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Quantum Particle Creation Near Strong

Curvature Regions

1 Introduction

One of the most important issues in current relativity theory is the occurenceof singular regions in the space-times. Singular regions, if naked, have farreaching implications. One of the very familiar example is the big bangsingularity in cosmological models. The study of gravitational collapse incases of imploding radiations, dust etc. have also pointed out the existance ofnaked singularities. Theoretically singularities(naked or covered) are boundto occur in relativity theory under certain general conditions imposed on thestress energy tensor. However, these classically reasonable conditions maynot hold if the matter field is quantized. In the late stages of gravitationalcollapse when regions of strong gravitational fields develop and the matter iscompactified in region of spacetime of the order of planck length, it becomesimportant to examine the quantum effects in the near singular regions of thespace-times. The Consideration and study of quantum effects, is thereforeimportant and it may lead to avoidance of singularities. An equally importantquestion is the nature of the spontaneous particle creation in the extremecurvature regions of the space-times.

2 Particle Creation in the Marginally Bound,

Self Similar Collapse of Inhomogeneous Dust

Let us consider a collapsing dust cloud. The field of a inhomogeneous dustcloud is described by the Tolman-Bondi metric given by *Tolman and Bondi.The metric is well known and given in comoving coordinates by

ds2 =dt2 R(t, r)2dr2 R2(t, r)d2 (1)The dust cloud is made up of concentric shells each labelled by r. R(t, r)isthe physical radius of such shells in the sense that the area of a shell labelled

by r is given by 4R2(t, r)).

R(t, r) denotes the derivative ofR(t, r)withrespect to r. The stress energy tensor describing the dust is

T=(t, r)0

0. (2)

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Whereua =deltaa0 represent the four velocity of dust particles.

*We are interested only in the marginally bound self similar collapsewhich is given by,

R(t, r) =r

1 3

2

t

r

23

(3)

The physical radius is seen to depend on one parameter, , (the mass pa-rameter). This parameter determines the total mass,M(r),lying within theshell labelled by ras 2GM(r) = r.The total mass of the dust is therefore2GM = = r0 where r0labels the outer boundary of the cloud. The ef-fective two dimensional metric giving the field of such a dust cloud is givenby

ds2 =dt2 R(t, r)2dr2 (4)where R is the physical radius of the collapsing dust cloud. Using the vari-ables (z, x) where z=lnr and x= t/r.

dr2 =r2dz2. (5)

dt2 =x2r2dz2 + 2xr2dxdz+r2dx2. (6)

Substituting equations (2)and (3) in equation (1), we get

ds2 =r2 dx2 + 2xdxdz+ (x2

R

2

(x))dz2 . (7)Now we will use the following,

=z+1

2(I+I+)=

1

2(I+I+), (8)

I =

dx

x R(9)

and equation (4) reads:

ds2 =r2

x2 R

d2 d2

. (10)

The physical radius Ris given by,

R = r

1 32

t

r

23

(11)

R = ez

1 32

t

r

23

(12)

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and the transformed collapsing dust radius is,

R=

dR

dr (13)

R=

(1 12

x)

(1 32

x)

1

3

(14)

We will evaluate the integrals of

I =

dx

x R(15)

The new transformed parameters of and becomes,

=z+

xdx

x2 R2

d=dz+ xdx

x2 R2

and and hence,

dz2 =

d

xdx

(x2

R

2

)

2

The value is defined as,

=1

2[I I+] (16)

and, we have

=

Rdxx2 R

2 (17)

and therefore,

d=

Rdx

x2 R2 (18)

and

dx=

x2 R

2d

R

(19)

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These values ofdz2 and dx2are substituted in equation (3)and it reads:

ds2 =r2

x2 R2

R2

d2 +R

2

d2

R

2

(20)

ds2 =r2

x2 R2

x2d2 R2

d2 2x2d2 +x2d2 +R2

d2

R

2

(21)

ds2

=r2

x2

R

2(d

2

d2

) (22)

For null coordinates such that in the limit as 0 these reduce to thestandard null coordinates in Minkowski space. Such coordinates are given

byu= +reI for x R >0region and u = reI for x R 0region and v= reI+ for x R

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From above equation dxand xare substituted in I+.

Then

I+ =3y2

a dy

1a3x

(1ax)13

(28)

I+ = 3y2

a dy

1y3

a +

2+y3

3

y

(29)

I+ = 9y3dy

3y

3y4 + 2a+ay3 (30)

I+ = 9y3dy

3y4 ay3 3y 2a (31)

I+ = 9

y3dy

3y4 ay3 3y 2a (32)

Similarly, for incoming case,

I =

dx

x R(33)

I = 3y

2

a dy

1y3

a (1a3x)

(1ax)13

(34)

I = 3y2

a dy

1y3

a

2+y3

3

y

(35)

I = 9y3dy

3y 3y4 2a ay3 (36)

I = 9

y3dy

3y4 ay3 3y 2a (37)

So the integral becomes

I= 9

y3dy

3y4 ay3 3y 2a (38)

and hence,f(y) = 3y

4 ay3 3y 2a (39)

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If we consider the coordinates with respect to f(y).

For region x R >0x R >0 (40)

(1 y3

a ) (1

12

x)

(1 32

x)

1

3

>0 (41)

(1 y3

a ) (1

a3

x)

(1 ax) 13>0 (42)

(1 y3

a ) [1

a3

(1y3

a )]

[1

a(1y3

a

)1

3 ]>0 (43)

3y 3y4 2a ay33ay

>0 (44)

3y4 ay3 + 3y 2a >0 (45)3y4 +ay3 3y+ 2a

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Therefore

f(y)> 0 (56)

Similarly for region x+R >0

(1 y3

a ) +

(1 12

x)

(1 32

x)

1

3

>0 (57)

(1 y3

a ) +

(1 a3

x)

(1 ax) 13>0 (58)

(1 y3a

) + [1 a3(1y3

a )][1 a(1y3

a )

1

3 ]>0 (59)

3y 3y4 + 2a+ay33ay

>0 (60)

3y4 +ay3 + 3y+ 2a >0 (61)3y4 ay3 3y 2a

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3y4

ay3

3y

2a >0 (70)

f+(y) = 3y4 ay3 3y 2a (71)Therefore

f+(y)> 0 (72)

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