1
$\begingroup$

I recall hearing something about there being a mechanical way that two neurons can communicate with each other in addition to the chemical and electrical synapse methods. Something about a certain frequency or physical pulse that can travel along the connexons across the gap junction that is received by the post-synaptic terminal.

Did I make this up?

$\endgroup$
2
  • $\begingroup$ Is it possible you have in mind sensory neuron/epithelia of the skin or cochlea? $\endgroup$ – Bryan Krause Sep 23 '20 at 16:35
  • $\begingroup$ Nope, definitely thinking on the level of the synapse. $\endgroup$ – Otherness Sep 29 '20 at 18:43
1
$\begingroup$

Short answer
Action potentials travelling through the axon are accompanied by a mechanical displacement of the axonal membrane.

Background
Multiple studies have shown that a mechanical displacement of the axonal membrane accompanies an action potential. Hady & Machta (2015) used a computer model to show that these mechanical displacements are generated by the storage of energy in the neuronal membrane when the action potential travels through the axon that alter the compressive electrostatic forces across the membrane. These forces lead to co-propagating mechanical displacements accompanying the action potentials, which the authors have dubbed 'action waves' (Hady & Machta, 2015).

It is important to realize that the action potential drives the action wave and communication between neurons. As far as I understand, the action wave has no known physiological function in neural communication, although it is a point of debate (Fox, 2018).

Reference
- Fox, Sci Am (2018), April issue
- Hady & Machta, Nat Comm (2015); 6: 6697

$\endgroup$
1
  • 1
    $\begingroup$ Ah yes, this must have been what I was thinking of. So cool, thank you! $\endgroup$ – Otherness Sep 29 '20 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.