The Gamma-Ray BurstControversy

Galactic or Cosmological

Shapley v. Curtis

Shapley argued thatthe spiral nebulaewere part of our ownGalaxy and the sundidn’t lie near thecenter of the Galaxy.

In 1920 two of the mosteminent astronomers of thetime argued over “The Scaleof the Universe.”Harlow

ShapleyHeber D.

CurtisCurtis argued that thespiral nebulae wereother galaxies like ourown and the sun liesnear the center of ourGalaxy.

Who resolved the debate?

Edwin Hubble observed Cephied variablestars in the Andromeda nebula. TheseCephieds are so much fainter than thosein our own galaxy that Andromeda mustlie outside Shapley’s idea of the Galaxy.Shapley was right about the side of theGalaxy and Curtis was right about the“island universes”.

What do GRBs look like?

GRBs are burstsof gamma rayslasting from lessthan one secondto over 100seconds.They exhibitstructure down totimescales of onemillisecond.Energies from 1keV to 18 GeV.1

Observational Tests (1)

IsotropyGRBs comefromeverywhereon the sky.There is nopreferenceto theGalacticcenter.

Distributions (1)

SNR

Distributions (2)

Observational Tests (2)

Let us imagine that each GRB has the same luminosityand they fill space uniformly. What would the fluxdistribution look like? Specifically, how many arebrighter than a particular threshold?

Also what would you expect the median value of to be?The observations give <V/Vmax>≈ 0.3 or equivalently thenumber counts increase more slowly than fmin

-3/2.N ~ f--1.4 for bright bursts and N ~ f--0.8 for weaker ones.

fmin =4ùd2

max

L, N =

3

4ùd3maxn → N ∝ fà3/2

f/fmin( )à3/2 = V/Vmax

Where could they be?

What other objects areisotropic on the sky andhave <(f/fmin)-3/2> < 1/2?

Galaxy Number Counts

The number of galaxiesbrighter than a particularthreshold falls short ofwhat you would expect fthey were standardcandles filling Euclideanspace.

Galaxies evolve.Galaxies don’t fill Euclideanspace - they lie atcosmological distances.

GRBs are cosmological orthe distribution ends.

Theoretical Issues (1)

Compactness problemThe minimumtimescale in GRBs isaround 1ms. Thisgives a size less than3 x 107 cm.If GRBs arecosmological, theyemit 1051 ergs of 1MeV photons (1057

photons).

t

t

Theoretical Issues (2)

Two photons above an energy of 511MeVcan produce an electron-positron pair.What is the optical depth?

If GRBs are indeed cosmological, thephotons that they produce should bethermal.The photon spectrum is non-thermal.

ü = ûTRn = ûTR4ùR3E

3L ø 1018L51Rà2

7

Theoretical Issues (3)

Energy Problem:The energy of cosmological GRBs has to bearound 1051 erg (one foe) in gamma rays.A supernova produces:

1051 erg of neutrinos1049 erg of kinetic energy1047 erg of light

How can GRBs produce such a large fractionof their energy as photons?

What to conclude? (1)

The observations indicate the the burstshave a cosmological origin.Theoretically it is difficult to account forthe energy and thermal spectrum ofcosmological bursts.

“One does not have to know much about the subject to realize that if there is onecorrect theory of the bursts then all but one are wrong. One may continue thisreasoning to note that if 99 out of 100 hundred published theories are wrong thenmost likely all 100 are wrong.” -- Bohdan Paczynsky

Galactic Evidence (1)

The key pieces of evidence are Highly magnetized neutron stars (I.e. thesoft-gamma repeaters) sometimes producegamma-ray bursts (e.g. the March 5 event).Some neutron stars are born with highvelocities and might form a galactic corona ata distance of about 100 kpc.

Galactic Evidence (2)

Additional supporting evidence includes:Observation of cyclotron lines in GRBsGRBs that apparently repeat on week-to-month timescales.A lack of bright galaxies in the error box ofbursts.To see bursts from Andromeda you wouldneed a much more sensitive instrument.

Galactic Evidence (3)

Who resolved the debate?

Who resolved the debate?

