Pseudo Tensor

 

Pseudotensors Math 1550 lecture notes, Prof. Anna Vainchtein

1

Prope Proper r and impr imprope oper r ortho orthogo gona nall trans transfo forrmations of bases

 {

}

Let e1 , e2 , e3  be an orthonormal basis in a Cartesian coordinate system and suppose we switch to another rectangular coordinate system with the e1 , ¯ e2 , ¯ e3 . Recall that the two bases are related orthonormal basis vectors ¯ ei e j : via an orthogonal matrix   Q  with components   Qij   =  ¯

 {

}

·

¯ ei  =  Q ij e j ,   ei  =  Q ji ¯ ei .

Let

 

∆ = detQ

(1)

 

±

(2)

and recall that ∆ = 1 because   Q  is orthogonal. If ∆ = 1, we say that the transformation is  proper orthogonal ; if ∆ = 1, it is an  improper orthogonal  transf tra nsform ormati ation. on. Not Notee tha thatt the hande handedne dness ss of the basis rema remains ins the sam samee under a proper orthogonal transformation and  changes   under an improper one. Indeed, ¯   = (¯ V  e1

−

× ¯e ) · ¯e  = (Q 2

3

1m

em

×Q

2n

·

en ) Q3l el  =  Q 1m Q2n Q3l (em

× e ) · e ,   (3) n

l

where the cross prod where product uct is tak taken en wit with h res respect pect to som somee und underl erlyin yingg rig right ht-handed han ded syst system, em, e.g. sta standa ndard rd basis basis.. Not Notee that this is a tri triple ple sum ov over er   m, n  and   l. Now, observe that the terms in the above some where ( m,n,l) is a cyclic (even) permutation of (1, 2, 3) all multiply   V   = (e1 e2 ) e3  because the scalar triple product is invariant under such permutations:

·

×

(e1

× e ) · e  = (e × e ) · e  = (e × e ) · e . 2

3

2

3

1

3

1

2

Meanwhile, terms where (m,n,l) is a non-cyclic (odd) permutations of (1, 2, 3) multiply V  , e.g e.g.. fo forr ( m,n,l) = (2, 1, 3) we have (e2 e1 ) e3   = (e1 e2 ) e3  = V  .   . All other terms in the sum in (3) (where two or more of the three indices (m,n,l) are the same) are zero (why?). So (3) yields

·

 − −

×

¯   =(Q11Q22Q33 +  Q13Q21Q32 +  Q 12Q23Q31 V 

−Q

12

Q21Q33

−Q

11

Q23Q32

1

−Q

13

Q22Q31)V,

·

 − ×

 

or, recalling (2) and the definition of the determinant, ¯   = ∆ V. V 

 

(4)

¯   and  V    hav havee the same sign (bases hav havee the same hande handedness) dness) if and Hence  V  only if ∆ = 1 (a proper orthogona orthogonall transform transformation ation). ). Othe Otherwis rwise, e, if ∆ = 1, a right-handed basis transforms into a left-handed one and vice versa.

 −

Example 1.   A counte counterclockwise rclockwise rotat rotation ion by the angle   θ  about  x 3 -axis is a

proper orthogonal transformation. Indeed, in this case [Qij

 cos ] =  − sin

θ   sin θ   0 θ   cos θ   0

0

0

1

 

.

which yields ∆ = cos2 θ  + sin2 θ   = 1. The hand handedn edness ess is pre preser serve ved d und under er such transformation (or any other pure rotation or combination thereof). Example 2.   Con Example Consid sider er now a reflec reflectio tion n in the (x2 , x3 ) pla plane ne:: x¯1   = x ¯2  =  x 2 , x¯3  =  x 3 . We have

[Qij

 −1 ]= 0 0

0 0 1 0 0 1

 

,   ∆=

 −x , 1

−1,

so the tran transfo sforma rmatio tion n is imp imprope roperr ort orthogo hogonal nal.. It is eas easy y to see tha thatt suc such h transformation reverses the handedness of the basis, as does any orthogonal transformation that involves an odd number of reflections.

