a simple theorem about basis vectors 7_12
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A short note by Phil dated 7.19.12 on how rotated Cartesian basis vectors relate to the original ones. Part (a) works the N=2 case with Rz(φ), giving a geometric proof and two index-based proofs. Part (b) proves the general N statement two ways, using the orthogonality relation R^-1 = R^T and dot products with the basis vectors.
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A simple theorem about basis vectors PhL 7.19.12
Just skip to part (b) where this theorem is quickly proven:
e'n = Ren => e'n = R-1nm em or = R-1
Contents:
(a) The Theorem for N = 2 1
(b) The Theorem for general N 3
(a) The Theorem for N = 2
Let's assume that two sets of basis vectors are related this way (N=2)
e'n = Rz(φ)en Rz(φ) = the usual active form =
The theorem is that you can write
= Rz(-φ) =
which is a shorthand notation for
e'n = Rz(-φ)nm em for n = 1,2
which is the same as
e'1 = cosφ e1 + sinφ e2
e'2 = - sinφ e1 + cosφ e2
Proof #1:
Write
e'1 = (e1 e'1) e1 + (e2 e'1) e2
e'2 = (e1 e'2) e1 + (e2 e'2) e2
The drawing shows that
(e1 e'1) = cosφ (e2 e'1) = cos(π/2-φ) = sinφ = picture shows positive
(e2 e'2) = cosφ (e1 e'2) = cos(π/2+φ) = -sinφ
Therefore
e'1 = cosφ e1 + sinφ e2
e'2 = -sinφ e1 + cosφ e2 QED.
Proof #2: Here is what we want to verify:
= Rz(-φ) (1)
and here is what we know
e'n = Rz(φ)en n = 1,2 (2)
Write out (1) this way
e'n = Rz(-φ)nm em (3)
Install (2) as the LHS of (3) to get
Rz(φ)en = Rz(-φ)nm em (4)
Take the ith component
Rz(φ)ij(en)j = Rz(-φ)nm (em)i (5)
But we know that (en)j = δn,j when evaluated in the black frame, so we then have
Rz(φ)ij δn,j = Rz(-φ)nm δm,i (6)
Rz(φ)in = Rz(-φ)ni (7)
But this is true since
Rz(-φ)ni = Rz(φ)-1ni = R(φ)Tni = R(φ)in (8)
Proof #3. Now let's reverse Proof 2 :
Rz(φ)in = Rz(-φ)ni (1)
Rz(φ)ij δn,j = Rz(-φ)nm δm,i (2)
Rz(φ)ij(en)j = Rz(-φ)nm (em)i (3)
Rz(φ)en = Rz(-φ)nm em (4)
e'n = Rz(-φ)nm em (5)
= Rz(-φ) (6)
(b) The Theorem for general N
The theorem is that, if this is true for how Cartesian basis vectors are related in two frames,
e'n = Ren n = 1,2....N , // (en)i = δn,i
then the following is true
= R-1
which is just a shorthand notation for
e'n = R-1nm em n = 1,2...N
Proof A:
Rin = R-1ni // since R-1ni = RTni = Rin (1)
Rij δn,j = R-1nm δm,i // rewrite (2)
Rij(en)j = R-1nm (em)i // ID the basis vectors (3)
Ren = R-1nm (em) // write above as vector equation (4)
e'n = R-1nm em // use known fact on LHS (5)
= R-1 // QED (6)
Proof B:
Since the en are a Cartesian basis, and since ab is a scalar under rotations, we may write
δn,k = en ek = e'n e'k (1)
Take the known true fact e'n = Ren and dot both sides into em to get
em e'n = em Ren = (em)i Rij(en)j = δm,i Rijδn,j = Rmn (2)
Now here is our claimed theorem statement :
e'n = R-1nm em n = 1,2...N (3)
To verify this is true, it suffices to show that it is true when dotted into all the e'k vectors since they form a complete basis. So we need to show that this is true
e'n e'k = R-1nm em e'k n = 1,2...N and k = 1,2...N (4)
But LHS is δn,k from (1), and we know that em e'k = Rmk from (2), so we get
δn,k = R-1nm Rmk = (R-1R)nk QED (5)