The classical conservation of kinetic energy in spherical coordinates within the framework of the thin layer approximation when the thermal effects are negligible is

1 2 M 0 ( r 0 ) v 0 2 = 1 2 M ( r ) v 2 , (1)

where M 0 ( r 0 ) and M ( r ) are the swept masses at r₀ and r, and v₀ and v are the velocities of the thin layer at r₀ and r; for further details, see [15].

In SR, the total energy of a particle is

E = M γ c 2 , (2)

where M is the rest mass, c is the light velocity, γ is the Lorentz factor 1 1 − β 2 , β = v c and v the velocity. The relativistic kinetic energy, E k , is

E k = M c 2 ( γ − 1 ) , (3)

where the rest energy has been subtracted from the total energy, see formula (23.1) in [16]. The relativistic conservation of kinetic energy in the thin layer approximation in two points ( r 0 , v 0 ) and ( r , v ) is

M 0 ( r 0 ) c 2 ( γ 0 − 1 ) = M ( r ) c 2 ( γ − 1 ) , (4)

where M 0 ( r 0 ) and M ( r ) are the swept masses at r₀ and r, respectively, γ 0 = 1 1 − β 0 2 and β 0 = v 0 c . A Taylor expansion about v = 0 and v 0 = 0 of

order three for the above relativistic conservation law gives the classic case given by Equation (1). This fact assures a smooth transition from relativistic to classical velocities.

When the ISM has a constant density, the Lorentz factor as function of the radius is

β ( r ; r 0 , β 0 ) = B N B D , (5)

where

B N = − r 0 3 / 2 [ − 2 ( ( ( r 3 − 1 2 r 0 3 ) β 0 2 − r 3 + r 0 3 ) C + ( − r 3 + r 0 3 ) β 0 2 + r 3 − r 0 3 )     × ( ( ( r 6 − 2   r 3 r 0 3 + r 0 6 ) β 0 2 − r 6 + 2 r 3 r 0 3 − 2   r 0 6 ) C   + ( 2 r 3 r 0 3 − 2 r 0 6 ) β 0 2 − 2 r 3 r 0 3 + 2 r 0 6 ) ] 1 2 , (6)

B D = C β 0 2 r 6 − 2 C β 0 2 r 3 r 0 3 + C β 0 2 r 0 6 + 2 β 0 2 r 3 r 0 3 − 2 β 0 2 r 0 6     − C r 6 + 2 C r 3 r 0 3 − 2 C r 0 6 − 2 r 3 r 0 3 + 2 r 0 6 , (7)

and

C = 1 − β 0 2 . (8)

The differential equation that regulates the motion can be obtained from the above equation by inserting d r d t = v = β × c and v 0 = β 0 × c ,

d r ( t ; r 0 , v 0 , c ) d t = C N C D , (9)

where

C N = r 0 3 / 2 4 [ ( ( ( − 1 2 c 2 + 1 2 r 0 2 ) ( r ( t ) ) 6 + ( c 2 r 0 3 − r 0 5 ) ( r ( t ) ) 3     − r 0 6 ( c 2 − 1 2 r 0 2 ) ) c 2 − r 0 2 + ( − r 0 3 c 3 + c r 0 5 ) ( r ( t ) ) 3 + r 0 6 c 3 − c r 0 8 )     × ( ( ( c 2 − r 0 2 ) ( r ( t ) ) 3 − r 0 3 ( c 2 − 1 2 r 0 2 ) ) c 2 − r 0 2   + ( − c 3 + c r 0 2 ) ( r ( t ) ) 3 + r 0 3 c 3 − c r 0 5 ) ] 1 2 c (10)

and

C D = c 2 − r 0 2 ( r ( t ) ) 6 c 2 − c 2 − r 0 2 ( r ( t ) ) 6 r 0 2 − 2 c 2 − r 0 2 ( r ( t ) ) 3 c 2 r 0 3     + 2 c 2 − r 0 2 ( r ( t ) ) 3 r 0 5 + 2 c 2 − r 0 2 c 2 r 0 6 − c 2 − r 0 2 r 0 8     + 2 ( r ( t ) ) 3 c 3 r 0 3 − 2 ( r ( t ) ) 3 c r 0 5 − 2 r 0 6 c 3 + 2 c r 0 8 . (11)

