In his lectures on Physics, Feynman illustrates a mathematical 'trick' in formulating the impedance of an infinite ladder, LC network. It basically counts on the assumption that adding one more 'rung' in an infinite ladder is just a drop in the bucket - you still have the same infinite ladder. I researched this question in our Physics Stack Exchange, and came upon these two questions that basically deal with the same trick:
Impedance of the infinitely long circuit
How uniquely determined is the impedance of an infinite-chain circuit?
But to the point of my question given the expression for the infinite ladder how can you apply it to a real physical system? (Feynman does not elaborate further).
For example, what physical values might one use for $L$, and $C$ in the infinite ladder given they are modeling infinitesimal elements? We first say the ladder is of infinite length, but in the practical world its length is finite, and the best you might have available are specs like Ohms/meter, Henries/meter, etc.
Answer
Using the "drop-in-a-bucket" trick, we find that an $LC$ laddar has impedance
\begin{align} Z &= Z_L + Z_C||Z \\ &= Z_L + \frac{Z_C Z}{Z + Z_C} \\ Z^2 - Z Z_L - Z_L Z_C &= 0 \\ Z &=\frac{1}{2} \left( Z_L \pm \sqrt{Z_L^2 + 4 Z_L Z_C} \right) \, . \end{align}
The impedance of an inductance $L$ is $Z_L = i \omega L$ and the impedance of a capacitor $C$ is $Z_C = 1 / i \omega C$. Plugging that in we get $$Z = \frac{1}{2} \left( i \omega L \pm \sqrt{-\omega^2 L^2 + 4 \frac{L}{C}} \right) \, .$$
Now suppose that each rung on the ladder has an inductance per length of $\mathcal{L}$ and a capacitance per length of $\mathcal{C}$. Suppose that each section of transmission line has length $\delta x$. Then we have $L = \mathcal{L}\delta x$ and $C = \mathcal{C} \delta x$, so
$$Z = \frac{1}{2} \left( i \omega \mathcal{L}\delta x \pm \sqrt{-(\omega \mathcal{L}\delta x)^2 + 4 \frac{\mathcal{L}}{\mathcal{C}}} \right) \, .$$
Now at this point you might hear people tell you to consider some sort of continuum limit by sending $\delta x$ to zero. While that does recover the usual result that $Z = \sqrt{\mathcal{L}/\mathcal{C}}$, it is totally bogus. You should never consider a limit as a quantity with physical dimensions goes to zero. Instead, we must identify a dimensionless quantity, i.e. a ratio of physical parameters, which tends to zero. Only then will we understand the physical limit in which the approximation applies.
Consider the limit in which the second term under the square root is much larger than the first:
\begin{align} \frac{\mathcal{L}}{\mathcal{C}} &\gg (\omega \mathcal{L} \, \delta x)^2 \\ 1 &\gg \omega^2 \mathcal{LC} \, \delta x^2 \\ 1 &\gg \left( \frac{\omega \, \delta x}{v} \right)^2 \end{align}
where $v \equiv 1 / \mathcal{LC}$ is a characteristic wave speed for the ladder. The usual relation between frequency $\omega$, wave speed $v$, and wave length $\lambda$,
$$v = \frac{\lambda \omega}{2 \pi}$$
brings us to
$$ 1 \gg \frac{\delta x}{\lambda} \, .$$
This equation says that the continuum approximation works when the wave length of the electrical signals put onto the ladder is much larger than the size of each rung.
This is precisely the case in real life. Ladder transmission line structures have a cutoff frequency above which their wavelength is on the same size scale as the ladder rungs. Above this frequency, the ladder doesn't act like a transmission line (they tend to become either open or short circuits).
This answer explains a few of the questions in the original post:
what physical values might one use for L, and C in the infinite ladder given they are modeling infitesimal elements?
You use the actual inductance and capacitance per length of the physical ladder. If each rung has length $l$ and inductance $L$ then $\mathcal{L} = L/l$.
We first say the ladder is of infinite length, but in the practical world its length is finite, and the best you might have available are specs like Ohms/meter, Henries/meter, etc.
You can certainly build a ladder out of discrete parts, and then figure out the inductance and capacitance per length form the physical extent of those parts. Again, this only yields the usual continuum limit expression if the wavelength is large enough (frequency is low enough).
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