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I'm trying to get a dimensionless form of the following expression:

\begin{equation} \delta (T) = a_{\delta} \left(\frac{1}{T}\right)^{b_\delta} + c_\delta ~, \end{equation}

so that the numerical values of the parameters ($a_{\delta}, {b_\delta}, c_\delta$) are independent of the selected system of units for temperature ($K$, $^o C$, etc.) [see Barenblatt, "Scaling", Cambridge Texts in Applied Mathematics, 1st ed. (2003)].

In fact, if I include a reference temperature, $T_0$, the expression above tends to be more consistent, but the first parameter $a_\delta$ still depends on the numerical value of the coefficient $(T_0 / T)$. The resulting expression will be:

\begin{equation} \delta (T) = a_{\delta} \left(\frac{T_0}{T}\right)^{b_\delta} + c_\delta ~. \end{equation}

Is better to perform a substraction of temperatures, given that a substraction between two temperatures in Celsius and Kelvin have both the same result? i.e.:

\begin{equation} \delta (T) = a_{\delta} \left(\frac{T_0}{T-T_0}\right)^{b_\delta} + c_\delta ~. \end{equation}

Note that $\delta$ is dimensionless.

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    $\begingroup$ What is the dimension of $\delta(T)$? * Note that the 2nd form, based on $\frac{T_0}{T - T_0}$, is not a correct choice, since it produces a (artificial) singularity at $T = T_0$ which was not in the original expression. $\endgroup$
    – AlQuemist
    Commented Aug 23, 2017 at 8:58
  • $\begingroup$ Excuse me for imprecise definition of the problem. $\delta (T)$ is the shape parameter of a Gumbel distribution function and is adimensional. In the problem, this parameter depends on temperature. It is absolutely correct the consideration about the singularity in $T - T_0$. Thanks for the answer. $\endgroup$ Commented Aug 23, 2017 at 9:00

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Let us assume that the $\delta$ has the dimension of $\delta_0$; i.e., $ [\![ \delta ]\!] = [\![ \delta_0 ]\!] $. Then, we can rewrite the relation (in simplified notation), $$ \delta(T) = a \, (\frac{1}{T})^b + c ~,$$ as $$ \delta_0 \, \frac{\delta(T)}{\delta_0} = a \, \frac{1}{T_0^b} \,(\frac{1}{T/T_0})^b + c ~, $$ or $$ \frac{1}{\delta_0} \,\delta(\frac{T}{T_0}) = \frac{a}{\delta_0 \, T_0^b} \,(\frac{1}{T/T_0})^b + \frac{c}{\delta_0} ~. $$

If we denote dimensionless variables by a tilde, $$\begin{align} \tilde{\delta} &:= \frac{\delta}{\delta_0} ~, \\ \tilde{a} &:= \frac{a}{\delta_0 \, T_0^b} ~, \\ \tilde{c} &:= \frac{c}{\delta_0} ~, \end{align} $$ we'll obtain the dimensionless form of the relation,

$$ \tilde{\delta}(\frac{T}{T_0}) = \tilde{a} \, (\frac{1}{T/T_0})^b + \tilde{c} ~. $$

Note that $b$ must be dimensionless itself, since it is a power.

Furthermore, the second suggestion, based on $\frac{T_0}{T - T_0}$, is not valid since it artificially produces a singularity at $T = T_0$ which did not exist in the original expression.

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  • $\begingroup$ @AdriánÁlvarez-Vázquez : Thanks for accepting the answer. For numerical calculations, it is usually better to rescale the parameters to some known relevant scale in the system; this makes them dimensionless and gives their magnitude a proper (relative) meaning. $\endgroup$
    – AlQuemist
    Commented Aug 23, 2017 at 10:30
  • $\begingroup$ Thanks newly for the suggestion. Is there any book on Scaling and Self-Similarities would you recommend? I usually use Barenblatt G. I. bibliography. $\endgroup$ Commented Aug 23, 2017 at 10:34
  • $\begingroup$ @AdriánÁlvarez-Vázquez : That book is good enough, I suppose; even far beyond what you'd need every day. $\endgroup$
    – AlQuemist
    Commented Aug 23, 2017 at 11:01

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