How to declare a 3D vector variable?

Question

How can I do vector calculations without telling Mathematica the vector entries?

I have very many arbitrary linear combinations in $\mathbb{R}^3$ which I want to perform some general calculations on (scalar and vector products) and want to use Mathematica to do this (especially for simplifying stuff like very long equations with scalar prodcuts of vector products).

So, I don't want to write for all my vectors stuff like v1={a1,b1,c1} ... vN={aN,bN,cN} and so on, but just want to say v1 ... vN are vectors.

How is this possible?

Answer 1

Assuming that we have three-dimensional real vectors :

$Assumptions = (u | v | w) ∈ Vectors[3, Reals];

we can use e.g. various tensor functions (new in ver. 9) e.g. TensorReduce to reduce (simplify) a tensor expression, e.g.

TensorReduce[ v.v + w.w - (v + w).(v + w) ]
TensorReduce[u \[Cross] (v \[Cross] w) ]

-2 v.w
-w u.v + v u.w

We can perform more interesting reductions, let's show e.g. the Jacobi identity:

TensorReduce[ u \[Cross] (v \[Cross] w) + v \[Cross] (w \[Cross] u) + 
              w \[Cross] (u \[Cross] v) ]

0

or write it in a traditional form:

Defer[   u \[Cross] (v \[Cross] w) + v \[Cross] (w \[Cross] u)
       + w \[Cross] (u \[Cross] v)] == 
TensorReduce[   u \[Cross] (v \[Cross] w) + v \[Cross] (w \[Cross] u)
              + w \[Cross] (u \[Cross] v) ] // TraditionalForm

Another common identity

TensorExpand[ (u \[Cross] v) \[Cross] (u \[Cross] w) ]
TensorExpand[ (u \[Cross] v).(u \[Cross] v) ]

u u \[Cross] v . w
-(u.v)^2 + u.u  v.v

Take a look at new differential operators:

Curl[ Curl[ f[x, y, z], {x, y, z}], {x, y, z}] == Laplacian[ f[x, y, z], {x, y, z}]

True

Answer 2

@Szabolcs is right, use Symbolic Tensors. But in that link it may be a bit confusing to find what you want. There are good examples on 3D vector operations. Read:

• Vectors

• TensorExpand

• Derive and Verify Vector Identities

For example, proving an identity:

a\[Cross](b\[Cross](c\[Cross]d)) == b a.(c\[Cross]d) - (a.b) c\[Cross]d // TensorExpand

True

Or expanding something very long:

((a\[Cross]b).c)^4 // TensorExpand

(a.c)^4 (b.b)^2 - 4 a.b (a.c)^3 b.b b.c + 4 (a.b)^2 (a.c)^2 (b.c)^2 + 2 a.a (a.c)^2 b.b (b.c)^2 - 4 a.a a.b a.c (b.c)^3 + (a.a)^2 (b.c)^4 + 2 (a.b)^2 (a.c)^2 b.b c.c - 2 a.a (a.c)^2 (b.b)^2 c.c - 4 (a.b)^3 a.c b.c c.c + 4 a.a a.b a.c b.b b.c c.c + 2 a.a (a.b)^2 (b.c)^2 c.c - 2 (a.a)^2 b.b (b.c)^2 c.c + (a.b)^4 (c.c)^2 - 2 a.a (a.b)^2 b.b (c.c)^2 + (a.a)^2 (b.b)^2 (c.c)^2