Schwarzchild Metric

Question 1

Schwarzchild Metric

Description

Answer 1

-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

Question 2

Schwarzchild Metric

Metric

Answer 2

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

Question 3

Schwarzchild Metric

External

Answer 3

-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

Question 4

The Schwarzchild Radius

Definition

Answer 4

Rs = 2GM/c²

-Rs varies with the mass you have

Question 5

The Schwarzchild Radius

Blackhole

Answer 5

-for a blackhole all of the mass, M, needs to be within the Schwarzchild radius

Question 6

When does the Schwarzchild metric apply?

Answer 6

• around a static star with spherical symmetry
• for a blackhole
• in empty space
• with time independent (‘stationary’) motion, around a rotating star

Question 7

Schwarzchild Metric

r > > Rs

Answer 7

-metric goes towards the weak field Schwarzchild metric

Question 8

Schwarzchild Metric

Rs/r -> 0

Answer 8

• 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)

Question 9

Schwarzchild Metric

Near Rs

Answer 9

-clocks run slower in a gravitatoinal potential:

dt > dτ

Question 10

Clocks Near Earth

Answer 10

• 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)

Question 11

Constants of Motion

Answer 11

-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

Question 12

Kepler’s Laws

Answer 12

-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

Question 13

Full Orbit Equation

Derivation

Answer 13

-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

Question 14

Full Orbit Equation

Word Equation

Answer 14

kinetic term + potential term = total energy

-a scaled energy equation per unit mass

Question 15

Full Orbit Equation

Potential

Answer 15

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

Question 16

Full Orbit Equation

Newtonian Comparison

Answer 16

Vn(r) = 1/2 h²/r² - GM/r
= centrifugal barrier term + real potential
-so GR differs by the addition of a 1/r³ term

Question 17

Schwarzchild Metric

Photons

Answer 17

• still have the conservation equations

- but now derivatives are with respect to λ not τ

Question 18

Full Orbit Equation

Potential for Photons

Answer 18

V = 1/2 h²/r² - GM/c² h²/r³

-no Newtonian 1/r term for photons

Question 19

Characteristics of Orbits

Circular Orbits

Answer 19

-circular orbit:
r’=0
AND
V’(r) = 0

Question 20

Characteristics of Orbits

Solutions for Massive Particles

Answer 20

• 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

Question 21

Plotting Orbits

Answer 21

• 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

Question 22

Plotting Orbits

GR

Answer 22

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

Question 23

Plotting Orbits

Newtonian

Answer 23

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

Question 24

Radial Trajectories - Photons

Equation

Answer 24

-for radial trajectory photons; ds²=0 and dθ=dφ=0
=>
dr/dt = ±c [1 - Rs/r]
-with + for outward and - for inward photons

Question 25

Radial Trajectories - Photons | Observers / Measuring Velocity

Answer 25

-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]

Question 26

Radial Trajectories - Massive Particles | r < Rs

Answer 26

-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

Question 27

Radial Trajectories - Massive Particles | Observers / Measuring Velocity

Answer 27

-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

Question 28

Gravitational Redshift | Overview

Answer 28

- 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

Question 29

Gravitational Redshift | Equation

Answer 29

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

Question 30

Gravitational Redshift | φ

Answer 30

-if ro, re > > Rs: νo/νe ~ 1 - (Δφ/c²) -if Δφ = φ(ro) - φ(re)

Question 31

The Event Horizon | Photons

Answer 31

- 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

Question 32

The Event Horizon | Massive Particles

Answer 32

- 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