Edwin Hubble again.On February 28, 1997 the satelliteBeppoSAX pinpointed the location of aGRB with sufficient accuracy to identify ahost galaxy.

K dwarf

Famous Bursts

970228 - firstafterglow970508 - firstredshift (0.835)971214 - z=3.24,E=0.16 MΧc2

990123 - 9thmagnitude opticaltransient (z=1.6)

Boom, Boom!

Let’s look at a big explosion. Neglectthe mass of the initial ejectathe internal energy of the surrounding matterasphericitiesradiation

What is left?Energy of explosion, EDensity of matter, ρRadius, r, and time, t

Similarity Solutions (1)

There is only one dimensionless numberthat combines these variables: .We actually don’t have to solve anythingto get the evolution.If we had made different assumptions, wewould have a different similarity variable.

If we neglect the mass of surroundingscompared to the ejecta, . This is a freeexpansion stage with constant velocity.

ú

E

r5t2

m

E

r2t2

Similarity Solutions (2)

Let’s make a variable that is proportionalto distance.Let’s take the fluid variables to befunctions of ξ. The radius of the blastwave is proportional to t 2/5.What this does is convert the fluidequations from partial differentialequations to ordinary differentialequations.

ø = ìrE

ú0ð ñ1/5

tà2/5

Similarity Solutions (3)

Because thesolution that weseek is a functionof ξ, the evolutionis self-similar, thefluid looks thesame at all times,just the scalechanges.

Relativistic Blast Wave (1)

In the relativistic problem there is anotherquantity, the speed of light so it is notstraightforward to make a similarityvariable.Let’s make the same assumptions andthink about the energetics.

Relativistic Blast Wave (2)

The total energy of the blast is E and it isshared with all the material behind theshock.The total energy of the material behindthe shock may be approximated by where w is the relativisticenthalpy. In the non-relativistic limit it isthe density of the material and we recoverthe Sedov-Taylor solution.

E = wΓ2ì2V

Relativistic Blast Wave (3)

Blandford and McKee found a similaritysolution that is valid in the limit of high Γ.

The radius of the blast wave increases ast while the value of Γ decreases as t -3/2.How far does the blast wave lag behindthe photons emitted at the start?

E = ûwΓ2ì2V = 8ùwt3Γ2/17

Relativistic Blast Wave (4)

The photons go a distance ct after a timet. The velocity of the blast wave variesso it goes a distance .Let’s take the difference of thesedistances.

d =R0

t ìcdt0

∆d =

Z0

t

(1 à ì)cdt0,

Γ2 = Atà3 =1 à ì2

1=

(1 à ì)(1 + ì)

1

∆d ùZ0

t

2Γ21

cdt =

Z0

t

2A

1(t0)3cdt0 = c

8A

t4

Relativistic Blast Wave (5)

Let’s say we receive thefirst photon from thegamma-ray burst at timet0. We will receivephotons emitted at time tin the blast at t0 + ∆d/c.Let’s say that we observethe blast wave to be atΓ=2 around one monthafter the burst, thenA=(4 x 106 s)3.

Log t

Log tobs

Γ=1401.8 day

This was of course a bit unrealisticbecause at such early times the mass ofthe ejecta are important, so you’re in thecoasting phase with constant Γ.The more realistic result is

Relativistic Blast Wave (6)

tobs =2Γ2

0

tt < tc GRB

2Γ2

0

tc+

8Γ2

0t3c

t4àt4ct > tc Afterglow

Relativistic Blast Wave (7)Γ0=100, tc=2 x 105 s=2.3 days

Shock Structure

Velocity Pressure

Does this work?

Does this work?

Yes and no. All of these similaritysolutions give a smooth lightcurve like asupernova but...

External v. Internal Shocks

The observed lightcurve reflects theactivity of the “innerengine”. Need TWOtime scales.To produce internalshocks the sourcemust be active andhighly variable over a“long” period.

D=ct

∆d=c∆t

The Whole Picture

InnerEngine

RelativisticWind

InternalShocks

γ-rays

ExternalShock

Afterglow

We see this.We don’t see this.

Next week

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