Download Advanced Electrodynamics Exercise 4

Advanced Electrodynamics Exercise 4 - metrics and
Lorentz transformations
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Signature of the metric
Consider an n-dimensional semi-Riemannian metric in an orthonormal basis.
Therefore it is diagonal and has k times −1 and n − k times +1 on its
diagonal. A special case is the Minkowski metric, diag(−1, 1, 1, 1). One can
wonder if we could find some other orthonormal basis which has a different
number of positive and negative entries. In this exercise you will show that
this is not the case. In other words, the signature of the metric is basis
independent.
a) Consider two orthonormal bases β = b1 , . . . , bn and β ′ = b′1 , . . . , b′n for
the metric G on the vector space V , with
(
(
−1, j ≤ l
−1, i ≤ k
(1)
and
b′j · b′j =
bi · bi =
+1, j > l.
+1, i > k
Now take the subspace X of V spanned by bk+1 , . . . , bn . The metric G
is positive definite, since for x 6= 0 ∈ X we have
G(x, x) = G(xi bk+i , xi bk+i ) =
n−k
X
(xi )2 > 0.
(2)
i=1
Now consider an arbitrary subspace W of V on which G is negative
definite. On the subspace W take a basis ω = w1 , . . . , wr . Show that the
vectors in the set
P = {w1 , . . . , w r , bk+1 , . . . , bn }
(3)
are linear independent. You might use the definiteness of the subspaces
X and W .
b) Now show that the linear independency of the set P implies that k = l,
i.e. that the signature of the metric is basis independent.
Hint: Use that dim(P ) ≤ dim(V ), since the vectors are linear independent. Further, notice that the vectors b′j with j ≤ l could be used as a
basis for W .
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The electromagnetic field tensor
During the lectures tensors were introduced and some emphasis has been put
on anti-symmetric tensors. A typical example of a rank 2 anti-symmetric
tensor is the electromagnetic field tensor, F µν . In Gauss units, this tensor
can be written in the electric and magnetic field components as


0 −Ex −Ey −Ez
Ex
0
−Bz By 
.
F µν = 
(4)
Ey Bz
0
−Bx 
Ez −By Bx
0
Since F µν is a tensor, it transforms under Lorentz transformations as
Feµν = Λµα Λν β F αβ .
(5)
In the case of a Lorentz boost in the x-direction, Λ has the following simple
form


γ
−βγ 0 0
−βγ
γ
0 0
.
Λµν = 
(6)
 0
0
1 0
0
0
0 1
a) The Lorentz transformations are special basis transformations. The have
the funny property that the inverse can be calculated by simply raising
and lowering the indices. Show that
(7)
Λ−1 µν = gµα gνβ Λαβ
directly follows from the fact that the components of the metric are
invariant under Lorentz transformations.
b) Show how the unit vectors ∂µ of a vector field transform under a Lorentz
boost in the x-direction (6). So give an expression for
∂
∂˜µ =
∂ x̃µ
(8)
in terms of ∂µ . It can be convenient to use the previous result.
c) Applying a Lorentz transformation in general mixes the electric and magnetic field components in the electromagnetic field tensor. Give an exe and B,
e in a
pression for the electric and magnetic field components, E
frame moving with a constant velocity in the x-direction in terms of the
original components E and B.
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d) If the electric and magnetic fields satisfy E and B satisfy the Maxwell
e and
equations, the electric and magnetic field in the moving frame, E
e
B, should also satisfy the Maxwell equations. Check this for the homogeneous Maxwell equations
∇ · B = 0,
(9a)
∇ × E + ∂0 B = 0.
(9b)
Later when the wedge product, exterior derivative and hodge-star operation are introduced, this becomes obvious for general Lorentz transformations.
e) Show that
Fµν F µν = 2 |B|2 − |E|2 .
(10)
f) The same relation holds for the components of Fe as well. By working
e and B,
e show that
out the E
Fµν F µν = Feµν Feµν ,
(11)
so is invariant under the x-boost (6).
g) It would be surprising if Fµν F µν is only invariant under a Lorentz boost in
the x-direction. Indeed, it is invariant under all Lorentz transformations.
Proof its invariance under general Lorentz transformations.
h) An other quantity invariant under proper orthochronous Lorentz transformations (det(Λ) = 1) is the completely anti-symmetric contraction
ǫαβγδ F αβ F γδ = 8 E · B ,
(12)
where the Levi-Civita tensor is defined as
(
sign(i1 , . . . , in ) all ij distinct
ǫi1 ...in =
0
otherwise.
(13)
Show that this quantity (12) is indeed invariant under general Lorentz
transformations. It can be useful to know that the determinant of a
matrix can be expressed as
det(A) = ǫi1 ...in A1i1 · · · Anin .
(14)
Each exercise is 1 point, so a maximum of 10 points.
Hand in Monday evening at latest 18:00 o’clock — mailbox of Robert van
Leeuwen, second floor Nanoscience Building
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Erratum exercise 3
The Schwarzschild metric got a bit messed up due to too many unit conversions. Here is the correct version in SI units
 2

0
0
0
c (1 − rrs )

0
−(1 − rrs )−1
0
0 

gµν (t, p) = 
(15)
2
2

0
0
−r sin (θ)
0 
0
0
0
−r 2
in spherical coordinates and rs = 2GM/c2 , where G is the gravitational
constant and M the mass of the object.
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