Notes on Neurons

Transporters, mainly sodium-potassium pumps, water channels, and gated ion channels, mainly the potassium leak channel, control the osmotic pressure of a cell and maintain the concentrations of sodium, potassium, and chloride at about 14, 140, and 14 mM inside the cell when it is immersed in a bath in which the Na+, K+, and Cl- concentrations are 140, 3, and 146 mM.   The resulting resting voltage across the membrane is
ΔV = V2 - V1 = Vinside - Voutside = -70 mV,
and the electric field E points into the cell.

A nerve cell, or neuron, consists of a cell body with branching dendrites and a long axon, which may range from 1 mm to 1 m in length, with its terminal branches which make synapses with target cells.  Axon diameters run from less than a micron to more than a millimeter.

The dendrites receive incoming signals; the axon and its branches carry signals to other cells.  

Nerve signals are called action potentials.   They are pulses of about 1 ms in which ΔV is + 40 mV instead of the resting potential of -60 or -70 mV.   Thus the action potential is a change in ΔV of about 100 mV.   These pulses travel with little change in shape at constant speeds in the range of 0.1 to 120 m/s.  

For about 1 ms after an action potential passes a given part of an axon, that part of the axon is hard to stimulate.   This pause is called the refractory period.   After the action potential passes a given part of an axon, ΔV recovers, overshoots, and then returns to its resting value in about 5 ms.   This behavior is called afterhyperpolarization .  

Neural contacts occur at synapses, most often when a branch of an axon touches a dendrite or the body of a target cell.

In this micrograph of the cell body of a motor neuron of a spinal cord, the axon terminals are red and the cytoskeleton is green:


The gap between the end of an axon branch and a dendrite (or a spine on a dendrite) of a target neuron is between about 10 and 30 nm; it is called the synaptic cleft.  Here are cartoons of a resting and an active synapse:

Neurotransmitters, such as acetylcholine, serotonin, GABA, glycine, and glutamate, can  open transmitter-gated sodium, potassium, and chloride channels.  The excitatory neurotransmitters open sodium channels, and the resulting influx of sodium depolarizes the membrane, making ΔV = V2 - V1 = Vinside - Voutside more positive and
possibly leading to an action potential.   The inhibitory
neurotransmitters open potassium or chloride channels.  The efflux of potassium further polarizes the membrane, making ΔV = V2 - V1  more negative.  The concentration of Cl- is 10 times higher outside the cell, and so when the chloride channels open, the influx of Cl- tends to further polarize the membrane.  So both potassium and chloride transmitter-gated channels tend to stop the generation of action potentials.

Mammalian brains add extra non-NMDA glutamate-receptor sodium channels after repeated entries of Ca2+ thru the NMDA receptor:

How mammals learn.

Long-term potentiation.
Here is a scanning electron micrograph of a neuromuscular junction in a frog:

Cartoon of neuromuscular junction.
Glial cells (Schwann cells and oligodendrocytes) wrap their plasma membranes about the axons of many of the neurons of vertebrates forming myelin sheaths .   The voltage-gated sodium channels are concentrated in gaps in the myelin, the nodes of Ranvier, about 1 mm apart. The action potential jumps from node to node, traveling faster than in unmyelated axons.   The disease multiple sclerosis destroys some of the myelin in the central nervous system, slowing down action potentials with devastating results.  

The human brain has about 100 billion (1011) neurons.   Here is a drawing by Golgi of a small part of the olfactory bulb (sense of smell) of a dog's brain: 

What happens if we electrically stimulate a nerve cell? If we slightly decrease ΔV throughout a small patch of the membrane of a nerve cell and then measure ΔV at points several mm away, we find a reduced decrease ΔV that fades with time and distance.   The cell resets its ΔV to - 60 mV, and at distances of 100 mm, the change in ΔV is very small.   A neuron recovers from a slight depolarization.   If we slightly increase Δ V in a small patch of membrane, then at points several mm away ΔV will show an increase that fades with distance and time.   The cell resets its ΔV to -60 or -70mV.   A neuron recovers from a slight hyperpolarization.   This passive spread of electrical signals is called electrotonus.   In both depolarization and hyperpolarization, the magnitude of the spreading change in ΔV is proportional to the stimulus: the response is graded.  