2

Pseu Pseudo dote tens nsor orss

Definition.   A  pseudotensor   pseudotensor    A  of order  n  in   R3 is a quantity with 3 n com-

ponents which under orthogonal transformations of coordinate system transform as follows: ¯i1 i2 ...i   = ∆ Qi1 j1 Qi2 j2 . . . Qi A

 j

n n

n

A j1 j2...j . n

In other words, pseudotensors transform the same way as ordinary tensors under  proper  orthogonal transformations but differently under the improper ones. 2

 

Example.   Consider ttwo wo vectors,  a  and   b, and let c  =  a

b

×

omputeed in the current current basis . The be defined as their cross product   ccomput hen n w wee have ci  =  a j bk ak b j ,   c¯i  = a ¯ j ¯bk a¯k¯b j ,   (5)

−

−

where for any   i, the indices   j   and   k  are selected so that ( i,j,k ) is the cyclic permutation of (1, 2, 3). Question: Ques tion: does   c  so defi defined ned tran transfo sform rm as a Car Cartes tesian ian tens tensor, or, i.e. do we have c¯i  =  Q ik ck  under the change of basis (1)? Let’s check: c¯i  = a ¯ j ¯bk

− a¯ ¯b   = Q

−Q

− a b ). Interchanging Interc hanging the summation indices m and n in the second term (−Q Q a b −Q Q a b ), we get c¯  = (Q Q − Q Q )a b .   (6) k  j

 jm am Qkn bn

kn an Q jm bm

 =  Q jm Qkn (am bn

n m

 jm

 jn

kn n m

km m n

i

 jm

kn

 jn

km

m n

Now, obs Now, observ ervee tha thatt the ter terms ms wit with h   m   =   n   in the above sum are zero, so only terms with   m =   n   rem remai ain. n. No Now, w, if   m   = 1 and   n   = 2, say, we get (Q j 1 Qk2 Q j 2 Qk1 )a1 b2, while m  = 2 and  n  = 1 yields (Q j2 Qk1 Q j1 Qk2)a2 b1 , so that the two terms combined give us

−

 

−

(Q j 1 Qk2

−Q

 j 2 Qk1

)(a1 b2

− a b ). 2 1

Similarly, the terms with ( m, n) = (2, 3) and (m, n) = (3, 2) yield (Q j 2 Qk3

 j 3 Qk2

−Q

)(a2 b3

−Q

)(a3 b1

3 2

−a b )

and the terms with (m, n) = (3, 1) and (m, n) = (1, 3) yield (Q j 3 Qk1

 j 1 Qk3

− a b ). 1 3

Combining these, we can see that (6) becomes c¯i  = (Q j 1 Qk2

+ (Q j2 Qk3 + (Q j3 Qk1

−Q −Q −Q

)(a1 b2  j 3 Qk2 )(a2 b3  j 1 Qk3 )(a3 b1  j 2 Qk1

3

−a b ) −a b ) − a b ), 2 1

3 2

1 3

(7)

 =

 

where we recall that (i,j ,k) is the cyclic permutation of (1, 2, 3). We now claim that Q jm Qkn

−Q

 jn Qkm

 = ∆Qil ,

 

(8)

where (i,j,k ) and (m,n,l) are both cyclic permutations of (1 , 2, 3). To see this, note that by (1), ¯ e j

× ¯e  =  Q

×e ,

 jm Qkn em

k

n

where both cross products are taken with respect to the same underlying right-handed basis, e.g. standard basis. Then e j (¯

× ¯e ) · e  =  Q k

 jm Qkn

l

(em

×e )·e. n

l

e2 ) e3  =  V 

en ) el ) = (e1

As in the earlier calculation, we note that ( em

×  ×· e ) · e   ×= −V ·   if it is

if (m,n,l) is a cyclic permutation of (1, 2, 3), (em non-cyclic and zero otherwise. This yields e j (¯

× ¯e ) · e  = (Q k

 jm Qkn

l

l

n

−Q

 jn Qkm

)V,

 

(9)

where (m,n,l) is a cyclic permutation of (1, 2, 3). But (i,j ,k) is also a cyclic permutation of (1, 2, 3), so we have ¯ e j

and hence e j (¯

× ¯e  =  V ¯ ¯e , k

i

× ¯e ) · e  =  V ¯ ( (¯e · e ) =  V¯ Q k

l

i

Combining (9) and (10), we get V¯ Q il   =  V  (  (Q jm Qkn

l

−Q

 jn Qkm

 

il .

