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Stefan-boltzmann solved for temperature?

  1. Jul 21, 2011 #1
    My physics is extremely rusty (I was bored to death back in school and had no practical application of the stuff to make it interesting and/or truly sink it.) Now, I have an application, and while I have a long way to go, it is pleasurable to be comprehending things as I go.

    To a point, and here is where I ask for some help and confirmations. Namely, I need to find the effective (surface?) temperature of a star (presumably main sequence and not extraordinary in mass) based on its mass.

    First, smash my face in if I miscombobulated the stefan-boltzmann law:

    L = 4[itex]\pi[/itex]R[itex]^{2}[/itex][itex]\sigma[/itex]T[itex]^{4}_{eff}[/itex]

    ... to look for T[itex]_{eff}[/itex]: (and turning 4[itex]\pi[/itex]R[itex]^{2}[/itex] into A for now)

    T[itex]_{eff}[/itex] = [itex]\sqrt[4]{L/A\sigma}[/itex]

    No?

    Walp, I've already plugged this into my code, and I'm getting absolutely absurd results for a near-solar-mass star: 4.23581197119e+23 (x10[itex]^{23}[/itex] for the noncoders out there) when I'm expecting to get roughly 5700-5800K. So I am forced to assume either that I flubbed my reconfiguring of S-BL or I am horribly misreading the S-B constant:

    [itex]\sigma[/itex] = 5.670373(21)×10[itex]^{−8}[/itex]Jm[itex]^{−2}[/itex]s[itex]^{−1}[/itex]K[itex]^{−4}[/itex].

    That's quite a unit... I'm almost certain I'm not reading it correctly. Tips?
     
    Last edited: Jul 21, 2011
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  3. Jul 21, 2011 #2

    George Jones

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    Yes, if you mean [itex]\sqrt[4]{L/\left(A\sigma\right)}[/itex] or [itex]\sqrt[4]{L/A/\sigma}[/itex], and I get the correct result using this.
    For troubleshooting, it would help if you posted [itex]\sqrt[4]{L/A\sigma}[/itex] exactly as you coded it, including the numerical values that you used.
     
  4. Jul 21, 2011 #3
    Yip yip, that's what I mean; good to hear.


    the code is simple python, accepting solar masses as an argument:

    star.mass = 1.98892e+30 (kg)
    star.luminosity = star.mass ** 3.5 (1.1095850642735086e+106)
    star.radius = (star.mass ** 0.8) * solar_radius (~695,500,000m)
    star.surface_area = (4 * pi * (self.radius ** 2)) (~6.078608e+18 m2
    star.effective_temperature = (self.luminosity / (self.surface_area * stefan_boltzmann_constant )) ** 0.25 (~4.2358e+23 wawahoozit units)

    The code seems fine to me (assuming I've taken accurate information from the solar-comparative equations peppering the web.) I think I am mangling the units. Probably starting with luminosity... looking at it in erg/s (on the order of 1033) I think I see a place to start
     
  5. Jul 21, 2011 #4

    George Jones

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    This can't be correct. I think it should be something like
    [tex]\frac{L}{L_{Sun}} = \left( \frac{M}{M_{Sun}} \right)^{3.5}.[/tex]
     
  6. Jul 21, 2011 #5
    pulled it from http://en.wikipedia.org/wiki/Mass–luminosity_relation ... I may be misusing the proportionality bit. Have we really not worked out how to model these based purely on mass yet? (i.e., independent of having to use the sun as a comparator.)
     
  7. Jul 21, 2011 #6
    In any event, I shall plug away it at again; converting my radius to cm, I was able to get closER (as (erg/s)/(cm2erg/scm2K4) cancelled down a lot easier)... at my next opportunity, I see what happens when I try that as you show it instead.
     
  8. Jul 23, 2011 #7
    Kay, so still doing something wrong... I'm getting 32486K for a sunmass star when I use what wikipedia claims. Am I misconstruing your equation when I flip it to

    [tex]L = \left(\left( \frac{M}{M_{Sun}} \right)^{3.5}\right)L_{Sun}.[/tex]

    Further, why do you say 3.5 and wikipedia claim 4?
     
  9. Jul 23, 2011 #8
    Perhaps I'm also asking the wrong question; I may still be blending my units poorly.

    [itex]T = \sqrt[4]{\frac{L}{A\sigma}}[/itex] and I try to read this in units:

    [itex]\frac{erg/s}{\frac{cm^{2} erg}{s cm^{2} K^{4}}}[/itex]

    If that isn't butchered, it does have a pleasing cascade of cancellations that lead me to

    [itex]\frac{1}{K^{4}}[/itex]

    That's either wrong or I don't know how to ~read~ that properly. Beware, I'm many years out of practice and didn't have anything by way of application at the time to let it truly sink in.
     
  10. Jul 25, 2011 #9
    Solved it; I was unwittingly trying to combine erg/s with watts
     
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