But if we depolarize the membrane by more than about 10 mV, then we can set in motion an action potential.  

The huge difference in the response is due to voltage-gated Na+ channels:  

How do we describe action potentials mathematically?   We note that the voltage drop across the membrane ΔV = V2 - V1 is the same for all three species, Na+, K+, and Cl-.   This voltage is the sum
ΔV = IiRi + VNi             (12.1)
of the Ohmic drop IiRi and the Nernst potential
ViN = - (kT/zie) ln(ci2/ci1) .             (11.1)
Here Ri = 1/(A gi) is the resistance, A the area, and gi the conductance, and Ii = jq,i A is the current of species i.  

Next, we will ignore the myelin and the sodium-potassium pumps.   Instead, we'll just assume the resting values of the bulk concentrations and of ΔV mentioned above. In any case, these values remain fairly constant for about 6 105 s after the pumps are stopped.   We'll denote the quasi-steady value of ΔV by V0.   If gi is the conductance of species i, then the current of that species will be (V0 - ViN)gi.   Since there is no build-up of charge inside the axon, the sum of these currents must vanish
i (V0 - ViN)gi = 0.             (12.2)
Let gtot denote the sum of the three conductances
gtot = ∑i gi.  
Then the quasi-steady membrane voltage is
V0 = ∑i (gi/gtot) VNi             (12.3)
which is called the chord conductance formula.   The relations gK+ = 25 gNa+ = 2 gCl- tell us that gCl- = 12.5 gNa+, and so that gtot = 38.5 gNa+.   The Nernst potentials from Table 11.1 are VNK+ = -75 mV, VNNa+ = 54 mV, and VNCl- = -59 mV.   So the chord conductance formula gives
V0 = (25/38.5)(-75) + (1/38.5) 54 + (12.5/38.5)(-59) mV = -66 mV
which is not very different from the value of -72 mV found in Eq.(11.12).  

The axon's membrane is a capacitor. When a voltage V is applied across it, opposite charges q and -q appear within a Debye length of the membrane.   If C is the capacitance of the membrane, then the charge q and the voltage V are related by
q = CV.             (12.4)
If the voltage is changing with time, then by differentiating the preceding equation, we find
(dq/dt) = I = C (dV/dt).             (12.5)
This current is called a capacitive current.   The capacitance C is proportional to the area A of the axon, C = A C.   Experimentally,
C = 10-2 F m-2 = 1 µ F cm-2.  

We still have not modeled the voltage-gated sodium channels, so we don't expect to get much of an action potential with the present model, but let's push ahead anyway.   Consider a cylindrical slice of radius a and thickness dx of an axon that points in the x-direction.   Let Ix(x) be the current flowing into the cylindrical slice at x.   Then -Ix(x+dx) is the current flowing into the cylindrical slice at x+dx.   The radial current flowing out is 2 π a dx jq,r plus the pile-up rate 2 π a dx C (dV/dt).   Since charge is conserved, we have
Ix(x) -Ix(x+dx) = - (dIx(x)/dx) = 2 π a (jq,r + C (dV/dt))dx.             (12.6)

We can relate the current Ix to the voltage difference
ΔVx = V(x+0.5dx) -V(x-0.5dx)
and the resistance of the axoplasm (the axon's cytoplasm)
R = dx/(π a2 κ)
by the formula
ΔVx = Ix R.  
Thus we get
Ix = - (V(x+0.5dx) -V(x-0.5dx))/R = - π a2 κ (dV/dx).             (12.6.5)

If we now put Eq.(12.6.5) into (12.6), then we find
π a2κ (d2V/dx2) = 2 π a ( jq,r + C (dV/dt) )             (12.7a)
aκ(d2V/dx2) = 2 ( jq,r + C (dV/dt) )             (12.7b)

which is the cable equation.  

