Myelin increases resistance across the cell membrane

PF Gold
P: 716
 Quote by atyy Yes, it is. But Prof C Koch is a nice guy, so he won't hit me, although he might punish me by discussing consciousness.
That's fine but there is another problem!
If the capacitance is effectively 50 time less with the wrapping of myelin, you forgot a crucial parameter in your computation: length => area.
if a node is 2.2 pF for 0.5 µm then an internode (1 mm) is (2.2/50)*2000= 88 pF.
Did I said Divergence?
Mentor
P: 17,329
 Quote by Moonbear I advised caution in comparing the concept of DEmyelinated to NONmyelinated, because I'm not sure that all else is equal if you remove the myelin, either via pathological or experimental mechanisms.
I agree.
PF Gold
P: 716
 Quote by DaleSpam Thanks for the info. They do appear to propagate. In number 5 they do mention varying degrees of demyleination, so I am not sure if it still propagates with complete demyleination, but there is at least some "extra" mylein than is strictly needed.
I agree.
http://www.physicsforums.com/showpos...8&postcount=19
Mentor
P: 17,329
It doesn't really matter as long as you are consistent. If you measure resistance in ohms*cm² and capacitance in farads/cm² then you have the same time constant as if you measure resistance in ohms and capacitance in farads. You just have to be consistent.
 PF Gold P: 716 I have some difficulty with your reply: A cylindrical resistance is proportional to volume => proportional to length. http://en.wikipedia.org/wiki/Electrical_resistance A plane capacitor is proportional to area => proportional to length. http://en.wikipedia.org/wiki/Capacitor#Capacitance See computations => http://butler.cc.tut.fi/~malmivuo/bem/bembook/21/21.htm I'm consistent. Where did you find a unit of ohm*cm²?
Mentor
P: 17,329
 Quote by somasimple A cylindrical resistance is proportional to volume
No, resistance is proportional to length/area. Volume is length*area. So, resistance is not proportional to volume.

 Quote by somasimple Where did you find a unit of ohm*cm²?
The "area specific resistance" is in ohm*cm². It is the appropriate "normalized" resistance for current through a membrane. To the best of my knowledge it is used primarily for characterizing fuel cell membranes and neuron membranes.

Note Gm listed in your last reference (http://butler.cc.tut.fi/~malmivuo/bem/bembook/21/21.htm). Conductance/unit area is simply the inverse of area specific resistance and is, IMO, more convenient.
P: 8,657
 Quote by somasimple That's fine but there is another problem! If the capacitance is effectively 50 time less with the wrapping of myelin, you forgot a crucial parameter in your computation: length => area. if a node is 2.2 pF for 0.5 µm then an internode (1 mm) is (2.2/50)*2000= 88 pF. Did I said Divergence?
Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.
PF Gold
P: 716
 Quote by atyy Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.
Atyy,
I bought the book of Koch. I'll see.
Edit:
I'm unable to draw the equivalent circuit since the capacitor must be oriented to the external milieu but connected to the internal one. The resistance must be longitudinal.
It gives a worst solution than before.
PF Gold
P: 716
 Quote by DaleSpam No, resistance is proportional to length/area. Volume is length*area. So, resistance is not proportional to volume.
You're right. So resistance is still proportional to length. So is Capacitance
PF Gold
P: 716
 Quote by atyy Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.
See table 2
http://www.pubmedcentral.nih.gov/pic...8&blobtype=pdf
http://www.ncbi.nlm.nih.gov/pubmed/1...ubmed_RVDocSum
The time constant is already 10 fold too high.
Mentor
P: 17,329
 Quote by somasimple You're right. So resistance is still proportional to length. So is Capacitance
Yes for resistance.

Capacitance is proportional to area/distance. So capacitance is inversely proportional to length if you wish to say it that way. (Although "distance" is a better description of the separation between plates than "length" since "length" connotes the largest dimension of an object and the separation between the plates is the smallest dimension)
 PF Gold P: 716 we are speaking about the area of the plates, here. the distance between the plates, d was already implied. two plates of an area of A and separated by a distance d. C is proportional to A and inversely proportional to d since C =e*A/d Thus if A is augmented (internode), even if d is augmented, C is augmented since a node has a length of L1= 0.5 µ => area = 2*pi*R*L1 => 2.2 pf the same plates where d is *50 => C/50 an internode is 2000 time longer 2*pi*R*L1*2000 => (2.2/50)*2000 = 88 pf and the computation is simplified since the perimeter augments with each wrap => C>88pf
Mentor
P: 17,329
 Quote by somasimple C is proportional to A and inversely proportional to d since C =e*A/d
Yes.

 Quote by somasimple Thus if A is augmented (internode), even if d is augmented, C is augmented
That depends entirely on which is augmented more. If they are both doubled then C is unchanged. If A is doubled and d is tripled then C is reduced to 2/3 of its original value. On the other hand if A is tripled and d is doubled then C is augmented to 3/2 of its original value.
 PF Gold P: 716 DaleSpam, Give the results in our example. 50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A).
Mentor
P: 17,329
 Quote by somasimple 50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A).
This is not correct. The presence or absence of the mylein doesn't change the length at all. It may slightly change A by a small increase in the circumference.
PF Gold
P: 716
 Quote by DaleSpam This is not correct. t may slightly change A by a small increase in the circumference.
Thats is not a result at all!

 Quote by DaleSpam The presence or absence of the mylein doesn't change the length at all.
Where did I said the length was modified?
Mentor
P: 17,329
 Quote by somasimple Thats is not a result at all! Please give us your result (and computation)?
What result and computation are you talking about?

 Quote by somasimple Where did I said the length was modified?
Your previous post where you said:
 Quote by somasimple 50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A).

It is very difficult for me to communicate with you. I know that a large part of that is a language barrier, so I am trying to be patient.
 P: 2,258

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