This differential does not have an analytical solution and therefore the solution should be derived in a numerical way, except about r = r 0 , where a third-order Taylor series expansion gives

r ( t ; r 0 , v 0 , t 0 ) = r 0 + v 0 ( t − t 0 )                                     + 3 2 ( c − v 0 ) ( c + v 0 ) ( c 2 − c c 2 − v 0 2 − v 0 2 ) ( t − t 0 ) 2 c c 2 − v 0 2 r 0 . (12)

The medium is supposed to scale as

ρ ( r ; r 0 ) = { ρ c   if   r ≤ r 0 ρ c ( r 0 r ) α   if   r > r 0 (13)

where ρ c is the density at r = 0 , r 0 is the radius after which the density starts to decrease and α > 0 .

The total mass swept, M ( r ; r 0 , ρ c , α ) , in the interval [0, r] is

M ( r ; r 0 , ρ c , α ) = 4 3 ρ c π r 0 3 − 4 r 3 ρ c π α − 3 ( r 0 r ) α + 4 ρ c π r 0 3 α − 3 .

The conservation of energy in SR gives the following differential equation

d r ( t ; r 0 , v 0 , c ) d t = P N P D , (14)

where

P N = − r 0 3 / 2 c [ − 54 ( ( 1 / 9 ( ( α 2 − 18 α + 72 ) c 2 + ( α 2 + 6 α − 54 ) v 0 2 ) r 0 6 + α ( r ( t ) ) − α + 3     + 2 / 3 ( α − 15 / 2 ) r 0 2 α + 3 ( c + v 0 ) ( c − v 0 ) ( r ( t ) ) − 2 α + 6 + r 0 3 α ( c 2 − v 0 2 ) ( r ( t ) ) − 3 α + 9     − 1 / 9 ( ( α 2 − 18 ) v 0 2 + ( α − 6 ) 2 c 2 ) r 0 9 ) ( c + v 0 ) ( c − v 0 ) c ( c 2 − v 0 2 ) − 1     − 1 / 9 ( ( α 2 − 18 α + 72 ) c 2 − 3 ( 6 + α ) v 0 2 ) ( c + v 0 ) ( c − v 0 ) r 0 6 + α ( r ( t ) ) − α + 3

  − 2 / 3 r 2 2 α + 3 ( ( α − 15 / 2 ) c 2 + 1 / 4 ( α + 15 ) v 0 2 ) ( c + v 0 ) ( c − v 0 ) ( r ( t ) ) − 2 α + 6   − r 0 3 α ( c − v 0 ) 2 ( c + v 0 ) 2 ( r ( t ) ) − 3 α + 9 + 1 / 9 r 0 9 ( ( α − 6 ) 2 c 4   − 1 / 6 ( α 3 − 3 α 2 − 36 α + 216 ) v 0 2 c 2 + 3 / 2 ( α + 3 ) v 0 4 ) ) ( α − 3 ) ] 1 2 , (15)

and

P D = − 6 c 2 c 2 − v 0 2 α c 2 r 0 6 + 6 c 2 c 2 − v 0 2 α r 0 6 v 0 2 − α 2 c 2 r 0 6 + 18 c 2 c 2 − v 0 2 c 2 r 0 6     − 18 c 2 c 2 − v 0 2 r 0 6 v 0 2 + 6 α c 2 r 0 6 − 18 c 2 r 0 6 + 9 r 0 6 v 0 2     + 6 c 2 c 2 − v 0 2 r 0 α + 3 ( r ( t ) ) − α + 3 α c 2 − 6 c 2 c 2 − v 0 2 r 0 α + 3 ( r ( t ) ) − α + 3 α v 0 2     − 18 c 2 c 2 − v 0 2 r 0 α + 3 ( r ( t ) ) − α + 3 c 2 + 18 c 2 c 2 − v 0 2 r 0 α + 3 ( r ( t ) ) − α + 3 v 0 2     + 18 r 0 α + 3 ( r ( t ) ) − α + 3 c 2 − 18 r 0 α + 3 ( r ( t ) ) − α + 3 v 0 2     − 9 r 0 2 α ( r ( t ) ) − 2 α + 6 c 2 + 9 r 0 2 α ( r ( t ) ) − 2 α + 6 v 0 2 . (16)

