Physics Asked by KvanteKaffe on July 13, 2021
In page 168 in Ref. [1], the authors search for a suitable order parameter for the nematic phase in liquid crystal. If $vec v^alpha$ is the direction of a single molecule, than due to the inversion symmetry, both $vec v^alpha$ and $-vec v^alpha$ contribute to order so the order parameter must be even in $vec v^alpha$. In addition, they want the order parameter to be $0$ in the disordered state. Since, I quote,
a symmetric traceless tensor will yield zero when averaged over all directions,
they define the following tensor
$$Q_{ij} = frac{V}{N} sum_{alpha} (v_i^alpha v_j^alpha – frac{1}{3} delta_{ij}) delta(vec x -vec x_{alpha}) .$$
Now, I understant that this tensor is symmetric and traceless. I can also insert $v_i^alpha = [cos phi sin theta, sin phi sin theta, cos theta]$ and see that the expression above is zero when averaged over all directions (but it does seem like magic).
So my question is, how general is the statement in the quote above?
For instance, in general, what does it mean to average a tensor over all directions? Does the quote require that the tensor has been defined in terms of quantity that has a direction in space (which is then the subject of averaging)? If this is the correct interpretation, is there an easy answer to why symmetric traceless tensors are zero when averaged over all directions?
[1] Chaikin, P. M., & Lubensky, T. C. (1995). Principles of condensed matter physics. Cambridge University Press.
The order parameter is a meso- or macroscopic quantity which is formed by adding together the tensor contributions from a bunch of different molecules: if each molecule $alpha$ has a tensor $T^{(alpha)}$, then the total order parameter is the sum of the $T^{(alpha)}$ from all the molecules in the relevant sample of material, i.e., $$ T = sum_alpha T^{(alpha)}, $$ or, when seen as a bilinear function with explicit arguments, $$ T(u,v) = sum_alpha T^{(alpha)}(u,v), $$ for arbitrary $u,vinmathbb R^3$.
The book's claim is that they want $T$ to be zero in the disordered state, in which the $T^{(alpha)}$ themselves are pointing in every direction: that is,
Here the integral over $mathrm{SO}(3)$ must be taken over the Haar measure of the group, which is the only definition of integration which is invariant w.r.t. the group action.
You might also be wondering what the rotated tensor "$R[T^{(0)}](u,v)$" means. In the abstract, it means that you rotate the tensor, but in practice, you can flip that "active transformation" view on its head into a "passive" one, by simply rotating the whole system ($R[T^{(0)}]$, $u$ and $v$) back to the original orientation, which gives you $$R[T^{(0)}](u,v) = T^{(0)}(R^{-1}u,R^{-1}v)$$ as a concrete recipe for evaluating $R[T^{(0)}]$.
The central claim here is the following:
Claim: this orientation averaging reduces the tensor to a multiple of the isotropic tensor $delta(u,v) = ucdot v$, with components proportional to the Kronecker delta $delta(e_i,e_j) = delta_{ij}$. Moreover, the coefficient is proportional to the trace of the reference tensor:
begin{align} T(u,v) & = int_mathrm{SO(3)} R[T^{(0)}](u,v) :mathrm dR & = frac13 mathrm{tr}(T^{(0)}) delta(u,v) . end{align}
(As pointed out by Valter Moretti, this is basically a whittled-down version of Schur's lemma, a central result in group representation theory which is well worth learning about if you feel like digging deeper.)
There are two main ways to understand this result: there is a fundamental core intuition that makes it work, but you can also prove it rigorously.
Intuition
The core of the result is that the orientation averaging means that $T(u,v)$ must be an isotropic tensor, i.e., it must be such that $$ T(Ru,Rv) = T(u,v) $$ for every $Rinmathrm{SO}(3)$. This is an extremely strong constraint: up to a constant multiple, there is only one isotropic tensor of rank two $-$ the Kronecker delta. This implies that $T(u,v) = lambda delta(u,v)$, and you just need to determine $lambda$. This is extremely simple to do, by just taking the trace: $$ mathrm{tr}(T) = lambda : mathrm{tr}(delta) = lambda sum_{i=1}^3 delta(e_i,e_i) = 3lambda, $$ so $lambda = frac 13 mathrm{tr}(T)$.
Proof
To complete the rigorous proof, thus, we only need to show that the Kronecker delta is the only isotropic tensor. As in V. Moretti's answer, the simplest way to do this is to consider the action of $T$ on a basis ${ e_i}$.
This completes the proof: $T(u,v)$ and $lambda:delta(u,v) = T(e_1,e_1)delta(u,v)$ coincide on a basis, so they must coincide everywhere.
