Electrodynamicx Fix

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Indigo5684
2025-09-30 13:19:42 -05:00
parent 99b9263649
commit 38a8ee0814

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@@ -524,25 +524,25 @@ $$V(z + \Delta z, t) - V(z, t) = -L' \Delta z \frac{dI(z, t)}{dt}$$
We can take the limit to see that
$$\frac{\partial V(z, t)}{\partial z} = -L' \frac{\partial I{z, t}}{\partial t}$$
$$\frac{\partial V(z, t)}{\partial z} = -L' \frac{\partial I(z, t)}{\partial t}$$
This gives is the two telegrapher's equations:
$$\begin{align}
\frac{\partial I(z, t)}{\partial z} &= -C' \frac{\partial V(z, t)}{\partial t} \\
\frac{\partial V(z, t)}{\partial z} &= -L' \frac{\partial I{z, t}}{\partial t}
\frac{\partial V(z, t)}{\partial z} &= -L' \frac{\partial I(z, t)}{\partial t}
\end{align}$$
We can then take the derivative of the first with respect to $t$ and the derivative of the second with respect to $z$, we see that
$$\begin{align}
\frac{\partial^2 I(z, t)}{\partial z^2} &= -C' \frac{\partial}{\partial z}(\frac{\partial V(z, t)}{\partial t}) \\
\frac{\partial}{\partial t}(\frac{\partial V(z, t)}{\partial z}) &= -L' \frac{\partial^2 I{z, t}}{\partial t^2}
\frac{\partial}{\partial t}(\frac{\partial V(z, t)}{\partial z}) &= -L' \frac{\partial^2 I(z, t)}{\partial t^2}
\end{align}$$
This tells us that
$$\frac{\partial^2 I(z, t)}{\partial z^2} = -C'(-L' \frac{\partial^2 I{z, t}}{\partial t^2}) = C'L' \frac{\partial^2 I{z, t}}{\partial t^2}$$
$$\frac{\partial^2 I(z, t)}{\partial z^2} = -C'(-L' \frac{\partial^2 I(z, t)}{\partial t^2}) = C'L' \frac{\partial^2 I(z, t)}{\partial t^2}$$
We can instead take the derivative of the first equation with respect to $z$ and the derivative of the second with respect to $t$ to see that