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Non-Equilibrium Continuum Physics TA session
TA: Michael Aldam 28.04.
Index Gymnastics: Gauss’ Theorem, Isotropic Tensors, NS Equations
The purpose of today’s TA session is to mess a bit with tensors and indices, which are a necessary tool for continuum theories and in particular for Solid Mechanics. We’ll see some simple examples and try to become comfortable with these mathematical tools. If time permits we’ll discuss the subject of dimensional analysis which, although very basic, is sometimes not understood well enough.
1 Isotropic tensors
A tensor is called isotropic if its coordinate representation is independent under coordi- nate rotation. Let’s look at all the possible forms of isotropic tensors of low ranks.
1.0 0 th^ rank tensors
A 0th^ rank tensor, a.k.a a scalar, does not change under rotations, therefore all scalars are isotropic (surprise!).
1.1 1 st^ rank tensors
A vector ~v is isotropic if for every rotation matrix Rij we have
Rij vj = vi. (1)
You can easily show that this condition is satisfied for arbitrary R only if ~v = 0. So the zero vector is the only isotropic vector (surprise #2!!).
1.2 2 nd^ rank tensors
Let’s hope we’re gonna get something a bit more interesting. A matrix A is isotropic if for every rotation matrix R we have Aij = RikRjlAkl, or in matrix notation:
RART^ = A. (2)
Let’s choose a specific rotation matrix, say a rotation of angle α around ˆz,
Rz^ (α) ≡
cos α sin α 0 − sin α cos α 0 0 0 1
The invariance equation now takes the form
A(0) = A(α) ≡ Rz^ (α)A Rz^ (α)T^ =
cos α sin α 0 − sin α cos α 0 0 0 1
A
cos α − sin α 0 sin α cos α 0 0 0 1
This is a complicated equation, with cosines and sines all mixed up in a very unpleas- ant manner. Luckily, we can find an equivalent condition that is significantly simpler. Differentiating with respect to α and plugging α = 0 gives
∂A(0)
∂α
∂A(α) ∂α
α=
∂Rz^ (α) ∂α
α=
A Rz^ (0) + Rz^ (0) A
∂Rz^ (α)T ∂α
α=
but since Rz^ (0) is the identity matrix, this reduces to the simple equation
≡Lz
A + A
Lz^ = ∂αR|α=0 is sometimes called “the generator of rotations around the z axis”, because Rz^ (α) = eαL z
. We see that equation A(0) = A(α) is equivalent to the much easier equation (notice the sign change)
A(0) = A(α) ⇐⇒ [A, Lz^ ] = 0. (7)
Explicitly calculating [A, Lz^ ] gives
[A, Lz^ ] =
−A 12 − A 21 A 11 − A 22 −A 23
A 11 − A 22 A 12 + A 21 A 13
−A 32 A 31 0
We see that commutation with Lz^ requires (a) A 13 = A 31 = A 23 = A 32 = 0 and (b) A 11 = A 22. Obviously, the choice of ˆz is arbitrary and isotropy means that A should also commute with Lx^ and Ly. If we repeat the above procedure for the other L’s, the analog of (a) will be that all off-diagonal elements must vanish, and the analog of (b) will be that all diagonal elements must be equal. That is,
Aij ∝ δij. (9)
I stress that this is true only in dimensions ≥ 3. In the HW you’ll see that in 2D there are isotropic tensors that are not proportional to the identity (can you already see how the above argument fails in 2D?).
1.3 3 rd^ rank tensors
Here we can use the same trick. A 3rd^ rank tensor A is isotropic iff for every rotation matrix Rij we have Riα Rjβ Rkγ Aαβγ = Aijk. (10)
You can imagine the mess that comes out of this if you plug in a real rotation matrix with sines and cosines and whatnot, and then start using trig identities. Phew, no thanks! So like before, we choose R = Rz^ (α), differentiate, and set α = 0. This gives
0 =
Lziα δjβ δkγ + δiα Lzjβ δkγ + δiα δjβ Lzkγ
Aαβγ
= Lziα Aαjk + Lzjβ Aiβk + Lzkγ Aijγ.
equation for ∂t~v as a function of ~v and its spatial derivatives. We take a perturbative approach, and expand ∂tv to second order in ~v and in its gradients:
∂tvi = Aij vj + Bijk ∂j vk + Dijkl vj ∂kvl + Eijkl ∂j ∂kvl + Fijk vj vk. (18)
Since ~v is a physical quantity (specifically, a 1st^ rank tensor), the dynamical equation for ∂t~v should be covariant under symmetries of the physical system in question. We’ll see what these symmetries impose on the form of the various tensors A, B, C, D, F. We begin with a Galilean transformation:
yi = xi − cit , (19) τ = t. (20)
Under this transformation, the velocity field now takes the form w~ = ~v − ~c. Also, by the chain rule:
∂xi =
∂yj ∂xi
∂yj +
∂τ ∂xi
∂τ = ∂yi , (21)
∂t =
∂τ ∂t
∂τ +
∂yj ∂t
∂yj = ∂τ − cj ∂yj. (22)
Applying this to Eq. (18) gives
∂twi − cj ∂j wi = Aij (wj + cj ) + Bijk∂j wk + Dijkl(wj + cj )∂kwl
- Eijkl∂j ∂kwl + Fijk(wj + cj )(wk + ck).
If we want the NS equation to be covariant, we need to impose that Eq. (23) will be equal, term by term, to Eq. (18), i.e.
