general relativity physics honours 2006
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General RelativityPhysics Honours 2006
A/Prof. Geraint F. LewisRm 557, A29gfl@physics.usyd.edu.au
Lecture Notes 8
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Astrophysical Black HolesUnfortunately, we don’t have time to cover astrophysical black holes in the lectures, but you should read through Chapter 12 of the textbook.
Chapter 13
You should ensure you read in detail Section 13.3 which explains how black holes can evaporate via Hawking raditiation.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Rotation
Chapter 14
So far, we have considered the motion of point like particles through spacetime. These travel along timelike (massive) or null (massless) geodesics;
This equation actually parallel transports the tangent vector to the geodesic (the 4-velocity) along the path. Clearly, in flat Minkowski spacetime, the Christoffel symbols are zero.
Other vectors are undergo parallel transport. The spin of a gyroscope can be represented as a spacelike spin 4-vector.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationIn our rest frame,
And the 4-spin and 4-velocity are orthogonal. However, this must hold in all frames. The 4-spin is parallel transported along the geodesic using
As we will see later, this is related to the covariant derivative, and the Christoffel symbols take into account the change in the coordinates over the manifold. In flat space, the spin of the gyroscope remains fixed. What about curved spacetime?
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationLet’s consider a gyroscope orbiting in the static Schwarzschild metric. We will see that General Relativity predicts that, relative to the distant stars, the orientation of the gyroscope will change with time. This is geodetic precession.If the gyroscope is on a circular orbit then
And so the 4-velocity is given by
(Remember we are orbiting in a plane where =/2)
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationWe can use the orthogonality to find
And solving we find
We can use the Christoffel symbols for the Schwarzschild metric and the gyroscope equation to find;
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationCombining these, we can derive the equation
This is just the usual wave equation with a frequency
Clearly, there is something weird here. The components of the spin vector change with time, but the frequency is not the same as the orbital frequency. So, the spin angle and the orbital frequency are out of phase.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationThe solution to these equations are quite straight-forward and
Where s.s=s2*. Let’s consider t=0 and align the spin along the
radial direction and we need to look at the orthonormal frame of the observer; as the metric is diagonal, we can define the radial orthonormal vector is
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationThe period of the orbit is P=2/ and therefore the component of spin in the radial direction after one orbit is
So, after each orbit, the direction of the spin vector has been rotated by
This is the shift for a stationary observer at this point in the orbit. We could also set up a comoving observer at this point (orbiting with the gyro) but the radial component of the spin vector would be the same.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
RotationFor small values of M/R the angular change is
For a gyroscope orbiting at the surface of the Earth, this corresponds to
While small, this will be measured by Gravity Probe B.
What would you expect if you were in an orbiting laboratory with windows that opened once per year?
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Slowly Rotating SpacetimeSo far, our massive objects have been static, but what if the body is rotating? We won’t have time to consider the derivation of the effect in general, but you can read it in Section 23.3. For a slowly rotating spherical mass
Where J is the angular momentum of the mass. Hence the form of the metric is deformed, but what is the effect of this distortion? Firstly, lets consider this metric in cartesian coordinates;
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Slowly Rotating SpacetimeLet’s drop a gyro down the rotational axis. We can approximate the Schwarzschild metric as being flat (as contributing terms would be c-5). For our motion down the z-axis (rotation axis) then
The non-zero Christoffel symbols for our flat + slow rotation metric are simply
When evaluated on the z-axis and retaining the highest order terms. From this, we can use the gyroscope equation to show;
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Slowly Rotating Spacetime
Again, we have a pair of equations that show that the direction of the spin vector change with time with a period of
This is Lense-Thirring precession. At the surface of the Earth this has a value of
While small, this will be observable by GP-B.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Rotating Black HolesUnfortunately, we will not get time in this course to look at general black holes. Here we will just review the basic properties. If a black hole is rotating, its spacetime geometry is described by the Kerr metric.
where
These are Boyer-Lindquist coordinates. This is asymptotically flat and the Schwarzschild metric when a=0.
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Rotating Black HolesWe can analyze particle motion in this metric using the same approach as we have used previously. The horizon is flattened and the singularity is distorted into a ring.
http://www.engr.mun.ca/~ggeorge/astron/shad/index.html
Lecture Notes 8 http://www.physics.usyd.edu.au/~gfl/
Lecture
Charged Black HolesThe only thing left that can influence the form of black holes is charge (black holes have no hair). The geometry of a charged black hole is described by the Reissner-Nordstrom metric.
Where is the charge of the black hole.
Again, when the charge is zero, we recover the Schwarzschild solution. Again, the metric results in weird structure not seen in the simple solution. If 2>m2, then we have a central singularity but no shielding horizon, it is naked.
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