the sedov solution - anu

The Sedov Solution

1 Similarity solutions

Equation of motion of pendulum:

(1)

mg

l

d2dt2---------

gl--- sin–=

Astrophysical Gas Dynamics: Bubbles 1/53

Suppose solution of the form:

(2)

Since f is a dimensionless function it cannot depend upon the di-mensions of t, l, or g.Hence we need to combine t, l and g intosome dimensionless combination, the only possible one being

and the solution is of the form

(3)

f 0 t l g =

tgl---

f 0 tgl---

=


In this problem the variable is a similarity variable. We con-

struct analogous variables in constructing useful solutions to theEuler equations.

tgl---


2 Spherically symmetric hydrodynamics

2.1General equations

Take the spherically symmetric Euler equations:

(4)

The last equation is equivalent to

(5)

t 1

r2-----

r

r2v + 0=

tv

vr

v 1---

rp

+ + 0=

t

p – vr

p – + 0=

p K s =dK s

dt--------------

t

K s vr

K s + 0= =


The independent variables in this problem are r and t. In generalwe need to solve partial differential equations in r and t to obtaina solution.

In the self-similar approach we look for solutions of one variable

, leading to ordinary differential equations. However, weneed some physical basis for choosing .

2.2 Point explosion in a uniform medium

Terrestrial application: Analysis of Atom Bombs

Astrophysical application: 2nd phase of supernova remnants

rt


We expect the general features of the explosion to be a sphericalexpansion preceded by a shock wave advancing into the undis-turbed gas.

EDensity 0= Shock


2.3Dimensional analysis

In order to find an appropriate similarity variable, we look at thedimensions of the parameters involved.

(6)

Thus an appropriate (dimensionless) similarity variable for thisproblem is

(7)

E V2 L3 =

E0------ L5T 2– LT 2 5/– 5= =

E0------ 1 5/

LT 2 5/–=

E0------ 1 5/–

rt 2 5/–


We aim to construct the similarity variable so that the radius ofthe shock produced by the explosion is at . We thereforetake

(8)

where is to be determined later.

It is not envisaged that all features of such an explosion are de-fined by taking the fluid variables to be functions of . However,what is envisaged is that after some time, the resultant flow willsettle down to a similarity solution in which the parameters , and the variables and are combined into one similarity varia-ble. This is a general feature of similarity solutions.

1=

E0------ 1 5/–

rt 2 5/–=

E r t


3 Self-similar form of the Euler equations

3.1Self similar forms of variables and derivatives

Guided by the dimensions of the dynamica variables, we take:

(9)

0G =

v Art--U =

cs2 B

r2

t2-----Z =


The constants A and B will be chosen to make the ensuing equa-tions as simple as possible. This form of the fluid variables ischosen in such a way that they are dimensionally correct. (Obvi-ously, there is a strong connection between dimensional analysisand self-similarity.)

The momentum equation is expresed in the following form:

(10)t

vv

rv

+1---

rp

–1---

p

r

– cs21---

r

– cs2

r

ln–= = = =


The final form of the Euler equations is:

(11)

Since

(12)

t

vr

r

v 2vr

---------+ + + 0=

tv

vr

vcs

2r ln+ + 0=

t---- v

r-----+

p – ln 0=

p – 1---p----- 1– – 1

---cs

2 1– –= =


then the last of the above equations can be written

(13)

We adopt the general self-similar variable:

(14)

where in the constant background density case dis-

cussed above. Other values of are relevant when the back-ground density is non-uniform.

t---- v

r-----+

cs2 1– ln–ln 0=

Crt=

25---–=


W e have to do the following preliminaries, since differentiationof functions of involves differentiation of with respect to and .

Derivatives of

(15)

tr

r

Ctr--= =

t Crt 1–

t--= =


Derivatives of density

(16)

Derivatives of velocity

(17)

r 0G

r--=

lnr

------------1r---G G

-----------------=

t 0

t

------G = lnt

------------t---G G

-----------------=

tv

Ar

t2---- U – U + =

rv A

t--- U U + =


Derivatives of sound speed:

Since

(18)

then

(19)

cs2ln constant ln 2 r

t--ln Zln+ +=

t---- cs

2ln1t--- 2– Z

Z ----------------+=

r----- cs

2ln1r--- 2

Z Z

----------------+=


3.2 Euler equations in self-similar form

3.2.1 Continuity equation

This is:

(20)

Substituting the expressions for , and derivatives:

(21)

t

vr

r

v2vr

------+ + + 0=

v

0t

------G Art--U 0G

r--+

0G At--- U U + 2

r---0G Ar

t--U + + 0=


The various terms have a common multiplicative factor of

, which cancels out. The terms can then be combined to

give the following equation:

(22)

which is, on dividing by

(23)

Note the appearance of terms such as

which often appear when we are dealing with self similar solu-tions.

