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5Characteristic acceleration timescale & rate for 2nd order Fermi

Acceleration as result of repeated collisions with "clouds":

Mean time between two collision = "scattering time": s

Average distance between two interactions (mean free path):

For energetic particle (v ' c) mean scattering time:

 

 

 

 

 

 

 

 

 

 

s '

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Characteristic acceleration timescale:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

E

 

 

'

3

 

 

 

 

3

 

 

 

c

 

2

 

 

 

 

< dE=dt >

' E

 

 

4 c2 c

 

 

4c

Vc

 

 

 

t

:=

 

 

 

 

 

 

 

 

 

 

s

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

acc

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Acceleration rate:

dt '

3 c2

,

 

dt

= tacc

 

 

3 c2E

tacc

= E

 

=

1

 

1

 

 

dE

4 c

 

 

 

 

 

dE

 

 

 

 

E

 

 

 

4c

 

 

 

11

= E
tacc

6Evolution of Particle Spectra - recap

Remember kinetic equation derived in lecture 6:

 

@n(E; t)

=

 

@

[ (E)n(E; t)] + Q(E; t)

 

@t

@E

 

 

 

n(E; t) dE number density of particle in interval dE at time t,

Q(E; t) injection term,

(E) := dEdt describing energy loss or gain (acceleration)

) for synchrotron losses (ignoring gains): dE=dt = c1E2

) for 2nd order Fermi (ignoring losses):

(E) = dEdt

12

7Particle Spectra with Escape

Evolution of particle distribution assuming particles to be accelerated by 2nd order Fermi and to escape on energy-independent timescale esc:

Kinetic equation with escape term:

@t

' @E

tacc n(E; t)

n( esc

@n(E; t)

@

 

E

 

E; t)

In steady-state @n(E; t)=@t = 0:

1

 

 

 

 

E dn(E)

 

n(E)

0 '

 

 

n(E)

 

 

 

 

 

 

tacc

tacc

 

dE

esc

) dE

' E 1 + esc

 

 

 

dn

 

n

 

 

tacc

 

Solution for particle spectrum is a power-law:

n(E) / E

with = 1 +

tacc

 

esc

 

13

Caution: The above treatment is only approximately valid (i.e., in the limit of small tacc= esc) and hides broadening/momentum dispersion < ( p)2= t > that accompanies the energy increase in this stochastic process. The latter can be correctly obtained within a Fokker-Planck approach:

14

Fokker-Planck Equation

Widely used for description of stochastic processes.

P (p;~ p~) := probability that particle with momentum p~ at time t changes momentum by p~ in time t.

f(~x; p;~ t):=phase space distribution function = probability to nd particle in phase space volume element d3xd3p.

Can write:

Z

f(p;~ t + t) = f(p~ p;~ t) P (p~ p;~ p~) d3( p)

Taylor expansion gives (to 2nd order in small p):

f(p;~ t) +

@f

t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

@t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

' Z f

@f

pi

+

1 @2f

 

pi pj P

@P

pi +

1 @2P

pi pj d3( p)

 

 

 

 

 

 

 

 

 

 

 

 

@pi

2

@pi@pj

@pi

2 @pi@pj

' Z f P

 

f P )

 

1

@2(f P )

pi pj::: d3( p)

 

 

 

 

@(

 

pi +

 

 

 

 

 

 

 

@pi

 

2

@pi@pj

 

 

 

 

15

Have from above:

 

 

 

 

 

 

 

f(p;~ t) +

@f

t

 

 

 

 

 

@t

 

 

 

 

 

 

 

 

 

 

 

 

 

' Z f P

f P )

 

1 @2(f P )

 

@(

pi +

 

 

 

pi

@pi

2 @pi@pj

pj::: d3( p)

Imposing normalization:

Z

P (p;~ p~) d3( p) = 1

and de ning "Fokker-Planck" coe cients:

Z

< pi > := P (p;~ p~) pi d3( p)

Z

< pi pj > := P (p;~ p~) pi pj d3( p)

) Fokker-Planck equation:

 

@f

 

@ < p

>

 

1

@2

 

< p p

j

>

 

 

 

=

 

 

i

 

 

f +

 

 

 

 

i

 

f

 

 

@t

@pi

t

 

 

2 @pi@pj

t

 

 

with 1st term (rhs) describing systematic energy gain/losses, and 2nd di u- sion part/dispersion/broadening.

