## Extracellular stimulation: Why dV/dx and not V as input

Anything that doesn't fit elsewhere.
calu
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### Extracellular stimulation: Why dV/dx and not V as input

Hello,

extracellular stimulation seems to have produced many questions in this forum and maybe this is an unneccessary one. But I can't make sense of the basic approach. I understand that the parameter set by segment._ref_e_extracellular is the field strength abs(E) = abs(dV/dx) (if the gradient is only non-zero in the x-direction). But in fact I would have assumed the actual potential to be the crucial parameter for stimulation, not the gradient. Because the changed membrane potential causes the actual spiking aktivity. (Cathode near nerve -> lower extracellular potential -> V_mem (= V_in - V_out) rises -> sodium channels fire.)

Why is the field strength the input? / Where is my mistake?

Thank you!
ted
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### Re: Extracellular stimulation: Why dV/dx and not V as input

calu wrote:I understand that the parameter set by segment._ref_e_extracellular is the field strength abs(E) = abs(dV/dx)
Where did you read that? That's not what the Programmer's Reference documentation of extracellular says.
calu
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### Re: Extracellular stimulation: Why dV/dx and not V as input

Oh... right. It is the voltage. Sorry, got that wrong. I am still unclear about dV/dx vs V in extracellular stimulation.

I do not understand (much like already discussed here viewtopic.php?f=15&t=1609) why a gradient will stimulate the cell, an arbitrary increase in extracellular potential will not. It seems to be the field strength only (as you pointed out here viewtopic.php?f=13&t=212&start=15) that is decisive. But why not the voltage itself? This is now more a question on biophysics than on the NEURON simulator. I'll probably simply read Rattay.
ted
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### Re: Extracellular stimulation: Why dV/dx and not V as input

calu wrote:I am still unclear about dV/dx vs V in extracellular stimulation.

I do not understand (much like already discussed here viewtopic.php?f=15&t=1609) why a gradient will stimulate the cell, an arbitrary increase in extracellular potential will not. It seems to be the field strength only . . . I'll probably simply read Rattay.

Suggest you instead perform this simple gedankenexperiment. Consider a circuit that consists of three resistors connected in series, but laid out like the upper case greek Pi. That is, two of the resistors are the vertical legs of the Pi, and the third resistor runs horizontally between their top ends. Attach the bottom ends of the vertical resistors to ground. There are no voltage or current sources in this circuit. What is the voltage drop across each of the vertical resistors? What is the difference between the voltages at the two ends of the horizontal resistor?

Now insert a 1 volt battery between the bottom end of each vertical resistor and ground, so that each battery's negative terminal is grounded and its positive terminal is attached to the bottom end of the corresponding vertical resistor. Now what is the voltage drop across each of the vertical resistors, and what is the difference between the voltages at the two ends of the horizontal resistor?

Finally, remove the battery on the right side of the Pi, and attach the bottom end of that vertical resistor to ground. Assuming that the three resistors have identical resistances, what is the voltage at the two ends of the horizontal resistor?
calu
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### Re: Extracellular stimulation: Why dV/dx and not V as input

Thank you very much for this answer.
calu wrote:Suggest you instead perform this simple gedankenexperiment. Consider a circuit that consists of three resistors connected in series, but laid out like the upper case greek Pi. That is, two of the resistors are the vertical legs of the Pi, and the third resistor runs horizontally between their top ends. Attach the bottom ends of the vertical resistors to ground. There are no voltage or current sources in this circuit. What is the voltage drop across each of the vertical resistors? What is the difference between the voltages at the two ends of the horizontal resistor?
Assuming there is no current or voltage source, the voltage drop is zero over any resistor.
calu wrote:Now insert a 1 volt battery between the bottom end of each vertical resistor and ground, so that each battery's negative terminal is grounded and its positive terminal is attached to the bottom end of the corresponding vertical resistor. Now what is the voltage drop across each of the vertical resistors, and what is the difference between the voltages at the two ends of the horizontal resistor?
True that there is still no voltage drop along the series of resistors. Ah, the potential of the intracellular medium will simply follow the extracellular with a difference equal to the membrane potential.
calu wrote:Finally, remove the battery on the right side of the Pi, and attach the bottom end of that vertical resistor to ground. Assuming that the three resistors have identical resistances, what is the voltage at the two ends of the horizontal resistor?
Now they are all at 1/3V. But I still think the voltage along the axon is not important for the stimulation, but the membrane-potential, right? Maybe what will cause the actual stimulation is the increased membrane voltage between the inner and outer medium along slightly diagonal paths like I drew in the picture underneath. When the gradient along the axon gets higher, the difference between phi e,2 and phi i,1 rises. ted
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### Re: Extracellular stimulation: Why dV/dx and not V as input

calu wrote:
Now insert a 1 volt battery between the bottom end of each vertical resistor and ground, so that each battery's negative terminal is grounded and its positive terminal is attached to the bottom end of the corresponding vertical resistor. Now what is the voltage drop across each of the vertical resistors, and what is the difference between the voltages at the two ends of the horizontal resistor?
True that there is still no voltage drop along the series of resistors. Ah, the potential of the intracellular medium will simply follow the extracellular with a difference equal to the membrane potential.
Yes, but what is the membrane potential?
Finally, remove the battery on the right side of the Pi, and attach the bottom end of that vertical resistor to ground. Assuming that the three resistors have identical resistances, what is the voltage at the two ends of the horizontal resistor?
Now they are all at 1/3V. But I still think the voltage along the axon is not important for the stimulation, but the membrane-potential, right?
Yes. Forget about diagonal paths. Just go ahead and calculate the membrane potentials (at any point in a cell it is the difference v_int - v_ext, which are the intra- and extracellular potentials, respectively; in this toy model, that would be v_topend - v_bottomend of a vertical resistor). Which leg of the Pi is depolarized (has a positive shift of membrane potential)--the one that is attached to the positive terminal of the battery, or the one that is grounded?

