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Theory of Planck Spacetime .pdf



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About spacetime in Planck scale.
Tomasz Kobierzycki
July 24, 2016

1

Contents
1 Spacetime quantized.

3

2 Gravity equation for one body.

4

3 Gravity equation for many body system.

5

4 Statistical recognition of measurement.

6

5 Energy density.

7

6 Statistical gravitational field.

8

7 Singularity-dynamic method of solving equations

9

8 Geometry - singularity and gravitational field .

10

9 Summary.

11

2

1 Spacetime quantized.
Space and time can be quantized by assuming that smallest quanta of space is Planck
length and smallest quanta of time is Plancka time.For start let consider body with no
acceleration, the Lorentz transfromation become:
"

t0
x(t + tp ) − x(t)
x =
x − ct
dt
tp

#!

0

t0
x(t + tp ) − x(t)
t =
x
t−
dt
lp

(1)

!

0

(2)

Where for photons, x0 = t0 = 0 what means that for photon all events are in present.
By that photon becomes non-local.For any body that have mass x0 > 0, t0 > 0.Change
in time (dt) can be written as:
dt = x(t + tp ) − x(t)

(3)

Time of event , start time for body that does not move is t0 ,it gives limit on how time can
t0
change- because dt
= γ maximally it can have ttp0 ,what means that smallest measuring
value is Planck time. It’s good to remember that rule that says about body without
acceleration. Speed is equal to :
x˙ =

x(t + tp ) − x(t)
tp

(4)

Derivative is mathematical sense can’t have valvue less than Planck time, examining
change in Planck time we derivative of position with respect to time. When describes
the movment in spacetime- i need variable by which funcion changes for the coordinate.
That variable is proper time:
xa (τ ) =

Z

mtp
X xa (t + tp ) − xa (t)
dt
dt
=
γ(t) t=0
t0

(5)

Where proper time can’t be for any event that last t0 less than t2p because increase in
time is minimum Planck time. Energy is related to proper time:
E a (τ ) =

~
xa (τ )

(6)

For photon probper time is related to wave length :
a

x (τ ) =

Z

mlp

X 1 xa (λ + lp ) − xa (lp )
1 dλ
=

c γ(t) λ=0 c
l0

What means that photon have energy, and proper time.

3

(7)

2 Gravity equation for one body.
When body accelerates it become more complicated. Let’s begin with vector r which is
equal to :

 1
x (τ )
x2 (τ )


(8)
ra = 

 ... 
xa (τ )
And tensor which after multiplied by vector gives distance squared:
Λab dr a = ds2b (τ )

(9)

ˆ
Where you can spread to product of energy tensor and operator R:
ˆ ab
Λab = RT

(10)

Which finaly gives for one body system:
ˆ ab dr a = ds2 (τ )
RT
b

(11)

ˆ it’s form is equal :
Starting with operator R
(

ˆ=
R

a=b

to

c

4 (x

a (τ ))3

~γ 2 (τ )

dr

a

a 6= b

;

to

2 (x

−c

a (τ ))3

~γ 2 (τ )

)

dr

a

(12)

While energy tensor is equal to:


Tab = a = b

to

~
2
c xa (τ )

;

a 6= b

~
a
x (τ )

to



(13)

Which allows to write down gravity equation for one body system:


1

2

2

2

a

2



c2 (xγ 2(τ(τ))) (dx1 (τ ))2 − c2 (xγ 2(τ(τ))) (dx2 (τ ))2 ... − c2 (xγ 2(τ(τ))) (dxa (τ ))2



...
ds2b (τ ) = 


a (τ ))2
2 (τ ))2
1 (τ ))2
(x
(x
(x
2
a
2
2
2
2
2
1
2
c γ 2 (τ ) (dx (τ )) ... − c γ 2 (τ ) (dx (τ )) − c γ 2 (τ ) (dx (τ ))

(14)

Where euqation is true for a = b, where a and b are number of dimensions. Laws are
equal for two dimensions and more. It’s gravity for one body system , for many body
system it becomes more complicated:
ˆ abcd..nm dr ac...n = ds2
RT
bd...m (τ )

4

(15)

3 Gravity equation for many body system.
To get equation for many body system i have to make transformation of romula from
last section. Let us multiply both side by d2 s2 d (τ ) :
bd...m

d
ˆ abcd..nm dr ac...n = 1
RT
d2 s2bd...m (τ )
Next by derivative

(16)

d2 s2bd...m (τ )
dr ac...n

ˆ abcd..nm
d2 s2bd...m (τ ) dRT

d2 s2bd...m (τ ) = 0
dr ac...n
ds2bd...m (τ )

(17)

Dividing both sides by d2 s2bd...m (τ ) it becomes:


ˆ abcd..nm
dRT
=0
d2 s2bd...m (τ )

(18)

dr ac...n −

d2 s2bd...m (τ )
=0
ˆ abcd..nm
dRT

(19)

1
dr ac...n
Rising both sides by power -1:

Pulling before bracket dr ac...n :
"

dr

ac...n

#

d2 s2bd...m (τ )
1−
=0
ˆ abcd..nm dr ac...n
dRT

(20)

