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Supernova Cas A ***

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Several parameters of the radio-bright supernova remnant Cas A, located approximately 3.4 kpc from Earth, are calculated in this worksheet. The supernova was observed in 1667, and its remnant has expanded to a diameter of 4 arcminutes. Relativistic electrons and protons are contained in the remnant, and the relativistic electrons are a source of synchrotron radiation.

Problem: (A) Find the total radio luminosity over the given frequency range in ergs; (B) find the minimum-energy magnetic field in Cas A; (C) find the minimum total energy from cosmic rays and magnetic fields, according to the radio emission of Cas A; (D) find the minimum energy in relativistic electrons; and (E) determine how often supernovae must occur in the Galaxy to maintain the observed synchrotron luminosity, assuming Cas A is a standard supernova remnant, that the total Galactic synchrotron radiation is approximately 1000 times that of Cas A,  and that all energy of cosmic-ray electrons converts to synchrotron radiation. (The problem is from Condon & Ransom (2006).)

NULL

Hint:

Perform the calculations in the order suggested. Output from one calculation is sometimes required as input for the following calculation.

 

Data:

 

#  d = 3.4 kpc  #  distance to Cas A

 

#  θ = 4 arcminutes  #  diameter of Cas A on the sky

NULL


NULL1 Jy = 10-23 erg s-1 cm-2 Hz-1:  #  jansky to erg conversion

 

#  η = up/ue = 40  #  The contribution of relativistic electrons to the cosmic-ray energy density in Cas A

 

Additional data appear below as required. Values for c12 and c13, needed in some of the equations, are given in the table at the end of the worksheet (Pacholczyk 1970).

 

Useful Equations:

 

NULL:  #  luminosity

#  Bmin = [4.5(1+η)c12L]2/7R-6/7  #  minimum-energy magnetic field strength 

 

#  
E[min](total) = c[13]*(1+eta)*L^(4/7)*R^(9/7)

 

#  E[min] = E[B]+E[CR]

 

#  E[e, min] = E[CR]/(1+eta)

 

#  tau = E[c]/L  time for Galaxy to produce one supernova if total synchrotron energy equals that of Cas A

 

 

Solutions:

 

(A) Find the total radio luminosity (L) over the given frequency range in ergs.

 

 

S = 2700*10^(-23)/((1/10^9)^.77*(nu/Hz)^.77)

S = 0.2298072703e-12/(nu/Hz)^.77

(1)

#     S = 2.298072703*10^(-13)*Unit('erg')/((nu/Hz)^.77*Unit('cm')^2*Unit('Hz'))

``

``

NULL

L = (4*3.14156)*((3.4*3.086)*10^21)^2*(int(2.30*10^(-13)/nu^.77, nu = 10^7 .. 10^11))

L = 0.4124069085e36

(2)

L = 4.12*10^35erg/s

``

NULL

(B) Find the minimum-energy magnetic field in Cas A.

 

The contribution of relativistic electrons to the cosmic-ray energy density in the Galaxy may be expressed as η = up/ue = 40. Assuming this to be the case in Cas A as well, the minimum-energy magnetic field of Cas A (Bmin) can be calculated as follows:

 

Find the radius, R.

 

R := evalf(3.086*((1/180)*(2*(1/60))*Pi*3.4)*10^21)

0.6104230882e19

(3)

Find the parameter c12 from Pacholczyk (1970, p. 233) (See table below.). For ν1 = 107 Hz, ν2 = 1011 Hz and α = 0.77, c12 = 3.57 * 107.

 

Bmin = [4.5(1+η)c12L]2/7R-6/7

 

B[min] := (4.12*(4.5*(1+40)*3.57)*10^7*10^35)^(2/7)*(6.1*10^18)^(-6/7)

0.7573669872e-3

(4)

B[min] = 757 μG.

NULL

 

(C) Find the minimum total energy from cosmic rays and magnetic fields, according to the radio emission of Cas A.

 

Use the following formula:

 

E[min](total) = c[13]*(1+eta)*L^(4/7)*R^(9/7)

 

where c13 = 0.921 * c[12]^(4/7) 

 

.921*(3.57*10^7)^(4/7)

19057.68040

(5)

E[min] := 19057.68040*(4.12*(1+40)*10^35)^(4/7)*(6.1*10^18)^(9/7)

0.5076888008e50

(6)

E[min](total) = 5.1*10^49*ergs

 

 

(D) Find the minimum energy in relativistic electrons.

 

E[min] = E[B]+E[CR]

E[min] = E[B]+E[CR]

(7)

E[min] = V*U[B]+E[CR]

E[min] = V*U[B]+E[CR]

(8)

E[min] = (1/6)*B^2*R^2+E[CR]

E[min] = 0.6210272443e37*B^2+E[CR]

(9)

NULL

 

E[min] = (1/6)*B[min]^2*R^3+E[CR]

E[min] = 0.2174474637e50+E[CR]

(10)

NULL

solve(E[min] = 2.174474636*10^49+E[CR], E[CR])

-0.2174474636e50+E[min]

(11)

NULL

E[e, min] = E[CR]/(1+eta)

E[e, min] = E[CR]/(1+eta)

(12)

NULL

E[e, min] = 2.902413372*10^49/(1+40)

E[e, min] = 0.7079057005e48

(13)

NULL

(E) Determine how often supernovae must occur in the Galaxy to maintain the observed synchrotron luminosity, assuming Cas A is a standard supernova remnant and all energy of cosmic-ray electrons converts to synchrotron radiation.

 

tau = E[c]/L

 

7.1*10^47/(4.2*10^35)

0.1690476190e13

(14)

or approximately 1.7 * 1012 s or about 54,000 y.

 

The total Galactic synchrotron luminosity is approximately 1000 times that of Cas A. Therefore, the supernova rate is approximately

 

1000*(1/54000)

1/54

(15)

or one supernova in 54 years.

 

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In the following table (Pacholczyk 1970), α is the spectral index, defined as

 

`≡`(alpha, `∂`(log)*S(nu)*(1/(`∂`(log)*nu)))

 

where S(ν) is radiative flux and ν is frequency.

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References

 

Condon, J. & Ransom, S. (2006). National Radio Astronomy Observatory. ASTR 534: Radio Astronomy. http://www.cv.nrao.edu/course/astr534/Homework/s07.pdf (accessed: 2016-01-09).

 

Pacholczyk. A. (1970). Radio Astrophysics: Nonthermal Processes in Galactic and Extragalactic Sources. New York: W. H. Freeman.

 

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