Introduction to Week 3 Lecture Notes Theoretical Cable Theory

Cable Theory

In this module we will derive the cable equation for cables of constant radius. The steady-state and time-dependent properties of this equation will then be analysed. The Rall ball-and-stick model of a soma coupled to a will then be introduced. Mathematical Physiology, Keener and Sneyd. Chapter 8 p249-265. Springer. Theoretical Neuroscience, Dayan and Abbott. Chapter 6 p203-220. MIT Press Spiking Models, Gerstner and Kistler. Chapter 2 p53-67. Cambridge Univeristy Press Methods in Neuronal Modelling, Koch and Segev. Chapter 2 p27-p92. MIT Press

• The current- relation in a cable Consider a short slice of dendrite that can be modelled as a cylindrical cable of radius a and length ∆. The are V (x)andV (x + ∆) at the left and right ends of the segment, respectively. The current flowing through the segment is I. From law we have:

V (x +∆)− V (x)=−IR (1) where the minus sign is because current flows down the potential gradient and R is the resistance of the short segment. On physical grounds it can be expected that the resistance is inversely proportional to the area, proportional to the length and also to some constant that has units of Ωm which we will call ra - the axial resistance:

∆ra R = . (2) πa2 Inserting this into equation (1) and re-arranging for I gives

πa2 V (x +∆)− V (x) πa2 ∂V I = − so in the limit ∆ → 0weget I(x)=− . (3) ra ∆ ra ∂x

• Derivation of the cable equation We now consider a section of cable which has leak channels, and where effects of the membrane are also accounted for. At position x of the cable, the current flowing in along the cable is I(x) and at position x + ∆ the current flowing out is I(x +∆). If C and gL are the capacitance and conductance per unit area, we get: dV IC =2πa∆C and IL =2πa∆gL(V − EL). (4) dt The total sum of currents must be zero, so

IC + IL = I(x) − I(x +∆). (5)

On inserting the forms for the current and dividing through by the surface area of the short cylinder we have ∂V 1 I(x +∆)− I(x) C = gL(EL − V ) − (6) ∂t 2πa ∆ which on the limit of ∆ → 0 yields ∂V 1 ∂I C = gL(EL − V ) − . (7) ∂t 2πa ∂x

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If a current Iapp were to be injected at x0 then by conservation of current:

I(x0 + ) − I(x0 − )=Iapp. (8) We now consider the equation for the voltage in a region of cable for which there is no injected current. Substituting in the axial current (3) and dividing by gL gives 2 ∂V 2 ∂ V τL = EL − V + λ (9) ∂t ∂x2 wherewehaveidentifiedaconstantλ with units of length that satisfies a λ2 = . (10) 2ragL The electrotonic length λ is an important quantity; it measures the extent to which voltage is attenuated in cable structures. The current voltage relation (3) can also be written in terms of this quantity: ∂V λ ∂V I = −(2πaλgL) λ = − (11) ∂x Rλ ∂x where a resistance Rλ =1/(2πaλgL) has been identified. This is the leak resistance of a section of cable λ long. It can be verified that it is also equivalent to the total axial resistance for a cable of length λ. To summarise, the equations that govern the electrical characteristics of a cable are: (8) if there is injected current; equation (9) in the bulk, away from any points of injection; equation (11) for relating voltage to current, as boundary conditions are sometimes on the current; and finally it can be noted that to prevent infinite currents there must be no jumps in the voltage V anywhere (though abrupt changes in the gradient are allowed, at the point of current injection). • Steady current injection on an infinite cable We now consider a section of cable that is infinitely long, with an applied current Iapp being injected constantly at x = 0. Imagine that the cable is horizontal, so we may say that everything to the left of the the point of injection is labelled by a 1 and everything to the right by a 2. We also measure the voltage from the baseline: v = V −EL. In the steady state the general solution of (9) is v = Ae±x/λ. (12) So, if the voltage is to remain finite when |x|→∞the solutions to the left and right of the point of current injection are x/λ −x/λ v1(x)=A1e and v2(x)=A2e (13) and because there can be no voltage jump at x =0itmustbethatA1 = A2 and we will call this quantity v0. We now write down the conservation-of-current equation where I1 is the current just before the point of injection and I2 just after. v0 v0 I1 + Iapp = I2 so that − + Iapp = (14) Rλ Rλ where the form for the current in terms of a voltage derivative has been used and the solutions (13) substituted. On solving for v0 we get