To understand it better i will start with two body in four dimension. Writing it mathematically i will get:
"
#
d2 s2bd (τ )
ac
dr
1−
=0
(21)
ˆ abcd dr ac
dRT
Where metric of four dimension spacetime can be defined as tensor product of coordinates (for two bodies), to do it i need to write metric in form of a matrix:
2

ds (τ ) =

4 h
X

ds2 (τ )

A



ds2 (τ )B

i=j=1

i

=

4 h
X

ds2 (τ )A (ds2 (τ )B )T

i1
2

(22)

i=j=1

Where down index means first body (A) and second one (B). For N bodies and m
dimensions, metric will be tensor product:
2

ds (τ ) =

m h
X

ds2 (τ )A ⊗ ds2 (τ )B ... ⊗ ds2 (τ )N

i=j=1

5

i

2
2N

(23)

4 Statistical recognition of measurement.
Single enery quanta can move in time t = tp in four possible directions for two dimension
spacetime. For four dimensions that number becomes sixteen possible directions, which
gives general formula for first movement of energy quanta :
Nm = 2am−1 ; N2 = 4

(24)

Where m is number of dimensions, when considering second movement , it has to assume
that quanta can move backwards, which gives two more movements to that number,
which gives four more movemnts for two dimensions. For four dimension it’s 4*2*2=16
more movements. Adding corection to equation:
Nm = 4am−1 ; N2 = 4

(25)

Where equation is not considering number of moves, because for third movement it’s
equal to Nm ∗ 2 wich gives series :
S(n, m) =

τ
X

2m−2 4n + 2n(m−2)

(26)

n=1

=By which it can be calculated probability of finding quanta in state n for m dimensions:
Z b
a

Pb

2m−2 4n + 2n(m−2)
≤1
m−2 4n + 2n(m−2)
n=0 2

Ψ(τ, m)dτ = Pn=a
τ

(27)

It’s good to remembert that in state n quanta can be in many possible states which
fallows that for time τ probability of finding quanta in one state is equal to:
Z τ

Ψ(τ, m)dτ =
τ −tp

τ
X
n=0



2m−2 4 + 2(m−2)



2m−2 4n + 2n(m−2)

≤1

(28)

Last variable in equation is energy, probability of finding quanta in energy state depends
on it’s position, for example if in first move energy is euqal to E0 (τ ) in second movement
more time flows , more chance we have still E0 (τ ), when we travel more close to Planck
scale , the chance that energy is equal to Emax (τ ) increases. Where maximum energy is
Plancka constant diveded by Planck time. Putting it together:
Z b
a

Rb
E0a (τ )
b
π
τ =a tp cos 2 1 − a Ψ(τ, m)dτ + tp
hP
i
Rn
E0a (τ )
n
π
τ =0 tp cos 2 (1 − τ =0 Ψ(τ, m)dτ + tp )


hP

Ψ(E a (τ ), m, τ )dτ =

i

(29)

Which means that energyy of quanta is not specified finally, more we get close to Planck
scale more chance that energy of quanta increases, the more we get out of Planck scale
more chance that energy is equal to ground energy of quanta.

6

5 Energy density.
Density of energy is very important value, when we try to measure gravity field. Let’s
start with gradient of energy multiplied bu vectpr r rac... (τ ):
1
∇Tabcd... rac... (τ ) = drbd... (τ )
~

(30)

Where ∇Tabcd... is equal to:
∂τ



∇Tabcd... = ~



∂xac... (τ )

τˆ

(31)

Because i have energy density i can put it in the main equation, what will simplify it,
but first it’s needed to made some mathematical transforms:
"

#

1
d2 s2bd...m (τ )
∇Tabcd... rac... (τ ) 1 −
=0
ˆ abcd..nm 1 ∇Tabcd... rac... (τ )
~
dRT
~

(32)

First i will move ~1 ∇Tabcd... rac... (τ ) to right side of equation:
#

"

1
d2 s2bd...m (τ )
= − ∇Tabcd... rac... (τ )
1−
1
ac...
ˆ
~
dRTabcd..nm ~ ∇Tabcd... r (τ )

(33)

After i will pull the front bracket ~1 ∇Tabcd... rac... (τ ):
1
∇Tabcd... rac... (τ )
~

"

#

d2 s2bd...m (τ )
1

= − ∇Tabcd... rac... (τ )
1
ac...
ˆ abcd..nm
~
(τ ) dRT
~ ∇Tabcd... r
1

(34)

Then i will divide both sides by ~1 ∇Tabcd... rac... (τ ):
d2 s2bd...m (τ )
= −1
ˆ abcd..nm
dRT

(35)

ˆ abcd..nm
dRT
1
∇Tabcd... rac... (τ ) − 2 2
= −1
~
d sbd...m (τ )

(36)

1
1
ac... (τ )
~ ∇Tabcd... r



And multiply by -1 power:

I will move to right side of equation
∇Tabcd... r

ac...

ˆ abcd..nm
dRT
d2 s2bd...m (τ )

and multiply both sides by ~:

ˆ abcd..nm
dRT
(τ ) = ~ −1 + 2 2
d sbd...m (τ )

!