A third-order Taylor series expansion gives

r ( t ; t 0 , r 0 , v 0 ) = r 0 + v 0 ( t − t 0 ) − 3 / 2 ( c − v 0 ) ( c + v 0 ) ( c − c 2 − v 0 2 ) ( t − t 0 ) 2 r 0 c . (17)

We assume that the medium around the SN scales with the piecewise dependence

ρ ( r ; r 0 ) = { ρ c   if   r ≤ r 0 ρ c ( exp − r b )   if   r > r 0   (18)

where ρ c is the density at r = 0 and r 0 is the radius after which the density starts to decrease. The total mass swept, M ( r ; r 0 , ρ c ) , in the interval [0, r] is

M ( r ; r 0 , ρ c , b ) = 4 3 ρ c π r 0 3 − 4 b ( 2 b 2 + 2 b r + r 2 ) ρ c e − r b π     + 4 b ( 2 b 2 + 2 b r 0 + r 0 2 ) ρ c e − r 0 b π . (19)

The conservation of energy in SR gives the following differential equation

d r ( t ; r 0 , v 0 , c , b ) d t = E N E D , (20)

where

E N = e 3 2 r ( t ) + r 0 b r 0 3 2 [ ( e r ( t ) + r 0 b r 0 3 v 0 2 + 12 ( ( − b 2 − b r 0 − 1 2 r 0 2 ) e r ( t ) b     + e r 0 b ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) ) b ( v 0 + c ) ( − v 0 + c ) ) c 2 − v 0 2     − 12 ( ( − b 2 − b r 0 − 1 2 r 0 2 ) e r ( t ) b + e r 0 b ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) )               × b ( v 0 + c ) ( − v 0 + c ) c ] 1 2 [ ( e 2 r ( t ) + 2 r 0 b c 2 r 0 6 − 72 b 2 ( v 0 + c )

  × ( − v 0 + c ) ( ( b 2 + b r 0 + 1 2 r 0 2 ) ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) e r ( t ) + r 0 b   − 1 2 e 2 r ( t ) b ( b 2 + b r 0 + 1 2 r 0 2 ) 2 − 1 2 ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) 2 e 2 r 0 b ) ) c 2 − v 0 2   − 12 b ( v 0 + c ) r 0 3 ( − v 0 + c ) c ( ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) e r ( t ) + 2 r 0 b   − e 2 r ( t ) + r 0 b ( b 2 + b r 0 + 1 2 r 0 2 ) ) ] 1 2 , (21)

and

E D = 1 c [ ( − 72 ( − v 0 + c ) ( b 2 + b r 0 + 1 2 r 0 2 ) ( v 0 + c ) b 2 ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 )     × e 2 r 0 + 2 r ( t ) b + e 3 r 0 + 3 r ( t ) b c 2 r 0 6 + 36 ( − v 0 + c ) ( e 3 r 0 + r ( t ) b ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) 2     + ( b 2 + b r 0 + 1 2 r 0 2 ) 2 e 3 r ( t ) + r 0 b ) ( v 0 + c ) b 2 ) c 2 − v 0 2 − 12 r 0 3 ( − v 0 + c ) ( v 0 + c )     × b ( ( − b 2 − b r 0 − 1 2 r 0 2 ) e 2 r 0 + 3 r ( t ) b + e 3 r 0 + 2 r ( t ) b ( b 2 + b r ( t ) + 1 2 ( r ( t ) ) 2 ) ) c ] . (22)

