# Hebbian learning / STDP -- what happens if one of the neurons do not fire?

I have read from the textbook that if neuron A is connected to neuron B with synapse C, then:

• if neuron A fires before neuron B fires, synapse C increases in strength roughly proportional to $$f(t) = 1/t$$ where $$t$$ is the time gap between the firing in the order of milliseconds
• if neuron A fires after neuron B fires, synapse C decreases in strength roughly proportional to $$f(t) = 1/t$$

I am trying to map out all the scenarios, so I have a few questions:

• what happens to synapse C when neuron A fires, but neuron B does not fire? As in, the signal from neuron A is not strong enough to activate the desired neural assembly? Is there a mathematical model that approximately describes the strength of C?
• what happens to synapse C when neuron A does not fire, but neuron B does fire? As in, the signal from neuron A is not relevant to the activated neural assembly? Is there a mathematical model that approximately describes the strength of C?

I have looked up various articles, one text book, and various academic papers on synaptic plasticity by searching for "Hebbian Learning" and "Spide-Timing Dependent Plasticity". For example, this chapter on Synaptic Plasticity by Gerstner and co: neuronaldynamics.epfl.ch/online/Ch19.S1.html. Where I got stuck is that they only mention cases where both neurons fire, but I dont recall any of these text mentioning what happens when only one of the two neurons fire or when neither fire.

• Stack Exchange requires questions to be specific, you shouldn't ask a bunch of different questions here at once, so I've simplified yours to just one question (well, two, but they're actually the same). I think your other questions can also be answered by looking at the same figure in my answer, though. Commented Jul 9 at 14:20

There's really no such thing as a neuron that does not fire.

All neurons fire at some rate, and there are homeostatic mechanisms to prevent neurons from never firing, just as there are homeostatic mechanisms to prevent neurons from firing too much: a neuron that never fires isn't useful.

So, knowing that both cells are going to fire at some point, we just need to talk about the relative timing of their firing. The closest scenario to the one you describe where one cell or the other does not fire would be that there is a big time difference (either before or after) in the spikes.

"Spike-Timing Dependent Plasticity" is indeed the best way to think about this, let's go find a figure from a review paper, here's one from Andrade-Talavera, Y., Fisahn, A., & Rodríguez-Moreno, A. (2023). Timing to be precise? An overview of spike timing-dependent plasticity, brain rhythmicity, and glial cells interplay within neuronal circuits. Molecular Psychiatry, 28(6), 2177-2188.:

Look at the x-axis: we're interested in long times, either positive or negative. What do the LTP and LTD curves do as we get to long times? They go towards zero. So, at least from this figure, we would say that there's really no plasticity (by the mechanisms depicted in the figure) if spike times are about 50 ms or more apart.

• This post answered my questions perfectly, thanks! Commented Jul 9 at 18:41
• Small follow-up question: "All neurons fire at some rate, and there are homeostatic mechanisms to prevent neurons from never firing" <-- would this explain why some neurons randomly fire and the spike timing charts seem stochastic rather than deterministic? These "random fires" should mostly be meaningless right? i.e. it shouldn't convey any real information to the brain? Commented Jul 9 at 18:45
• @user3667125 That might be a partial explanation at some level of understanding for some cells, but generally no, I wouldn't take that by itself as an explanation of "noise" in the brain. Commented Jul 9 at 19:07