0t 1–

AU + G AG U 3AU G + + 0=

G

AU + G G

----------------- AU 3AU + + 0=

G G

-----------------d G ln

d ln---------------------=


3.2.2 Momentum equation

We can treat the momentum equation similarly.

tv

vr

vcs

2r ln+ + 0=

Ar

t2---- U– U' + A

rt--U A

t--- U U' + +

Br2

t2-----Z

r--

G' G -------------+ 0=


A factor of cancels and we can then collect terms in ,

and to obtain the final result:

(24)

3.2.3 Entropy equation

Similarly, the entropy equation becomes:

(25)

r

t2---- U'

U G'

A AU + U AU AU 1– +

BZ G G

-----------------+ 0=

Z Z

---------------- 1– G G

-----------------–2 AU 1– AU +

---------------------------------+ 0=


3.3Final form of self-similar equations

The set of self-similar equations is:

AU + G AG U 3AU G + + 0=

A AU + U AU AU 1– BZ G G

-----------------+ +

0=

Z Z

---------------- 1– G G

-----------------–2 AU 1– AU +

---------------------------------+ 0=


For the point explosion problem in a constant density back-

ground, we have . Therefore, 25---–=

25---– AU + G AG U 3AU G + + 0=

A25---– AU + U 2

5---U 2

5---U 1– BZ G

G -----------------+ +

0=

Z Z

---------------- 1– G G

-----------------–2 AU 1–

25---– AU +

---------------------------------+ 0=


The above equations are simplified, if we take

(26)

The final self-similar form is:

(27)

A25---= B

25--- 2 4

25------= =

U 1– d Glnd ln-------------

dUd ln----------- 3U+ + 0 (Continuity)=

U 1– dUd ln----------- U U

52---– Z

d Glnd ln-------------+ + 0 (Momentum)=

d Zlnd ln------------ 1– d Gln

d ln-------------–

2 U 5 2– U 1–

----------------------------+ 0 (Entropy)=


4 Conditions at outgoing shock

4.1 Development of boundary conditions

The surface is to represent the outgoing shock wave.Since

(28)

(29)

1=

E0------ 1 5/–

= rt 2 5/–

Radius of shock Rs 1– E0------ 1 5/

t2 5/= =

Velocity of shockdRsdt

---------25--- 1– E

0------ 1 5/

t 3 5/– 25---

Rst

------= = =


Within the constraints of this self-similar solution the only sortof shock that we can contemplate is an infinitely strong shock. Ifwe were to introduce another parameter into the solution corre-sponding to say, the strength of the shock, then our initial as-sumption that the solution only depends upon E, , r and t

would be invalid and the self-similar solution could no longer besatisfied.

0

v1'v2'

In frame of shock

v1'drsdt

--------=v1' v2'–

Frame of stationary gas

rsrs


We use the relationships for a strong shock:

(30)

In view of the self-similar relationship between v and U, viz.

, this boundary condition becomes:

(31)

v2 1–

1+-----------v1

=

v rs v1 v2

–=

2 1+-----------v1

2 1+-----------

25---

rt--

= =

v25---

rt--U =

U 1 2 1+-----------=


Using the density ratio for a strong shock, the density just insidethe shock wave is given by:

(32)

In the present notation the relationship between shock velocity,pressure and density is:

(33)

rs 0G 1 1+ 1–-----------0= =

G 1 1+ 1–-----------=

v1 2 1+

2-----------

p rs

0------------= p rs 2

1+-----------0

drsdt--------

2=


Hence, the sound speed just inside the shock is given by:

(34)

