Neuron equilibrium potential
Before finishing up the final step in the detection process (generating an output), we will need to use the concept of the equilibrium membrane potential, which is the value of \(V_m\) that the neuron will settle into and stay at, given a fixed set of excitatory and inhibitory input conductances (if these aren’t steady, then the \(V_m\) will likely be constantly changing as they change). This equilibrium value is interesting because it tells us more clearly how the tug-of-war process inside the neuron actually balances out in the end. Also, we will see in the next section that it is useful mathematically.
To compute the equilibrium membrane potential (\(V_m^{eq}\)), we can use an important mathematical technique: set the change in membrane potential (according to the iterative \(V_m\) updating equation from above) to 0, and then solve the equation for the value of \(V_m\) under this condition. In other words, if we want to find out what the equilibrium state is, we simply compute what the numbers need to be such that \(V_m\) is no longer changing (i.e., its rate of change is 0). Here are the mathematical steps that do this:
\[ V_m(t) = V_m(t-1) + dt_{vm} \left[ g_e (E_e-V_m) + g_i (E_i-V_m) + g_l (E_l-V_m) \right] \]
This is the part that is driving the changes (time constant omitted as we are looking for equilibrium):
\[ \Delta V_m = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) \]
which we set to zero to find when it stops changing:
\[ 0 = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) \]
and then do some algebra to solve for \(V_m\):
\[ V_m = \frac{g_e}{g_e + g_i + g_l} E_e + \frac{g_i}{g_e + g_i + g_l} E_i + \frac{g_l}{g_e + g_i + g_l} E_l \]
The detailed math is shown below in Derivation.
In words, this says that the excitatory drive \(E_e\) contributes to the overall \(V_m\) as a function of the proportion of the excitatory conductance \(g_e\) relative to the sum of all the conductances (\(g_e + g_i + g_l\)). And the same for each of the others (inhibition, leak). This is just what we expect from the tug-of-war picture: if we ignore \(g_l\), then the \(V_m\) “flag” is positioned as a function of the relative balance between \(g_e\) and \(g_i\) — if they are equal, then \(g_e / (g_e + g_i)\) is .5 (e.g., just put a “1” in for each of the g’s — 1⁄2 = .5), which means that the \(V_m\) flag is half-way between \(E_i\) and \(E_e\).
So, all this math just to rediscover what we knew already intuitively! (Actually, that is the best way to do math — if you draw the right picture, it should tell you the answers before you do all the algebra). But we’ll see that this math will come in handy next.
Here is a version with the conductance terms explicitly broken out into the “g-bar” constants and the time-varying “g(t)” parts:
\[ V_m = \frac{\overline{g}_e g_e(t)}{\overline{g}_e g_e(t) + \overline{g}_i g_i(t) + \overline{g}_l} E_e \frac{\overline{g}_i g_i(t)}{\overline{g}_e g_e(t) + \overline{g}_i g_i(t) + \overline{g}_l} E_i + \frac{\overline{g}_l}{\overline{g}_e g_e(t) + \overline{g}_i g_i(t) + \overline{g}_l} E_l \]
For those who really like math, the equilibrium membrane potential equation can be shown to be a Bayesian optimal detector.
Derivation
This shows all the algebra to derive the equilibrium membrane potential from the update equation — it will only be viewable on the PDF version.
iterative Vm update equation:
\[ V_m(t) = V_m(t-1) + dt_{vm} \left[ g_e (E_e-V_m) + g_i (E_i-V_m) + g_l (E_l-V_m) \right] \]
just the change part:
\[ \Delta Vm = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) \]
set it to zero:
\[ 0 = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) \]
solve for Vm: (multiply all the g’s through)
\[ 0 = g_e E_e - g_e V_m + g_i E_i - g_i V_m + g_l E_l - g_l V_m \]
solve for Vm: (gather Vm terms on other side)
\[ g_e V_m + g_i V_m + g_l V_m = g_e E_e + g_i E_i + g_l E_l \]
solve for Vm: (get only one Vm guy, then divide each side by g’s to get..)
