## Lecture 12: The Stress Tensor and the Relativistic Stress-Energy Tensor

12. The Stress Tensor and the Relativistic Stress-Energy Tensor

Classical Stress Tensor

The classical stress tensor measures the internal forces that parts of a medium-such as a fluid or the interior of a star-exert on other parts (even though there may be zero net force at each point, as in the case of a fluid at equilibrium).

This is how you measure it: if S is an element of surface in the medium, then the material on each side of this interface is exerting a force on the other side. (In equilibrium, these forces will cancel out.) To measure it physically, pretend that all the material on one side is suddenly removed. Then the force that would be experienced is the force we are talking about. (It can go in either direction: for a liquid under pressure, it will push out, whereas for a stretched medium, it will tend to contract in.)

To make this more precise, we need to distinguish one side of the surface S from the other, and for this we replace S by a vector S = nS whose magnitude is S and whose direction is normal to the surface element (n is a unit normal). Then associated to that surface element there is a vector F representing the force exerted by the fluid behind the surface (on the side opposite the direction of the vector S) on the fluid on the other side of the interface.

Since we this force is clearly effected by the magnitude S, we use instead the force per unit area (the pressure) given by

 T(n) = limS0 FS .

Note that T is a function only of the direction n (as well as being a function of the point in space at which we are doing the slicing of the medium); specifying n at some point in turn specifies an interface (the surface normal to n at that point) and hence we can define T.

One last adjustment: why insist that n be a unit vector? If we replace n by an arbitrary vector v, still normal to S, we can still define T(v) by multiplying T(v/|v|) by |v|. Thus, for general v normal to S,

 T(v) = limS0 FS .|v|.

We now find that T has this rather interesting algebraic property: T operates on vector fields to give new vector fields. If is were a linear operator, it would therefore be a tensor, and we could define its coordinates by

Tab = T(eb)a,

the a-component of stress on the b-interface. In fact, we have

 Proposition 12.1 (Linearity and Symmetry) T is a symmetric tensor, called the stress tensor.

Sketch of Proof To show it's a tensor, we need to establish linearity. By definition, we already have

T(v) = T(v)

for any constant . Thus, all we need show is that if a, b and c are three vectors whose sum is zero, that

T(a) + T(b) + T(c) = 0.

Further, we can assume that the first two vectors are at right angles. Why? Since all three vectors are coplanar, we can think of the three forces above as stresses on the faces of a prism as shown in the figure. (Note that the vector c in the figure is meant to be at right angles to the bottom face, pointing downwards, and coplanar with a and b.)

If we take a prism that is much longer that it is thick, we can ignore the forces on the ends. It now follows from Pythagoras' theorem that the areas in this prism are proportional to the three vectors. Therefore, multiplying through by a constant reduces the equation to one about actual forces on the faces of the prism, with T(a) + T(b) + T(c) the resultant force (since the lengths of the vectors a, b and c are equal to the respective areas). If this force was not zero, then there would be a resultant force F on the prism, and hence an acceleration of its material. The trouble is, if we cut all the areas in half by scaling all linear dimensions down by a factor , then the areas scale down by a factor of 2, whereas the volume (and hence mass) scales down by a factor 3. In other words,

T(2a) + T(2b) + T(2c) = 2F

is the resultant force on the scaled version of the prism, whereas its mass is proportional to 3. Thus its acceleration is proportional to 1/ (using Newton's law). This means that, as becomes small (and hence the prism shrinks) the acceleration becomes infinite -- hardly a likely proposition.

The argument that the resulting tensor is symmetric follows by a similar argument applied to a square prism; the asymmetry results in a rotational force on the prism, and its angular acceleration would become infinite if this were not zero.

The Relativistic Stress-Energy Tensor

Now we would like to generalize the stress tensor to 4-dimensional space. First we set the scenario for our discussion:

We now work in a 4-manifold M whose metric has signature (1, 1, 1, -1).

We have already call such a manifold a locally Minkowskian 4-manifold. (All this means is that we be using different units for time in our inertial frames and MCRFs.)

Example 12.2
Let M be Minkowski space, where one unit of time is defined to be the time it takes light to travel one spacial unit. (For example, if units are measured in meters, then a unit of time would be approximately 0.000 000 003 3 seconds.) In these units, c = 1, so the metric does have this form.