(10)

),

which by (4) yields (8). Now No w obse observ rvee that that (8 (8)) wi with th (m, n) = ( 1, 2) (a (and nd th thus us   l   = 3) yi yiel elds ds Similarly arly,, with (m, n) = (2, 3) we get   Q j 2 Qk3 Q j 1 Qk2 Q j 2 Qk1   = ∆Qi3 . Simil Q j 3 Qk2  = ∆Qi1  and with (m, n) = (3, 1) we have   Q j 3 Qk1 Q j 1 Qk3   = ∆Qi2 . Substituting these in (7), we get

−



−

−

−

−

−

c¯i  = ∆ Qi3 (a1b2 a2 b1 )+Qi1 (a2 b3 a3 b3 )+Qi2 (a3 b1 a1 b3 )

4



−

= ∆Qil (am bn an bm ),

 

where again (m,n,l) is a cyclic permut permutation ation of (1, 2, 3). But this implies that c¯i  = ∆Qil cl ,

so that   c  defined as the cross product in the current basis is indeed a pseudotensor of first order (a  pseudovector , or  axial vector ) rather than a regular vector. vec tor. It transfor transforms ms as a regular vecto vectorr under proper ortho orthogonal gonal transf transforormations (∆ = 1), c¯i   =   Qil cl , but not under improper ones (∆ = 1) where we have c¯i  = Qil cl . Note that if we just compute   c  =  a b  in some basis and then consider the transformation of the resulting quantity under a change to another basis (without  recomputing the cross product with respect to the new basis), then of course course the qua quant ntit ity y wil willl tra transf nsform orm as a reg regula ularr ve vecto ctor, r, c¯i   =   Qil cl . But the point is that the new components will   not   be components of the cross product  a b as computed  in the new basis  unless   unless  Q  is proper orthogonal orthogonal.. So

 −

−

×

×

  the cross product operation  it really gives us a pseudovector because weisuse the convention (right hand rule that in a right-handed basis and left hand rule in a left left-hande -handed d one). The conv convent ention ion allows us to compute the cross product the same way,

a

 {

×b

  =  

e1   e2   e3 a1   a2   a3 b1   b2   b3

}

 

,

in   any   orthonormal basis e1 , e2 , e3 , regardless of its handedness, but the price we pay is that it gives us a pseudovector if we compute the cross product in the current basis. Remark.   In gener general al bases, the ccross ross produc productt is give given n by

a

×b

  =  

e1   e2   e3 a1 a2 a3 b1 b2 b3

 √     =  G

e1 e2 e3 a1   a2   a3 b1   b2   b3

  √ 1

G

,

where   G   =   V   2 = det[gik ] is the determin determinan antt of the met metric ric ten tensor sor.. In the orthonormal case   G   = 1 and the covariant and contravariant components coincide, as do regular and reciprocal basis vectors.

