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Units, Constants and Useful Formulas

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Last modified: June 8, 2022 ✓

The Cauchy Stress Tensor ------------------------ Consider the stress, σ, inside a material. The corresponding stress tensor is: - - | σxx σxy σxz | σ = | σyx σyy σyz | | σzx σzy σzz | - - Stress is a vector defined as σ = F/A which is the same formula used to determine pressure. However, stress and pressure, though related, are different. Pressure is a scalar since it is not dependent on direction and is a force applied externally to an object. Pressure causes stress inside of the object, so stress is an internal force. The pressure is the average of the 3 normal stresses at the point, p = (1/3)Tr(σ). The off-diagonal components represent shear stresses. Stress-Energy-Momentum Tensor, Tμν --------------------------------- The stress–energy-momentum tensor is a tensor that describes the density and flux of energy and momentum in spacetime. It is an attribute of matter, radiation, and non-gravitational force fields. It is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity. The stress-energy-momentum tensor is symmetric. Therefore, Tμν = Tνμ. The covariant form of T is obtained from: Tμν = Tαβgαμgβν Continuity Equation ------------------- The continuity equation describes the transport of a conserved quantity. Continuity equations are the local form of conservation laws. Basically, the total amount (of the conserved quantity) inside any region can only change by the amount that passes in or out of the region through the boundary. A conserved quantity cannot increase or decrease, it can only move from place to place. First, consider the situation for charge. Divergence corresponds to the flow. An outward flow corresponds to a negative divergence. For a unit volume: jμ = (cρ,jx,jy,jz) - the 4-current density. ∂ρ/∂t + ∇.j = 0 ∂ρ/∂t + ∂jx/∂x + ∂jy/∂y + ∂jz/∂z = 0 Write ∂ρ/∂t as ∂j0/∂xo. Therefore, ∂jμ/∂xμ = 0 By analogy with the conservation of charge we can also specify an energy density and a momentum density and a flow of both in and out of a small region. The momentum 4-vector is: pμ = (E/c,p1,p2,p3). Momentum flow would be the momentum passing through a unit area per unit time which has the same units as pressure. Physically, one can think of a gas at constant pressure in a box. If a hole of unit area is opened in the side of the box, the pressure would be the amount of momentum escaping per unit time. Pressure = F/A where F = dp/dt = d(mv)/dt = ma. = ma/A [ma/A] = kg.(m/s2).(1/m2) = kg/ms2 = 1 pascal We can write a continuity equation for each row of Tμν: ∂T00/∂t + ∂T01/∂x + ∂T02/∂y + ∂T03/∂z = 0 or, ∂ρe/∂ct + ∇.Tti = 0 Therefore, energy is conserved. ∂T10/∂ct + ∂T11/∂x + ∂T12/∂y + ∂T13/∂z = 0 Therefore, x-momentum is conserved. Likewise for the y and z directions. In shorthand: DμTμν = 0 Where Dμ is the covariant derivative. Note: Conservation of energy-momentum only applies locally (at an infinitesimal point). In large regions there is no law. For example, redshifted light has a lower energy than the source. This energy is lost - not conserved. Tμν for a Perfect Fluid ----------------------- Homogenious and isotropic spacetime can be considered to be a perfect fluid. Perfect fluids have no shear stresses, viscosity, or heat conduction. The Stress-Energy tensor of a perfect fluid can be written in the form: Tμν = (p + ρe)UμUν + pgμν Where U is the 4-velocity vector field of the fluid. If the fluid is only changing with time and not moving spacially then U = (1,0,0,0). This gives: - - - - - -    | p + ρe 0 0 0 | | -1 0 0 0 | | ρe 0 0 0 | Tμν = | 0   0 0 0 | + p| 0 1 0 0 | = | 0  p 0 0 |    | 0   0 0 0 | | 0 0 1 0 | | 0  0 p 0 |    | 0   0 0 0 | | 0 0 0 1 | | 0  0 0 p | - - - - - - Where ρe = ρmc2 Tμν for Dust ------------ A collection of particles not exerting pressure on each other is often employed in cosmology as a model of a toy universe, in which the dust particles are considered as idealized models of galaxies, clusters, or superclusters. Tμν = ρeUμUν - -     | UtUt UtUx UtUy UtUz | Tμν = ρe| UxUt UxUx UxUy UxUz |     | UyUt UyUx UyUy UyUz |     | UzUt UzUx UzUy UzUz | - - If the dust particles are only changing with time and not moving spacially then U = (1,0,0,0). This gives: - -    | ρe 0 0 0 | Tμν = | 0  0 0 0 |    | 0  0 0 0 |    | 0  0 0 0 | - - Where again ρe = ρmc2