From the moment you short a battery, how long before the current BEGINS to flow. Does a virtual photon need to travel the path of the wire? I was thinking it would be the length of the wire divided by the speed of light. Or maybe it has to do with the distance of the battery to the spot where it was shorted like if there was a switch on the wire that shorts the battery. I have literally no idea tho
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Ignoring resistance, the Telegrapher's equations help to understand this question.
The voltage along the circuit should behave as
$\frac{d^{2}V}{dx^{2}} = u^{2}\frac{d^2V}{dt^{2}}$.
This is a wave equation with speed $u = \frac{1}{\sqrt{LC}}$ with inductance-per-length L and capacitance-per-length C. This speed is the speed of light "for transmission lines made of parallel perfect conductors with vacuum between them."
When shorting the circuit, a step difference in the potential will excite waves according to this wave equation, traveling at speed $u$. So current will already be flowing in parts of the wire, due to the voltage differences that the battery has already established on both sides of the short. These waves will reach the battery after time:
$t = l/u$
where $l$ is the closest distance from the short to the battery.
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$\begingroup$ Your equation for u is not a speed, it is a frequency in natural units? $\endgroup$ Nov 19, 2021 at 5:51
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1$\begingroup$ Good catch jamie, both values are per-length which makes it a speed. I edited my answer accordingly. $\endgroup$– AlwinNov 19, 2021 at 6:02