(37)

finaly divide both sides by rac... (τ ):
∇Tabcd... = −

~
rac... (τ )

ˆ

+

~dRTabcd..nm
2
d s2bd...m (τ )rac... (τ )

Which gives the maximum energy equal to − t~p + ~.

7

(38)

6 Statistical gravitational field.
To describe gravity as field, it’s needed to take into account that it is field when there is
probability to be in some state. There are no definitive states, it comes from equation of
theory, in respect to number of dimensions and states there are many solutions. Because
using equation from four chapter , i want to have not only probability finding quanta but
energy that is related to that probability, which gives possibility to calculate possible
position.
Z b

Ψ(E(τ ), m, τ, ∇E(τ ))dτ =

Z

Z

Ψ(E(τ ), m, τ )dτ
Vn

a

∇E(τ )dτ

(39)

Vn

Because integral can be divided into single episodes with planck time , this can be
simplified to simple sum:
Z

mtp

Z

Vn

X

∇E(τ )dτ =

Ψ(E(τ ), m, τ )dτ
Vn

Ψ(E(τ ), m, τ )∇E(τ )

(40)

τ =0

This equation is not very strict, it’s good to remember that in equation there is energy
tensor not a function of energy, i can write full equation:

Vn

mtp

Z

Z

Ψ(Tabcd... (τ ), m, τ )dτ

Vn

∇Tabcd... (τ )dτ =

X

Ψ(Tabcd... (τ ), m, τ )∇Tabcd... (τ ) (41)

τ =0

Last part is to designate trajectories for any particle, it’s not easy, because we have sum
of many moves. Starting from one quanta example , equation is simplified to:

Vn

mtp

Z

Z

Ψ(Tab (τ ), m, τ )dτ

Vn

X

∇Tab (τ )dτ =

Ψ(Tab (τ ), m, τ )∇Tab (τ )

(42)

τ =0

Putting it all together i will get:
Rb
E0a (τ )
b
π
τ =a tp cos 2 1 − a Ψ(τ, m)dτ + tp
hP
i
Rn
E0a (τ )
n
π
τ =0
τ =0 tp cos 2 (1 − τ =0 Ψ(τ, m)dτ + tp )
mtp

hP



i

X

ˆ ab
~dRT
~
− a
+ 2 2
r (τ ) d sb (τ )ra (τ )

!

(43)

To get position i need to multiply energy gradient by trajectories and calculate integral:
Z
Vn

∇Tab (τ )ra (τ )dτ =

Z

dra (τ )dτ

(44)

Vn

Where integral is equal to:
Z
Vn

mtp

dra (τ )dτ =

X
τ =0

!

ˆ ab
~dRT
~
+ 2 2
ra (τ )
− a
r (τ ) d sb (τ )ra (τ )

8

(45)

7 Singularity-dynamic method of solving equations
There are to ways of solving equation in this theory. First is geometric, second is
dynamic. I will show here first a dynamic one. Writing main equation again :
"

dr

ac...n

#

d2 s2bd...m (τ )
1−
=0
ˆ abcd..nm dr ac...n
dRT

(46)

Which leads to equation:
ˆ abcd..nm dr ac...n
d2 s2bd...m (τ ) = dRT

(47)

ˆ differential of that operator is equal to:
Where here important is change of operator, dR
(

ˆ=
dR

a=b

(dxa (τ ))3 a
dr
c4
~dγ 2 (τ )

to

;

a 6= b

(dxa (τ ))3 a
− c2
dr
~dγ 2 (τ )

to

)

(48)

Taking into account differential of energy operator: :


dTab = a = b

to

~
2
c dxa (τ )

a 6= b

;

to

~
a
dx (τ )



(49)

I will get:
n

ˆ ab dr a = a = b
d2 s2b (τ ) = dRT

− c2 (aa (t))2 d2 t2
(50)
Which gives me dynamic solution of that equation with acceleration. When i put it for
one body system i will get distance matrix:
to

c2 (aa (t))2 d2 t2

;

a 6= b

to





c2 (a1 (t))2 d2 t2 − c2 (a2 (t))2 d2 t2 ... − c2 (aa (t))2 d2 t2


...
d2 s2b (τ ) = 

c2 (aa (t))2 d2 t2 − c2 (aa−1 (t))2 d2 t2 ... − c2 (a1 (t))2 d2 t2

(51)

It means that i can measure not only differential of distance but second differential of
2
it. This differential is connected to acceleration. It’s in units [ m
] what is understood
s2
as spacetime acceleration .
Energy density depends on acceleration oraz speed of time change which means time
acceleration. In case of liniar movement (without acceleration) it simplifies to first chapter equations. But when it considered that body is accelerating energy can be more or
less dense, which creates gravitacional field. For black hole (center of it) acceleration is
always equal to t1p it is because when i put Planck time everywhere i will get :
c2 (a1 (t))2 d2 t2 = c2 (

t2p 2
)t = c2
t4p p

(52)

WHat means that inside a black hole time is accelerating with speed of light same with
space.

9

o


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