Classical Electromagnetism, Fall 2013 Stress-Tensor Notes

Physics 112: Classical Electromagnetism, Fall 2013 Stress-Tensor Notes

Please read through these notes. There are a few bulleted questions in the discussion which you should include in your solutions for next seminar. There are also a few questions at the end of these notes which you should solve.

1 Introduction to the stress-tensor

Press your hands together in the y − z plane and feel the force that one hand exerts on the other across a tiny area A– say, one square millimeter of your hands’ palms. That force, of course, is a vector F~ and has three independent components: it has a normal component (along the x- direction); it also has a tangential component: if you try to slide your hands past each other, you feel a component of force along the surface– a “shear” force in the y and z directions. The normal and tangential (i.e. shear) directions are the natural directions into which we decompose F~ .

Not only is the force F~ a vector, but we can associate a vector with the surface itself: A~ = Axˆ. Where the direction of the surface element points perpendicularly to the surface itself. In this case we have chosen the negative side of the surface to be the −x side and the positive side to be +x. With this choice, the force F~ is that which the negative hand, on the −x side, exerts on the positive hand. Now, I assert that the force F~ is a linear function of the surface A~. This is another way of saying that if I double the magnitude of A~ then the force F~ also doubles. This is clearly true if the force per unit area is a constant, and we assume that the area over which the force acts is small enough we can take the force per unit area as constant (what is another word for ‘force per unit area’?). It should seem obvious that two vectors which are linearly related to each other fall under the purview of linear algebra. In particular, in general there exists a matrix which relates them. I am going to work in index notation in these notes using the Einstein summation notation. In this notation the force Fi and the surface Ai are related by a matrix Tij:

Fi = TijAj. (1)

The matrix Tij is also known as a tensor– and in particular is the stress tensor. It is the set of coeﬃcients which takes a small surface and reports the force that is exerted across that surface.

1 Let us write out in words what Tij is: i−component of force per unit area Tij = (2) across a surface perpendicular toe ˆj  i−component of momentum that crosses a unit area  =  which is perpendicular toe ˆj per unit time,  (3) with the crossing pointing from − eˆj to +e ˆj Notice the fact that both of these descriptions are identical since force is just the rate of change of momentum. However, we will see that the second description will better describe the situation we have with the electromagnetic ﬁelds.

The stresses inside a table with a heavy weight on it are described by the stress tensor Tij, as are the stress in a flowing fluid or plasma, in the electromagnetic field, and in any other physical medium. Accordingly, we shall use the stress tensor as an important (really central) mathematical tool in our study of force balance in kinetic theory, elasticity theory, fluid mechanics, plasma physics, general relativity, electromagnetism, or any other field theory.

Figure 1: Two shear forces acting on a cube of length L.

It is not obvious from its deﬁnition, but the stress tensor Tij is always symmetric in its two indices. To see this, consider small cube with side L in any medium (or ﬁeld). The medium outside the cube exerts forces, and therefore also exerts torques, on the cube’s faces. The z-component of the torque is produced by the shear forces on the front and back faces and on the left and right– see Fig. 1. Let us use the stress-tensor in order to calculate the torque about the z-axis. In particular, the force acting on the negative y surface (with a normal pointing in the −yˆ-direction) is 2 Fi = −TiyL . (4) The torque caused by this force about thez ˆ-axis is then: L L3 N = − (ˆy × F~ ) → N = T . (5) z 2 z z 2 zyk ky

2 We can use the properties of the Levi-Civita symbol to show that the only component of the stress-tensor which contributes to this component of the torque is Txy: L3 N = T . (6) z 2 zyx xy

Finally, we know that xyz = +1 and that each subsequent permutation ﬂips the sign– therefore zyx = −1– so we have L3 N = − T . (7) z 2 xy The shear forces on the other three faces of the cube also contribute to the torque aboutz ˆ. • Work out the torque on the remaining three faces and show that the total torque about thez ˆ axis is 3 Nz = L (Tyx − Txy). (8)

Next note that the moment of inertia of a uniform cube about its center is I = (1/12)ρL5 and we see that the cube has an angular acceleration about thez ˆ-direction: 1 dΩ L3(T − T ) = ρL5 z . (9) yx xy 12 dt

Now imagine taking the size of the cube to zero– L → 0– if Tyx − Txy remains non-zero, then what happens to the angular acceleration? Using this result, answer the question:

• Why do we require Tyx = Txy? Convince yourself that the same argument works for the other faces of the cube so that we conclude the stress-tensor must be symmetric.

