General Astronomy/Fluence and Extragalactic Nature

From Wikibooks, open books for an open world
Jump to: navigation, search

Fluence[edit]

The distances to sources of gamma-ray bursts (GBRs) and their origins were not known until late 1990s. In particular, it was not clear whether GBRs originated in our solar system, in our galaxy, or in distant galaxies. In addition, the light spectrum of GRBs has multiple peaks and a complex structure. This causes changes in the energy received per second. Without knowing our distance from the progenitors, and since through the course of the GRB the energy randomly changes, the energy output must be given in terms of total energy received per unit area of detector surface. The can be done by integrating the energy flux over the duration of the burst. This quantity is called the fluence (S) with units of J/m2. Gamma-ray bursts typically have fluences between 10-12Jm-2 to 10-7Jm-2. [1] Data for flux and fluences for GRBs can be seen for yourself at http://www.batse.msfc.nasa.gov/batse/grb/catalog/4b/tables/4br_grossc.flux.

In order to investigate possible distance to the progenitor of a GRB we need to derive an equation that relates Fluence with distances. Consider a progenitor that has a luminosity of L and is at a distance r from earth, and assume an isotropic burst. After the light has traveled a distance of r, the luminosity will be spread out equally over the surface of a sphere with radius r, with the progenitor at the center of this sphere. The surface area of this sphere will be


A =  4 \pi r^2


The flux (in W/m2) that earth receives will be equal to the luminosity divided by the surface area of a sphere with radius r,

F =  \tfrac{L}{4 \pi r^2}

This is the well known inverse square law of radiation. The fluences will be the amount of energy received over the time duration of the GRB. This can be found by integrating the flux with respect to time.

\int F\,dt \, ,
 =\int \tfrac{L}{4 \pi r^2} dt
 = \tfrac{1}{4 \pi r^2}\int L\, dt
 = \tfrac{1}{4 \pi r^2} E
S =  \tfrac{E}{4 \pi r^2}

Where E is the energy of the gamma-ray burst and r is the distance the source is from earth. S is fluences and has units of J/m2.

Enlightening Example[edit]

If we assume the GRB to be in a galaxy 1 Gpc way, what amount of energy would be needed to produce a 10-7Jm-2 fluences, which is the upper limit of fluences detected in a GRB.

E = (4 \pi r^2) \cdot S
 E = 4 \pi \cdot (1 Gpc)^2 (10^{-7} Jm^{-2})
 E = 1 \cdot 10^{45} J

This is the average energy of a GRB given earlier in the introduction. [2] Now, in contrast, let's assume that the GRB source is within our galaxy. The extended corona around our Galaxy is about 100 kpc away.

E = (4 \pi r^2) \cdot S
 E = 4 \pi \cdot (100 kpc)^2 (10^{-7} Jm^{-2})
 E = 1 \cdot 10^{37} J

This is the total energy emitted by a GRB if it were much closer, such as in our Galaxy. This is smaller than the energy given off by a type IA supernova which gives off about 1044J. [3] This smaller energy seemed more reasonable to astronomers when GRBs were first being studied. However further evidence shows that the larger of these two numbers is more accurate.

A long time ago, In a galaxy far far away[edit]

Wikimedia commons image
The isotropic angular distribution of GRB seen by the BATSE detector on board the CGRO

The Compton Gamma Ray Observatory satellite on average observed one GRB every 25 hours. Measured positions from the dataset are shown in the figure to the right. The figure shows that there was no statistically significant deviation from an isotropic distribution. The figure also shows that the distribution is not homogeneous. This is significant because this means that the sources of the GRBs can’t be neutron stars in the Galactic Disk. The origin must lie outside of our galaxy. From the earlier example it can be seen that the greater distances have a dramatic effect on the energy required to produce GRBs. One big question about GRBs is understanding what could generate such a large amount of energy in such a short time.

To determine whether there is an edge to the distribution a brightness distribution argument using fluence can be used. Assume that a gamma ray source has an energy E and a distance r from earth, its fluence would be

S =  \tfrac{E}{4 \pi r^2}.

Solving for r provides,

r(S) =  (\tfrac{E}{4 \pi S})^{1/2}.

We assume that all burst sources are isotropic and have the same intrinsic energy, E. For each value of S0 the source will be within a sphere of radius r(S0). If there are n burst sources per unit volume, then the number of sources with fluence of S0 is

N(S_{o}) = n \tfrac{4}{3} \pi (r(S_{o}))^3.


N(S_{o}) =  \tfrac{4}{3} \pi n (\tfrac{E}{4 \pi S_{o}})^{3/2}.


This shows that if gamma-ray burst are uniformly distributed throughout space, then the number of bursts observed with fluence greater than some value S0 is proportional to S0^{-3/2). The data from the Compton Gamma Ray Observatory satellite show that this proportionality is not upheld when S is small enough to included very distant galaxies. This implies an edge to the distribution. This could be the edge of the observable universe. Thus we are must likely at the center of an spherically symmetric distribution of gamma ray burst sources.

References[edit]

  1. Dale A. Ostlie, and. Carroll, Bradley W. Introduction to Modern Stellar Astrophysics. Boston: Addison Wesley Company, 2006.
  2. PETROSIAN, VAHÉ, and THEODORE T. LEE. "The Fluence Distribution of Gamma-Ray Bursts." THE ASTROPHYSICAL JOURNAL (1996). Center for Space Science and Astrophysics, Stanford University. <http://www.iop.org/EJ/article/1538-4357/467/1/L29/5195.text.html>.
  3. Khokhlov, A.; Mueller, E.; Hoeflich, P. (1993). "Light curves of Type IA supernova models with different explosion mechanisms". Astronomy and Astrophysics 270 (1–2): 223–248. http://adsabs.harvard.edu/cgi-bin/bib_query?1993A%26A...270..223K. Retrieved on 2007-05-22.