5

 

3

Prope Propert rtie iess of pseu pseudo dote tens nsor orss 1. The sum of two pseu pseudote dotensors nsors of order  n  is a pseudotensor of order   n. Indeed, Indee d, if   C i1 ...i   =  A i1 ...i  +  Bi1...i   and n

n

¯i1...i   = ∆Qi1 j1 . . . Qi A n

n

 j

n n

¯i1 ...i   = ∆ Qi1 j1 . . . Qi A j1...j ,   B n

n

 j

n n

B j1 ...j , n

¯i1 ...i   = ∆ Qi1 j1 . . . Qi  j C  j1 ...j , so   C  =  A + B  is a pseudotensor. then  C  n

n n

n

2. The produc productt of a pseud pseudotens otensor or of order m  and a regular tensor of order n  is a pseudotensor of order   m + n. Let   A  be a pseudotensor of order   m   and   B  be a regular tensor of or¯k1 ...k   = der   n. The Then n we ha have ve  A¯i1...i   = ∆Qi1 j1 . . . Qi  j A j1...j   and  B Qk1l1 . . . Qk l Bl1 ...l . Thus fo forr   C i1 ...i k1 ...k   =   Ai1...i Bk1 ...k   we have ¯i1 ...i k1 ...k   = ∆Qi1 j1 . . . Qi  j Qk1l1 . . . Qk l C  j1 ...j l1 ...l , so  C  is a pseuC  m

n n

m

m m

n

m

n

m

n

m m

n

m

n n

m

n

n

dotensor of order   m + n. 3. The produc productt of tw twoo pse pseudo udoten tensor sorss of ord order erss   m   and   n  is an ordinary (Cartesian) tensor of order   m + n. Proved the same way as above, except now we get ∆ for both   A   and B, so we obtain ∆2 = 1 in the transformation law for   C.

 ≥

4. Con Contract traction ion of a pseudotens pseudotensor or of order   n  2 gives a pseudotensor of  order   n 2. Indeed, if  A¯i1i2...i   = ∆Qi1 j1 Qi2 j2 . . . Qi  j A j1 j2 ...j , then its contraction in the  k th and  m th index (replace i m  by  i k  and sum over these) satisfies

−

n

n

n n

¯ Ai1 ...i

k

...ik ...in

But since   Qi  j

k k

¯i1 ...i A

k

...ik ...i n

k k

k m

n n

  = ∆ Qi1 j1 . . . Qi  j  . . . Qi  j  . . . Qi  j A j1 ...j Qi  j   =  δ   jj  j   this yields k m

k

...jm ...jn .

k m

 =

∆Qi1 j1 . . . Qi

 j

so   Ai1...i

 are components of a pseudotensor of order   n

k−1 k−1

k

...i k ...i n

Qi

 j

k+1 k+1

 . . . Qi

 j

m−1 m−1

6

Qi

 j

m+1 m+1

 . . . Qi

 j

n n

A j1 ...j

k

...j k ...j n ,

− 2.

 

4

Levi Levi-C -Civ ivit ita a sym symbol bol

Consider εijk   = (ei e j ) ek ,   (11) where   the cross product is taken with respect to the orthonormal basis e1 , e2 , e3 . Then we have

× ·

 {

εijk

}

+1   if ( ) is a cyclic (even) permutation of (1 2 3)  = −1   if ( (odd) permutation of (1 2 3) 0   otherwise,) isi.e.a non-cyclic whwen any two two indices coincide ,

i,j,k

,

i,j,k

, ,

, ,

(12)

,

Note that there are   only six nonzero components : ε123 =  ε 231  =  ε 312  = 1,

ε132 =  ε 321  =  ε 213  =

−1

and cise). observe that any matrix   A, detA  =   εijk A1i A2 j A3k  (show this as an exer exercise) . Obse Observe rvefor also that εmnk   =  ε nkm  =  ε kmn ,

 

(13)

i.e. the components are invariant under the cyclic permutation of the indices. We claim that   εijk   is a   third order pseudotensor . Ind Indee eed, d, in a new ba basis sis ¯ e1 , ¯ e2 , ¯ e3  we have ei ¯ ¯ e j ) ¯ ek , ε¯ijk   = ( ¯

{

}

× ·

where the bar above the cross product sign indicates that we are taking the cross product  with respect to the new basis . Recalling (1), we have ε¯ijk  = (Qim em ¯ Q jn en ) Qkl el  =  Q im Q jn Qkl (em ¯ en ) el ,

×

·

× ·

where again we are taking the cross product in the new coordinate system. But we know that the cross products taken in the new and old systems coincide, (em ¯ en ) el  = (em en ) el  if the two bases have the same handedness (and thus   Q  is a proper orthogonal transformation, ∆ = 1) and have opposite signs, (em ¯ en ) el   = (em en ) el   if they have different handedness and thus the transformation is improper (∆ = 1). This means that

× · × · × ·  − ×

·

(em ¯ en ) el  = ∆(em

× ·

−

× e ) · e  = ∆ε n

l

and thus we have ε¯ijk  = ∆Qim Q jn Qkl εmnl ,

7

mnl ,

 

proving that   εijk  is a third order pseudotensor. Observe that if   c  =  a b, we can now write

×

ci  =  ε ijk a j bk

 

(14)

in any Cartesian coordinate system. Indeed, c1  =  ε 1 jk a j bk  =  ε 123a2 b3 +  ε132a3b2  =  a 2 b3

−a b , 3 2

where we have used the fact that by (12) the only nonzero components   ε1 jk are   ε123   = 1 and   ε132   = 1. Co Comp mpar arin ingg th this is to the fir first st equa equali litty in (5 (5)) with (i,j ,k) = (1, 2, 3), we see that this is indeed the first component of the cross product. Similarly, one can show that (14) yields   c2  =  a 3 b1 a1 b3   and c3   =   a1 b2 a2 b1 , which are the second and third components of the cross product according to the first equality in (5) taken with (i,j ,k) = (2, 3, 1) and (i,j,k ) = (3, 1, 2), respectively respectively.. φ  is a scalar, Recalling the properties of pseudotensors, one can see that if   φ then  ε ijk φ  are components of a third order pseudotensor, and if   T iijj  are components of a second order tensor, then  ε ijk T  jk  are compone components nts of a first order pseudotensor, or pseudovector (obtained by the double contraction of a fifth order pseudotensor pseudotensor). ). For examp example, le,   T  jk   =   a j bk   are components of a regular Cartensian tensor, ans so   ci   =   εijk T  jk   =   εijk a j bk  are components of a pseudovector dov ector.. If   T iijk jk   are components of a third order tensor, then   εijk T iijk jk   is a pseudoscalar   (meaning (meaning that it changes sign as we change from a right handed system sys tem to a lef leftt han handed ded or vic vicee ve versa rsa). ). For examp example, le,   T iijk jk   =   ai b j ck , we get the pseudoscalar

 −

−

 −

d  =  ε ijk T iijk jk   =  ε ijk ai b j ck   =  a (b

c).

· ×

Meanwhile,   εijk εmnl  is a sixth order regular Cartesian tensor, which satisfies the following identity: εijk εmnl

  =  

δ iim m   δ iin n   δ iill δ  jm   δ  jn   δ  jl δ km km   δ kn kn   δ k kll

 

.

In particular, expanding this determinant about the third row and setting l  =  k  (which means summing over   k ), we obtain the fourth order tensor εijk εmnk   =  δ iim m δ  jn

8

− δ 

 jm δ iin n.

 

(15)

 

Indeed, the expansion gives us ε

δ iim m   δ iin n

 δ 

ε ijk mnl

 =

kl

Setting   l  =  k , we get εijk εmnk  =  δ kkkk



δ  jm

 

δ iim m   δ iin n δ  jm   δ  jn

 

δ  jn

δ iim m   δ iill

δ  kn

−   −

δ kknn

 

δ  jm

 

δ  jl

δ iim m   δ iik k δ  jm   δ  jk

δ iinn   δ iill

 δ  km

+   +

 δ kkm m

 

δ  jn

 

δ  jl

δ iinn   δ iikk δ  jn   δ  jk

.

  

.

