Computing a value of radius for a typical white dwarf

In summary, computing the value of radius for a typical white dwarf involves using equations and data from observational studies to determine the mass, density, and temperature of the star. This information is then used to calculate the radius, which is typically around 0.01 times the radius of the sun. The process also takes into account the degeneracy pressure and electron degeneracy of the core, which play a crucial role in determining the size of a white dwarf. Overall, accurately computing the radius of a white dwarf is essential for understanding the properties and behavior of these fascinating celestial objects.
  • #1
fab13
312
6
I try to get a correct result for the radius of a standard white dwark (roughly 10000 km).

I just want the order of magnitude , i.e with the common values of a solar mass into Earth radius sphere.

From http://www.astro.umontreal.ca/~bergeron/CoolingModels/Synthetic_Calibration.pdf page 1223, I took the following (typical ?) values to compute this radius :

##\Phi_{\text{bolometric}}=10^{-9}\,\text{erg}.\text{cm}^{-2}.\text{s}^{-1}##

##T_{\text{surface}}=10000 K##, ##\text{D}=140## parsec,

and by using the following formula :

##R=\sqrt{\dfrac{ \Phi_{\text{bolometric}} D^{2}}{\sigma T^{4}}}##

Then I calculate the radius with all these values in (SI) units ##(1\,erg.cm^{-2} = 10^{-7} 10^{4}\,J.m^{-2}=10^{-3}\,J.m^{-2}##

$$Radius = \sqrt{\dfrac{(10^{-9}*10^{-7}*10^4*(140*3.26*3600*24*365*3*10^8)^2)}{(5.67*10^{-8}*(10000^4)}}$$

and I get ##Radius = 1.8134\,10^8 \,\text{meters} = 1.8134\,10^{5} \,\text{kilometers}##

That's too high as result, I expect rather a scale ##10^{3}## km < Radius < ##10^{4}## km.

If someone could tell me where is my error to get a standard value of radius for a white dwarf ?

UPDATE :

I get expected results with a flux equal to : ##\Phi_{\text{bolometric}}=10^{-12}## erg.s^-1.cm^-2

such that :

##Radius = \sqrt{\dfrac{(10^{-12}*10^{-7}*10^4*(140*3.26*3600*24*365*3*10^8)^2)}{(5.67*10^{-8}*(10000^4)}}##
= 5.7343 10^3 km

Anyone could confirm me the typical value of a bolometric flux (apparent brightness) equal to ##10^{-12}##erg.s^-1.cm^-2 for a white dwarf distant from 140 pc ?

I have difficulties to do the link between the monochromatic spectral flux (exprimed in erg.cm^-2.s^-1.Angstrom^-1) and the total flux ( in erg.cm^-2.s^-1, i.e the monochromatic spectral flux integrated on all wavelength), like for example in this figure :

W1UmO.png


But from this figure, I can only get the total flux between 1300 and 1600 Angstrom, not the total flux over all wavelengths (##\Phi_{\text{bolometric}}(1300<\lambda<1600)\,\approx\,8*10^{-12}*300\,\approx\,2.4*10^{-9}\, erg.cm^-2.s^-1##)

Any help or suggestion is welcome, regards
 

Attachments

  • W1UmO.png
    W1UmO.png
    27.2 KB · Views: 453
Astronomy news on Phys.org
  • #2
That is a lot to plug into a calculator.

Parsec ≈ 31 petameters. 3.1 x 1016

square parsec is 931 square petameters, or 9.6 x 1032m

10-12 x 1402 x 9.6 x 1032 = 1.88 x 1025

1.88 x 1025/(5.67 x 108)= 3.3 x 1016

√(3.3 x 1012) = 181659021 ≈ 1.8 x 108

That looks O.K.

Should your equation have a 4π next to σ?

The table on page 1223 of your link has values for Vega magnitude. You should be calculating Vega's radius.
 
  • #3
stefan r said:
That is a lot to plug into a calculator.

Parsec ≈ 31 petameters. 3.1 x 1016

square parsec is 931 square petameters, or 9.6 x 1032m

10-12 x 1402 x 9.6 x 1032 = 1.88 x 1025

1.88 x 1025/(5.67 x 108)= 3.3 x 1016

√(3.3 x 1012) = 181659021 ≈ 1.8 x 108

That looks O.K.

Should your equation have a 4π next to σ?

The table on page 1223 of your link has values for Vega magnitude. You should be calculating Vega's radius.

@stefan r

It seems that you have not used the conversion between erg and SI units since ##\sigma = 5.67 10^{-8} \,W.m^{-2}##, haven't you ?

Regards
 
  • #4
fab13 said:
@stefan r

It seems that you have not used the conversion between erg and SI units since ##\sigma = 5.67 10^{-8} \,W.m^{-2}##, haven't you ?

Regards
I used the same numbers you had. 5.67 x 10-8 x (104)4 = 5.67 x 108

σ has units W m-2s-1T-4 but that works fine in the equation in your first post.
 
  • #5
@stefan r

Sorry, I must precise that in the calculation :

##Radius = \sqrt{\dfrac{(10^{-12}*10^{-7}*10^4*(140*3.26*3600*24*365*3*10^8)^2)}{(5.67*10^{-8}*(10000^4)}}##

I apply the factor "##10^{-7}*10^{4}##" since the bolometric flux is expressed as : ##\Phi_{\text{bolometric}}=10^{-12}\,\text{erg.cm^-2.s^-1} = 10^{-12}\,10^{-7}\,\text{J.cm^-2.s^-1} = 10^{-12}\,10^{-7}\,10^{4}\,\text{W.m^-2}=10^{-15}\,\text{W.m^-2}##

I hope you understand
 
  • #6
In real life, I stop giving out partial credit when the numbers go in. This thread is an example why.
 

Related to Computing a value of radius for a typical white dwarf

1. What is a white dwarf?

A white dwarf is a star that has reached the end of its life and has collapsed down to a small size, around the size of Earth. It is composed mainly of carbon and oxygen and is extremely dense.

2. How is the radius of a white dwarf calculated?

The radius of a white dwarf can be calculated using the Chandrasekhar limit, which is the maximum mass that a white dwarf can have before it collapses into a neutron star. This limit is dependent on the mass of the white dwarf and the properties of the material it is composed of.

3. What is the typical radius of a white dwarf?

The typical radius of a white dwarf is around 0.01 to 0.02 times the radius of the Sun, or approximately the size of Earth. However, this can vary depending on the mass of the white dwarf.

4. How does the radius of a white dwarf compare to other types of stars?

White dwarfs are much smaller in size compared to other types of stars, such as main sequence stars or red giants. This is because they have reached the end of their life and have undergone gravitational collapse.

5. Can the radius of a white dwarf change over time?

Yes, the radius of a white dwarf can change over time as it cools down and loses heat. This process can take billions of years and the white dwarf will gradually become smaller and smaller. However, the radius will eventually reach a minimum size and remain constant.

Similar threads

  • Astronomy and Astrophysics
Replies
9
Views
2K
  • Astronomy and Astrophysics
Replies
7
Views
5K
  • Introductory Physics Homework Help
Replies
17
Views
240
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Astronomy and Astrophysics
Replies
5
Views
5K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Special and General Relativity
Replies
5
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
3K
  • Astronomy and Astrophysics
Replies
23
Views
4K
  • Astronomy and Astrophysics
Replies
2
Views
3K
Back
Top