Geodesic Deviation and Spacetime CurvaturePreviously we talked about geodesics, the paths of freely falling particles. We alsoindicated early on that the only “force” that gravity can exert on a particle is a tidal force.This is also another way to characterize the curvature of spacetime. In flat spacetime, twoparticles infinitesimally close to each other that are initially moving freely along neighboringpaths will not deviate from each other. Similarly, two lines that are initially parallel in aplane remain parallel indefinitely. However, we know that tidal effects separate two particlesin a gravitational field, in the same way that in a two-dimensional surface with curvature(e.g., a sphere or a hyperboloid), the separation between two initially parallel paths willchange. It is this observation that initiated the study of non-Euclidean geometry in theearly 1800s. We can therefore use geodesic deviation to study curvature (see Figure 1 for apictorial representation of geodesic deviation on a sphere).For our purposes, we don’t need to go into details. Let’s just say that one can definea fourth-rank (!) tensor called the Riemann curvature tensor, which describes the way thatinitially nearby geodesics deviate from each other. This tensor can then be manipulatedto form a second-rank tensor called the Einstein curvature tensor, which in a mathematicalsense is divergence-free and symmetric. This is written G, or Gαβin contravariant (up-index)form or Gαβin covariant (down-index) component form. This tensor involves derivatives ofthe metric tensor and, unfortunately, is nonlinear in them (that is, it involves products of thegs), which means that in very strong gravity there are significant complications. The tensorG, which depends on the curvature of spacetime, is what must be related to the presence ofmatter or energy, which curves spacetime. To do this, we introduce the stress-energy tensor.The Stress-Energy TensorSo far we have concentrated on the motion of test particles in curved spacetime. But“matter tells space how to curve”, so we need a machine to quantify that as well. We’llexamine this by defining the stress-energy tensor and looking at its components, then doinga couple of examples of what the stress-energy tensor is in a particular circumstance.That machine is the stress-energy tensor, sometimes called the energy-momentum tensor.It is a symmetric second-rank tensor written T, or in component form Tαβ. At a givenlocation, the meaning of the components is as follows. Consider an observer with four-velocity uα. That observer will see a density of four-momentum (i.e., four-momentum perunit of three-dimensional volume), ofdpα/dV = −Tαβuβ. (1)This can also be thought of as inserting the four-velocity into one of the slots of the stress-energy tensor: T(u, . . .) = T(. . . , u). That means that the nαcomponent of the four-Fig. 1.— Because a sphere’s’ surface is curved, nearby geodesics deviate from each other. Fromhttp://upload.wikimedia.org/wikipedia/commons/thumb/2/24/Earth geo.gif/250px-Earth geo.gifmomentum density is n · dp/dV = −Tαβnαuβ. Inserting the four-velocity into both slotsgives the density of mass-energy measured in that Lorentz frame:T(u, u) = Tαβuαuβ. (2)Finally, suppose we pick a particular Lorentz frame and choose two spacelike basis vectors ejand ek. Then T(ej, ek) = Tjkis the j, k component of the stress as measured in that Lorentzframe. That is, Tjkis the j-component of force per unit area acting across a surface with anormal in the k direction, from xk− ² to xk+ ². Symmetrically, it is also the k-componentof force per unit area acting across a surface with a normal in the j direction, from xj− ² toxj+². This means that the diagonal components (j = k) are the components of the pressureas measured in that Lorentz frame, and the off-diagonal components are the shear stresses.Suppose we pick a particular observer with a particular Lorentz frame. Then what do thecomponents mean? Here we’ll use the notation (fairly widespread) that the “0” componentis the time component.T00= −T00= T00is the density of mass-energy measured in that frame. Tj0= T0jis thevolume density of the j-component of momentum, measured in that frame. Alternatively(and equivalently), T0kis the k-component of the energy flux. Finally, Tjkis as definedbefore, which can also be thought of as the k-component of flux of the j-component ofmomentum.Symmetry of the Stress-Energy TensorThe stress-energy tensor must be symmetric. For a clever partial proof of this (involvingthe space components only), we can use an argument also used in Newtonian theory. Thinkof a very small cube, with a mass-energy density T00and dimension L. The moment of inertiaof the cube is I ∼ ML2∼ T00L5. The torque on the cube from various stresses is simplythe sum of the forces times the lever arms. For example, the z component of the torque isNz= (−Tyx)L2(L/2) + (TyxL2)(−L/2) − (−TxyL2)(L/2) − (TxyL2)(−L/2) . (3)That is, “the torque is the y-component of force on the +x face times the lever arm to the+x face, plus the y-component of the force on the −x face times the lever arm to the −xface, minus the x-component of the force on the +y face times the lever arm to the +y faceminus the x-component of the force on the −y face times the lever arm to the −y face”.Summed, this givesNz= (Txy− Tyx)L3. (4)Ask class: what would this (plus the moment of inertia) imply for the angular velocityinduced as L −→ 0? The ratio N/I −→ ∞, so an infinitesimal cube would be set spinninginfinitely fast unless Txy= Tyx. Therefore, the spatial components are symmetric. So arethe space-time components.Example 1: Perfect FluidLet’s think about a perfect fluid. This is defined as a fluid that has no shear stresses(e.g., no viscosity). In the “rest frame” of the fluid, where fluid motions are isotropic, thereare therefore no off-diagonal components of the stress-energy tensor. Calling the mass-energydensity in this frame ρ, we know that T00= ρ. In addition, since fluid motions are isotropicin this frame, the pressure is the same for all three spacelike diagonal components; call itTxx= Tyy= Tzz= p. This gives the stress-energy tensor in that special frame, but whatabout in general? The key trick here is to write a general tensorial equation that is valid inthe special frame for which the calculation is easy. In this case, let the four-velocity of