IappRλ IappRλ −|x|/λ v0 = giving the full solution V (x)=EL + e . (15) 2 2

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It can be noted that the input resistance, defined by the voltage at the point of current injection divided by the current injected, is Rin = Rλ/2 which is consistent with the two branches behaving as parallel resistors to ground with resistance Rλ each: 1 1 1 = + . (16) Rin Rλ Rλ It can be noted that the injecting electrode has an effect that decays with a length constant λ and so any experimental measurements probe only a distance λ from the recording electrode. • Boundary Effects There are two kinds of boundary that can be considered. The first is a short-circuit where the end of the cable is open to the extracellular media (for example if the dendrite was cut) so that at this point the absolute voltage is grounded V = 0mV. If such a cut boundary is located at the LHS of a semi-infinite cable, the boundary conditions require that the voltage distribution looks like −x/λ V = EL 1 − e . (17)

This is not a case which is important for the usual function of . A second more relevant case is a closed end of a dendrite. In this case it is the current at the boundary that is zero I = 0 (assuming negligible effects from channels sitting on the cylinder end). This requires that λ ∂V ∂V I = − = 0 which is a gradient condition on V so =0. (18) Rλ ∂x ∂x We will now examine the consequences of this condition. • Semi-Infinite Cable with a Closed Boundary We will now derive the distribution for a semi-infinite cable x ≥ 0 with a closed boundary at x = 0 and a steady current injection at x = x0. The solution below x0 we will call v1 and above v2, where the voltage is measured from the baseline EL. Hence

(x−x0)/λ −(x−x0)/λ −(x−x0)/λ v1 = Ae + Be and v2 = Ce . (19)

At the closed boundary at x = 0 no current flows so

Ae−2x0/λ = B. (20)

At the point of current injection the voltage does not jump and the total currents must add up to zero, where I1 is the current just to the left of the point of injection and I2 the current just to the right

A + B = C and I1 + Iapp = I2 so that A − B + C = IappRλ (21)

The three equations in (20) and (21) can be solved to give IappRλ IappRλ IappRλ A = ,B= e−2x0/λ and C = 1+e−2x0/λ . (22) 2 2 2

On insertion into the solution, and using modulus signs to write equations v1 and v2 together, we have

IappRλ IappRλ v = e−|x−x0|/λ + e−|x+x0|/λ. (23) 2 2

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This solution is similar to the case of the infinite cable, except that there appears to be also an effect of an identical strength input at x = −x0, which of course is past the end of the cable. This is the mirror image of the voltage distribution coming from the true point of injection reflected in the closed boundary at x =0.Itisanexampleofthemethod of images that can be used to solve linear equations with closed (reflecting) boundaries, once the solution on the infinite line is known. The physical interpretation is that the current “reflects” from the closed boundary. • Finite Cable with one Closed Boundary We now consider a finite cable of length L with a closed boundary at L and a voltage v0 due to a current input Iapp at x = 0. The general solution is the double exponential

v = Aex/λ + Be−x/λ (24) with the boundary conditions

2L/λ A + B = v0 at x =0 and Ae = B at x = L from the IL = 0 condition. (25)

On solving for A and B we get L−x v0 (x−L)/λ (L−x)/λ cosh λ v = e + e = v0 . (26) eL/λ + e−L/λ L cosh λ

This can be related to the current input at x = 0 and the input resistance at x =0forthe closed length of cable R0 = v0/Iapp. v0 L L I0 = tanh so that R0 = Rλ coth . (27) Rλ λ λ

We will refer to the input resistance of this structure as RC and the conductance as GC =1/RC . • The Rall Soma and Dendrite Model We now consider a model neuron made from a isopotential soma Vs attached to a single dendritic cylinder with a spatially dependent voltage Vd(x)withtheendatx = 0 attached to the soma andtheendatx = L closed. An experimentalist is injecting an applied current Iapp at the soma to measure the total input resistance of the combined structure:

∂Vs τL = EL − Vs +(Iapp − I0) Rs soma (28) ∂t ∂Vd 2 ∂Vd τL = EL − Vd + λ dendrite (29) ∂t ∂x2 where by continuity Vs = Vd(0) = V0 and where I0 is the current flowing from the soma into the dendrite,

λ ∂Vd I0 = − . (30) Rλ ∂x

As before we measure all voltages from EL. In the steady state the somatic voltage obeys

vs = v0 =(Iapp − I0)Rs (31)

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and the current I0 flowing into a closed cylinder was given in equation (27) in terms of v0. Inserting this current into equation (31) gives Rs L v0 1+ tanh = IappRs. (32) Rλ λ

So the input resistance Rin = v0/Iapp of the entire neuronal structure is

Rs Rin = . (33) Rs L 1+ Rλ tanh λ

This is consistent with the somatic and dendritic resistance being in parallel 1 1 1 = + . (34) Rin Rs RC where RC is the cylinder input resistance given in equation (27). • Conductance of Complex Dendritic Structures Dendritic structures do not contain closed loops and hence their total conductance can be calculated by starting at the ends of the dendritic branches and adding up the conductances as they join together towards the soma. For example, imagine two closed (at the end of part of the dendritic tree) that join to a third parent dendrite. The conductance at the end of the parent dendrite (where it joins the two end dendrites) is

GE = GC1 + GC2. (35)

We will consider that this is at x = L of the parent cylinder. The conductance into this cylinder is G0. If we can calculate the input conductance as a function of the conductance at the end, then this process could be continued recursively until the conductance of the whole structure is calculated.