The use of MCRFs allows us to define new physical scalar fields as follows: If we are, say, in the interior of a star (which we think of as a continuous fluid) we can measure the pressure at a point by hitching a ride on a small solid object moving with the fluid. Since this should be a smooth function, we consider the pressure, so measured, to be a scalar field. Mathematically, we are defining the field by specifying its value on MCRFs. Note that there is a question here about ambiguity: MCRFs are not unique except for the time direction: once we have specified the time direction, the other axes might be "spinning" about the path-it is hard to prescribe directions for the remaining axes in a convoluted twisting path. However, since we are using a small solid object, we can choose directions for the other axes at proper time 0, and then the "solid-ness" hypothesis guarantees (by definition of solid-ness!) that the other axes remain at right angles; that is, that we continue to have an MCRF after applying a time shear as in Lecture 11.

Now, we would like to measure a 4-space analogue of the force exerted across a plane, except this time, the only way we can divide 4-space is by using a hyperplane; the span of three vectors in some frame of reference. Thus, we seek a 4-dimensional analogue of the quantity nS. By coincidence, we just happen to have such a gizmo lying around: the Levi-Civita tensor. Namely, if a, b, and c are any three vectors in 4-space, then we can define an analogue of nS to be ijklaibkcl, where is the Levi-Civita tensor. (See the exercises.)

Next, we want to measure stress by generalizing the classical formula

 stress = T(n) = FS

for such a surface element. Hopefully, the space-coordinates of the stress will continue to measure force. The first step is to get rid of all mention of unit vectors -- they just dont arise in Minkowski space (recall that vectors can be time-like, space-like, or null...). We first rewrite the formula as

T(nS) = F,

the total force across the area element S. Now multiply both sides by a time coordinate increment:

T(nSx4) = Fx4 = p,

where p is the 3-momentum (classically, force is the time rate of change of momentum). This is fine for three of the dimensions. In other words,

 T(nV) = p, or T(n) = pV ...   (I)

where V is volume in Euclidean 4-space, and where we take the limit as V0.

But now, generalizing to 4-space is forced on us: first replace momentum by the 4-momentum P, and then, noting that nSx4 is a 3-volume element in 4-space (because it is a product of three coordinate invrements), replace it by the correct analogue for Minkowski space,

(V)i = ijklxjykzl,

getting

T(V) = P,

where P is 4-momentum exerted on the positive side of the 3-volume V ("positive" being given by the direction of V) by the opposite side. But, there is a catch: the quantity V has to be really small (in terms of coordinates) for this formula to be accurate. Thus, we rewrite the above formula in differential form:

T(dV) = T(ndV) = dP

This describes T as a function which converts the covariant vector dV into a contravariant field (P), and thus suggests a type (2, 0) tensor. To get an honest tensor, we must define T on arbitrary covariant vectors (not just those of the form V). However, every covariant vector Y* defines a 3-volume as follows.

Recall that a one-form at a point p is a linear real-valued function on the tangent space Tp at that point. If it is non-zero, then its kernel, which consists of all vectors which map to zero, is a three-dimensional subspace of Tp. This describes (locally) a (hyper-)surface. (In the special case that the one-form is the gradient of a scalar field , that surface coincides with the level surface of passing through p.) If we choose a basis {v, w, u} for this subspace of Tp, then we can recover the one-form at p (up to constant multiples) by forming ijklvjwkul. (Indeed, all you have to check is that the covariant vector ijklvjwkul has u, w, and v in its kernel. But that is immediate from the anti-symmetric properties of the Levi-Civita tensor.) This gives us the following formal definition of the tensor T at a point:

Definition 12.3 (The Stress Energy Tensor) For an arbitrary covariant vector Y at p, we choose a basis {v, w, u} for its kernel, scaled so that Yi = ijklvjwkul, and define T(Y) as follows: Form the parallelepiped V = {r1v + rrw + r3u | 0 ri 1} in the tangent space, and compute the total 4-momentum P exerted on the positive side of the volume element V on the positive side of this volume element by the negative side. Call this quantity P(1). More generally, define
 P() = total 4-momentum P exerted on the positive side of the (scaled) volume element 3V on the positive side of this volume element by the negative side.

Then define

 T(Y) = lim0 P()3 .

Note Of course, physical reality intervenes here: how do you measure momentum across volume elements in the tangent space? Well, you do all your measurements in a locally intertial frame. Proposition 9.6 then guarnatees that you get the same physical measurements near the origin regardless of the inertial frame you use (we are, after all, letting approach zero).

To evaluate its coordinates on an orthonormal (Lorentz) frame, we define

Tab = T(eb)a,

so that we can take u, w, and v to be the other three basis vectors. This permits us to use the simpler formula (I) to obtain the coordinates. Of interest to us is a more usable form -- in terms of quantities that can be measured. For this, we need to move into an MCRF, and look at an example.

Note It can be shown, by an argument similar to the one we used at the beginning of this section, that T is a symmetric tensor.

