac{a_o}{\lambda}\) is approximately zero, so the form factor ends up being 1.

The problem, however, also asks for both of the limiting cases if the form factor ends up depending on \(\theta\). I'm not sure why this would be asked for each case if one of them didn't end up with a \(\theta\) dependence...
So I'm asking for some advice, would it make sense to use a taylor expansion on either or both of these limiting cases? Why or why not?

Thanks for your attention!

Hi skatenerd!

Can it be that something like the following is intended? (Wondering)
\begin{array}{c|c|c|}
& a_0 \gg \lambda & a_0 \ll \lambda \\
\hline
\theta=0 & f = 1 & f=1\\
\hline
\theta = \pm \frac \pi 2 & f = 0 & f = 1\\
\hline
\end{array}
 

FAQ: Does this problem warrant a taylor expansion? (solid state physics)

What is a Taylor expansion and how is it used in solid state physics?

A Taylor expansion is a mathematical tool used to approximate a function by breaking it down into a sum of simpler functions. In solid state physics, it is often used to analyze the behavior of a system near a specific point or within a small region.

How do I determine if a problem in solid state physics warrants a Taylor expansion?

A problem in solid state physics may warrant a Taylor expansion if the function being analyzed is sufficiently smooth and the desired level of accuracy can be achieved by using only a few terms in the expansion. Additionally, the problem may warrant a Taylor expansion if the behavior of the system can be described by a known mathematical function.

What are the limitations of using a Taylor expansion in solid state physics?

One limitation of using a Taylor expansion in solid state physics is that it is only accurate within a certain range or region. If the function being analyzed has sharp discontinuities or behaves differently outside of the region, the results may not be accurate. Additionally, using too few terms in the expansion can also lead to inaccurate results.

Can a Taylor expansion be used for any type of problem in solid state physics?

No, a Taylor expansion is not suitable for all problems in solid state physics. It is most commonly used for problems that involve analyzing the behavior of a system near a specific point or within a small region. Problems that involve complex or non-linear behavior may not be suitable for a Taylor expansion.

Are there alternative methods to a Taylor expansion for solving problems in solid state physics?

Yes, there are alternative methods such as perturbation theory, numerical methods, and other mathematical techniques that can be used to solve problems in solid state physics. The choice of method will depend on the specific problem and the desired level of accuracy.

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