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White dwarf mass-radius relation

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Compute the numerical constant C for an electron gas (take Z = 6 and A = 12) and determine the radius of a white dwarf whose mass is 0.6 solar masses.

    [itex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ Jh\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s\ s[/itex]

    [itex]m_{e}\ =\ 9.10938188(72)\ \times\ 10^{-31}\ kg[/itex]

    [itex]G\ =\ 6.673(10)\ \times\ 10^{-11}G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}\ m^{3} kg^{-1} s^{-2}[/itex]

    Mass of the white dwarf: [itex]\ 1.2 \times\ 10^{30}kg[/itex]

    2. Relevant equations

    M = [itex]\frac{f}{R^{3}}[/itex]

    f = [itex]\frac{π}{3}[/itex][itex]\left(\frac{15C}{2πG} \right)^{3}[/itex]

    [itex]\frac{N}{V}[/itex] = [itex]\frac{ρN_{0}}{2}[/itex], since Z/A = 1/2

    [itex]P=Cρ^{\frac{5}{3}}[/itex] = [itex]\left(\frac{N}{V}\right)^{\frac{5}{3}}[/itex] [itex]\left(\frac{3h^{3}}{8π}\right)^{\frac{2}{3}}[/itex][itex]\frac{1}{5m}[/itex]

    3. The attempt at a solution

    C = [itex]\frac{1.064\times10^{-67}}{5m}[/itex][itex]\left(\frac{N_{0}}{2}\right)^{\frac{5}{3}}[/itex]

    I took [itex]N_{0}[/itex] = 6.02 x [itex]10^{23}[/itex] so C = 31.57

    [itex]\frac{15C}{2πG}=\frac{473.53}{4.19\times 10^{-10}}=1.13\times10^{12}[/itex]

    Putting that into f, we get f = [itex]1.511\times10^{36}[/itex]

    Now, [itex]1.2\times10^{30}[/itex] = [itex]\frac{1.511\times10^{36}}{R^{3}}[/itex]

    and finally R = 107.98m, which doesn't make any sense to me. The only place where I think I might have it wrong is the value of [itex]N_{0}[/itex].
     
  2. jcsd
  3. Dec 15, 2012 #2

    mfb

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    Working with units would help to spot the error.
    What is N0? If it is the avogadro constant, why do you get it as factor between a density and another density? Which units do you use for ρ?
     
  4. Dec 15, 2012 #3
    I hate the textbook and the class notes but, per class notes, I assume ρ is in g/cc, and N/V is in particles per cubic meter.
     
  5. Dec 15, 2012 #4

    mfb

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    cc? cm3? Then you need an additional factor of 10^6 to convert between m3 and cm3.
     
  6. Dec 15, 2012 #5
    But then I'd have to divide by 1,000 to go from grams to kilograms... so I'd need an additional factor of 1,000 rather than 10^6.
     
  7. Dec 15, 2012 #6

    mfb

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    As you can see, it would be useful to work with units everywhere. It is easier to spot (or avoid) those prefactors if you know the units of your parameters.
     
  8. Dec 15, 2012 #7
    And now it gives a more sensible size (somewhat larger than Earth in volume, yet on the same order of magnitude); I find that the Coulomb pressure wouldn't affect the radius very much.
     
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