A third-order Taylor series expansion gives

r ( t ; t 0 , r 0 , v 0 , b ) = r 0 + v 0 ( t − t 0 ) + 3 / 2 ( − v 0 + c ) ( v 0 + c ) ( c 2 − c c 2 − v 0 2 − v 0 2 ) ( t − t 0 ) 2 c c 2 − v 0 2 r 0 e − r 0 b     + v 0 ( v 0 + c ) ( − v 0 + c ) ( t − t 0 ) 3 b r 0 2 c 2 ( ( c 2 − v 0 2 − c ) ( b − 1 2 r 0 ) c e r 0 b         + 6   b ( c c 2 − v 0 2 − c 2 + 3 / 4 v 0 2 ) ) 1 e 4 r 0 b . (23)

We assume that the medium around the SN scales as a n = 5 Emden profile, [13] [17] [18],

ρ ( r ; r 0 ) = { ρ c   if   r ≤ r 0 ρ c 1 ( 1 + 1 3 r 2 b 2 ) 5 / 2   if   r > r 0 (24)

where ρ c is the density at r = 0 and b is the scale.

The total mass swept, M ( r ; r 0 , ρ c ) , in the interval [0, r] is

M ( r ; r 0 , ρ c , b ) = 4 3 ρ c π r 0 3 + 4 b 3 r 3 ρ c 3 π ( 3   b 2 + r 2 ) 3 / 2 − 4 b 3 r 0 3 ρ c 3 π ( 3   b 2 + r 0 2 ) 3 / 2 . (25)

The conservation of energy in SR gives the following differential equation

d r ( t ; r 0 , v 0 , c , b ) d t = D N D D , (26)

where

D N = 243   c r 0 3 / 2 [ ( b 2 + 1 3 ( r ( t ) ) 2 ) ( 1 3 ( v 0 + c ) r 0 3 ( ( r ( t ) ) 3 ( b 2 + 1 3 r 0 2 ) 2     × 3 b 2 + ( r ( t ) ) 2 − r 0 3 ( b 2 + 1 3 ( r ( t ) ) 2 ) 2 3 b 2 + r 0 2 ) ( b 2 + 1 3 ( r ( t ) ) 2 )     × b 3 ( b 2 + 1 3 r 0 2 ) c ( − v 0 + c ) 3 ( c 2 − v 0 2 ) − 1 − 1 3 ( r ( t ) ) 3 ( v 0 + c ) r 0 3     × ( b 2 + 1 / 3 ( r ( t ) ) 2 ) b 6 ( b 2 + 1 3 r 0 2 ) ( − v 0 + c ) 3 b 2 + r 0 2 3 b 2 + ( r ( t ) ) 2     + ( ( 1 2 c 2 − 1 2 v 0 2 ) b 12 + r 0 2 ( 1 2 c 2 − 1 2 v 0 2 ) b 10 + r 0 4 ( 1 / 6 c 2 − 1 / 6 v 0 2 ) b 8

  + 1 / 18 ( c 2 − 2 / 3 v 0 2 ) r 0 6 b 6 + b 4 c 2 r 0 8 54 + b 2 c 2 r 0 10 162 + c 2 r 0 12 1458 ) ( r ( t ) ) 6   + 1 3 ( ( c 2 − 1 2 v 0 2 ) b 6 + 1 2 b 4 c 2 r 0 2 + 1 / 6 b 2 c 2 r 0 4 + c 2 r 0 6 54 ) r 0 6 b 2 ( r ( t ) ) 4   + ( ( c 2 − 1 2 v 0 2 ) b 6 + 1 2 b 4 c 2 r 0 2 + 1 / 6 b 2 c 2 r 0 4 + c 2 r 0 6 54 ) r 0 6 b 4 ( r ( t ) ) 2   + ( ( c 2 − 1 2 v 0 2 ) b 6 + 1 2 b 4 c 2 r 0 2 + 1 / 6 b 2 c 2 r 0 4 + c 2 r 0 6 54 ) r 0 6 b 6 )   × ( 2 / 3 ( v 0 + c ) ( r ( t ) ) 3 ( b 2 + 1 3 r 0 2 ) 2 3 b 2 + ( r ( t ) ) 2