Now V0 is the quasi-steady voltage across the membrane — the voltage that gives no net radial flux jq,r.   So the outward radial current flux is
jq,r = (V - V0)gtot .  

Now let
v(x,t) = V(x,t) -V0 .  
And let's define the axon's space constant as
λ = (aκ/2gtot)1/2             (12.8a)
and its time constant as
τ = C/gtot.             (12.8b)
With these definitions and assumptions, the cable equation becomes the linear cable equation
λ2 (d2v/dx2) -τ (dv/dt) = v.             (12.9)

Letting v(x,t) = exp(-t/τ) w(x,t), we find that (12.9) becomes
(λ2) (d2w/dx2) = dw/dx
which is the diffusion equation we studied in Sec.(4.6.5).   Using its solutions, we have with  N a normalization constant
v(x,t) = N exp(-t/τ) t-1/2 exp(-x2/(42/τ)             (12.10)
which is the passive-spread solution.   With a = 0.5 mm, gtot = 5 m-2Ω-1, C =10-2F m-2, and κ = 3 Ω-1 m-1, we get
λ 12 mm            (12.11a)
and τ = 2 ms.             (12.11b)

Here is a cartoon that describes both voltage-gated sodium channels
(slightly better) voltage-gated potassium channels:

Cartoons of voltage-gated sodium channels.

When the membrane is depolarized by more than about 10 mV,  the voltage-gated sodium channels open.  When enough of them are open on a given patch of the axon's membrane, the sodium conductance momentarily jumps from the value of Table 11.1, which was gNa+   = 0.04 gKto
gK+ = 0.05 gNa+ = 2gCl-          (12.14)

that is, the sodium conductance jumps from 

gNa+   = 0.04 gKto gNa+   = 20 gK+

which is to say by a factor of 25 x 20 = 500.   Sodium ions come flooding in when the sodium channels are open.  

The voltage-gated potassium channels are slower than the voltage-gated sodium channels to respond to changes in the voltage across the membrane.   They open only after this voltage V = V2 - V1 has reached its maximum of about +40 mV.

The action potentials of some muscle, egg, and endocrine cells depend upon voltage-gated calcium channels.

We seek a traveling-wave solution, that is, one of the form
V(x,t) = V(t - x/s)   
where s is the speed of the wave.   The text has a tilde on the second V and uses a Greek theta for the speed, but I am not good enough at html to be so fancy.  Now the spatial derivative of this
traveling wave is
V(x,t)/dx = dV(t - x/s)/dx = (-1/s) dV(t - x/s)/dt      (12.145)
and its second
spatial derivative is
d2V(x,t)/dx=  (1/s2) d2V(t - x/s)/dt2   .     (12.146)

If we put this
traveling-wave solution into the cable equation (12.7b), then we find

(aκ/(2s2)) d2V/dt2 jq,r + C (dV/dt)             (12.15)

which is an ordinary, not a partial, differential equation.

Our simplified voltage-gating hypothesis is that the outward radial current  jq,r is

   jq,r  = i ( V - ViN ) gi(V).               (12.17)

Now we need to model the sodium conductance somehow.  The text uses the abbreviation v = V - V0 where V0 is the resting potential and models gNa+ (v) by means of the quadratic formula
gNa+ (v) = g0NaB v2  .                       (12.19)

Nelson then finds an action potential that rises from 0 to a voltage he calls v2
But an action potential rises and then falls slightly below 0.  Also, the voltage-gated sodium channels stop the sodium current once the voltage v has risen to about 100 mV (V rises from -60 mV to + 40 mV).   So  Eq.(12.19)  is a very bad approximation. 

