Question about relation between two time constants I am not a chemist, but a neuroscientist, so bear with me. I have struggled with this problem for over a week now and have realized it's a chemical question. I'll try to explain it so no neuroscience background is needed.

I am trying to fit a model to experimental data. The data describes how a certain channel opens and closes in response to a change in membrane potential of the cell the channel sits in. This is captured in two datasets:


The fraction of open channels at equilibrium for different values of the membrane potential (an example of a plot of this can be found here. This fraction I call G/G$_{max}$
The time constant describing the time it takes for the fraction to come to its equilibrium. This time constant I call $\tau_{G/G_{max}}$


It is thought that for each open channel, 4 gates have to be in the 'open' state. That is, it is thought that:

G/G$_{max}$ = n$_{\infty}^4$

The gate n, too, can be described by its fraction of open-state/total. 
The rates of n moving between these two states are described by $\alpha_n$ and $\beta_n$. Once there are four gates in the open-state, the channel is instantly open.
 If I try to describe this in a chemical reaction, I would guess something like:

$$\ce{C_n &lt;=&gt;[\alpha_n][\beta_n] O_n}$$

where $C_n$ and $O_n$ depict 'closed' and 'open' n-gates respectively. And

$$\ce{4\cdot O_n &lt;=&gt;[instant][instant] O_{ch} }$$,
where $O_{ch}$ describes the whole channel being open.

Now is the data I've got for G/G$_{max}$(V) and $\tau_{G/G_{max}}$ (this is the only thing the experimentalists can measure). However, to model this behavior, I need $\alpha$(V) and $\beta$(V). I already know that 

$n_{\infty} = (G/G_{max})^{1/4}$

but I cannot figure out how $\tau_n$ and $\tau_{G/G_{max}}$ are related to each other. Even though the reaction 4O --> open channel is instant, the shape of approaching the equilibrium differs and thus the point that they cross 1-1/e would as well. n will reach its equilibrium with a single exponential, while G/G$_{max}$ reaches it with an exponential to the fourth power. 

I think the answer should be quite simple, and should only depend on the 4'th power.

I hope my question is clear, but if there are more details that will help solve this I am more than happy to clarify certain points.

Edit: Here is a figure that shows how the 'measured' time constant and the one for the differential variable n differ. Also, it shows that there is not a linear relationship.



Just to be clear about what I did:

The differential system describing this model looks like:
G/G$_{max}$ = n$^4$
with
$\frac{dn}{dt} = \alpha_n(V)(1 - n_{\infty}(V)) - \beta(V) n_{\infty}(V)$

As one of the answers pointed out (sort of), a solution to this system with initial condition V1 is:

$n(t) = n_{\infty}(V) \cdot e^{-(\alpha_n(V) + \beta_n(V))} + n_{\infty}(V) \cdot \Big(1 - e^{-(\alpha_n(V) + \beta_n(V))}\Big)$

and thus

$G/G_{max}(t) = n(t)^4$

To find the $\tau$ of either of these traces (the time constant being the time it takes to reach $1-1/e \cdot$(final value) (see link), I plotted:

$\frac{n - min(n)}{max(n) - min(n)}$

and

$\frac{n^4 - min(n^4)}{max(n^4) - min(n^4)}$
I think I understood your question now. I will reformulate it in more 'chemical' terms and give you my solution. Not nicely formatted for now, I will do that later if I find the time.

You have a cell with $G_{max}$ gates on it. They can open and close reversibly, depending on a voltage V. Closed gates open with rate constant $\alpha(V)$, and open gates close with rate constant $\beta(V)$.

This implies one differential equation. If $G_o$ is the number of open gates at time t, and $G_c$ is the number of closed gates at time t, then:

$\frac {dG_o}{dt} = \alpha G_c - \beta G_o$

And as the total number of gates is $G_{max} = G_o + G_c$:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o$

At steady-state, i.e. when $G_o$ stops changing, you have:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o = 0$

which, solved for $G_o/G_{max}$, yields:

$\frac {G_o}{G_{max}} = \frac {\alpha}{\alpha + \beta}$

The ratio $G_o/G_{max}$ is the probability that a single gate is open. As you say, a 'channel' is open when 4 of its gates are open. If we consider each gate being open or closed independently, the probability that 4 gates in a channel are open is indeed $(G_o/G_{max})^4$

What you measure is the time it takes, when the voltage is abruptly changed from an initial value to a final value, for the number of open channels to go from an initial value to 'equilibrium'.
This time is in theory infinite, but in practice it is probably the time it takes for the number of open gates to go from its steady-state value at voltage $V_1$, to a value sufficiently close to the new steady-state value at voltage $V_2$.

If so, it is sufficient to solve the above differential equation for $G_o$ with initial conditions $(t=0,G_o=G_{o,V_1})$. The solution is:

$\frac {G_o(t)}{G_{max}} = \frac {G_{o,V_1}}{G_{max}} \cdot e^{-{(\alpha+\beta)t}} + \frac {G_{o,V_2}}{G_{max}} \cdot [1-e^{-{(\alpha+\beta)t}}]$

where:

$\frac {G_{o,V_i}}{G_{max}} = \frac {\alpha(V_i)}{\alpha(V_i) + \beta(V_i)}$

The difficulty I see is the arbitrary nature of the 'time' you consider sufficient to reach steady-state. I would rather fit time points, i.e. number of open gates vs time. And my earlier suggestion to use numerical fitting may still be valid, actually.