General Relativity Physics Honours 2006

General RelativityPhysics Honours 2006 A/Prof. Geraint F. Lewis Rm 557, A29 gfl@physics.usyd.edu.au Lecture Notes 6

Gravitational Lensing There are many astrophysical applications using what we have already learnt. Here we will examine two; gravitational lensing and relativistic accretion disks. Given that massive objects can deflect the path of a beam of light, this means that objects can behave as a gravitational lens. Such lenses resemble glass lenses in many ways, giving multiple images which can be magnified. Chapter 11

Gravitational Lensing How “strong” is a gravitational lens? Remember that the deflection angle is given by It is straightforward to define a lens equation of the form; Given astronomical distance scales, and the deflection angle being small, then

Gravitational Lensing The gravitational lensing equation for a point mass can then be written as; where This is known as Einstein radius (or angle). A source at =0 will result in a circular image on the sky, with a radius of =E. But how big is E?

Gravitational Lensing Let’s consider a star in our Galaxy, with the mass of the Sun. All of the distances are ~10kpc = 1017m. For this But for galaxies (M~1041 kg and D~1025 m) then And for clusters of galaxies, then

Gravitational Lensing Given the lens equation, we can find the image positions to be For the mathematically minded, the lensing equation is one-to-many (one source -> several images); this makes gravitational lensing tricky. If the source is not a point, then it is magnified (stretched). It should be clear that the angular (polar) extent is the same for the images as the source.

Gravational Lensing We can differentiate  and so So it is straightforward to calculate the area of an image. How bright an image? We can use the fact that surface brightness is conserved through lensing (be it a glass lens or gravitational lens). Therefore the brightness of the image compared to the source is simply the ratio of the areas;

Gravitational Lensing Hence we can write the magnification of an image as Hence, we have two images, the outer one brighter than the inner one. The total magnification is simply

Microlensing We know that our galaxy is surrounded by a halo of dark matter, but we don’t know what it is. If it is black holes or dead stars, then every-so-often, one will pass in front of a distant star, microlensing it. The resultant images are 10-3-10-6 arcsecs apart, and are unresolvable. We should just expect to see the distant star brighten and fade with a characteristic shape. Also, the variations should be achromatic.

Microlensing

Multiple Lenses What if we have more than one microlens? We can write the lens equation as Where ,  &  are all 2-dimensional vectors For a particular source position, solving for  becomes very tricky as there will be several images (and the number changes as you change ). When the number of lenses exceeds 2, then numerical methods are used.

Microlensing

Smooth Matter Lenses It is straightforward to go from many individual lenses to smooth matter distributions. The deflection angle becomes Where  is the surface mass density.

Multiply Imaged Quasars Quasars can be multiply imaged and exhibit the Shapiro time delay.

Accretion Disks Accretion disks are important in astrophysics as they efficiently transform gravitational potential energy into radiation. Accretion disks are seen around stars, but the most extreme disks are seen at the centre of quasars. These orbit black holes with masses of ~106-9 M¯, and radiate up to 1014 L¯, outshining all of the stars in the host galaxy. If we assume the black hole is not rotating, we can describe its spacetime with the Schwarzschild metric and make predictions for what we expect to see.

Accretion Disks Suppose we have radiating material orbiting a black hole. The photons leave the disk and are received by a distant observer. How much redshifting of the photons does the observer see?

Accretion Disks How fast is the material orbiting? Assuming circular orbits So we write the 4-velocity of the emitting matter to be And using the normalization of the 4-velocty

Accretion Disks We can now us the Killing vector conserved quantities; Remember, the conserved quantities and the impact parameter are related via

Accretion Disks Remember we are in a plane where =/2. What redshift do we see? Let’s consider the matter moving transversely to the line-of-sight. For these, b=0 and so So, there is no Doppler component. What about the matter moving directly towards or away from the observer? Now we explicitly need to consider the value of b

Accretion Disks Remembering that photons are null And so The resultant frequency ratio is

Accretion Disks For small values of M/r then where

Accretion Disks Accretion disks are hot an emit in highly ionized species. Remembering that the radius for the last stable orbit occurs at r=6M, the maximum redshift for an edge-on disk is For a face-on disk, this is ~0.7. Above is the X-ray spectrum of the 6.4keV Fe line in a Seyfert galaxy. The solid line is a fit to the data using a relativistic accretion disk seen at 30o. The extent of the redshift reveals this must be a black hole.

General Relativity Physics Honours 2006