Schwarzchild Metric Flashcards
Schwarzchild Metric
Description
-spherical symmetry - angular part of the metric:
r²(dθ² + sin²θ dφ²)
-not homogeneous - gravitational field varies with position so dt ≠ dτ
-use static spacetimes:
1) no time derivatives of gμν
2) the geometry should be the same if t -> -t
Schwarzchild Metric
Metric
ds² = goo c² dt² + grr dr² + r²(dθ² + sin²θ dφ²)
-with goo and grr functions of r only
-often written as:
ds² = - e^2α c²dt² + e^2β dr² + r²(dθ² + sin²θ dφ²)
-where α and β are functions of r only
Schwarzchild Metric
External
-for the space between stars => T^αβ = 0 and T = 0 -hence R00 = R11 = R22 = R33 = 0 -can use these conditions to find α and β
The Schwarzchild Radius
Definition
Rs = 2GM/c²
-Rs varies with the mass you have
The Schwarzchild Radius
Blackhole
-for a blackhole all of the mass, M, needs to be within the Schwarzchild radius
When does the Schwarzchild metric apply?
- around a static star with spherical symmetry
- for a blackhole
- in empty space
- with time independent (‘stationary’) motion, around a rotating star
Schwarzchild Metric
r > > Rs
-metric goes towards the weak field Schwarzchild metric
Schwarzchild Metric
Rs/r -> 0
- recover the flat space metric of special relativity
- spacetime far away from a star is described as being ‘asymptotically flat’
- this is also proper time: dτ -> dt(r=0)
Schwarzchild Metric
Near Rs
-clocks run slower in a gravitatoinal potential:
dt > dτ
Clocks Near Earth
- the clock at the pole experiences only the gravitational effect
- the clock on the equator sees the same gravitational effect but also experiences time dilation due to its motion
- the clock in orbit (geostationary) sees less of a gravitational effect because its further away but a larger effect due to its motion (satellite has to move faster than surface to maintain same relative position because its at a greater radius)
Constants of Motion
-from the Euler Lagrange Equations:
r²φ’ = const. = h
-conservation of angular momentum, where h is the angular momentum per unit mass
t² (1 - Rs/r) = const. = K
-conservation of energy where K is the energy per unit rest mass
Kepler’s Laws
-found from the Geodesic equation in r with dr=0 (circular motion) , θ’=0 and θ=π/2
=>
(dφ/dt)² = GM/r³
-this is true for circular obits even when r is not greater than Rs
Full Orbit Equation
Derivation
-use:
U^μ Uμ = -c² = gαβ U^α U^β = gαβ x^α’ x^β’
-with θ = π/2 and λ=τ for massive particles
-sub in t’ and φ’ from the conservation equations
-rearrange and multiply by 1/2
Full Orbit Equation
Word Equation
kinetic term + potential term = total energy
-a scaled energy equation per unit mass
Full Orbit Equation
Potential
V(r) = - 1/2 Rs/r c² + 1/2 h²/r² - 1/2 Rs h²/r³
= - GM/r + 1/2 h²/r² - GM/c² h²/r³
Full Orbit Equation
Newtonian Comparison
Vn(r) = 1/2 h²/r² - GM/r
= centrifugal barrier term + real potential
-so GR differs by the addition of a 1/r³ term
Schwarzchild Metric
Photons
- still have the conservation equations
- but now derivatives are with respect to λ not τ
Full Orbit Equation
Potential for Photons
V = 1/2 h²/r² - GM/c² h²/r³
-no Newtonian 1/r term for photons
Characteristics of Orbits
Circular Orbits
-circular orbit:
r’=0
AND
V’(r) = 0
Characteristics of Orbits
Solutions for Massive Particles
- case h < √3 Rs c - no solution so there are no circular orbits, at low angular momentum particles MUST fall towards the centre
- case h = √3 Rs c - minimum radius of circular orbit, inflection point so not stable
- case h > √3 Rs c - looking at the limit h > > √3 Rs c gives two solutions
- a maximum of V so unstable and a minimum which is potentially stable
- -it goes to increasingly large radii as h increases
Plotting Orbits
- in the units used, 2V/c² is dimensionless
- parameterise h in terms of Rs c, i.e. h = m Rs c with m a dimensionless number
Plotting Orbits
GR
massive particles
- potential barrier
- from unstable orbit:
- -could be knocked outwards and drift away
- -of fall in
- there is a stable orbit radius
photons
-only unstable orbits for photons
Plotting Orbits
Newtonian
massive particles
- infinite potential barrier, if you have any angular momentum at all you can never hit the central source
- all orbits are stable
photons
- also infinite potential barrier but no orbits
- photons can be deflected but will always continue on past
Radial Trajectories - Photons
Equation
-for radial trajectory photons; ds²=0 and dθ=dφ=0
=>
dr/dt = ±c [1 - Rs/r]
-with + for outward and - for inward photons
Radial Trajectories - Photons
Observers / Measuring Velocity
-r is not a proper length unless measured at r=∞, the same is true for t
-an observer experiences a time dτ and sees proper lengths dl so a measurable velocity for an observer is dl/dτ = c
-i.e. to a local observer light appears to still travel at c
-for a observer at r=∞ they see light as if its moving in a medium of refractive index:
1 / √[1 - Rs/r]
Radial Trajectories - Massive Particles
r < Rs
-massive particles are timelike: ds² < 0
-if r < Rs and dr²=0 then ds² > 0 -> contradiction
-so we must have dr² ≠ 0
=>
-no stationary observers for r < Rs - everything falls towards the centre
-OR the only way to have non-stationary observers within the event horizon is for them to be moving faster than c which is not possible
Radial Trajectories - Massive Particles
Observers / Measuring Velocity
-find dr/dτ by subbing in the constant of motion equation with K
-a fixed observer at radius r instead sees:
v = dl/dτ = …
-as r -> Rs, v -> c as it should at the event horizon
-if the observer is at ∞ instead they measure the rate of change of dl with respect to their clock dτ(r=∞)=dt which gives v -> 0 as r -> Rs
Gravitational Redshift
Overview
- imagine firing off a signal from a stationary emitter to a statoinary observer then doing the same again after a coordinate time difference Δt
- the observer receives these separated by the same coordinate time difference Δt, but they will have different clock times
Gravitational Redshift
Equation
νo/νe = Δτe/Δτo = √[1 - Rs/re] / √[1 - Rs/ro]
- if ro > re then νo < νe
- ir re -> Rs then νo -> 0
- can’t see anything ar r=Rs if we are at larger r
Gravitational Redshift
φ
-if ro, re > > Rs:
νo/νe ~ 1 - (Δφ/c²)
-if Δφ = φ(ro) - φ(re)
The Event Horizon
Photons
- consider a radially orbitting photon
- light cones at r > > Rs have gradient ±1 i.e. 45 degree cones
- they close up as r -> Rs (the gradient -> ∞)
- for r < Rs the sign of the gradient flips and as r -> 0 the gradient -> 0 and the cones are completely closed and looking at the singularity at r=0
The Event Horizon
Massive Particles
- starting at rest near r=∞ (so k²-1=0 since there is no kinetic or potential energy)
- the proper time that elapses in going from r1 -> r2 is finite regardless of where r2 is, even if r2=0
- a particle falling in from any finite distance, BY ITS OWN CLOCK will experience a finite time to reach the central singularity and no boundary exists at r=Rs
- an observer from outside sees the particle slowing down (as it approaches Rs) and never crossing the event horizon
