Question

t5.14. Let x and y be two different coordinate patches for part of a surface M Let X Xx, X, and Y = Y'x Y'y. be two vector fields. Define symbols Z* and Z by χι ΣΓ/Υιχ | and Prove that 2 EZ*(dv|8u*). (Hint: Problem 4.11.) This proves thatZx £ Zy, defines a vector field Z = VxY, called the covariant derivative of Y with respect to X. This is one of the most fundamental concepts of modern differential geometry. It is due to Levi-Civita (1917) and will be used in Chapter 7 4.11. Let Fdenote the Christoffel symbols for a coordinate system U, and f:Uua coordinate transformation. Use Gauss's formulas to prove This formula is the first example of a transformation law that is no well behaved (because of the term (dudv d))

Answer 1

Answer:

In 3-dimensions, we define the Levi-Civita tensor, εijk, to be totally antisymmetric, so we get a minus sign under interchange of any pair of indices. We work throughout in Cartesian coordinate. This means that most of the 27 components are zero, since, for example,

ε212 = −ε212

if we imagine interchanging the two 2s. This means that the only nonzero components are the ones for which i, j and k all take different value. There are only six of these, and all of their values are determined once we choose any one of them. Define

ε123 ≡ 1

Then by antisymmetry it follows that

ε123 = ε231 = ε312 = +1

ε132 = ε213 = ε321 = −1

All other components are zero.

Using εijk we can write index expressions for the cross product and curl. The i th component of the cross product is given by

[u × v] i = εijkujvk

as we check by simply writing out the sums for each value of i,

[u × v] 1 = ε1jkujvk

= ε123u2v3 + ε132u3v2 + (all other terms are zero)

= u2v3 − u3v2

[u × v] 2 = ε2jkujvk

= ε231u3v1 + ε213u1v3

= u3v1 − u1v3

[u × v] 3 = ε3jkujvk

= u1v2 − u2v1

We get the curl simply by replacing ui by ∇i = ∂ ∂xi ,

[∇ × v] i = εijk∇jvk

If we sum these expressions with basis vectors ei , where e1 = i, e2 = j, e3 = k, we may write these as vectors:

u × v = [u × v] i ei

= εijkujvkei

∇ × v = εijkei∇jvk

There are useful identities involving pairs of Levi-Civita tensors. The most general is

εijkεlmn = δilδjmδkn + δimδjnδkl + δinδjlδkm − δilδjnδkm − δinδjmδkl − δimδjlδkn

To check this, first notice that the right side is antisymmetric in i, j, k and antisymmetric in l, m, n. For example, if we interchange i and j, we get

εjikεlmn = δjlδimδkn + δjmδinδkl + δjnδilδkm − δjlδinδkm − δjnδimδkl − δjmδilδkn

Now interchange the first pair of Kronecker deltas in each term, to get i, j, k in the original order, then rearrange terms, then pull out an overall sign,

εjikεlmn = δimδjlδkn + δinδjmδkl + δilδjnδkm − δinδjlδkm − δimδjnδkl − δilδjmδkn

= −δilδjmδkn − δimδjnδkl − δinδjlδkm + δilδjnδkm + δinδjmδkl + δimδjlδkn

= − (δilδjmδkn + δimδjnδkl + δinδjlδkm − δilδjnδkm − δinδjmδkl − δimδjlδkn)

= −εijkεlmn

Total antisymmetry means that if we know one component, the others are all determined uniquely. Therefore, set i = l = 1, j = m = 2, k = n = 3, to see that

ε123ε123 = δ11δ22δ33 + δ12δ23δ31 + δ13δ21δ32 − δ11δ23δ32 − δ13δ22δ31 − δ12δ21δ33

= δ11δ22δ33

= 1

Check one more case. Let i = 1, j = 2, k = 3 again, but take l = 3, m = 2, n = 1. Then we have

ε123ε321 = δ13δ22δ31 + δ12δ21δ33 + δ11δ23δ32 − δ13δ21δ32 − δ11δ22δ33 − δ12δ23δ31

= −δ11δ22δ33

= −1

as expected.

We get a second identity by setting n = k and summing,

εijkεlmk = δilδjmδkk + δimδjkδkl + δikδjlδkm − δilδjkδkm − δikδjmδkl − δimδjlδkk

= 3δilδjm + δimδjl + δimδjl − δilδjm − δilδjm − 3δimδjl

= (3 − 1 − 1) δilδjm − (3 − 1 − 1) δimδjl

= δilδjm − δimδjl

so we have a much simpler, and very useful, relation

εijkεlmk = δilδjm − δimδjl

A second sum gives another identity. Setting m = j and summing again,

εijkεljk = δilδmm − δimδml

= 3δil − δil

= 2δil

Setting the last two indices equal and summing provides a check on our normalization

εijkεijk = 2δii = 6

This is correct, since there are only six nonzero components and we are summing their squares.

Collecting these results,

εijkεlmn = δilδjmδkn + δimδjnδkl + δinδjlδkm − δilδjnδkm − δinδjmδkl − δimδjlδkn

εijkεlmk = δilδjm − δimδjl

εijkεljk = 2δil

εijkεijk = 6

Now we use these properties to prove some vector identities. First, consider the triple product,

u · (v × w) = ui [v × w] i

= uiεijkvjwk

= εijkuivjwk

Because εijk = εkij = εjki, we may write this in two other ways,

u · (v × w) = εijkuivjwk

= εkijuivjwk

= wkεkijuivj

= wi [u × v] i

= w · (u × v)

and

u · (v × w) = εijkuivjwk

= εjkiuivjwk

= vj [w × u] j

= v · (w × u)

so that we have established

u · (v × w) = w · (u × v) = v · (w × u)

and we get the negative permutations by interchanging the order of the vectors in the cross products. Next, consider a double cross product:

[u × (v × w)]i = εijkuj [v × w]k

= εijkuj εklmvlwm

= εijkεklmujvlwm

= εijkεlmkujvlwm

= (δilδjm − δimδjl) ujvlwm

= δilδjmujvlwm − δimδjlujvlwm

= (δilvl) (δjmujwm) − (δjlujvl) (δimwm)

= vi (umwm) − (ujvj ) wi Returning to vector notation, this is the BAC − CAB rule,

u × (v × w) = (u · w) v − (u · v) w

Finally, look at the curl of a cross product,

[∇ × (v × w)]i = εijk∇j [v × w]k 3

= εijk∇j (εklmvlwm)

= εijkεklm ((∇jvl) wm + vl∇jwm)

= (δilδjm − δimδjl) ((∇jvl) wm + vl∇jwm)

= δilδjm ((∇jvl) wm + vl∇jwm) − δimδjl ((∇jvl) wm + vl∇jwm)

= (∇mvi) wm + vi∇mwm − (∇jvj ) wi − vj∇jwi

Restoring the vector notation, we have

∇ × (v × w) = (w · ∇) v + (∇ · w) v − (∇ · v) w − (v · ∇) w

If you doubt the advantages here, try to prove these identities by explicitly writing out all of the components!