If you had to generalize, would you say that driving extracellular potential in a positive direction at some location A along a cell will depolarize or hyperpolarize the local membrane?

Up to this point, membrane capacitance has been ignored, and changing extracellular potential at a point affects membrane potential everywhere immediately. What happens if membrane capacitance is not ignored?
calu
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### Re: Extracellular stimulation: Why dV/dx and not V as input

Hi,

calu wrote:Which leg of the Pi is depolarized (has a positive shift of membrane potential)--the one that is attached to the positive terminal of the battery, or the one that is grounded?
The grounded one is depolarized. It's v.m = v.int - v.ext = 1/3V - 0V = 1/3V whereas for the one at 1V its v.m = v.int - v.ext = 2/3V - 1V = -2/3V.
calu wrote:If you had to generalize, would you say that driving extracellular potential in a positive direction at some location A along a cell will depolarize or hyperpolarize the local membrane?
If we have a chain of such pi-elements with a constant gradient of voltage along the chain, the influence on the 'batteries' between the lower contacts on the potentials of adjacent inner nodes compensate each other. Only the first and the last node will be influenced by the applied voltage gradient. In order to have an excitation of all segments, the voltage gradient needs to cange over distance. That is probably why the second derivative of the voltage is decisive for stimulation efficiency. (Rattay, right?)
calu wrote:Up to this point, membrane capacitance has been ignored, and changing extracellular potential at a point affects membrane potential everywhere immediately. What happens if membrane capacitance is not ignored?
The time course of the stimulation signal has an impact. Voltage over capacitors can't change abruptly. Considering a step from 0V to 1V of the 1V-node. At first, the membrane potential will not change, as the capacitors are not charged, then the same situation as described above will arise.
ted
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### Re: Extracellular stimulation: Why dV/dx and not V as input

Your answers were fine as long as you stuck to the narrowest focus of my questions. Where you drifted away from that focus, things started to go a bit wobbly. The activating function approach seems to have given you some false impressions.
calu wrote:
Which leg of the Pi is depolarized (has a positive shift of membrane potential)--the one that is attached to the positive terminal of the battery, or the one that is grounded?
The grounded one is depolarized. It's v.m = v.int - v.ext = 1/3V - 0V = 1/3V whereas for the one at 1V its v.m = v.int - v.ext = 2/3V - 1V = -2/3V
Yes, the grounded one is depolarized. Incidentally, the one that is driven by the extracelllar positivity is hyperpolarized by 1/3 volt, not by 2/3 volt.
calu wrote:
If you had to generalize, would you say that driving extracellular potential in a positive direction at some location A along a cell will depolarize or hyperpolarize the local membrane?
Extracellular positivity at location A will hyperpolarize membrane potential at location A.
If we have a chain of such pi-elements with a constant gradient of voltage along the chain, the influence on the 'batteries' between the lower contacts on the potentials of adjacent inner nodes compensate each other. Only the first and the last node will be influenced by the applied voltage gradient.
That's what the "activating function" approach predicts, but it's not what would happen in the physical system or in an analytic or numerical solution to the cable equation that describes that physical system. In the steady state, membrane potential will vary continuously with position. For a cylindrical cable in a field with a constant potential gradient, maximum polarization will be at the opposite ends of the cable. Membrane potential will be unchanged midway between the two ends. Derivative of membrane potential relative to distance along the cable will be nonzero everywhere, but steepest near the two ends (this is related to what has been called the "sealed end effect" in other contexts).
In order to have an excitation of all segments, the voltage gradient needs to cange over distance.
First, all segments will not be excited, ever. If an extracellular stimulus depolarizes some part of a neuron, there must be some other part of that neuron that is hyperpolarized by that same stimulus. Remember that positive charge can't accumulate in some part of a cell unless it entered through the membrane in some other part of the cell. And the latter membrane is necessarily hyperpolarized.

Second, the extracellular potential gradient does not have to vary in order for it to affect a cell. Not at all. That misconception comes to you thanks to the activating function approach. The activating function approach predicts that the extracellular potential gradient must have a nonzero second derivative, because one of its key assumptions is that the neurite is so long that effects at the ends can be ignored). But, as many have observed, it's not what you don't know that gets you into trouble--it's what you know for sure that ain't so.

"Well, if the activating function approach is all that bad, what use is it?"

It's most useful in situations in which there is a relatively localized disturbance of extracellular potential, especially when the spatial scale of that disturbance is small compared to the overall length of the neurite that is being stimulated. Good examples would be focal extracellular stimulation of a periperhal nerve many cm away from the origin or termination of that nerve (e.g. in the clinical setting of, say, performing nerve conduction velocity testing of the radial nerve at the wrist, or stimulation of the dorsal columns of the spinal cord, or maybe even deep brain stimulation of a pathway that is a few cm long by a bipolar electrode that is located within a mm or so of the axons). In such cases it can be quite useful.
calu wrote:
Up to this point, membrane capacitance has been ignored, and changing extracellular potential at a point affects membrane potential everywhere immediately. What happens if membrane capacitance is not ignored
The time course of the stimulation signal has an impact. Voltage over capacitors can't change abruptly.
True. The time derivative of membrane potential changes abruptly, but charge must accumulate on membrane capacitance, and that process requires the passage of time.