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UMD ASTR 680 - Orthonormal Tetrad

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Orthonormal Tetrads, continuedHere’s another example, that combines local frame calculations with more globalanalysis. Suppose you have a particle at rest at infinity, and you drop it radially into aSchwarzschild black hole. What is the radial velocity as seen by a local static observer atradius r? The particle being at rest at infinity means that its total energy is ut= −1.Radial motion has θ and φ components zero, so u2= −1 meansurur+ utut= −1grr(ur)2+ gttu2t= −1(ur)2/(1 − 2M/r) − 1/(1 − 2M/r) = −1(ur)2= 2M/r .(1)Therefore, ur= dr/dτ = 2M/r. The radial velocity seen by a local static observer isvˆr= uˆr/uˆt= −uˆr/uˆt= −eˆrrur/(etˆtut)= −(1 − 2M/r)−1/2ur/[(1 − 2M/r)−1/2ut]=q2M/r .(2)Therefore, the locally measured radial velocity is just the same as the Newtonian expression,when Schwarzschild coordinates are used. By comparison, the radial velocity as measuredat infinity isvr=drdt= ur/ut= ur/(gttut) = (1 − 2M/r)ur= (1 − 2M/r)q2M/r . (3)This drops to zero at the horizon. Note that there is one factor of (1 − 2M/r)1/2from theredshift and one from the change in the radial coordinate.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.Here we introduce one more bit of terminology: D/dλ = ∇u, that is, D/dλ isthe directional derivative along the four-velocity. In component notation, for example,DAα/dλ = Aα;µuµ. Then, consider two nearby geodesics with separation vector n. The“equation of geodesic deviation”, which encodes tidal effects in general relativity, is (incomponent notation)D2nαdλ2+ Rαβγδuβnγuδ= 0 . (4)This equation defines the “Riemann curvature tensor”, which can also be written asRαβγδ= Γαβδ,γ− Γαβγ,δ+ ΓαµγΓµβδ− ΓαµδΓµβγ. (5)Here the Γs are the connection coefficients, as defined before. Ask class: what are thecomponents of R in flat spacetime? Since Γαβγ= 0 in flat spacetime, Rαβγδ= 0.From the Riemann curvature tensor one can form other tensors by contraction. Forexample, the Ricci curvature tensor is Rµν= Rαµανand the scalar curvature is R = Rµµ.Incidentally, the scalar curvature measures the local presence of matter. That is, in avacuum, even if that vacuum is just outside a star, R = 0. There is another contraction ofthe Riemann tensor (the Weyl tensor) that encodes “background” curvature.The most important second-rank tensor one can form from the Riemann tensor is theEinstein curvature tensorGµν= Rµν−12δµνR , (6)where δ is the Kronecker delta. The Einstein curvature tensor has deep geometricalsignificance; in particular, it satisfies the “contracted Bianchi identities”Gµν;ν= 0 . (7)Stated in coordinate-free form, since this is the divergence of G, ∇ · G = 0, it says that G isa divergence-free tensor. It is also symmetric (Gµν= Gνµ). These two properties are linkedto the utility of G in the Einstein field equation, which links the curvature of space to thepresence of matter. To understand this better we have to find a way to express the amountof matter and stresses in tensorial form.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-momentumtensor. It is a symmetric second-rank tensor written T, or in component form Tαβ. Ata given location, the meaning of the components is as follows. Consider an observer withfour-velocity uα. That observer will see a density of four-momentum (i.e., four-momentumper unit of three-dimensional volume), ofdpα/dV = −Tαβuβ. (8)This can also be thought of as inserting the four-velocity into one of the slots of thestress-energy tensor: T(u, . . .) = T(. . . , u). That means that the nαcomponent of thefour-momentum density is n · dp/dV = −Tαβnαuβ. Inserting the four-velocity into bothslots gives the density of mass-energy measured in that Lorentz frame:T(u, u) = Tαβuαuβ. (9)Finally, suppose we pick a particular Lorentz frame and choose two spacelike basis vectorsejand ek. Then T(ej, ek) = Tjkis the j, k component of the stress as measured in thatLorentz frame. That is, Tjkis the j-component of force per unit area acting across asurface with a normal in the k direction, from xk− ² to xk+ ². Symmetrically, it is also thek-component of force per unit area acting across a surface with a normal in the j direction,from xj− ² to xj+ ². This means that the diagonal components (j = k) are the componentsof the pressure as measured in that Lorentz frame, and the off-diagonal components are theshear stresses.Suppose we pick a particular observer with a particular Lorentz frame. Then whatdo the components mean? Here we’ll use the notation (fairly widespread) that the “0”component is 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


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