 Definition 12.4 Classically, a fluid has no viscosity if its stress tensor is diagonal in an MCFR (viscosity is a force parallel to the interfaces).

Thus, for a viscosity-free fluid, the top 33 portion of matrix should be diagonal in all MCRFs (independent of spacial axes). This forces it to be a constant multiple of the identity (since every vector is an eigenvector implies that all the eigenvalues are equalÉ). This single eigenvector measures the force at right-angles to the interface, and is called the pressure, p.

Question Why the pressure?

Answer Let us calculate T11 (in an MCRF). It is given by

 T11 = T(e1)1 = P1V ,

where the 4-momentum is obtained physically by suddenly removing all material on the positive side of the x1-axis, and then measuring 1-component of the 4-momentum at the origin. Since we are in an MCRF, we can use the SR 4-velocity formula:

 P = m0(v1, v2, v3, 1)/(1-v2/c2)1/2.

At the instant the material is removed, the velocity is zero in the MCRF, so

P(t=0) = m0(0, 0, 0, 1).

After an interval t in this frame, the 4-momentum changes to

P(t=1) = m0(v, 0, 0, 1)/ (1-(v)2/c2)1/2,

since there is no viscosity (we must take v2 = v3 = 0 or else we will get off-diagonal spatial terms in the stress tensor). Thus,

P = m0(v, 0, 0, 1) / (1-(v)2/c2)1/2.

This gives

 (P)1 = m0v(1-(v)2/c2)1/2 = mv (m is the apparent mass) = (mv) = Change of measured momentum

Thus,

 P1V = (mv)yzt = Fyz (force = rate of change of momentum)

and we interpret force per unit area as pressure.

What about the fourth coordinate? The 4th coordinate of the 4-momentum is the energy. A component of the form T4,1 measures energy-flow per unit time, per unit area, in the direction of the x1-axis. In a perfect fluid, we insist that, in addition to zero viscosity, we also have zero heat conduction. This forces all these off-diagonal terms to be zero as well. Finally, T44 measures energy per unit volume in the direction of the time-axis. This is the total energy density, . Think of is as the "energy being transferred from the past to the future."

This gives the stress-energy tensor in a comoving frame of the particle as

 p 0 0 0 0 p 0 0 0 0 p 0 0 0 0
.

What about other frames? To do this, all we need do is express T as a tensor whose coordinates in a the comoving frame happen to be as above. To help us, we recall from above that the coordinates of the 4-velocity in the particle's frame are

u = [0   0   0   1]      (just set v = 0 in the 4-velocity).

(It follows that

uaub=
 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

in this frame.) We can use that, together with the metric tensor,

g=  1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -1
,

to express T as

Tab = ( + p)uaub + pgab.

 Stress-Energy Tensor for Perfect Fluid The stress-energy tensor of a perfect fluid (no viscosity and no heat conduction) is given at a point m Ž M by Tab = ( + p)uaub + pgab, where: is the mass energy density of the fluid p is the pressure ui is its 4-velocity.

Note that the scalars in this definition are their physical magnitudes as measured in an MCRF.

Conservation Laws

Let us now go back to the general formulation of T (not necessarily in a perfect fluid), work in an MCRF, and calculate some covariant derivatives of T. Consider a little cube with each side of length l, oriented along the axes (in the MCRF). We saw above that T41 measures energy-flow per unit time, per unit area, in the direction of the x1-axis. Thus, the quantity

T41,1l

is the approximate increase of that quantity (per unit area per unit time). Thus, the increase of outflowing energy per unit time in the little cube is

T41,1(l)3

due to energy flow in the x1-direction. Adding the corresponding quantities for the other directions gives

 - Et = T41,1(l)3 + T42,2(l)3 + T43,3(l)3,

which is an expression of the law of conservation of energy. Since E is given by T44(l)3, and t = x4, we therefore get

- T44,4 (l)3 = (T41,1 + T42,2 + T43,3)(l)3,

giving

T41,1 + T42,2 + T43,3 + T44,4 = 0

A similar argument using each of the three components of momentum instead of energy now gives us the law of conservation of momentum (3 coordinates):

Ta1,1 + Ta2,2 + Ta3,3 + Ta4,4 = 0

for a = 1, 2, 3. Combining all of these and reverting to an arbitrary frame now gives us:

 Einstein's Conservation Law .T = 0 where .T is the contravariant vector given by (.T)j = Tjk|k.

This law combines both energy conservation and momentum conservation into a single elegant law.

Exercise Set 11

1. If a, b, and c are any three vector fields in locally Minkowskain 4-manifold, show that the field ijklaibkcl is orthogonal to a, b, and c.( is the Levi-Civita tensor.)

Last Updated: January, 2002