  − r 0 3 ( b 2 + 1 3 ( r ( t ) ) 2 ) 2 3   b 2 + r 0 2 ) b 3 c ( − v 0 + c ) 3 ( c 2 − v 0 2 ) − 1   − 2 / 3 ( r ( t ) ) 3 ( v 0 + c ) b 3 ( b 2 + 1 3 r 0 2 ) 2 ( − v 0 + c ) 9 b 2 + 3 ( r ( t ) ) 2   + r 0 3 ( b 2 + 1 3 ( r ( t ) ) 2 ) 2 ( 2 / 3 b 3 ( − v 0 + c ) ( v 0 + c ) 9 b 2 + 3 r 0 2       + ( b 2 + 1 3 r 0 2 ) 2 v 0 2 ) ) ] 1 2 18   b 2 + 6 r 0 2 , (27)

and

D N = − 486 ( v 0 + c ) r 0 3 [ ( ( r ( t ) ) 3 ( b 2 + 1 3 r 0 2 ) 2 3   b 2 + ( r ( t ) ) 2     − r 0 3 ( b 2 + 1 3 ( r ( t ) ) 2 ) 2 3 b 2 + r 0 2 ) ( b 2 + 1 3 ( r ( t ) ) 2 ) b 3 ( b 2 + 1 3 r 0 2 )     × c ( − v 0 + c ) 3 ( c 2 − v 0 2 ) − 1 + 486 ( r ( t ) ) 3 ( v 0 + c ) r 0 3 ( b 2 + 1 3 ( r ( t ) ) 2 )     × b 6 ( b 2 + 1 3 r 0 2 ) ( − v 0 + c ) 3 b 2 + r 0 2 3 b 2 + ( r ( t ) ) 2     + ( ( − 729 c 2 + 729 v 0 2 ) b 12 + ( − 729 c 2 + 729 v 0 2 ) r 0 2 b 10 + ( − 243 c 2 + 243 v 0 2 ) r 0 4 b 8

  + ( − 81 c 2 + 54 v 0 2 ) r 0 6 b 6 − 27 b 4 c 2 r 0 8 − 9 b 2 c 2 r 0 10 − c 2 r 0 12 ) ( r ( t ) ) 6   + ( ( − 486 c 2 + 243 v 0 2 ) r 0 6 b 8 − 243 b 6 c 2 r 0 8 − 81 b 4 c 2 r 0 10 − 9 b 2 c 2 r 0 12 ) ( r ( t ) ) 4   − 1458 ( ( c 2 − 1 2 v 0 2 ) b 6 + 1 2 b 4 c 2 r 0 2 + 1 / 6 b 2 c 2 r 0 4 + c 2 r 0 6 54 ) r 0 6 b 4 ( r ( t ) ) 2   − 1458 ( ( c 2 − 1 2 v 0 2 ) b 6 + 1 2 b 4 c 2 r 0 2 + 1 / 6 b 2 c 2 r 0 4 + c 2 r 0 6 54 ) ] r 0 6 b 6 . (28)

A third-order Taylor series expansion gives

r ( t ; t 0 , r 0 , v 0 , b ) = r 0 + v 0 ( t − t 0 ) − ( − 27 v 0 + 27 c ) ( v 0 + c ) ( c c 2 − v 0 2 − c 2 + v 0 2 ) 3 b 5 ( t − t 0 ) 2 2   ( 3 b 2 + r 0 2 ) 5 / 2 c 2 − v 0 2 r 0 c . (29)

A test for the quality of the fits is represented by the merit function χ 2

χ 2 = ∑ j ( r t h − r o b s ) 2 σ o b s 2 ,

where r t h , r o b s and r o b s are the theoretical radius, the observed radius and the observed uncertainty, respectively. A fit can be done by assuming a power law dependence of the type

r ( t ) = r p t α p ,

where the two parameters r p and α p , as well as their uncertainties can be found using the recipes suggested in [12]. Figure 1 reports the power law fit to the data.

In the case of a constant profile of density, we present a numerical solution as given by the numerical integration of the differential Equation (11). Figure 2 displays the theoretical model versus the astronomical data. Figure 3 presents the Taylor approximation of the trajectory as given by (12) in the restricted range of time [0.001 yr - 0.02 yr].

In the case of a power law profile for density, we present a numerical solution as given by the numerical integration of the differential Equation (14). Figure 4 displays the theoretical model.

