Peter Latham, February 6, 2018 1 Dendrites and axons 1 Dendrites Our goal is to write down an equation describing the membrane potential in a dendrite, and then solve it. The derivation is mainly physics and the solution is mainly math. The solution is important: it will give us intuition about how dendrites work; combined with some experimental facts, it explains the size of dendritic trees. We'll initially consider a dendrite to be an infinitely long cylinder. At the end we'll talk about more realistic geometries. 1.1 The cable equation A short section of a dendrite is shown in Fig. 1 (see the figure caption for details). We're going to use only two equations: V = IR and Q = CV , the latter implying CdV=dt = I. Here V and I are voltage and current, and C and R are resistance and capacitance. We'll start with CdV=dt = I. Treating each section as equipotential, we have @V (x; t) C = I (x − dx=2; t) − I (x + dx=2; t) − I (x; t) + I (x; t) (1) m @t L L m e where IL is the longitudinal current, Im is the current due to membrane channels, Ie is the injected current, Cm is the membrane capacitance and RL is the longitudinal resistance. And now we use V = IR, V (x − dx; t) − V (x) I(x − dx=2; t) = (2) RL where RL is the longitudinal resistance; a similar equation applies to I(x+dx=2; t). Inserting Eq. (2) into (1), we have @V (x; t) V (x − dx; t) − V (x; t) V (x; t) − V (x + dx; t) Cm = − − Im(x; t) + Ie(x; t): (3) @t RL RL Taylor expanding membrane potential to second order in dx gives us @V (x; t) dx2 @2V (x; t) Cm = 2 − Im(x; t) + Ie(x; t): (4) @t RL @x That's the easy part. The hard(er) part is taking the limit dx ! 0. Basically, we have to figure out how Cm and RL scale with dx. We'll start with RL. It's a physics fact that resistance scales linearly with the length of a material and inversely with its area, with a scale factor that's an intrinsic property of the material. For the longitudinal resistance, we'll use rL for that intrinsic property, giving us dx R = r : (5) L L πa2 Peter Latham, February 6, 2018 2 Im(x) Ie(x) a IL(x-dx/2,t) IL(x+dx/2,t) V(x-dx,t) V(x,t) V(x+dx,t) x-dx x-dx/2 x x+dx/2 x+dx Figure 1: A short section of a cylindrical dendrite with radius a. The vertical bars are spaced by dx, which we'll eventually take to zero, so the drawing is slightly misleading: the bars should be very close together (think stacks of pancakes), but then there would be no room to show the currents. There are three kinds of currents: longitudinal current, IL, which flows along the dendrite; current due to membrane channels, Im (which is outward by convention); and injected current, Ie (which is inward by convention). Here dx is the length and πa2 is the area (remember the dendrite is cylindrical). Capacitance also depends on area. To see how, we need a physics fact about membrane potential: inside the dendrite, at any particular value of x, the membrane potential is con- stant. That means V (x) is the change in potential across the membrane, and that change is caused by charge building up on the inside and outside of the membrane. It follows (with a little thinking) that for fixed membrane potential the total charge scales linearly with area, with a scale factor the depends on the membrane. We'll use cm for that scale factor, giving us Cm = cm2πadx: (6) Inserting Eqs. (5) and (6) into Eq. (4), and performing a small amount of algebra, we arrive at @V (x; t) a @2V (x; t) cm = 2 − im(x; t) + ie(x; t) (7) @t 2rL @x where im and ie are current densities, I (x; t) i (x; t) ≡ m (8a) m 2πadx I (x; t) i (x; t) ≡ e : (8b) e 2πadx Peter Latham, February 6, 2018 3 Equation (7) is independent of dx, as desired. For dendrites (but, as we'll see in the next section, not for axons), it's convenient to do one more thing. Equation (5) is a general expression that relates total resistance to geometry and intrinsic properties. For something with fixed thickness { like the membrane of a neuron or dendrite { we can think of