2 Examples of stress-tensors

An example will make the stress-tensor more concrete.

Inside a static perfect ﬂuid there is an isotropic pressure P , so Txx = Tyy = Tzz = P (the normal forces per unit area across surfaces in the y − z plane, the z − x plane and the x − y plane are all equal to P . Using index notation the stress-tensor takes the form

Tij = P δij. (10)

To check this result, consider a surface A~ = Anˆ with area A oriented perpendicular to some arbitrary unit vectorn ˆ. The force that the ﬂuid exerts across A~, in index notation, is

Fi = TikAk = P δikAnk = P Anj, (11) so we can see, as expected, it is a normal force with a magnitude equal to the ﬂuid pressure P times the surface area A. This is what it should be!

3 Momentum conservation and the stress-tensor

The stress tensor plays a central role in the Newtonian law of momentum conservation. Recall the physical interpretation of Tij as the i-component of momentum that crosses a unit area perpendicular

3 to the j-direction per unit time. Apply this definition to the little cube in Fig. 1– and now imagine that the cube is not a solid object, but is a volume into which and out of which momentum can flow. 2 The momentum that flows into the cube in unit time across the front face (at y = 0) is Tjy(0)L , 2 and across the back face (at y = L) is −Tjy(L)L ; and their sum is ∂T T (0)L2 − T (L)L2 ' jy L3. (12) jy jy ∂y

• Argue a similar relationship is true for the other faces and that the rate of change of momentum- density inside of the cube is given by ∂(momentum density) ∂p i + ∇ T = i + ∇ T = 0, (13) ∂t j ij ∂t j ij where ~p is the momentum density. This has the standard form for any local conservation law: the time derivative of the density of some quantity (here momentum), plus the divergence of the ﬂux of that quantity (here the momentum ﬂux is the stress-tensor), is zero.

4 So, what is the stress-tensor?

As we’ve seen in these notes, the stress tensor can be ‘defined’ in at least three, equivalent, ways (though we did not prove the equivalence explicitly!): I. The stress-tensor is a matrix which tells us the linear relationship between force acting on a ~ ~ surface, A, and the surface A itself: Fi = TijAj = TijAnˆj. II. The stress-tensor tells us the momentum flux that crosses a unit area, A~. III. The stress-tensor is a tensor whose negative divergence is equal to the time rate of change of momentum-density. It is important to realize that all three of these ‘definitions’ are equivalent and each may be useful in different circumstances.

5 Problems

You don’t have to hand in solutions, but you should sketch out the solutions for yourself! 1. Equations of motion for a perfect fluid (a) Consider a perfect fluid that moves with a velocity ~v that varies in space and time– ~v(x, y, z, t). Explain why the fluid’s momentum density is ρ~v, and explain why its momentum flux (i.e., stress- tensor) is Tij = P δij + ρvivj. (14)

(b) Explain why the law of mass conservation for this ﬂuid is ∂ρ + ∇~ · (ρ~v) = 0. (15) ∂t

4 (c) Explain why the total-time derivative d ∂ ≡ + ~v · ∇~ , (16) dt ∂t describes the rate of change as measured by an observer who moves locally with the fluid, i.e. with velocity ~v. This is called the fluid’s advective time derivative. (d) Use the advective time derivative to rewrite mass conservation as 1 dρ = −∇~ · ~v. (17) ρ dt In words this says that the divergence of the fluid’s velocity field is minus the fractional rate of change of its density, as measured in the fluid’s local rest frame. If the fluid is incompressible what is the divergence of the fluid’s velocity field? (e) Show that the law of momentum conservation for the fluid can be written d~v ρ = −∇~ P. (18) dt This is called the fluid’s Euler equation. Explain why this Euler equation is Newton’s second law of motion, F~ = m~a, written on a per unit volume basis. We also saw this equation in seminar 2 when we considered the conditions for hydrostatic equilibrium of a star. 2. Electromagnetic Stress-Tensor Now we get the the electromagnetic stress-tensor. In problem 1 we considered a fluid, and that is a bit easier for us to imagine. However, as we have seen throughout our discussion of electromagnetism, the electric and magnetic fields carry, energy, momentum, and exert pressure. (a) The Griffiths’ discussion just jumps in and solves the full problem. Lets go a bit slower here to make sure that we understand what is going on. In this part we will consider a static electric field and derive the stress-tensor associated with it. First, assume that all of the charges are free– i.e., we are not in a dielectric medium. The force density experienced by the charges is f~ = ρE.~ (19) We can use Gauss’ law to write ~ ~ ~ ~ f = 0(∇ · E)E. (20) Now, when we say ‘force-density’ we are really saying ‘rate of change of momentum-density’. There- fore this equation can be re-stated as ∂~p = (∇~ · E~ )E,~ (21) ∂t 0 where ~p is the momentum density. Looking at definition III of the stress-tensor, we want to find a tensor which satisfies the equation ~ ~ ∇iTij = −0Ej(∇ · E). (22)

