Re: Need help to a beginner.
- To: mathgroup at smc.vnet.net
- Subject: [mg9704] Re: [mg9642] Need help to a beginner.
- From: David Withoff <withoff>
- Date: Fri, 21 Nov 1997 01:31:52 -0500
- Sender: owner-wri-mathgroup at wolfram.com
> Hi, all. I am quite new to Mathematica, and am using the older version,
> v. 2.2. I hope someone will help me. My problem is nothing to do with
> homework of a math class, or any sort. I am a physics graduate student,
> and for my research project am trying to have an algebraic expression
> of potential energy of certain ionic crystalline lattice. The
> electrostatic (i.e. Coulomb) energy has the expression of 1/r, where r
> is the distance between two charges.
> I am still far from getting the answer I want because of this problem I
> am facing in the very first stage where I am supposed to be used to
> Mathematica.
>
> First of all, let me explain my problem. It is known, if you expand
> 1/Sqrt[1-x], provided that x^2<<1, you have
> 1+(1/2)*x+(3/*)*x^2+(5/16)*x^3+.... Now, let's have a similar
> expression to this: 1/Sqrt[(a-x)^2+y^2+z^2].
> Suppose x, y and z are cartesian coordinates, and r^2=x^2+y^2+z^2. And a
> is some positive constant, and it satisfies a>>r. So factoring the
> denominator by a^2 and kicking it out of the Sqrt as a, I can continue
> this algebra by my hand, and I should end up with:
> (1/a)*(1+x/a-r^2/(2*a^2)+(3*x^2)/(2*a^2)+....) in which I only keep up
> to the 2nd order of r/a (and x/a).
>
> I would like to be able to do this by Mathematica. That is, given this
> sort of a reciprocal of a Sqrt of quadratic expression with x, y and z,
> I'd like it to expand "approximately" so that a resultant expression
> only contain the terms up to a specified order of r/a (thus, x/a, y/a,
> and z/a). How can I do this? This kind of expansion goes on forever,
> but I don't need many higher order terms. How can I specify the maximum
> order that I want to have?
>
> I will greatly appreciate someone's help to this problem. Meanwhile, I
> am trying to find a solution in the manual by myself. If you don't
> mind, please send your solution to my email address: shink at iastate.edu
>
> Thank you for your attention.
If I am understanding this correctly, then implicit in this calculation
is the assumption that x/a and r/a are of the same order, and the
objective is to do an expansion in powers of the common scale of these
variables, rather than a two-variable expansion. This can be done by
rewriting the express in terms of a variable that represents that
scale, and then doing the expansion.
In[1]:= expr = 1/Sqrt[Expand[(a-x)^2+y^2+z^2]] /.
x^2 + y^2 + z^2 -> r^2
1
Out[1]= ---------------------
2 2
Sqrt[a + r - 2 a x]
In[2]:= expr = PowerExpand[Factor /@ (expr /. {r -> t a ra, x -> t a
xa})]
1
Out[2]= ---------------------------
2 2
a Sqrt[1 + ra t - 2 t xa]
In[3]:= ser = Series[expr, {t, 0, 2}]
2 2 2
1 xa t (-ra + 3 xa ) t 3 Out[3]= - + ---- +
----------------- + O[t]
a a 2 a
after which
In[4]:= Normal[ser]/. {t -> 1, ra -> r/a, xa -> x/a} //Expand
2 2
1 r x 3 x
Out[4]= - - ---- + -- + ----
a 3 2 3
2 a a 2 a
gives the terms to second order in x/a and r/a in terms of the original
variables.
Dave Withoff
Wolfram Research