So let's try a more general expression for the sodium conductance:
gNa+ (v) = g0Naf(v) .                                           (A.1)
Then the outward radial current is from (12.17)
jq,r  = i ( V - ViN ) gi0(V)  +  ( V - VNa+N ) f(v)                 (A.2)
 or by (12.2-3)
 jq,r  = ( V - V0 ) gtot( V - V- (VNa+N - V0) ) f(v)         (A.3)

      = v
gtot +  (v - H )  f(v       (A.4)
where H = VNa+N - V0 = 54 -(-66) mV = 120 mVSo the cable equation (12.15) now is
(aκ/(2s2)) d2V/dt2 =  v gtot +  (v - H f(v + C (dV/dt)                (A.5)
or with v(t) = V(t - x/s) - V0 more simply
(aκ/(2s2)) d2v/dt2 =  v gtot +  (v - H f(v + C (dv/dt) .               (A.6)
Now  v = V - V0 is between zero and about 100 mV, so the second term on the right-hand side of this equation is negative and represents the flow of sodium
thru the voltage-gated sodium channels into the axon.  

Let's simplify (A.6).  First, we drop the first term
(aκ/(2s2)) d2v/dt2, which is reasonable when the  radius a of the axon is half a micron, as in mammals, rather than half a millimeter, as in giant squids. 

Next, we take
the term  f(v) to be
f(v) =  f θ(v - 0.01) θ(v - 0.1)         (A.7)  
in which the Heaviside function
θ(x) = 1 if x is positive and 0 otherwise and f is a constant, which we'll take to be 
 f = gNa+ = 20 gK+ = 20 x 25 x gtot/38.5 = 65 m-2 Ω-1  .     (A.8)
So we now have
0 =
v gtot +  (v - H ) f θ(v - v1 ) θ(v2 - v) + C (dv/dt) .               (A.9)

Now we imagine that at time t
= 0 the voltage v = v1 which we take to be just enough to open the voltage-gated sodium channels, roughly 10 mV.  While these channels are open,  Eq.(A.9) is
C (dv/dt) = f - ( gtot  f ) v .       (A.10)
We write this first-order, linear, inhomogeneous differential equation as
dv/[ f - ( gtot  f ) v ]  = dt/C    (A.11)
which we may integrate to
- (1/(
gtot + f )) ln {[ Hf - ( gtot + f ) v ]/[ Hf - ( gtot + f ) v1 ]} = t /(A.12)
[ Hf - ( gtot + f ) v ]/[ Hf - ( gtot + f ) v1 ] = e -(gtot + f ) t /C .    (A.12)
A little high-school algebra now gives
  v(t) = Hf [ 1 - e -(gtot + f ) t /C ]/( gtot + f ) + v1 e -(gtot + f ) t /C .    (A.13)
This function starts at
v1 at t = 0 and then rises to Hf /( gtot + f ).
v(t) approaches Hf /( gtot + f ), we assume that the sodium channels close at t = t2 after which time the differential equation is simply
(dv/dt) =
- gtot v .       (A.14)
To integrate this simpler equation, we write it as
dv/v = - gtot dt / C .       (A.15)
Integrating both sides, we find
ln (
v/v2 ) = - gtot ( t - t2 )/ C .       (A.16)
Thus we find
v(t) = v2 e - gtot ( t - t2 )/ C  for t > t2 .       (A.17)
Here's a plot of v(t) as given by (A.13) followed by (A.17):
Capacitive action potential

We may improve this model by including the voltage-gated potassium channels.  After the
voltage-gated sodium channels close and the voltage-gated potassium channels open, instead of (A.1), we have
gK+ (v) = g0K+  k(v) .                                           (A.18)
And instead of (A.4), we have
jq,r = v gtot +  (v - Kk(v)         (A.19)
in which K = VK+N - V0 = -75 -(-66) mV = -9 mV.   For simplicity, we will ignore the dependence of  k(v)  upon the voltage v and simply set k(v) = k a constant after the sodium channels close and while the potassium channels remain open.  Then in place of (A.9), we have
  0 =  v (gtot + k)   - k K  + C (dv/dt)                    (A.20)
which we may integrate as we did
(A.9).  We get
  v(t) k K [ 1 - e -(gtot + k ) (t-t0)/C ]/( gtot + k ) + v1 e -(gtot + k ) (t-t0) /C .    (A.21)
Here's a plot of the resulting action potential (A.13, 21, & 17):