• Argue that from this equation we know that the electromagnetic stress-tensor will be quadratic in the electric ﬁeld.

5 Lets proceed by guessing– there are only a few ways to write down the product of two electric ﬁeld components which is symmetric! Lets try the solution Tij = aEiEj, where a is a constant which we will determine: ∇iTij = a∇i(EiEj) = a(Ej∇iEi + Ei∇iEj). (23) Well, it didn’t work out.

2 Instead, lets try Tij = aEiEj + bδijE :

2 ∇iTij = a(Ej∇iEi + Ei∇iEj) + bδij∇iE . (24)

Using the fact that ∇~ × E~ = 0 in electrostatics we have (from vector identity 4):

2 ∇jE = 2Ei∇iEj, (25)

so that if we choose a = −0 and b = 0/2 then we have T static electric ﬁeld = 0 E2δ − E E . (26) ij 2 ij 0 i j

• Following the same procedure as before, use the equations of magnetostatics to show that the stress-tensor for a static magnetic ﬁeld takes the form

static magnetic ﬁeld 1 2 1 Tij = B δij − BiBj. (27) 2µ0 µ0

Note: Here we’ve derived the stress-tensor by considering the forces acting on the charge or current within the static electric or magnetic field– we’ve used definition III of a stress-tensor. As outlined in Griffiths’ Sec. 8.2.2, we can play the same game within electrodynamics and we have

0 2 1 2 1 Tij = E δij − 0EiEj + B δij − BiBj. (28) 2 2µ0 µ0

You will note that this is the negative of Griffiths’ Eq. (8.17)– this is because we have written our equation for the conservation of momentum in the same form as the continuity equation for charge conservation where in a charge and current free region ∂p i + ∇ T = 0, (29) ∂t j ij ~ ~ ~ and ~p = 0µ0S = 0(E × B). Compare this to Griffiths’ Eq. (8.4) and the sign of the divergence. • Note that we have derived the form for the stress-tensor using only electro and magneto statics– however, its agrees with Eq. 8.17 (up to a sign) which was derived in the context of electrodynamics! Although this may seem very odd, note that we have said nothing related to the Poynting vector. It is the time variation of the Poynting vector which encapsulates the effects of time varying electric and magnetic fields.

6 Generalizations

• The ﬂow of a scalar (like charge density ρ) is described through a vector (like current density J~).

6 • The flow of a vector (like momentum density P~ ) is described through a tensor (like the negative of the Maxwell stress tensor Tij ). What mathematical tool would one use to describe the flow of a tensor? I ask this question not to make your brain hurt, but to open your mind to more and richer possibilities. The tensor that weve discussed, namely the Maxwell stress tensor, is an example of a “rank-2 tensor”. In three dimensions, a rank-2 tensor can be described using 9 components, which are conveniently presented in a 3 × 3 matrix. The flow of a rank-2 tensor is described through a “rank-3 tensor”. In three dimensions, a rank-3 tensor can be described using 27 components, and theres no real convenient way to present one on flat paper. (Sometimes I use a stack of three index cards, on each of which I write a 3 × 3 matrix. But even this is not really effective.) From this point of view, a vector is a rank-1 tensor and a scalar is a rank-0 tensor. In general, a rank-r tensor in d dimensions is specified through dr components.