In the page that led upto this we looked at two categories of algebra that help us express physical representations in a way that is independent of the coordinate system used:
Both of these require leaning a lot of new ideas and techniques which is not easy. So on this page we will try to get a flavor of the different approaches by using simple examples to help you decide which helps your intuitive understanding.
3D rotations
We don't need this extra complexity to handle 3D rotations, we can use quaternions or matrices (which we can think of as specific examples of Clifford and tensor algebras respectively) as described on this page. However it may help to start with something relatively familiar.
Using Clifford algebras
We use the even subalgebra based on 3D vectors which gives us quantities containing a scalar and a 3 dimensional bivector. The rotation transform is represented by:
v' = q * v * q^{1}
where:
 v',v = output and input vectors
 q = multivector containing scalar and 3D bivector (quaternion).
q is four dimensional but it is normalised to give 3 degrees of freedom.
However here we want to take into account any changes in the coordinate system so we need to define v and q in terms of their coordinates as follows:
v = vx e1 + vy e2 + vz e3
q = qw + qx e2e3 + qy e3e1 + qz e1e2
which gives the formula for the rotation in terms of the basis vectors as:
v' = (qw + qx e2e3 + qy e3e1 + qz e1e2) * (vx e1 + vy e2 + vz e3) * (qw + qx e2e3 + qy e3e1 + qz e1e2) ^{1}
Using Tensor algebra
We can represent the rotation by a grade 2 tensor (matrix) with two subscripts, one raised and one lowered:
v^{j}=R_{i}^{j} * v^{i}
In the 3D case R_{i}^{j} will be a square 3×3 matrix but we dont allow any linear transform it is constrained to be orthogonal.
The above transformed the coordinates but we could have got the same result by transforming the basis vectors in the opposite direction.
e_{j}= e _{i} * R_{j}^{i}
In 3D with orthogonal coordinates this distinction between transforming the coordinates and transforming the basis vectors has minimal effect on the actual computations but when things get more complicated, for example if we are using curvilinear coordinates then we need to be much more careful about these things.
Inertia Tensor
At first sight this would appear to be a 3D linear transform similar to the rotation above, for instance it can be used to relate torque and angular acceleration (or L=[I]ω):
T= [I]α
where:
 T= Torque
 [I] = Inertia Tensor
 α = angular acceleration
However what we are interested in here is how it changes with a change in coordinates which is where it is very different. Here we will see that continuously rotational quantities like torque and angular acceleration vary with the reciprocal basis.
Using Clifford algebras
The inertia tensor maps bivectors to bivectors (for example torque to angular acceleration)
T= i_{1} α_{1} e_{2} e_{3} + i_{2} α_{2} e_{3} e_{1} + i_{3} α_{3} e_{1} e_{2}
where:
 i_{1},i_{2},i_{3} = eigen vectors (symmetry axes)
 α_{1},α_{2},α_{3} = components of angular acceleration
 e_{1},e_{2},e_{3} = basis vectors
If the basis vector is e then :
I_{e} = e•Ie
where:
I = pseudoscalar
Using Tensor algebra
How do we represent rotational quantities like ω=v×r and α=a×r in tensor algebra? Since they can be formed from cross products we can use
v×r = ε_{ijk}v^{j}r^{k}
where:
 ε_{ijk} = permutation symbol
An alternative way to investigate this is to convert the rotational quantity as a antisymmetric grade 2 tensor:
[~w]= 

Yet another approach is to use the determinant:
v×r= det 

where:
 e_{1},e_{2},e_{3} = basis vectors.
So what about the inertia tensor itself?
_{N}  
I_{jk} =  Σ  m_{k}(r_{k}δ_{ij}  r_{ki} r_{kj}) 
^{k=1} 
where:
δ_{ij} = Kronecker delta
I_{e} = [l m n] 