\[ V_m (g_e + g_i + g_l ) = g_e E_e + g_i E_i + g_l E_l \]
solution!
\[ V_m = \frac{g_e E_e + g_i E_i + g_l E_l}{g_e + g_i + g_l} \]
Another way of writing this solution, which makes its meaning a bit clearer, is:
\[ V_m = \frac{g_e}{g_e + g_i + g_l} E_e + \frac{g_i}{g_e + g_i + g_l} E_i + \frac{g_l}{g_e + g_i + g_l} E_l \]
In the Adaptive Exponential function, there is an adaptive factor \(\omega\) (greek omega) that enters into the membrane update equation, which we can include in our equilibrium calculation:
iterative Vm update equation:
\[ V_m(t) = V_m(t-1) + dt_{vm} \left[ g_e (E_e-V_m) + g_i (E_i-V_m) + g_l (E_l-V_m) - \omega \right] \]
just the change part:
\[ \Delta Vm = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) - \omega \]
set it to zero:
\[ 0 = g_e \left(E_e-V_m\right) + g_i (E_i-V_m) + g_l (E_l-V_m) - \omega \]
solve for Vm: (multiply all the g’s through)
\[ 0 = g_e E_e - g_e V_m + g_i E_i - g_i V_m + g_l E_l - g_l V_m - \omega \]
solve for Vm: (gather Vm terms on other side)
\[ g_e V_m + g_i V_m + g_l V_m = g_e E_e + g_i E_i + g_l E_l - \omega \]
solve for Vm: (get only one Vm guy, then divide each side by g’s to get..)
\[ V_m (g_e + g_i + g_l ) = g_e E_e + g_i E_i + g_l E_l - \omega \]
solution:
\[ V_m = \frac{g_e E_e + g_i E_i + g_l E_l - \omega}{g_e + g_i + g_l} \]
And here is the derivation of the equation for \(g_e^{\Theta}\):
equilibrium Vm at threshold:
\[ \Theta = \frac{g_e^{\Theta} E_e + g_i E_i + g_l E_l}{g_e^{\Theta} + g_i + g_l} \]
solve for g_e:
\[ \Theta (g_e^{\Theta} + g_i + g_l) = g_e^{\Theta} E_e + g_i E_i + g_l E_l \]
(multiply both sides by g’s), then solve for g_e:
\[ \Theta g_e^{\Theta} = g_e^{\Theta} E_e + g_i E_i + g_l E_l - \Theta g_i - \Theta g_l \]
(bring non-g_e’s back to other side), then solve for g_e:
\[ g_e^{\Theta} \Theta - g_e^{\Theta} E_e = g_i (E_i - \Theta) + g_l (E_l - \Theta) \]
(bring other g_e over and consolidate other side, then divide both sides to isolate g_e to get), solution:
\[ g_e^{\Theta} = \frac{g_i (E_i - \Theta) + g_l (E_l - \Theta)}{\Theta - E_e} \]
In the AdEx function, there is an adaptive factor \(\omega\) (greek omega) that enters into the membrane update equation, which we can include in our equilibrium calculation as above:
equilibrium Vm at threshold:
\[ \Theta = \frac{g_e^{\Theta} E_e + g_i E_i + g_l E_l - \omega}{g_e^{\Theta} + g_i + g_l} \]
solve for g_e:
\[ \Theta (g_e^{\Theta} + g_i + g_l) = g_e^{\Theta} E_e + g_i E_i + g_l E_l - \omega \]
(multiply both sides by g’s), then solve for g_e:
\[ \Theta g_e^{\Theta} = g_e^{\Theta} E_e + g_i E_i + g_l E_l - \Theta g_i - \Theta g_l - \omega\]
(bring non-g_e’s back to other side), then solve for g_e:
\[ g_e^{\Theta} \Theta - g_e^{\Theta} E_e = g_i (E_i - \Theta) + g_l (E_l - \Theta) - \omega \]
(bring other g_e over and consolidate other side, then divide both sides to isolate g_e to get), solution:
\[ g_e^{\Theta} = \frac{g_i (E_i - \Theta) + g_l (E_l - \Theta) - \omega}{\Theta - E_e} \]