Aij cj = 0 , (24) −cj ∂j wi = Dijkl cj ∂kwl, (25) Fijk(wj + cj )(wk + ck) = Fijkwj wk. (26)
All these should hold for arbitrary ~c, ~w. The first constraint clearly means A = 0. For the third one, choose for example w~ = −~c, and get that Fijkwj wk = 0 for arbitrary w~. Note that this is exactly the last term in Eq. (18), so we see that it vanishes identically. The constraint Eq. (25) may be written as
−δilδjk cj ∂kwl = Dijkl cj ∂kwl.
Since ~c, ~w are arbitrary, Dijkl = −δilδkj , and Eq. (18) can be written as
∂tvi + vj ∂j vi = Bijk∂j vk + Eijkl∂j ∂kvl. (27)
You have to admit that this is a very big improvement... Now let’s look at rotations yj = Rij xj. Demanding Eq. (27) to be invariant means that the tensors B, E are isotropic.
We’ve just seen that the only 3rd^ rank isotropic tensor is the Levi-Civita tensor, so the B term is proportional to ∇ ×~ ~v and thus is forbidden by reflection symmetry. It’s too bad that we know already that the A and F terms are gone, because they would also be forbidden by rotational symmetry. For example, the F term must be proportional to ~v × ~v and therefore vanishes identically (note that we didn’t show that F = 0, but only that it gives zero when it acts on the same vector in its two slots). As for E, we know that we have exactly three choices, given in Eqs. (14),(15),(16). These give, respectively,
δij δkl ∂j ∂k vl = ∂i∂j vj = ∇~ (∇ · ~v) = grad (div ~v) , (28) δil δjk ∂j ∂k vl = ∂j ∂j vi = ∇^2 ~v = div (grad ~v) , (29) δik δjl ∂j ∂k vl = ∂i∂j vj = same as Eq. (28) , (30)
so the third option is redundant. Note that if we wanted to use Eq. (17) we’d get
Eijα Eαkl∂j ∂k vl = Eijα ∂j
∇ ×~ ~v
α
= ∇ ×~
∇ ×~ ~v
which is also redundant because of the vector calculus identity which you all know by heart: ∇ × (∇ × A) = ∇(∇ · A) − ∇^2 A. To sum up, we see that the only form of ∂t~v which is covariant under rotations, reflections and Galilean symmetries is
( ∂t + ~v · ∇~
~v = η∇^2 ~v + μ∇~ (∇ · ~v) , (31)
where η and μ are two scalars. In incompressible flows, (∇ · ~v), there’s only one η, as the μ term vanishes. Lastly, note that there’s another term that clearly does not violate any symmetries: ∇^ ~P where P is some scalar function.
2.1 An historical note about the power of symmetries in con-
tinuum theories
Euler’s equation (∂t + vj ∂j )vi = ∂iP , regarding inviscid incompressible flows, was derived sometime around 1750. It took the scientific community almost 80 years (!!) to under- stand how to incorporate viscosity into the business. Mind you, some of the greatest minds of the time were devoted to the problem, including Cauchy, Poisson, d’Alembert, Bernoulli, and of course, Navier and Stokes. Not exactly Elitzur Ra’anana, if you see what I mean. So what took them so long? The answer, very very roughly, is that they tried to model viscosity on a molecular level: to understand the dissipation mechanisms, stress-transfer mechanisms, and what- not. One of the great strengths of continuum theory is that measly insignificant mortals like us were able to do here in 45 minutes a derivation that the primordial gods needed 80 years to do. Moreover, we did that without caring even the slightest bit about the underlying physics. In fact, this is the crux of the matter – the use of symmetries allows us to state very powerful statements about the functional form of the viscosity term, without having to
√g ` τ^. This means that this combination has to be constant, therefore
τ = c
`
g
which is a remarkable result, considering that we know almost nothing about the system and no equations were solved. Fully formulating and solving the problem would of course tell us that c = 2π, but consider that even if we would not know how to do so, we can measure just one pendulum, find the value of c, and know it happily ever after for any other pendelum. What happens if we relax the requirement of small deflections? Then we introduce a new parameter, α, the initial angle. This means that we now have two dimensionless combinations, so we have one function of one parameter
√ g `
τ = f (α) ⇒ τ =
`
g
f (α). (34)
We actually know f , which is an elliptical integral, but it’s beyond the point right now. Consider the power of this technique. All you have to do is to measure enough values of the function f for one pendulum, and again you know enough about all pendelums in the world. As a final example I’ll use an historical anecdote. During the early days of atomic testing, the American government published a series of photos from a nuclear test, but kept the details of the explosion classified. British physicist Taylor looked at the images and realised that it gave him enough data to calculate the amount of energy released in the blast!
Figure 1: The experimental points determined by Taylor from the movie film lay on a single straight line with slope unity in the coordinates log t, (5/2) log rf. Taylor was thus able to determine the energy of the explosion from the series of photographs.
Let’s get to it. The quantities we need to consider are the radius at a specific time r, the time since detonation t, the energy released E, and the initial air density ρ. Looking at all the possible combinations Π = rργ^ Eβ^ tα^ we find that the only possible dimensionless combination is
α = −
, β = −
, γ =
ρ^1 /^5 r E^1 /^5 t^2 /^5
which as before, being the only dimensionless quantity, has to be constant, so
r =
cE^1 /^5 ρ^1 /^5
t^2 /^5. (36)
Thus, in a Log Log plot, we should expect to see a straight line, as can be seen in Fig. 1. From the y axis intersection a value for cE^1 /^5 ρ−^1 /^5 can be found, and knowing ρ and assuming c to be unity^3 Taylor got quite a good estimate for E. At the time, Taylor’s publication of this value (which turned out to be approximately 1021 erg) caused, in his words, “much embarrassment” in American government circles: this figure was considered top secret, even though the film was not classified.
(^3) A similar problem in gas dynamics has shown c to be about unity.