Figure 5 presents the Taylor approximation of the trajectory as given by (3.2) in the restricted range of time [7.2 × 10⁻⁵ yr - 2.2 × 10⁻⁴ yr].

In the case of an exponential profile for density, we present a numerical solution as given by the numerical integration of the differential Equation (20); Figure 6 displays the theoretical model.

Figure 7 presents the Taylor approximation of the trajectory as given by (23) in the restricted range of time [10⁻³ yr - 8 × 10⁻³ yr].

In the case of an Emden profile for density, we present a numerical solution as given by the numerical integration of the differential Equation (26). Figure 8 displays the theoretical model.

Figure 9 presents the Taylor approximation of the trajectory as given by (29) in the restricted range of time [10⁻³ yr - 2 × 10⁻² yr].

For an observer who moves on the expanding shell, the proper time τ * is

τ * = ∫ t 0 t d t γ = ∫ t 0 t 1 − β 2 d t ,

see [19]. Let us take the example of an Emden profile with the initial trajectory characterized by the Taylor expansion given by Equation (29). The value of β as given by the Taylor expansion is

β ( t ; r 0 , β 0 , t 0 , b ) = 1 c d d t r ( t ; t 0 , r 0 , v 0 , b ) = B E N ( 3 b 2 + r 0 2 ) 5 / 2 − β 0 2 + 1 r 0 , (30)

where

B E N = 9   − β 0 2 + 1 r 0 ( b 2 + 1 / 3 r 0 2 ) 2 β 0 3   b 2 + r 0 2     + 27 b 5 c 3 ( β 0 − 1 ) ( β 0 + 1 ) ( β 0 2 + − β 0 2 + 1 − 1 ) ( t − t 0 ) . (31)

The time dilation can be evaluated once the following integral is done

F ( t ; r 0 , β 0 , t 0 , b ) = ∫ 1 − β ( t ; r 0 , β 0 , t 0 , b ) 2 d t ,

which is

F ( t ; r 0 , β 0 , t 0 , b ) = F A 3 3 F B + t , (32)

where

F A = − 27 ( − β 0 c + c ) ( β 0 c + c ) ( c − β 0 2 c 2 + c 2 − c 2 + β 0 2 c 2 ) 3 b 5 t ( 3   b 2 + r 0 2 ) 5 / 2 − β 0 2 c 2 + c 2 r 0 c + β 0 c     + 27 ( − β 0 c + c ) ( β 0 c + c ) ( c − β 0 2 c 2 + c 2 − c 2 + β 0 2 c 2 ) 3 b 5 t 0 ( 3   b 2 + r 0 2 ) 5 / 2 − β 0 2 c 2 + c 2 r 0 c , (33)

and

F B = − 243 c 4 b 5 ( β 0 − 1 ) ( β 0 + 1 ) ( β 0 2 + − β 0 2 + 1 − 1 ) ( 3   b 2 + r 0 2 ) 5 / 2 − β 0 2 + 1 r 0 . (34)

The time dilation is therefore

τ * = F ( t ; r 0 , β 0 , t 0 , b ) − F ( t 0 ; r 0 , β 0 , t 0 , b ) .

A measure of the time dilation is

D = τ * t − t 0 ,

with 0 < D < 1 . The time dilation is displayed as function of the time in Figure 10 and as a function of time and scaling in Figure 11.

The decay of a radioactive isotope is modeled by the following law for particles, which are in the laboratory frame

N ( t ) = N 0 e − t − t 0 τ ,

where τ is the proper lifetime, N 0 is the number of nuclei at t = t 0 and the half-life is T 1 / 2 = ln ( 2 ) τ . In a frame that is moving with the shell, the decay law is

N ( t ) = N 0 e − τ * τ .

Let us analyze the isotope ⁵⁶Ni for which τ = 8.757   d or T 1 / 2 = 6.07   d . We now express the proper lifetime in yr (1 yr = 365.24219 d), τ = 0.024 and Table 1 reports the number of nuclei that have survived at a given time. From the above table, it is evident that the number of nuclei that are embedded in the moving layer is bigger when the time dilation is considered.