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White dwarf mass-radius relationship

  1. Oct 15, 2014 #1
    The mass-radius relationship for a white dwarf star is defined by :

    R= (9pi)^0.66 /8 * h^2/m1 * 1/(Gm2^1.66*M^.333),

    where m1= electron mass, m2=proton mass, G=grav. constant, h=planck's constant
    I want to take a proportion with the solar mass and solar radius, which would involve a division where I think everything should cancel out. But then for one solar mass, a white dwarf would have a radius equal to that of the Sun. I've found that the following is the correct relationship:

    R/R(solar)=0.010(M(solar)/M)^.333

    But where in the world does the factor of 0.010 come from???

    If anything needs to be explained more clearly, please let me know.
    Thank you very much!!!
     
    Last edited: Oct 15, 2014
  2. jcsd
  3. Oct 16, 2014 #2

    Drakkith

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  4. Oct 16, 2014 #3
    Yeah I guess there are variations to the equation out there. Would the constants not also divide out in the equation that you have mentioned, though?
     
  5. Oct 16, 2014 #4

    Vanadium 50

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    White dwarfs are smaller than main sequence stars. The 0.01 tells you how much smaller. (And it's made up of h's and c's and pi's and the like)
     
  6. Oct 16, 2014 #5

    Ken G

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    Your formula includes various physical constants and numerical factors, you are supposed to plug those in and see what you get. That's where the 0.01 comes from, there's no reason to expect those physical constants would yield a solar-radius white dwarf if it has a solar mass, because the physics of the Sun is very different from the physics of a white dwarf. In particular, a white dwarf has lost a whole lot of net heat, relative to the Sun, and that's why it is so much smaller. So the answer to your question is, the 0.01 comes from all that net heat that the white dwarf had to lose to get to a white dwarf.
     
  7. Oct 19, 2014 #6

    Matterwave

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    All you need to know to figure out a relation like R/R(solar)=X*(M(solar)/M)^(1/3) is to know that ##R\sim M^{1/3}## and how large a solar mass white dwarf is. I think this is perhaps easier than trying to figure out how all the constants work out. Once you know how big a solar-mass white dwarf is, you can plug in 1 solar mass to the right hand side of the equation and get that R/R(solar)=X. That's where the .01 comes from. A solar mass white dwarf will have a radius that is 1/100th the radius of the Sun (or about 6900km). Of course, this relationship fails quite dramatically as M reaches 1.4 solar masses, and so it's not like the range of validity of this relationship is all that broad in the first place.
     
  8. Oct 22, 2014 #7
    Hi guys,
    A white dwarf is a polytrope with index n=3 since it is relativistic and degenerate.
    For all the polytropes, the mass radius relation is: M~R(n-3)/(n-1). So for a white dwarf the mass is independent of the radius.
     
  9. Oct 22, 2014 #8

    Ken G

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    White dwarfs are generally not particularly relativistic. You are talking about what happens as the mass approaches the "Chandrasekhar mass." That mass is independent of the radius because it is just one mass, generally about 1.4 solar masses, where the white dwarf goes highly relativistic, and that is also where it collapses into a neutron star. In the opposite limit of a lower-mass white dwarf, it is nonrelativistic so has a polytrope index of n=3/2, and a mass-radius relationship that radius scales like mass to the -1/3 power, as above.
     
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