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I read in Wikipedia how the cable equation was derived (here) and had a specific problem regarding one of its equation: At the start of the derivation it states that we first need to pretend that the membrane is perfectly sealed (Rm=∞) and that the capacitance of the membrane equal zero. Then we can say that the change in voltage over distance depends only on the resistance of the axoplasm, by the equation: and from that, the current equal:

But later when they derive the last equation, , they set as in the equation above, which seems to me incorrect, since the resistance of the membrane also affect . What did I get wrong?

Thanks!

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The part you are quoting is about helping you conceptualize the cable equation by starting from some approximations that let you remove most of the terms. See the sentence beginning (emphasis mine):

To better understand how the cable equation is derived, first simplify the theoretical neuron even further and pretend...

They then walk you through how, because these approximations are just pretend, one needs to include terms to account for the actual permeability and capacitance of the membrane. This means that you do not have a constant internal current with distance as you travel along; instead, the current decays with distance because of leak through the membrane and membrane capacitance.

This is what the last equation you quote from says: the change in internal current over distance x (written as dil/dx) is equal to the current lost to the membrane, which includes two parts: leak through the membrane (V/rm) and charging the membrane capacitor (cm*dV/dt).

Another correction that might be tripping you up: you write

change in voltage over time

in your question, but that is not what the equations are representing, they are representing change in voltage over distance x.

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  • $\begingroup$ Thanks for the answer. I also corrected another mistake I wrote (I mean I(i) instead of I(m) at the end). But I didn't understand yet how they can set I(i) (as was derive at the beginning) to the I(i) at the last equation, because when they set I(i) in that equation (the second-order partial differential equation) its the equation which don't take into account the R(m). $\endgroup$
    – Idop11
    Apr 10, 2020 at 16:45
  • $\begingroup$ @Idop11 Did you miss equations (7) and (8) from the Wiki page? Note that the definitions are changing throughout, this is not a derivation per se, where you can connect different steps algebraically. Don't confuse symbols that are based on certain assumptions with the same symbols once those assumptions are changed. $\endgroup$
    – Bryan Krause ♦
    Apr 10, 2020 at 16:58

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