Observations of Supernova 1987a R. D. Gehrz University of MInnesota

Theories Confirmed by Observations of Supernova 1987a Neutrino pulse: right size and duration Infrared line emission and visual light curve decay rate: reveal formation of heavyelements Spectrum of precursor star: blue supergiant Thermal IR emission: formation of dust grains Light echoes: illuminated shells ejected by the precursor star and also illuminated a “phantom” nebula caused by reflections of the explosion off interstellar clouds Central remnant: Is a pulsar or black hole left over in the center of the system?

The Supernova Explosion The core collapses until the neutrons touch: The collapse stops because of neutron degeneracy pressure A shock wave rebounds through the outer layers of the star: some of the neutrons fly out through the ejecta and make heavy elements by 56 Fe + Nn  56+N X The core continues to collapse: it either halts as a stable neutron star or becomes a black hole depending upon the mass remaining in the core after the rebound

Core Collapse of a Massive Star Formation of the iron core: nuclei lighter than iron give off energy when they are formed by fusion, nuclei heavier than iron absorb energy when they are formed by fusion. Iron is on the dividing line between these processes. Photo-disintegration of iron and helium:  + 56 Fe  13 4 He + 4n and  + 4 He  4n Neutronization: p + + e -  n + Formation of a neutron star or a black hole: A neutron star forms if M c < 1.44 Solar Masses (the Chandrasekhar mass limit) A black hole forms if M exceeds M c R Black Hole = 2GM/c 2

The Escape Velocity Derivation When an object is thrown up into the air there is a battle between the object's kinetic energy and the gravitational pull on the object (its gravitational potential energy). If our object is to escape the gravitational pull of the Earth, its kinetic energy must balance the gravitational force acting upon it. The kinetic energy = ½ mv 2, where m is the mass of the object and v is its velocity. The gravitational potential energy of an object = GmM/r, where G is the gravitational constant, m is the mass of the object, M is the mass of the Earth and r is the Earth's radius. For a balance between kinetic energy and gravitational potential energy we must have: ½ mv 2 = GmM/r _______ Doing simple algebra, we obtain our escape velocity formula: V esc = √ 2MG/r

The Black Hole Equation Recall the escape velocity formula: ________ V esc = √ 2MG/r To trap light, we must have V esc = c Therefore: R Black Hole = 2GM/c 2

Black Hole’s of Various Masses R Black Hole = 2GM/c 2 : For M = 1M sun, R Black Hole = 3 km For M = 10M sun, R Black Hole = 30 km For M = 10M earth, R Black Hole = 1 cm For M = 10 6 M sun, R Black Hole = 3x10 6 km For M = 10 8 M sun, R Black Hole = 3x10 8 km

Neutron Stars For a star spinning at the limit of rotational stability: