When I(inward) = I(outward) dV/dt = 0 right?

[Bill Connelly]

Hi.

I made a simple soma with Cm and HH. I added an AlphaSynapse. Then I summed all the currents and compared this with the soma.v. I was assuming that when the total current was outward, soma.v would be depolarizing. When the total current was inward, soma.v would be hyperpolarizing and when the total current was zero soma.v would be stable (i.e. the peak of the EPSP).

However, this was not the case.

The currents I summed were
soma.i_cap(0.5)
soma.ik(0.5)
soma.il_hh(0.5)
soma.ina(0.5)
AlphaSynapse[0].i

In hindsight, the reason for this effect is that all those 'currents' are actually in mA/cm^2 apart from AlphaSynapse right? So how do I convert those currents into whole cell currents? (A way that didn't just use 4 . pi . r^2 would be good)

Thanks.

[ted]

Seems easier to convert the single current that is in "absolute" units into a value in "density"
units. See the Programmer's Reference via its link at
http://www.neuron.yale.edu/neuron/docs
for documentation of the area function, which returns the area of the currently accessed
segment in um^2.

[Bill Connelly]

So a soma with a length of 10, and a diam of 10 has an area of 314?
That means that neuron is modelling it as a cylinder without the ends. For a section where Length = radius doesn't that mean neuron is 50% out when calculating currents given in current/cm^2?

[ted]

doesn't that mean neuron is 50% out when calculating currents

No.

NEURON assumes a "sealed end" boundary condition (longitudinal current == 0 and
dv/dx == 0) at the 0 or 1 end of any section that is not attached to another section at that
point. If you want a different boundary condition, either attach a section to the point of
interest, or attach a point process that implements the desired boundary condition (e.g.
an IClamp, SEClamp, ExpSyn, etc.).