Since

(35)

then the boundary condition on is

(36)

cs2 rs

p rs

rs ---------------

2 1– 1+ 2

----------------------drsdt--------

2 2 1– 1+ 2

----------------------425------

rs2

t2-----= = =

cs2 4

25------

r2

t2-----Z =

Z

Z 1 2 1– 1+ 2

----------------------=


4.2Summary of boundary conditions

(37)

U 1 2 1+-----------=

G 1 1+ 1–-----------=

Z 1 2 1– 1+ 2

----------------------=


5 First integral via energy equation

5.1Conservation laws for a moving surface

When considering conservation laws earlier, we took the surface

enclosing a given volume to be stationary. Let us now considerthe case when the enclosing surface is moving. We consider the

S

V

ni UiVi


generic conservation law:

(38)

corresponding to conservation of a density f with correspondingflux density .

tf Fi

xi--------+ 0=

t---- fd3x

V FinidS

S+ 0=

Fi


When the surface S is moving with velocity it “sweeps up” f

and the corresponding flux into S is . Hence the corre-

sponding conservation law is:

(39)

5.2 Application to the Sedov solution

In our development of the Sedov solution we have the movingsurfaces .

Ui

fUini

t---- fd3x

V FinidS

S+ fUini Sd

S=

t---- fd3x

V Fi fUi– nidS

S+ 0=

constant=


Energy within :

(40)

1 2 3

E0------ 1 5/–

rt 2 5/–=

1=Shock

constant=

E 4 12---V2

---+

r2dr0

r

=


Now

(41)

(42)

---

1 1–-----------

P---

1 1– -------------------

P

------1

1– -------------------

P

------1

1– -------------------cs

2= = = =

E 4 12---V2

cs2

1– -------------------+

r2dr0

r

=


We have

(43)

r 1– E0------ 1 5/

t2 5/=

dr 1– E0------ 1 5/

t2 5/ d=

v2 r t 425------

r2

t2-----U2 =

cs2 4

25------

r2

t2-----Z =


Therefore the energy becomes:

(44)

The terms explicitly involving r and t combine to give:

(45)

so that the energy is

(46)

E 4 0G 12---

425------

r2

t2-----U2 1

1– -------------------

425------

r2

t2-----Z +

0

=

3– E0------ 3 5/

t6 5/ 2d

r2t 4 5/– 2– 2 E0------ 2 5/

=

E 1625

---------= 5– E 4G 12---U2 1

1– -------------------Z + d

0

One important aspect of this result is that is independent ofthe time t. We shall use this result shortly. The other importantresult is that it gives the way of calculating . The energy within

must be equal to the total energy of the blast, so that and

(47)

The parameter can be calculated from this once expressionsare obtained for and .

E

1=E 1 E=

5 1625

--------- 4G 12---U2 1

1– -------------------Z + d

0

1

=

U Z


5.3 First integral

The conservation of energy applied to the time evolving volumeenclosed by is

(48)

constant=

t---- 1

2---V2+

d3x0

r

V12---V2 h+ 1

2---V2+

dr dt

-------------– r2d+ 0=


Since the energy is independent of time, the surface integralmust be zero. The terms in the integrand are independent of an-gle and therefore the integrand itself must be zero. Thus,

(49)

v12---v2 h+ 1

2---v2+

dr dt

-------------=

25---

rt--U 1

2---v2 h+ 2

5---

rt-- 1

2---v2+=

U 12---v2

cs2

1–-----------+

---

12---v2+

cs2

1– -------------------

12---v2+= =


Using the self-similar substitutions for v and gives

(50)

This is easily solved for Z to give:

(51)

This is the energy integral!

cs2

U 12---U2 1

1–-----------Z +

Z 1– -------------------

12---U2 +=

Z 1– 2

-------------------U2 1 U –

U 1– ------------------------------------------=


One can see that this integral automatically satisfies the bound-ary conditions

(52)

(exercise).

U 1 2 1+-----------=

Z 1 2 1– 1+ 2

----------------------=


6 Final solution

Since we have first integral of the equations, the number of in-dependent equations is now reduced from three to two. The eas-iest equation to drop out of the previous set of 3 is themomentum equation and the most convenient form of the equa-tions to derive the final solution is

(53)

1 U– –d Glnd ln-------------

dUd ln----------- 3U+ + 0 (Continuity)=

d Zlnd ln------------ 1– d Gln

d ln-------------–

2 U 5 2– U 1–

----------------------------+ 0 (Entropy)=

Z 1– 2

-------------------U2 1 U –

U 1– ------------------------------------------ (First integral)=


Plan

• Eliminate from continuity and entropy equations giving

a differential equation connecting U and Z.• Use the first integral also connecting Z and U to solve for as a function of V.

