Download Advanced Electrodynamics Exercise 5

Advanced Electrodynamics Exercise 5 - Lorentz
Force and Hodge Star
This exercise will consider the Hodge star operation ? and its importance
in treating physical problems in different coordinate systems. Consider for
example the Lorentz force on a particle
~
F~ = q ~v × B
(1)
~ in the Euclidean coordiwith F~ = (F1 , F2 , F3 ) and similarly for ~v and B
nate system. Of course this equation should also hold if we go to another
coordinate system. So with the transformation OF~ 0 = F~ we should have
~0 .
(2)
OF~ 0 = q O~v 0 × OB
In terms of the basis vectors, transforming according to ei =
we have for example
e1 × e2 =
3
X
O3j f j =
j=1
3
X
O1k O2l f k × f l
P3
j=1 Oij f
j,
(3)
k,l=1
It is not easy to see that this is true in general. We will now use Eqn. (173)
from the fourth lecture notes, which relates the cross-product to the Hodge
star operation ?, i.e.
~
?(v ∧ B) = ~v × B.
(4)
We see that we actually consider the components of a one-form. Now we
want to show that not only the above equality is true but that the Hodge
star is basis independent under orthogonal transformations.
The Hodge star operation for an n-dimensional vector space V maps Ωp (V ) →
Ωn−p (V ) and is defined by acting on a dual orthogonal basis as (see Eqn.
(143) and the discussion leading to Eqn. (39) in the lecture notes 4)
p
? ei1 ∧ ... ∧ eip = i1 ...in g i1 i1 ...g ip ip |g| eip+1 ∧ ... ∧ ein .
(5)
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Here |g| = det(gij ) and (ip+1 ...in ) = (1...n) − (i1 ...ip ) as defined in the
lecture. We then start by assuming an orthogonal transformation O from
an orthonormal basis {ei } to an other orthonormal basis {fi } by
fj =
n
X
ei Oij .
(6)
i=1
We choose here two orthonormal
basis sets
p
p because then we do not have to
care about the factors |g(ei , ej )| and |g(fi , fj )| in the definition of the
Hodge product. Further we then have for the metric


λ1

...
g=λ=
(7)
λn
where λi = ±1. Already in the first exercise we had introduced the idea of
an orthonormal matrix (leaving a given metric invariant) by
OT gO = g.
(8)
We will further assume that the basis transformation does not change the
orientation, i.e. det(O) = 1. Examples would be the matrices of the proper
orthochronous Lorentz group.
a) Show first that from (6) we find for the dual basis
ei =
n
X
Oij f j .
(9)
j=1
(1 point)
Using this result we can now consider the invariance of the Hodge star
operation under basis transformation. Our claim is using (5), that


n
n
X
X
? ei1 ∧ ... ∧ eip = ? 
Oi1 j1 f j1 ∧ ... ∧
O ip jp f jp 
(10)
j1 =1

= i1 ...in λi1 ...λip 
n
X
jp =1
Oip+1 jp+1 f jp+1 ∧ ... ∧
jp+1 =1
n
X

Oin jn f jn  ,
jn =1
i.e. the Hodge star is independent whether we do a transformation before
or after we have performed the operation. In order for this equality to be
true we will need to derive equations connecting the sub-determinants of an
orthogonal matrix. So in order to simplify matters we introduce the matrix
O(i1 ...ip ,j1 ...jp )
2
(11)
consisting of the rows (i1 ...ip ) and the columns (j1 ...jp ) of the orthogonal
matrix O. For example we have for


O11 O12 O13
(12)
O = O21 O22 O23 
O31 O32 O33
the sub-matrix
O(23,12) =
O21 O22
.
O31 O32
(13)
In accordance we define the determinant of the sub-matrix O(i1 ...ip ,j1 ...jp ) as
the sub-determinant |O|(i1 ...ip ,j1 ...jp ) of the matrix O.
b) Derive now from Eqn. (10) the equality connecting the two sub-determinants
n
|O|(i1 ...ip ,j1 ...jp ) = λi1 ...λip λj1 ...λjp δji11 ...i
...jn |O|(ip+1 ...in ,jp+1 ...jn ) ,
(14)
n
where we have defined δji11 ...i
...jn as in Eqn. (38) in the lecture notes 4.
(3 points)
So, if the Hodge star operation is basis independent, then any two subdeterminants connected by (ip+1 ...in ) = (1...n) − (i1 ...ip ) and (jp+1 ...jn ) =
(1...n) − (j1 ...jp ) in the above sense should have the same value up to a sign.
On a first glance this seems strange.
c) Take the matrix of the Lorentz-transformation consisting of a boost
in z-direction by the constant velocity v and a rotation in the spatial
part around the z-axis by an angle Θ. Take two random pairs of subdeterminants of the above form, e.g. one pair could be |Λ|(134,134) and
|Λ|(2,2) , and verify that the absolute values are the same.
(1 point)
In a next step we will reorder the original matrix O such that the sub-matrix
O(i1 ...ip ,j1 ...jp ) becomes the left upper block and O(ip+1 ...in ,jp+1 ...jn ) becomes
the right lower block, i.e.
A B
Õ =
.
(15)
C D
where A = O(i1 ...ip ,j1 ...jp ) and D = O(ip+1 ...in ,jp+1 ...jn ) . This will help us
to finally relate both sub-determinants by just considering an appropriate
matrix equation. If we define the reordering matrix
(ΠI )p,q = δp,iq
(16)
where I = (i1 ...in ) we can see (check with a simple example) that (ΠI )T O
reorders the rows (1...n) → (i1 ...in ) and OΠJ reorders the columns (1...n) →
(j1 ...jn ).
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d) Show now with the definition of ΠI in (16) that
Õ = (ΠI )T OΠJ .
(17)
Further show that
det(ΠJ ) = j1 ...jn
and
n
det(Õ) = δji11 ...i
...jn det(O).
(18)
Show that the reordered metric λ̃I = (ΠI )T λΠI where λ is defined in (7)
obeys
λ̃I,1
0
T
.
(19)
λ̃J = Õ λ̃I Õ =
0 λ̃I,2
In what follows we will use the reordering (1...n) → (i1 ...ip ip+1 ...in ) so
that I, 1 = (i1 ...ip ) and (1...n) → (j1 ...jp jp+1 ...jn ) so that J, 1 = (j1 ...jp ).
(2 points)
Now we will take the last step to show the above claim. To do so we define
T
A
CT
˜I Õ.
(20)
R=
0
1
From Eqn. (19) we can deduce by explicitly calculating the matrix components that
AT λ̃I,1 A + C T λ̃I,2 C = λ̃J,1
T
T
A λ̃I,1 B + C λ̃I,2 D = 0.
(21)
(22)
e) Show now that
det(D) det(λ̃J,1 ) = det(A) det(λ̃I,1 ) det(Õ)
(23)
and deduce from this Eqn. (14), which finishes our proof.
(3 points)
Maximal points: 10
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