Na-K Action Potential

The voltage v(t) rests at 0 for the first 0.5 ms; then the sodium channels open and (A.13) raises v above 0.1 V in
0.5 ms; then the sodium channels close, and the potassium channels open, and (A.21) drops v drops, overshooting; then the potassium channels close, and (A.17) returns v to its resting value of 0.   The input values were: k = 50 m-2 Ω-1,  C = 0.01 F m-2, and K = - 9 mV.

Let's now return to the full Eq.(A.6) with the Heaviside representation (A.7) of the flow thru the voltage-gated sodium channels but without the potassium channels.   Before depolarization, the voltage v is lying at 0, and the channels are closed, so (A.6) is
  (aκ/(2s2)) d2v/dt2 =  v gtot + C (dv/dt) .               (A.22)
This is a linear, homogeneous, second-order differential equation. 
We may solve it by letting v = exp(pt).   Since
dv/dt = p v, and
d2v/dt2 = p2v,  we get
(aκ/(2s2))  p2v =  v gtot + C pv               (A.23)
which is a quadratic equation for p
   (aκ/(2s2))  p2 - C p   - gtot =  0.               (A.24)

There are two solutions
    p+ = s2 (  C + ( C2 + 2 gtot/s2 )1/2  ) /(aκ)    >  0    (A.25)
   p- = s2 (  C - ( C2 + 2 gtot /s2 )1/2  ) /(aκ)    < 0 .    (A.26)
The general solution to Eq.(A.22) is then
  v = c1 exp(
p+t ) + c2 exp( p-t )    .         (A.27)
There are three time zones here:  before the v-g sodium
channels open, while they are open,
and after they close.  Since the voltage v relaxes to 0 in the far past and in the far future, the solution
before the v-g sodium channels open must be
  v(t) =
c1 exp( p+t )  for t < t1         (A.28)
  v(t) = c2 exp( p-t )  for t > t2        (A.29)
after they close.

While they are open,  Eqs.(A.6&7) give
  (aκ/(2s2)) d2v/dt2 =  v (gtot + f)   - H + C (dv/dt)               (A.30)
which is a linear, second-order, inhomogeneous DEQ.
It is easy to find a particular solution of the
inhomogeneous equation: the constant
  v(t) = H f/(
gtot+ f )               (A.31)
will do.  To find the general solution of the homogeneous equation, we try v(t) = exp(qt).
Inserting this into
(A.26), we find that q must satisfy the quadratic equation
(aκ/(2s2)) q2 - Cq - (gtot + f)   =  0   .     (A.32)
The solutions are
    q+ = s2 (  C + ( C2 + 2 aκ(gtot+ f ) /s2 )1/2  ) /(aκ)    >  0  (A.33)
   q- = s2 (  C - ( C2 + 2 aκ(gtot + f )/s2 )1/2  ) /(aκ)    < 0 . (A.34)
So the general solution to the
inhomogeneous equation (A.26) is
v(t)  =  
d1 exp( q+t ) + d2 exp( q-t )   +  H f/( gtot+ f )    for   t<  t < t2  .         (A.35)

The voltage-gated sodium channels open when v = v1
  and  close when v = v2 .
So the solutions (A.24&25) must be
  v(t) = v1 exp( p+(t - t1 ) )     for t < t1         (A.36)
 v(t) = v2 exp( p-(t - t2 ) )       for t > t2      .  (A.37)

Let's allow a discontinuity in the time derivative
dv/dt of the voltage
at t =
due to an applied stimulating voltage or to the sudden opening of the transmitter-gated sodium channels.  So we will write (A.31) as
v(t)  =  
d1 exp( q+(t - t1 )) + d2 exp( q-(t - t1 ))   +  H f/( gtot+ f )    for   t<  t < t2           (A.38)
with coefficients that satisfy
  d1 + d2   H f/( gtot+ f )  = v1   .     (A.39)
Here is a plot of (A.33-35) for s = 10 m/s, t1 = 0.0001 s, t2 = 0.0005 s, and d1 = 0.01 V:

Action potential

This plot has two defects:  the time-derivative dv/dt is discontinuous at t2 = 0.0005 s and the voltage remains positive; there is no overshoot due to an efflux of potassium ions.  To try to remedy these defects, we model the voltage-gated potassium channels. 

