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Thread: transcendental equation

  1. #1 transcendental equation 
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    Hi,
    I am doing my PhD in Physics (electrical Physics) and I need to solve a transcendental equation (first time I met such a problem).
    This equation comes from a physics article writen in 1974.

    Here is the equation:

    exp(E/kT) = s/B * (kT^2)/E

    - k and s are some physcal constants (Boltzmann constant etc..)
    - B is the temperature rate, given in "degre Kelvin/seconds" (known value)
    - T is the temperature (in Kelvin)
    - E is the energy

    and from that I need to plot E versus T...?!
    In the article, the author doesn't really say how he gets the graph but the solution/graph he shows is a linear line such as
    E = CT - D
    where C and D are some constants that he determines from the slope and the intercept with the y-axis.
    So I would like to know how do I solve this equation and get this graph ? Do I need to use any program/algorythm..?
    Thanks for your help !
    Marc


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  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Well, the way you wrote that equation is a little ambiguous. I read it as: exp(E/(k*t)) = s/B * ((k*T)^2)/E. (Correct me if I'm wrong.)

    With that equation, you can substitute u=E/(k*T) and v=s/B * k*T and get exp(u) = v/u or u*exp(u) = v. This can be solved with the Lambert W function. After that, it's easy to solve for E.


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  4. #3  
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    sorry for the writing.
    It is:

    exp (E / (k*T)) = s/B * (k*T^2)/E

    (T^2 in the right hand part)

    So if I do as you recommand, u=E/(k*T) and v=s/B * k*T, I end up with:

    v = k*u*exp(u) ( k is a physical constant)

    I checked on the link you gave for the Lambert W function but I am not familiar with this type of calculation...
    could you develop a bit more ? how to use this lambert function in order to solve the pb ?
    thanks a lot for your help.
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  5. #4  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Well, if that's the case, then just rewrite v = s/B * T.

    As stated on the Lambert W page, if u*exp(u) = v, then u = W(v). (It uses X and Y, but, same thing.) Undo the substitutions and solve for E and you get something like E = k*T*W(s/B * T). (You should double-check that though.) This certainly isn't linear, but if the range of temperatures is right, it might could be estimated by a linear function.
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  6. #5  
    Forum Professor river_rat's Avatar
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    If you are feeling adventurous you can to a power series expansion of E in terms of T and see where that gets you. Else use the implicit function theorem and see what the first two approximations will be about some point that is in the region you are interested.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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