Suppose a black body at temperature $T_0$ which is subject to a permanent and constant energy influx $J_{in}$. The outgoing energy flux is described by Stefan-Boltzmann law: $J_{out} = \sigma\cdot T^4$. For $T_0 \neq T_e \equiv\left(\frac{J_{in}}{\sigma} \right)^{\frac{1}{4}}$ the temperature of the black body will tend to the equilibrium temperature $T_e$.

Now I would like to know the black body's temperature as a function of time $T(t)$.

For an ideal gas (or as this answer states) the internal energy of the system is proportional to its temperature: $U \propto T$. The energy change equals the net power emerging from the ingoing and outgoing energy flux:

$$ \frac{dT(t)}{dt} = \frac{1}{k}\frac{dU(t)}{dt} = k^{-1}A\cdot \left(J_{in} - J_{out}\right) = k^{-1}AJ_{in} - k^{-1}A\sigma T^4 $$

where $A$ is the surface of the system. This is a first order non-linear differential equation of the form

$$ \dot{x} + ax^4 + b = 0 $$

where $x(t) \equiv T(t)$, $a \equiv k^{-1}A\sigma$ and $b \equiv -k^{-1}AJ_{in}$. Wolfram Alpha gives this rather cumbersome solution for the ODE from which I didn't manage to deduce $x(t)$ (i.e. the temperature as a function of time).

ODE solution by Wolfram Alpha

Does anyone know how I can express the temperature as a function of time and obtain $T(t)$?

  • $\begingroup$ A couple of things, 1) define $c^4 = -a/b$ (assuming $J_{in} > 0$). This will factor out 1 variable and get rid of most of the roots in the solution so it looks much cleaner. 2) use the relation $\tan^{-1} x = \frac{\imath}{2} \ln\left(\frac{1-\imath x}{1+\imath x}\right)$ (don't worry all the imaginary numbers must cancel out in the end). This will let you get everything into a single logarithm, which you can then exponentiate. $\endgroup$ – By Symmetry Jan 29 '18 at 15:35
  • $\begingroup$ ...and I think the $A$s are not completely correct. I think it should be $k^{-1}(J_\mathrm{in}-J_\mathrm{out}) $ which is $k^{-1}(J_\mathrm{in}-A\sigma T^4) $ $\endgroup$ – mikuszefski Jan 29 '18 at 15:40
  • $\begingroup$ @BySymmetry I tried replacing $a \rightarrow a^4$, $b \rightarrow b^4$ as well as used the $\tan^{-1}$ formula you suggested and ended up with a term $i\cdot\log\left(\frac{(b - i\sqrt{2}ax)^2 + b^2}{(b + i\sqrt{2}ax)^2 + b^2}\right)$ from which I didn't manage to eliminate the imaginary terms. Also I cannot join this term with the other log terms in order to exponentiate (well I could have an imaginary exponent, but I don't think this is helpful). $\endgroup$ – a_guest Jan 29 '18 at 23:03
  • $\begingroup$ You can make your life a lot simpler (and this is a general technique) by setting $x=x_0y$ and $t=t_0\tau$ and solving $\frac{dy}{d\tau}+y^4+1=0$ by choosing appropriate values for $x_0$ and $t_0$ $\endgroup$ – StephenG Jan 29 '18 at 23:31
  • $\begingroup$ @StephenG Thanks, I tried that version using Wolfram Alpha but still I don't see how to deduce $y(t)$ from the resulting solution (it's still a combination of $log$s and $tan^{-1}$s). $\endgroup$ – a_guest Jan 30 '18 at 0:20

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