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I have a function that outputs a large expression containing dot products of vectors. None of the vector components are known, so all dot products are symbolic. For instance, part of the output may look like2(p1.q1)(p3.p2) + (p1.p2)^2.

I know what some of the dot products evaluate to. For example, I know things like:

p1.p1 = m
p1.p2 = 0

I want mathematica to simplify the expression as much as possible, making use of the known dot products and being sure to simplify all possible cancellation.

What is a good way to do this?

The way I am currently doing it is clunky and doesn't always work well. What I did was define a dot product function d[x,y] and then explicitly specified some of the dot products like d[p1,p2] = m and d[p1,p2] = 0. This became cumbersome because I constantly had to explicitly input commutitivity for each dot product (i.e. d[p1,p2] = d[p2,p1] = 0), and even when I did this, mathematica was not fully taking commutitivity into account when simplifying (via FullSimplify) my expression. Is there a better way?

Carl Woll
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    To help you with your d function, you could SetAttributes[d,Orderless]. Then all you need to define is d[p1,p2]=m (shouldn't the m be squared?). Then d[p1,p2] and d[p2,p1] will automatically be replaced with m. – QuantumDot Nov 29 '14 at 23:00
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    I also suggest that you investigate the option TransformationFuntions that is given to Simplify to teach it to handle expressions involving your d function. – QuantumDot Nov 29 '14 at 23:02
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    The orderless attribute simplifies the code a lot, and Mathematica seems to have an easier time simplifying expressions after I added that. –  Nov 30 '14 at 20:15

2 Answers2

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TensorExpand knows about commutativity of vectors. So, one approach is to define UpValues for your known relations (in canonical order):

p1 /: p1.p1 = m;
p2 /: p1.p2 = 0;

Then, use TensorExpand on your expression with the assumption that your variables are vectors:

TensorExpand[
    2 (p1.q1) (p3.p2) + (p1.p2)^2,
    Assumptions->(p1|p2|p3|q1) ∈ Vectors[d]
]

2 p1.q1 p2.p3

Carl Woll
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2

You might formulate a list of rules that take into account the existing relations, such as the following. Assume there are 2 known relations p1.p2=mand p1.q1=s. The rules are as follows:

rules = {p1.p2 -> m, p2.p1 -> m, p1.q1 -> s, q1.p1 -> s};

Their application is straightforard:

    2 (p1.q1) (p3.p2) + (p1.p2)^2 /. rules

(*  m^2 + 2 s p3.p2   *)

Have fun!

Alexei Boulbitch
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