Timeline for How is the electrical information within a biological system captured using simple wires?
Current License: CC BY-SA 4.0
22 events
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| Nov 21, 2024 at 21:51 | comment | added | S.C. | Your final comment about "very very tiny excess of electrons" is what I needed to help with my internal model...because i think this implies that the charge density of each wire is effectively different (assuming one wire tip sees more Na+ ions than the other wire tip). | |
| Nov 21, 2024 at 21:43 | comment | added | S.C. | Awesome - thank you for all of the help and attention. You've given me enough to write a more precise question, which I will supply in a separate post in the upcoming days. Cheers~ | |
| Nov 21, 2024 at 21:41 | vote | accept | S.C. | ||
| Nov 21, 2024 at 21:21 | comment | added | Neil_UK | @S.C. tissue doesn't magically 'share a ground' with the equipment, unless you add one, a leg electrode for instance. Electrons get pulled off, and pushed into, the relevant connection wires, as Na+ and Na metal get converted, 'ground' is irrelevant. If ground is connected, then it's just another wire that also has these reactions going on at its interface. A very very tiny excess of electrons in one wire with respect to the other is what constitutes the voltage differenc ebetwen them. Creating this difference is called 'charging up', and is a very brief flow of current. | |
| Nov 21, 2024 at 20:29 | comment | added | S.C. | Also, you write, "...wires will charge up very quickly." What do you mean by 'charge up'? What particular electrical phenomenon do you intend on conveying with that phrasing? My inexperience with terminology has led to false conclusions previously haha. | |
| Nov 21, 2024 at 20:18 | comment | added | S.C. | I have read your responses; thank you for confirming several of my conceptualizations. I don't want to take much more of your time, but, just to be really certain: is it correct that the biological tissue shares a common ground with the electrical equipment on the opposite side of the wires? If true, then it would be correct to say that the electrons that 'jump off' of the tip of the wires to make non-ionic Na are being replaced by electrons being 'pulled' from ground on the equipment end, right? | |
| Nov 21, 2024 at 20:00 | comment | added | Neil_UK | @S.C. I was misunderstanding the depth of your misunderstanding, sorry for misleading you initially, I have updated my answer substantially. | |
| Nov 21, 2024 at 20:00 | history | edited | Neil_UK | CC BY-SA 4.0 | added 1111 characters in body |
| Nov 21, 2024 at 19:45 | history | edited | Neil_UK | CC BY-SA 4.0 | added 1111 characters in body |
| Nov 21, 2024 at 19:32 | comment | added | Neil_UK | @S.C. I was using confine laterally for the wires, not moving in and out of the insulation. At the tips, and the opamp, they of course move in and out of the ends of the wire to complete the circuit if there is current flowing. Such little charge needs to move to charge the wires to the voltage across the tip locations that we can ignore that flow. When a current flows at the contact with the aqeuous ions, there are reduction and oxidation reactions that change between Na+ and Na(metal), with the exchange of electrons to or from the wire. 'Jump off' (or on) is a reasonable shorthand for that. | |
| Nov 21, 2024 at 17:21 | comment | added | S.C. | Regarding your third comment, I agree that the illustrations may be confusing about what SPECIFICALLY my question is. I really just wanted to know whether or not this whole ensemble is actually a circuit where electrons flow (regardless of the op-amp). Your choice in the word 'confine' gave me one impression, but some of your subsequent comments gave me a different impression. Should I think of the biological tissue as a voltage-varying battery that has a common ground with the amplification equipment (or whatever other circuit components) that is present on the opposite side of the wire tip? | |
| Nov 21, 2024 at 17:14 | comment | added | S.C. | You stated previously that the electrons are confined to the wire, which I interpreted as, UNLIKE in the battery case, the electrons in this biology context will not 'jump off' of the wire tips and onto the cations (e.g. Na+). Sorry again, I'm sure communicating with a novice is frustrating hah. | |
| Nov 21, 2024 at 17:14 | comment | added | S.C. | Thank you for the insightful responses. Regarding your first (of the three most recent) response, let us tackle the non-ideal case where there is, indeed, current flow into the op-amp. In the event of direct current...what is 'replenishing' the electrons of the wire? In a circuit setup (with a battery) I understand that the cathode and anode are essentially enabling the transit of electrons through the wire (I apologize for the original post language of 'consume' and 'supply'). In the biology context, though...if electrons move into the op-amp...what is replacing the electrons of the wire? | |
| Nov 21, 2024 at 17:09 | comment | added | Neil_UK | @S.C. I've just looked at your second illustration. The electrodes will pick up the difference in the Na+ concentration between the two tips, not the overall concentration as your diagram appears to show (it's difficult to interpret what it's actually meant to show, but the total concentration appears to be the significant change between the top and bottom diagrams, not the difference between the two tips). As such, the distance between the tips will change the sensitivity to a concentration gradient of Na+. | |
| Nov 21, 2024 at 17:03 | comment | added | Neil_UK | @S.C. You might want to specify whether you are measuring the full DC component, or whether you can AC couple to measure only the changes, the former is more difficult and subject to long-term drifts. | |
| Nov 21, 2024 at 17:02 | comment | added | Neil_UK | @S.C. See my comment on that poster's post. With an op-amp as the load, it's very high impedance, so for an ideal op-amp, no, no current flows. Your electrodes need to have no current flow otherwise they will polarise. Polarisation is a change of voltage due to a current-induced change of ionic concentration, which will change the very thing you are trying to measure. If the op-amp is replaced with a low value resistor, then yes, a current flows. Current will flow into a non-ideal op-amp, just not a significant amount, you can easily find op-amps with nA or even pA of bias current specified. | |
| Nov 21, 2024 at 15:56 | comment | added | S.C. | Thank you for the follow up. The other poster's remarks have given me pause. Is current actually flowing through the wires (in any meaningful sense)? | |
| Nov 21, 2024 at 15:42 | comment | added | Neil_UK | @S.C. Yes, it's flat, to all intents and purposes. The wire maintains the potential along its length constant, just like a pipe full of water, lying flat, would do. If there's a current flowing, then the potential (pressure or voltage) along the pipe or wire would vary slightly from end to end. Electrons physics is such that their concentration does not change meaningfully even when current flowing. While electron concentration, electric field, voltage etc are all related, it's not intuitively quantitive, and it's not necessary. Circuit theory just needs volts and amps, job done! | |
| Nov 21, 2024 at 15:39 | history | edited | Neil_UK | CC BY-SA 4.0 | added 723 characters in body |
| Nov 21, 2024 at 15:38 | comment | added | S.C. | Thank you for updating your answer. If you are so inclined, could you please tell me what the electron distribution graph should look like? Is it just a uniform distribution? | |
| Nov 21, 2024 at 15:31 | history | edited | Neil_UK | CC BY-SA 4.0 | added 723 characters in body |
| Nov 21, 2024 at 15:27 | history | answered | Neil_UK | CC BY-SA 4.0 |