First, we use (A.18 & 19) to model the
voltage-gated potassium channels after the voltage-gated sodium channels have shut down.     So instead of (A.1), we have
gK+ (v) = g0K+  k(v) .                                           (A.40)
And instead of (A.4), we have
jq,r = v gtot +  (v - Kk(v)         (A.41)
in which K = VK+N - V0 = -75 -(-66) mV = -9 mV.   For simplicity, we will ignore the dependence of  k(v)  upon the voltage v and simply set k(v) = k a constant after the sodium channels close and while the potassium channels remain open.  Then in place of (A.30), we have
  (aκ/(2s2)) d2v/dt2 =  v (gtot + k)   - k K  + C (dv/dt)     .               (A.42)
The general solution v(t) to this linear, inhomogeneous, second-order ordinary DEQ is the sum of a particular solution to the
inhomogeneous equation, such as
  v = k K/(
gtot + k)                                                              (A.43)
and the general solution of the
homogeneous equation
  (aκ/(2s2)) d2v/dt2 =  v (gtot + k)  + C (dv/dt)     .               (A.44)
The general solution to this linear,
homogeneous equation is
  v(t) =  e1 exp(r+ t) + e2 exp(r- t)   
in which
s2 (  C + ( C2 + 2 aκ(gtot+ k ) /s2 )1/2  ) /(aκ)    >  0     (A.46)
  r-s2 (  C - ( C2 + 2 aκ(gtot+ k ) /s2 )1/2  ) /(aκ)    <  0 .    (A.47)
So the
general solution v(t) to Eq.(A.44) is
v(t) = e1 exp(r+ t) + e2 exp(r- t)  +  k K/(gtot + k)  .     (A.48)

We now stitch together the solution v(t) = 0 while the neuron is resting, followed by the solution
(A.35) for 0.5 ms while the voltage-gated sodium channels are open (and the K ones closed), followed by the solution (A.48) for 0.5 ms while the voltage-gated potassium channels are open (and the Na ones closed), followed by the solution (A.29) for 1 ms after all the voltage-gated ion channels are closed.  We match the coefficients to keep the time-derivative dv/dt continuous except at t = 0.0005 s when the voltage v jumps due to the opening of the transmitter-gated sodium channels, which start the action potential.  It is interesting that for k = 50 m-2 Ω-1K = - 9 mV, a = 0.5 mm,  andκ= 3.0 m-1 Ω-1, the resulting solution v(t) behaves weirdly unless the speed s lies within the interval 67 ≤    77 m/s.

Here is a plot for speed s = 75 m/s:

Action Potential for s = 75 m/s

The voltage v stays at zero until the initial depolarization opens the sodium channels at t = 0.5 ms.  The influx of Na+ ions increases v until around v = 0.1 V the sodium channels close at t = 1 ms, and the potassium channels open.  Then the voltage v drops below zero until the potassium channels close at 2 ms.  Then it slowly recovers to its resting value of zero, which the sodium-potassium pumps maintain.  In real life, the potassium channels are more sluggish.

In Eq.(A.42), the speed s enters only thru the combination
(aκ/(2s2)) and so s must be proportional to the square-root of aκ.  So if we reduce the radius a by a factor of 4, then the allowed speed s should drop by a factor of 2.    Here's a plot for a = 0.000125 and s = 37.5 m/s with all the other parameters unchanged:

Action Potential for Thinner Axon