1
$\begingroup$

I am dealing with first-order perturbative theory, that is, any variable may be decomposed as

$a = a + \delta a$

and the same would occur with several different variables, let's say, $b$ and $c$. However any term that is accompanied by a $\delta$ (i.e. $\delta a$, $\delta b$ and $\delta c$) are called "perturbative terms", that is, any $\delta$-term multiplied by any other term with $\delta$ are zero (using a different lingo: just keeping perturbation to the first order):

$\delta a ^2 =0$;

and

$\delta b \delta c = 0$.

In my case these terms are used in several different calculations throughout my program, so I would like to discriminate these conditions as a global assumption (actually any other idea on how to do this would be gladly accepted) in the beginning of the program. So I have tried something like:

$Assumptions = $\delta a^2 == 0$ && $\delta b \delta c == 0$

and henceforth for every different combination of variables.

Obviously, it does not work out. I have tried several different ways to do it and, in the end of my big calculations, I just to simplify equations using Simplify or FullSimplify, get results in which second order perturbative terms do not appear.

Edit:

In my case, the perturbative terms are dependent of two different variables:

$\delta a = \delta a (t,r)$.

So it is really common to appear derivative terms multiplying between the $\delta$-terms, in which must be set to zero. Examples of this would be:

$\partial_r \delta a \partial_r \delta b = 0$, or $\partial_t \delta a \partial_r \delta a = 0$, or $(\partial_t \delta c)^2 = 0$.

How should I add rules to take care of these derivative terms?

$\endgroup$
  • 1
    $\begingroup$ You can introduce scaling of the variables, for instance $a=a_0+\alpha x$, $b=b_0+\beta x$, and $c=c_0+\gamma x$. Next you do your calculations and at the end perform series expansion with respect to $x$ and keep only terms up to the 1st order in $x$, that is Series[f[x],{x,0,1}]//Normal. $\endgroup$ – yarchik Sep 13 at 15:06
1
$\begingroup$

You could introduce the perturbative part as δ[a] and define an extra rule how multiplication works with those (δ can be entered quickly via EscdeltaEsc or via \[Delta]):

δ /: Times[___, _δ, _δ, ___] = 0;

Now every time at least two δ[_] symbols are multiplied they will be simplified to zero automatically. For example in

(* Input *)
(a + δ[a]) (b + δ[b])
% // Expand

(* Output *)
(a + δ[a]) (b + δ[b])
a b + b δ[a] + a δ[b]

after expanding the expression, the δ[a]δ[b] part was replaced by zero automatically.

We have to be a bit careful because this rule doesn't catch powers of δ[_] so we should add another rule for that:

δ /: Power[_δ, n_Integer?(# >= 2 &)] = 0;

Now we can for example do

(* Input *)
Table[δ[a]^k, {k, 0, 3}]

(* Output *)
{1, δ[a], 0, 0}

I'm not sure what should happen for negative powers with absolute value of at least two, but you can modify/add another rule similarly to the two above.

$\endgroup$
  • 1
    $\begingroup$ Probably safer to use UpValues rather than to redefine something as fundamental as Times: \[Delta] /: Times[_\[Delta], _\[Delta]] = 0;. $\endgroup$ – march Sep 13 at 16:25
  • $\begingroup$ @march Great idea! i'll update my answer to use UpValues instead. $\endgroup$ – Thies Heidecke Sep 13 at 16:29
  • 1
    $\begingroup$ I suppose that δ /: Times[___, _δ,___, _δ, ___] = 0; is good to make sure that it always works. However, since Times is Orderless, I'm pretty sure the pattern matcher will just check to see if there are two \[delta][_]'s that are multiplied, regardless of whether there are other quantities there or not. I think that \[Delta] /: Times[_\[Delta], _\[Delta]] = 0 on it's own should work. $\endgroup$ – march Sep 13 at 20:01
  • $\begingroup$ Ok, great answer. It seems like it is working flawlessly. However, if I may ask, lots of derivative terms also appear like $\partial_r \delta a \partial_r \delta b$ and even with different variables, $\partial_t \delta a \partial_r \delta c$. They also should turn out to be zero (since anything with two $\delta$ must become zero). Any easy way to work around this? $\endgroup$ – Edison Santos Sep 14 at 13:35
  • $\begingroup$ @march Yes, originally i had the extra ___ in the pattern, when i had the same realisation about the Orderless property, after which i removed it. Both patterns should work i guess. $\endgroup$ – Thies Heidecke Sep 15 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.