Correct answer by Emilio Pisanty on July 13, 2021
Symmetry requirement is not necessary. Let us take an orthonormal basis $e_1,e_2,e_3$ and consider all the unit vectors $n in S^2$. $$int_{S^2} T(n,n) d n = sum_{i,j=1}^3T(e_i,e_j) int_{S^2} n^i n^j dn:.$$ $dn$ is the standard rotation-invariant measure on $S^2$ with total value $4pi$, i.e., referring to standard spherical coordinates $$dn = sin theta dtheta dphi:.$$ Now, for every choice of $i=1,2,3$ and $j=1,2,3$ with $ineq j$ there is a rotation $R$ such that $(Rn)^i=-n^i$ but $(Rn)^j = n^j$. Since $dn = dRn$ we immediately have that $$int_{S^2} n^i n^j dn= -int_{S^2} n^i n^j dn= 0 $$ if $ineq j$. Therefore begin{align} sum_{i,j=1}^3T(e_i,e_j) int_{S^2} n^i n^i dn & = sum_{i=1}^3T(e_i,e_i) int_{S^2} n^i n^i dn & = T(e_1,e_1)int_{S^2} (n^1)^2 dn + T(e_2,e_2)int_{S^2} (n^2)^2 dn & qquad quad + T(e_3,e_3)int_{S^2} (n^3)^2 dn:. end{align} To conclude, observe that rotational invariance (as above) $$int_{S^2} n^i n^i dn = d quad mbox{independently of $i=1,2,3$}$$ and thus $$4pi = int_{S^2} sum_{i=1}^3 (n^i)^2 dn = 3d$$ which implies $$d = frac{4pi}{3}:.$$ Inserting in the formula above begin{align} int_{S^2} T(n,n) d n & = sum_{i,j=1}^3T(e_i,e_j) int_{S^2} n^i n^j dn frac{4pi}{3} sum_{i=1}^3 T(e_i,e_i) & = frac{4pi}{3} mathrm{tr}(T):. end{align} In other words, averaging over all directions we obtain the trace up to the factor $1/3$: $$langle Trangle_{S^2} := frac{1}{4pi}int_{S^2} T(n,n) d n = frac{1}{3} mathrm{tr}(T):.$$ Your hypothesis also requires $mathrm{tr}(T)=0$ concluding the proof.
Answered by Valter Moretti on July 13, 2021
OK, another proof according to the Remark by Emilio that made clear the interpretation of the question. We have to average the tensor (as a bilinear map) in the space of the rotations. The only way to do it, as Emilio wrote, is to take an integral with respect to the normalized (left-invariant) Haar measure of $SO(3)$: $$langle Trangle_{SO(3)}(u,v) := int_{SO(3)} T(R^tu,R^tv) dR$$ Since $dB^tR = dR$ for every $Bin SO(3)$, $$langle Trangle_{SO(3)}(Bu,Bv) = int_{SO(3)} T(R^tBu,R^tBv) dR = int_{SO(3)} T((B^tR)^tu,(B^tR)^tv) dR $$ $$ = int_{SO(3)} T((B^tR)^tu,(B^tR)^tv) dB^{t}R = int_{SO(3)} T(R'^tu,R'^tv) dR' = langle Trangle_{SO(3)}(u,v):.$$ In matricial form, writing $R$ for $B$, $$u^tR langle Trangle_{SO(3)} R^tv = u^tlangle Trangle_{SO(3)}v:. $$ Since $u,v$ are arbitrary, $$R langle Trangle_{SO(3)} R^t = langle Trangle_{SO(3)}:. $$ That is $$R langle Trangle_{SO(3)}= langle Trangle_{SO(3)}R quad forall R in SO(3):.$$ Since the fundamental representation of $SO(3)$ is irreducible, Schur's lemma implies that, for some constant, $tin mathbb{R}$ $$ langle Trangle_{SO(3)} = tI:.$$ In particular, since every unit vector can be transformed to any other unit vector with a rotation, an orthonormal a basis is transformed into an orthonormal basis, and the trace does not depend on the basis, $$3t = sum_{i=1}^3langle Trangle_{SO(3)}(e_i,e_i)$$ $$ = trlangle Trangle_{SO(3)} = int_{SO(3)} sum_{i=1}^3T(R^te_i,R^te_i) dR = int_{SO(3)} tr(T) dR = tr(T):.$$ In summary $$ (langle Trangle_{SO(3)})_{ab} = frac{1}{3}tr(T)delta_{ab}:.$$
Answered by Valter Moretti on July 13, 2021
The author here maybe concentrate on how to construct a tensor by a vector, so the average over direction means the average over direction of vector,namely $$langle T_{ij}rangle=frac{1}{4pi}int T_{ij}dOmega $$ where $dOmega$ is the solid angle
The argument
a symmetric traceless tensor will yield zero when averaged over all directions
has some background on group theory -- the representation of SO(3) group. We could consider two vector $U,V$ and their tensor product $U_iV_j$ the problem is $U_iV_j$ could be divided into several part obeying different rules under rotation (that is to say, this is a reducible representation)
$$U_iV_j=frac{1}{3}Ucdot Vdelta_{ij}+frac{1}{2}(U_iV_j-U_jV_i)+(frac{U_iV_j-U_jV_i}{2}-frac{Ucdot V}{3}delta_{ij})$$
where the number of independent variable is $3times3=1+3+5$
the first part is a scalar-- it's invariant under rotation. The second part is $(Utimes V)_kepsilon_{ijk}$ which transform like vector. In fact, the tensor is divided into terms called spherical tensor. The last part ,not that simple, tranform as l=2 sphere harmonic function. From the orthogonality of nonequivalent irreducible representation you could find that the integral (interpreted as the inner product of the l=2 with l=0 component) is zero.
Spherical tensor will be useful if you want to discuss items concerning angular momentum ,though they seems do little thing with what you ask. You may refer to books like Modern Quantum Mechanics J.J Sakurai if you're interested in this topic.
Answered by Black Monolith on July 13, 2021
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