16

"Principle of detailed balance": if scattering process is reversible in the sense that:

P (p;~ p~) = P (p~ p;~ p~)

Taylor expand rhs:

 

 

 

 

@2P

P (p;~ p~) ' P (p;~ p~) pi

@P

+

1

pi pj

 

 

 

 

@pi

2

@pi@pj

Integrating over d3( p) gives (treating p and p as independent variables):

@pi

< pi > 2 @pj

< pi pj >

= 0

@

 

 

 

1 @

 

 

 

 

 

 

 

 

 

 

 

 

 

)

< pi > 1

@ < pi pj >

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

t

2

@pj

 

t

 

 

Fokker-Planck eq. then reduces to a di usion equation in momentum space:

 

@f

=

@

Dij

@f

 

 

 

 

 

 

@t

@pi

@pj

where

Dij := < pi pj >

2 t are the components of the di usion tensor.

17

Example: Stochastic 2nd order Fermi for isotropic di usion in momentum space with f(p~) ! f(p) (i.e., f(p~)d~p = 4 p2f(p)dp; use spherical coordinates], and with constant escape:

@f

1 @

@f

 

f

 

=

 

 

 

p2Dp

 

 

 

@t

p2

@p

@p

esc

with Dp(p) = < p p> =: D0p2 momentum space di usion coe cient.

2 t

Thus, to properly incorporate di usion in kinetic eq. before (Sec. 7), an additional term / @E@ E2 @E@n needs to be added.

Steady-state solution assuming power law ansatz f(p) = f0 p s:

1 @

 

 

 

@f

 

@f

 

@2f

 

 

f

 

 

 

 

 

p2Dp

 

 

= 4D0p

 

 

+ D0p2

 

 

=

 

 

 

 

 

 

p2 @p

@p

@p

@p2

esc

)

 

4D0f0( s) + D0f0( s)( s 1) =

 

 

f0

 

 

esc

s2 3s D0 esc

= 0

)

 

s = 2

+ 2r

 

 

 

 

 

 

1 +

9 D0 esc

 

 

 

 

 

1

 

 

 

 

3

3

 

 

 

4

1

 

(choose positive solution to ensure nite particle number).

18

We have < E= t >= (4=3)(c= ) c2E for relativistic particles (E ' pc). Using "detailed balance"-relation for Fokker-Planck coe cients:

 

 

 

p

1

@

 

 

 

( p)2

 

 

 

 

 

in spherical coordinates, D t E =

 

 

 

p2

D

 

 

E

 

 

 

2p2

 

 

@p

 

t

 

 

 

)

Dp(p) :=

2 t

= p2 Z

p2

 

 

 

t

 

 

dp = p2 Z

p2

3 c2p dp =

3 c2p2

 

 

< p p > 1

 

 

< p >

 

1

 

 

4 c

1 c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

)D0 = 13 c c2

Power law index for f(p) = f0p s becomes:

s = 2

+ 2r

1 +

9 D0 esc

= 2

+ 2r

1 +

9 esc

 

3

3

 

4 1

 

3

3

 

 

 

16tacc

noting that (e.g. sec. 5) tacc = E= < dE=dt >= 34c

1

 

= 1=(4 D0).

2

 

 

 

 

 

 

 

 

c

 

 

Di erential particle distribution: n(p) / p2f(p) / p2 s / p with

= s 2 ' (

3

2) +

3

1 +

8 tacc

= 1 +

4 tacc

 

 

 

 

 

 

 

 

 

 

2

2

9 esc

3 esc

with expansion for large D0 esc. This is close to our estimate from the kinetic equation.

19

82nd order Fermi acceleration and its limits

E cient Acceleration of CR by 2nd order Fermi in the Galaxy?

Slow process: Acceleration time too long

) ISM clouds < Vc > 10 km/s, i.e. c 10 4, so < E=E >' c2 10 8 ) Distance or mean free path L 1 pc, so s L=c a few years.

) Acceleration timescale:

tacc =

s

108 yr > tCR 107 yr

< E=E >

tCR=typical CR con nement time (leaky box) in the Galaxy.

Injection Problem:

) Coulomb losses independent of E, dominate at low energy

) Protons have to be injected with kinetic energy of a few 100 MeV ) need pre-acceleration/injection mechanism.

Non-universal index: power law index s sensitive to tacc and esc ) tacc depends on c

) esc depends on density of scattering clouds

20

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