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Time-integral of a voltage function?

  1. Nov 10, 2007 #1
    Hi, this is my first post. I've not taken a calculus-based physics course yet so these may be rather naive questions. I'm looking at field potential recordings in rat hippocampi. My aim is to characterize the large-amplitude spontaneous waves seen in the data. I was hoping someone with a stronger physics background would explain 1) the meaning of the time integral of a voltage function and 2) the meaning of the second derivative of a voltage function.
    As to the integral, it seems to me that this would be a measurement of ion flux across the membranes of the population of neurons in vicinity of the recording electrode. Also, I'm pretty sure that the first derivative of a voltage function is current but is there any meaning to its second derivative. Thank you for your help,

  2. jcsd
  3. Nov 11, 2007 #2


    Staff: Mentor

    Are the voltage measurements transmembrane potentials or are they more like EEG lead voltages? I took a 1 semester neurophysiology course a long time ago, but I only remember the transmembrane stuff, not the more macroscopic stuff. And even the transmembrane stuff is a little hazy.

    If you can assume a constant resistance then the voltage is proportional to current and the time integral of current is charge. The resistance of a membrane is definitely not constant, when the membrane ion channels are closed the resistance is very high but when they are open it is relatively low. But maybe it is a reasonable approximation for EEG leads.

    Similarly, if you have a constant capacitance then the first derivative of your voltage is proportional to the current. I think that constant capacitance is probably a better approximation than constant resistance, but I am not certain.

    I don't know of any significance to the second time derivative of the voltage. The frequencies involved are low enough that I think it is safe to neglect the magnetic fields and EM radiation generated by these voltages. In any case, it would be a really noisy value, so probably of limited usefulness even if there were some great physical significance.
  4. Nov 11, 2007 #3
    Thanks for replying Dale.

    We record local field potentials in brain slices; the electrodynamic principles are probably similar to EEG recordings but I don't know for sure. The waves I'm interested in reflect the synchronous discharging of a whole population of neurons. That they're negative-going reflects the increased Na+ influx. This means that membrane conductance, resistance as well as capacitance are variable.

    Other groups have reported using an estimate of the second derivative as a means of identifying "significant" waves in the data. Maybe the only significance is that it's a measure of the magnitude of the slope at a given time?

    As to the time integral of current measuring charge, that's makes sense since current is a rate but resistance changes in our case. The electrochemical gradient across the membrane is the potential energy that drives ion flux. As you said, when ion channels open it's like discharging a capacitor and ions flow down their potential energy gradients (changing direction at their reversal potentials). Is the time integral of the wave then equivalent to amount of energy dissipated in the duration of the wave? I'm confused!
  5. Nov 12, 2007 #4


    Staff: Mentor

    Since resistance and capacitance are time varying I do not think you can read anything physical into the integrals or derivatives of the voltage. If you really need a physical interpretation then you will need to measure current as well.
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