Why is t-distribution independent of true variance? Why is the t-distribution independent of the true variance of the underlying function? Intuitively I would not expect this to be true. Can anyone explain why this is true, using words rather than mathematical formulas?

Here is an example of exactly what I mean:
Consider 2 functions F and G.
F returns values which are normally distributed around mean 0 with variance 1.
G returns values which are normally distributed around mean 0 with variance 5.
Consider taking a sample of size 20 from F or G.
If we take multiple samples like this from F, the sampling distribution will be Student's t-distribution with 20-1 degrees of freedom.
If we take multiple samples like this from G, the sampling distribution will be Student's t-distribution with 20-1 degrees of freedom.
Intuitively I would expect to get a different distribution.
 A: The first thing to note is that the sample mean and sample variance are independent. The proof of this is a bit long to include here but can be found online pretty easily. The sample mean has distribution $$\bar{X} -\mu \sim N(0, \sigma^{2} /n)$$ and the sample variance has dist $$(n-1)S^{2} / \sigma^{2} \sim \chi^{2}_{n-1}$$
With abuse of notation
$$\frac{\bar{X}-\mu}{S / \sqrt{n}}= \frac{\sigma Z / \sqrt{n}}{\sigma \sqrt{\chi^{2}_{n-1}/(n-1)}/\sqrt{n}}=\frac{Z}{\sqrt{\chi^{2}_{n-1}/(n-1)}}=t_{n-1}$$
Here the Z is a standard normal, and is independent of the Chi squared term, and the final equality comes from the definition of a t random variable.
A: By definition (at least my definition), the Student's T distribution is the distribution of the variable
$$T=\sqrt n \frac {\overline{X_n} - \mu} {\sqrt{\frac 1 {n-1}\sum(X_i-\overline{X_n})^2}}$$
Where the $X_i$ are independent $N(\mu, \sigma^2)$ variables. The question is: why is this definition actually a definition? Why does the distribution of $T$ not depend on $\mu$ and $\sigma^2$?
A simple way to prove that the distribution is well-defined is to do the following algebraic manipulations. First, rewrite the sample variance by adding and removing $\mu$:
$$\sum(X_i-\overline{X_n})^2 = \sum((X_i-\mu)-(\overline{X_n}-\mu))^2$$
And now divide both parts of the fraction by $\sigma$:
$$T=\sqrt n \frac {(\overline{X_n} - \mu)/\sigma} {\sqrt{\frac 1 {n-1}\sum(\frac{{X_i-\mu}}\sigma-\frac{\overline{X_n}-\mu}\sigma)^2}}$$
The point is, we've gotten an expression of the form:
$$T=f\left(\frac{X_1-\mu}{\sigma}, ..., \frac{X_n-\mu}{\sigma}, \frac{\overline{X_n}-\mu}\sigma\right)$$
That is, we have expressed $T$ as a function of the variables $Y_i=\frac{X_i-\mu}{\sigma}$ and of $\frac{\overline{X_n}-\mu}\sigma$, which are all normally distributed variables. In fact, since $\frac{\overline{X_n}-\mu}\sigma$ can itself be expressed in terms of the $Y_i$, we can even write, for some complicated function $f$:
$$T=f(Y_1, ..., Y_n)$$
Since the joint distribution of the vector $(Y_1, ..., Y_n)$ doesn't depend on $\mu$ or $\sigma^2$ (it is $N(0, I_n)$), neither does the distribution of $T$.
A: Keep in mind that the statistic is divided by the sample standard deviation so in a way you "use" this value as a (roughly speaking) standardisation tool.
Thus, if you take 20 sample values from either f then both the nominator and the denominator will be lower (most probably) compared to if you do the same from g. This will make both statistics to follow the t-distribution.