To do this we consider a cylinder for which: at x =0wehavev = v0, G0:andatx = L we have GE. The general solution is most conveniently written L − x L − x v = A cosh + B sinh (36) λ λ at v(L)=vL so A = vL and the current flowing at L is IL, so after differentiating we get B = RλIL = vLGE/Gλ. The full solution is therefore L − x GE L − x v = vL cosh + sinh . (37) λ Gλ λ

To calculate the input conductance we need G0 = I0/v0 andsothecurrentatx =0mustbe calculated. This takes the form L GE L I0 = vLGλ sinh + cosh (38) λ Gλ λ and so L I0 GE + Gλ tanh λ G0 = = . (39) v0 GE L 1+ Gλ tanh λ

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This can be used to calculate the input conductance for the entire dendritic tree, and can be applied even when each dendrite has a different λ and Rλ. • Dynamics in Passive Structures Again we measure voltage from the resting potential EL. The cable equation is

2 ∂v 2 ∂ v τL = −v + λ . (40) ∂t ∂x2 It is convenient to introduce φ where v = φe−t/τL . On insertion into the cable equation we get

2 ∂φ 2 ∂ φ τL = λ (41) ∂t ∂x2 which is in the familiar form of the diffusion equation. The general solution can be found using a variable-separable method

2 2 τL ∂T λ ∂ X φ = T (t)X(x)sothat = = −k2 (42) T ∂t X ∂x2 where the form −k2 is chosen so that the modes will decay in time

2 −k t/τL ±ikx/λ Tk = e and Xk = e . (43)

The general solution can be written

2 −k t/τL ikx/λ −ikx/λ φ = e φk where φk = Ake + Bke (44) k where the sum over k could be an integral if the spectrum is continuous. We now consider the solution of this equation for a closed cable, to examine transients, and for an infinite cable, to examine the velocity of the signal in a passive structure. • Transients in a Closed Structure Consider a cable of length L that is closed on both ends and with some initial distribution of voltage at time t = 0. The closure requires that the gradient is zero at x =0sothatAk = Bk and at the other boundary, the zero gradient condition is λ 1=ei2Lk/λ so that k = πn . where n =0, 1, 2, ··· (45) L Hence the full general solution for this case is ∞ 2 − t 1+ πnλ πnx τ ( L ) v = an cos e L . (46) n=0 L where the an constants depend on the form of the initial voltage. The transient time constants are

τL τn = 2 . (47) πnλ 1+ L

The n = 0 mode decays slowest with a τL and with a spatial form that is a constant along the cable. Hence, it can be said that the transients τn with n>1 act to equilibriate the

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voltage throughout the neuron, whereas the zero-order mode with τ0 = τL governs the charge leaving the entire structure. Together with the second mode, the late-time voltage takes the form −t/τ −t/τ1 πx v(x, t)  a0e L + a1e cos . (48) L

The two time constants τL and τ1 can be measured experimentally and can then be used to yield an effective λ/L. • Signal Velocity in a Long Passive Dendrite The solution of the cable equation on an infinite dendrite, due to an initial brief injection of charge Q can be written − x2 QRλ exp 4λ2t/τ v(x, t)= e−t/τL L . (49) τL 4πt/τL

We now follow Dayan and Abbott and consider the time when the voltage at a distance x reaches its maximum value, before decaying to zero. This can be found by taking the derivative of (49) with respect to time, and setting this to zero. This yields the equation

2 − v − v vx 0= + 2 2 (50) τL 2t 4λ t /τ which results in a quadratic equation for t. Taking the positive root of this equation gives   2 t 1 − x  = 1+ 1+4 2 . (51) τL 4 λ

For x  λ the relation is t/τL  x/2λ whichcorrespondstoavelocity

dx 2λ  . (52) dt τL

Typically λ ∼ 100µmandτL = 20ms. This yields a velocity that is of the order of 1cm/s which is very slow (given the size of a human, for example). Moreover, the signal is very strongly attenuated, by virtue of the exponential decay in time. Hence, passive cables described by equation (40) cannot be used for signalling over long distances.

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