Elimination of from continuity and entropy equations

gives

(54)

d Glnd ln-------------

d Glnd ln-------------

dUd ln-----------

1 U– 1–

------------------d Zlnd ln------------–

5 3 1– U– 1–

--------------------------------------– 0=


Multiply this through by gives

(55)

Now we eliminate from this equation using the firstintegral which can be expressed as:

(56)

d lndU

-----------

11 U– 1–

------------------d ZlndU

------------–5 3 1– U–

1–--------------------------------------

d lndU

-----------– 0=

d Zln dU

Zln 1– 2

-------------------ln 1 U– ln 2 Uln U 1– ln–+ +=

d ZlndU

------------ 1–1 U–-------------

2U----

U 1–----------------–+=


Substitution of this expression into the above yields, after somesimplification:

(57)

All that remains to do at this stage is to expand the factors andintegrate giving

(58)

d lndU

-----------

5 3 1– U–---------------------------------

2 1 U– U 5 3 1– U– ------------------------------------------–=

1 U– U 1– 5 3 1– U–

-----------------------------------------------------------+

5 ln132 7– 12+ 2 1+ 3 1–

---------------------------------------- 5 3 1– U– ln–=

2 Uln–5 1– 2 1+

------------------- U 1– ln constant+ +


The constant is determined by the condition

(59)

The result is:

(60)

U 1 2 1+-----------=

5 1+2

-----------U2– 1+

7 –----------- 5 3 1– U–

1 1+ 1–----------- U 1–

2=

1132 7– 12+ 2 1+ 3 1–

----------------------------------------–=

25 1– 2 1+

-------------------=


Solution for G

Using the continuity equation:

(61)

1 U– –d Glnd ln-------------

dUd ln----------- 3U+ + 0=

d GlndU

-------------1

1 U–-------------

3U1 U–-------------

d lndU

-----------+=


We can express in terms of U using eqn. (57) and then in-

tegrate to obtain . The result is:

(62)

d lndU

-----------

G U

G 1+ 1–-----------

1+ 1–----------- U 1–

3 1+7 –----------- 5 3 1– U–

4=

1+ 1–----------- 1 U–

5


where

(63)

33

2 1+---------------=

4

12 –-----------–

132 2– 12+3 1– 2 – 2 1+

--------------------------------------------------------= =

52

2 –-----------–=


7 Determination of

Change the variable of integration of the integral for from to. Note that corresponds to

(64)

The last integral can be easily integrated numerically given all of

the previous expressions for , etc. (One point to note

is that the integral contains a removable singularity at )

U 0= U 1 =

5 1625

--------- 4G 12---U2 1

1– -------------------Z + d

0

1

=

1625

--------- 5G U 12---U2 1

1– -------------------Z U +

d lndU

-----------dU1

2 1+

=

5 U d lndU

-----------

U 1 =


Typical results are

(65)

8 Application of Sedov solution to supernova remnants

Take relationship between

(66)

The locus of the outgoing blast wave is given by

1.4= 0.968=

53---= 0.868=

r

r E0------ 1 5/

t2 5/=

1=


Knowing the radius of the blast wave and the time we can calcu-late the energy released by the supernova.

(67)

Take ISM number density

(68)

E 0

rBW

t2 5/-------------

5

=

05– rBW

5 t 2–=

n 1 cm 3– 106 m 3–

E nm 5– rBW5 t 2–

5.84210 J

n

106--------- rBW

pc------- 5 t

100 yrs ------------------- 2–

=


e.g. Crab nebula

(69)

Velocity of expansion

(70)

rBW 3pc = t 1992 1054– 938 yrs= =

E 1.64310 J

drdt-----

25---

rt--=

Vsh 3900 km/sr

pc----- t

100 yrs------------------ 1–

=

1250 km/s for Crab=


Velocity of shocked gas

For a strong shock . This

velocity is typical of the expansion velocity of the Crab nebulafilaments.

V1V2V1 V2– V 0=

V214---V1=

V1

V1 V2–34---V1 940 km/s for Crab= =


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