# How can TMS induce enough depolarization to cause an action potential?

Creating an action potential needs at least tens of mV potential difference across the membrane to occur. But the membrane is really thin, surely less then 0.01µm (some reference give 25A).

Then the potential gradient, that is to say the electric field E, must be very strong : The order of magnitude of the needed electric field is much greater than 10mV/0.01µm = $10^{6}$ V/m

But TMS produces E of the order of 200 V/m on the scalp, which is even less inside the brain. Then how can TMS induces neurons depolarization ?

• Do you have a reference for those TMS numbers? I found here that computed field is even less, on the order of 1 V/m – aaaaa says reinstate Monica Feb 28 '15 at 2:18
• cntrics.ucdavis.edu/c2meeting1/talks/Lisanby.pdf – agemO Feb 28 '15 at 10:38
• my guess is that TMS is used to depolarize myelinated axons (white matter). They are thicker (up to 20um) so that required field is on order of 100s $V/m$, not $10^6V/m$. And here TMS fields are noted at 500V/m level – aaaaa says reinstate Monica Feb 28 '15 at 11:13
• That would make sense – agemO Feb 28 '15 at 14:15
• And by "depolarize" I mean that you don't explicitly cause action potential to fire (like cortical electrodes do), but rather increase excitability of the region. Does that make sense? – aaaaa says reinstate Monica Feb 28 '15 at 21:25

## 1 Answer

I think my mistake was to believe that the electric field has to be somewhat perpendicular to the membrane, so that it can directly apply a voltage difference across the membrane, and that the resistance inside and outside the axon was the same, so that a tangential field was useless.

In fact these are false : depolarization and action potential creation happen where the axon is parallel to the electric field and where the gradient of the tangent field is the bigger.

Here you have a complete explanation but I think we can make a simplified one : http://stbb.nichd.nih.gov/pdf/Nerve_Fiber_Stimulation_Model.pdf

Let's look at a membrane model (Hodgkin-Huxley) : As we said if we try to apply an electric field $\vec{E}$ along $\vec{y}$ , it should be incredibly strong to produce a trans-membrane potential $V$ strong enough to create an action potential.

In between the inside an the outside of the membrane you have all the channels and the capacitance of the membrane. but I think the crux of the matter here are the resistances inside and outside : the small surface of a section of axon limits the conductivity, whereas the outside is supposed to be bigger so that its resistance is negligible.

As a result if you apply a gradient $\frac{dE_{x}}{dx}$ you can create trans-membrane current and potential: because of the gradient, the fields $E_{1} = E(x)$ and $E_{2} = E(x+\Delta x)$ are different : $E_{2} = E_{1} + \frac{dE_{x}}{dx}\Delta x$. Then using electricity laws : $$i_{m} = I_{1} - I_{2}$$ $$i_{1} = -\frac{ E_{1} \Delta x}{R}$$ $$i_{2} = -\frac{ E_{2} \Delta x}{R}$$

then :

$$i_{m} = \frac{\Delta x}{R} (E_{2}-E_{1}) = \frac{\Delta^{2} x}{R} \frac{dE_{x}}{dx}$$

This trans-membrane current will produce a potential across the membrane which will be proportional to $\frac{dE_{x}}{dx}$, electromagnetic stimulations shows that TMS can easily produce gradients strong enough to create actions potential.