But   δ kkkk   =   δ 1111 +  δ 2222 +  δ 3333  = 1 + 1 + 1 = 3, and due to the Kronecker Deltas in front of the determinants in the second and third terms, we can simplify this further as

      − +  = 3       Observing that interchaging the columns in the last determinant changes its sign to the opposite, we get        = 2   = − εijk εmnk

δ iim m   δ iin n δ  jm   δ  jn

δ iim m   δ iin n δ  jm   δ  jn

δ in in   δ iim m δ  jn   δ  jm

εijk εmnk

δ iim m   δ iin n δ  jm   δ  jn

δ iim m   δ iin n δ  jm   δ  jn

δ iim m   δ iin n δ  jm   δ  jn

,

which yields (15). Contracting (15) further, we obtain the second order tensor εink εmnk   = 2δ iim m.

Indeed, setting   j   =  n  in (15) (and thus summing over   n), we get   εink εmnk   = im δ nn δ  2δ iim m.

nm δ in

− δ 

. But   δ nn  = 3 (see above), so we get   εink εmnk   = 3 δ im

In particular, we can use (15) to prove a

im

− δ 

 =

× (b × c) = b(a · c) − c(a · b).

× ×

Note that   a (b c) is a regular Cartesian vector (not a pseudovector) because we apply the cross product tw twice. ice. To prove the ident identity ity,, let   d   = a (b c). Then by (14)

× ×

di  =  ε ijk a j (b

× c)   = ε k

9

ijk a j εkmn bm cn

 

But by (13)   εkmn  =  ε mnk , so that

−

−

di  =  ε ijk εmnk a j bm cn  = (δ iim m δ   jn δ   jm δ in in )a j bm cn  =  δ iim m δ  jn a j bm cn δ   jm δ iin n a j bm cn ,

where we used (15) to get the first equality. Hence di  =  a n bi cn

−a

m bm ci

 =  b i (an cn )

− c (a

m bm

i

· − c (a · b).

) =  b i (a c)

i

Thus, a

proving the identity.

5

× (b × c) = d  =  b (a · c) − c(a · b),

Antisy tisymm mmet etri ric c seco second nd orde order r te tens nsor orss an and d pseudovectors

  Aij  be components of an antisymmetric second order tensor, i.e.   A ji   = Let A . The matrix representation of the tensor is

−

ij

[Aij

 0 ]= −

 

A12   A13   0   A23 A23   0

A12 A13

−  −

 

,

so tha thatt the there re are onl only y six nonze nonzero ro compone component nts. s. No Now w obs observ ervee tha thatt we can write Aij   =  ε ijk ak , or

 0  −  ]= −   0   −   0    a3

[Aij

a3

a2   a1

.

a2 a1 Indeed,   A23   =   ε231a1   =   a1 ,   A12   =   ε123a3   =   a3 ,   A31   =   ε312a2   =   a2 , while A32   =   ε321a1   = a1 ,   A21   =   ε213a3   = a3   and   A13   =   ε132a2   = a2 . The diagonal components are all zero because   εiik  = 0 (no sum). It follows that

 −

 −

ai  =

 −

 1 εijk A jk . 2

 

(16)

Indeed, A 1  =   12 (ε123A23 +ε132A32) =   12 (A23 A32), a 2  =   21 (ε231A31 +ε213A13) = 1 (A31 A13) and   a3  =   12 (ε312A12 +  ε321A21) =   21 (A12 A21). We write 2

−

−

a  =

 1 A  =  vecA. 2 ×

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−

 

Since   a   is obtained by a double contraction of a second order tensor and a third order pseudotensor, the result must be a pseudovector (recall the properties of pseudotensors). And indeed, under a change of variables, a ¯i  =

 1  1 ε¯ijk  ¯ A jk   = ∆Qim Q jn Qkl εmnl Q jp Qkq A pq . 2 2

But   Q jn Q jp  =  δ np np   and   Qkl Qkq   =  δ lq lq , so we have a ¯i  =

 1  1 ∆Qim εmnl δ np ∆Qim εmnl Anl   = ∆ Qim am , np δ lq lq A pq   = 2 2

since by (16)   εmnl Anl   = 2am . So   a, whose components are related to the components of the antisymmetric tensor   A  via (16), transforms as a pseudovector, or an axial vector.

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