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Hi,

I am modelling a soma with nseg=1 and a dendrite with nseg=2.
I know from reading the neuron textbook that the nseg governs
how the internals nodes are placed. Let's also assume that
the soma and dendrites in my model are represented by simple
RC circuits.

(i)Since the soma has nseg=1, the nodes are placed at 0,1/2 and 1?

(ii) Figure 3.5A in the NEURON textbook looks to me like it has nseg=2,
however the caption in the diagram says it has 3 nodes. If nseg=2,
should there not be 4 nodes ie at 0,0.25,0.75 and 1?

(ii) Equations 3.29a and 3.29b in the textbook can be used to solve
for v1(t) and v2(t). Assuming that Figure 3.5A represents a dendrite (dend[0])
with nseg=2, then
would dend[0].v(0.25) be v1(t) ?? (Eq 3.30a)
and dend[0].v(0.75) be v2(t) ?? (Eq 3.30b)

Can you please advise?

Thanks,
Meena

Good questions.

Meena wrote:I am modelling a soma with nseg=1 and a dendrite with nseg=2.

It's not a good idea to make nseg even. v(0.5) will report v at one of the two
adjacent internal nodes (in this case, v(0.25) or v(0.75)), depending on
roundoff error. It is never good for one's fate to be at the mercy of roundoff error.

(i)Since the soma has nseg=1, the nodes are placed at 0,1/2 and 1?

True.

(ii) Figure 3.5A in the NEURON textbook looks to me like it has nseg=2,
however the caption in the diagram says it has 3 nodes.

It has three nodes. Two are ungrounded (one labeled V1, the other labeled V2),
and the third node is grounded. Also see sentence 2 of paragraph 2 on page 45,
which states that the circuit in Fig. 3.4 has "two nodes, but one is grounded so
its potential is 0."

In the context of continuous system solutions on a digital computer, the ground
node is of no particular interest because its potential is always 0, i.e. no need
to solve for it. One is concerned only with the locations at which the values of
time- and space-continuous variables are unknown and must therefore be
calculated. One discretizes a continuous system in order to sample a large
extent of space and time in sufficient detail (fine enough spatial grid, small
enough time steps) in order to generate solutions that approximate the
behavior of the continuous system with acceptable error. The points in space
that are sampled are called nodes.

Returning to NEURON--if, for a particular section, one assigns nseg = 1, then
NEURON will solve a set of ODEs that describe that section's continuous
variables at one point in space, so one says that "the section has one internal
node."

If nseg=2, should there not be 4 nodes ie at 0,0.25,0.75 and 1?

There are. Try However, only the internal nodes will be associated with ODEs. Potential at 0
and 1 will be calculated as the weighted average of the potentials at adjacent
internal nodes
v(0 or 1) = SUMMA (vadjacent / rax) / SUMMA (1/rax)
where the SUMMA are over all adjacent internal nodes, and the rax are the
axial resistances that separate the 0 (or 1) end from the corresponding
internal node. And potential at the ground node (which always exists, even if
we don't talk about it) is always 0.

(ii) Equations 3.29a and 3.29b in the textbook can be used to solve
for v1(t) and v2(t). Assuming that Figure 3.5A represents a dendrite (dend[0])
with nseg=2, then
would dend[0].v(0.25) be v1(t) ?? (Eq 3.30a)
and dend[0].v(0.75) be v2(t) ?? (Eq 3.30b)

True.

At the risk of muddying the waters, I should mention that inserting extracellular
causes a whole bunch of new nodes to be created, which sit between the section
and ground--for a diagram, see the Programmers' Reference documentation
about extracellular.