rL×thickness as the intrinsic membrane property, which we'll called rm. This gives us r R = m (9) m Area where Rm is the actual resistance. We're going to multiply both sides of Eq. (7) by rm. This gives us a factor of rmcm on the left hand side. To interpret this factor, we note, as discussed above, that capacitance is proportional to area. And so we may write r R C = m c Area = r c (10) m m Area m m m where Cm is the total capacitance. In general, RmCm is the membrane time constant; thus, so is rmcm. If dendrites have similar properties to cell somas, which, it turns out, they do, then rmcm is on the order of 10 ms. Multiplying both sides of Eq. (7) by rm, and defining τm ≡ rmcm; (11) we have @V (x; t) @2V (x; t) τ = λ2 − r i (x; t) + r i (x; t) (12) m @t @x2 m m m e where λ, which is known as the electrotonic length, is given by ar λ2 ≡ m : (13) 2rL Equation (12) is the cable equation. In real dendrites, the current density, im, consists of both passive and active channels (the active channels are just like the ones we saw in the Hodgkin Huxley equations). Here, however, we'll consider only passive channels, V − E 2πadx Im = = (V − E); (14) Rm rm comparing this to Eq. (8a), we see that rmim = V − E; (15) yielding @V (x; t) @2V (x; t) τ = λ2 − (V (x; t) − E) + r i (x; t): (16) m @t @x2 m e This is the famous passive cable equation. We're going to solve it first in steady state (V (x; t) is independent of time); then we'll tackle the time-dependent case. In either case, Peter Latham, February 6, 2018 4 we can simplify our equations by working with voltage relative to E; we do that by defining a new variable, u ≡ V − E; (17) giving us the equation @u(x; t) @2u(x; t) τ = λ2 − u(x; t) + r i (x; t): (18) m @t @x2 m e 1.2 The response of a dendrite to steady injected current To compute the membrane potential in steady state, we need to specify the current density, ie(x; t). We'll assume that the current is injected at one point, so it's something like I0=2πadx for x within dx=2 of zero and zero otherwise. In the limit that dx ! 0, the factor of 1=dx turns into a delta-function, leading to the equation @u(x; t) @2u(x; t) r I τ = λ2 − u + m 0 δ(x) (19) m @t @x2 2πa where δ(x) is the Dirac delta-function; it has the property that it is zero when x 6= 0 and it integrates to 1: so long as a and b are positive, then Z b dxδ(x) = 1: (20) −a The delta-function is one of the most useful functions in the world; it's a good idea to know about it. If we let the cable equation evolve for a long time, eventually we'll reach steady state { meaning @u(x; t)=@t = 0. Thus, in the limit t ! 1, our cable equation becomes @2u(x; t) λ2 − u = −R I λδ(x) (21) @x2 λ 0 where r r λ R ≡ m = L : (22) λ 2πaλ πa2 As an aside, Rλ has a very natural interpretation: Recall that total resistance of a material is equal to resistivity, rL times the length of the material divided by its area. Thus, Rλ is the resistance of a dendrite with radius a and length λ. Which, in hind sight, is not especially surprising. But it's kind of cool. Solving equations with delta-functions is often kind of easy, and that's the case here. That's because when x 6= 0, we have a simple linear ODE. The only problem is to figure out what happens at x = 0. For that we just have to take derivatives of discontinuous functions. If you know that, or you don't know it and don't care, you can skip the next couple of paragraphs. However, you'll need it to understand the solution to Eq. (21). Consider a function g(x) that's continuous everywhere except at x = x0. For instance, 2 2 we might have g(x) = x if x < x0 and g(x) = x + 4 if x > x0. We'll define g0 and g1 to be Peter Latham, February 6, 2018 5 the values of g(x) when approached from the below and above x0, respectively. In the above 2 2 example, g0 = x0 and g1 = x0 + 4. To compute a derivative, we'll use the usual expression, dg(x) g(x + /2) − g(x − /2) = lim : (23) dx !0 Most of the time this just gives us the derivative.