Deriving Wien's displacement law (Zettili)According to Zettili, we can derive Wien's displacement law from Planck's energy density u ~ ( λ , T ) = 8 π h c λ 5 1 e h c / k T λ − 1 where λ is the wavelength and T is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets α λ = 5 ( 1 − e − α / λ )where α = h c / ( k T ). To solve this, Zettili goes on to say that we can write α λ = 5 − ϵso that the previously mentioned equation becomes 5 − ϵ = 5 − 5 e − 5 + ϵ Apparently, ϵ ≈ 5 e − 5 is a good answer. My first question is why do we use 5 − ϵ to rewrite the equation? What method is this? I've never seen that before and other resources just say that they got the values by solving numerically. My second question is why ϵ ≈ 5 e − 5 when there is an extra factor of e ϵ in the equation? Why can we ignore that factor?

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Deriving Wien&#039;s displacement law (Zettili)According to Zettili, we can derive Wien&#039;s displacement law from Planck&#039;s energy density            u      ~        (  λ  ,  T  )  =            8      π      h      c              λ              5                  1                  e                  h          c                      /                    k          T          λ                    −      1      where   λ is the wavelength and   T is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets      α    λ    =  5  (  1  −      e          −      α              /            λ        )where   α  =  h  c      /    (  k  T  ). To solve this, Zettili goes on to say that we can write      α    λ    =  5  −  ϵso that the previously mentioned equation becomes  5  −  ϵ  =  5  −  5      e          −      5      +      ϵ      Apparently,   ϵ  ≈  5      e          −      5       is a good answer. My first question is why do we use   5  −  ϵ to rewrite the equation? What method is this? I&#039;ve never seen that before and other resources just say that they got the values by solving numerically. My second question is why   ϵ  ≈  5      e          −      5       when there is an extra factor of       e          ϵ       in the equation? Why can we ignore that factor?

Isla Preston · Accepted Answer

Well this is sort of a Taylor expansion. Let us assume we do not take 5 from the beginning, what is the method behind? We want to solve something like:  x  =  5  (  1  −            e              −      x        )Ok then, let us assume the solution is close to some value       x    0   that we might guess, but only differs by a small   ϵ  ≪  1, i.e.   x  =      x    0    −  ϵ. We then have:      x    0    −  ϵ  =  5  (  1  −            e              −              x        0            +      ϵ        )We then obviously have  ϵ  =      x    0    −  5  +  5            e              −              x        0                        e        ϵ  with             e        ϵ    ≈  1. We need   ϵ to be small, because that was our assumption in the first place.The value of       e          −              x        0             is probably a small number, but what about       x    0    −  5? Well this is small if       x    0    =  5, as in this case this is zero. The result then must provide self consistently that   ϵ is small. Otherwise the ansatz was wrong, but this is OK here. So it works.Note that this requires a large factor for the parenthesis. 5 is OK, it would have been less precise for, e.g., 3 and would completely fail for 1.To continue the idea let&#039;s see what we can do in case of 3. The True numerical solution for   x  =  3  (  1  −            e              −      x        ) is   x  =  2.82144... but the simple idea from above gives   ϵ  =  0.149..., i.e.   x  =  3  −  0.149...  =  2.8506... so we are notably off. Our mistake is the assumption that             e        ϵ    =  1 but with   0.149 this is   1.16..., i.e.   16  % off. So lets make a Taylor expansion of this.             e        ϵ    ≈  1  +  ϵ. Plugging this into the above for a factor of 3 we get      x    0    −  ϵ  =  3  −  3            e              −              x        0              (  1  +  ϵ  )as before (but now with 3) setting       x    0    =  3 and solving for   ϵ we get  ϵ  =            3                        e                          −          3                            1      −      3                        e                          −          3                      =  0.175...and   3  −  ϵ  =  2.8244, reducing the error from   1.2  % to   0.4  %And just because it is sort of fun, let me even improve. Let&#039;s approximate             e              ϵ        ≈  1  +  ϵ  +      ϵ    2        /    2. This results in a quadratic equation for   ϵ which gives us:  ϵ  =      1    a    −  1  −      (          1      a        −    1          )      2        −    2    ;    a  =  3            e              −      3      The above result is actually an approximation of this as (just to practice Taylor series a little bit)                    ϵ                            =                    (                  1          a                −        1        )        −                  (                      1            a                    −          1                      )            2                    −          2                                                  =                    (                  1          a                −        1        )        −        (                  1          a                −        1        )                  1          −                      2                          (                              1                a                            −              1                              )                2                                                                                      =                    (                  1          a                −        1        )                  (          1          −                      1            −                          2                              (                                  1                  a                                −                1                                  )                  2                                                              )                                                  ≈                    (                  1          a                −        1        )                  (          1          −                      (            1            −                          1                              (                                  1                  a                                −                1                                  )                  2                                                      )                    )                                                  =                              1                      (                          1              a                        −            1            )                          =                  a                      1            −            a                              In any case, plugging in the value of   3, the quadratic approximation gives   3  −  ϵ  =  2.82162, which is   0.006  %If you would have done this with the factor   5 you&#039;d get for the exact solution   x  =  4.96511423... and for the approximation   x  ≈  4.96511448... providing a relative error of   5  ⋅      10          −      8

Deriving Wien's displacement law (Zettili) According to Zettili, we can derive Wien's displacement

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