Energy reflected by a mismatched line termination can be entirely separated from the forward-travelling wave, and then be dissipated in a temperature calibrated resistor, and accurately measured as I^2R heat.
This is more or less correct, with a couple of caveats.
First, it is possible to mostly, but not entirely separate the reflected wave. This is done a directional coupler. Practical directional couplers have isolation error, which causes a small portion of the input signal to appear at the measurement port, in addition to the reflected signal that is intended to be measured.
Second, the measurement is not typically done by heating a resistive element. This can be done and it is called a bolometric power sensor. However it's more common in my experience to use an rf detector based on a diode. The nonlinear response of the diode converts some of the rf energy to a dc voltage, which is read with a voltmeter.
Bolometric sensors might be used for very high power conditions, or when calibration to a non-electrical standard is required (e.g. a thermometer).
Edit Replying to your comment, "the generator supplies only the actual power that is transmitted to the load."
This depends a lot on the details of the generator. You refer to a white paper that suggests the following scenario:
Suppose a lossless line is terminated by a pure open circuit, and suppose the the line is exactly one wavelength long at the operating frequency. In this case the current at the generator will be zero, and so the current in its internal impedance will be zero, so there is no power dissipated in it.
This is correct if the generator is actually a perfect voltage source with a 50-ohm series resistance. But an actual benchtop generator might contain other circuits like a levelling circuit or power monitor between the actual generator and the front panel port. Also you rarely know the actual line length to the load --- maybe there is some internal transmission line between the actual source and its front-panel port. If you don't know you have perfectly tuned the transmission line length, then the reflected power is the power you should be prepared to absorb at the generator, even if you don't have to absorb that much in every case.
Also, the case of an open circuit termination and half-wavelength line means that the generator sees an effective open-circuit load (that's why the current is 0). But not every type of generator is designed to work correctly with an open circuit load. A practical circuit could end up demanding more power from other elements within it, or generating more harmonic content when incorrectly terminated. This could still damage the generator even if the ideal components view of the circuit says there's no power transferred in the standing wave.
Finally, if you did insert a directional coupler into this scenario, you would transfer power through the coupled port and into whatever terminates that port (assuming it's not a perfect open or short). This means you would have "separated the forward and reverse waves" as suggested by the author you quoted, even though you did it in a system that was not transferring power before you inserted the directional coupler.
non-real characteristic impedance transmission lines exist? Are they typical?
Yes.
An ideal transmission line has R = 0 and G = 0. This gives a real characteristic impedance. But this is an idealization. These numbers are never actually 0 in reality, so the imaginary part of characteristic impedance is never truly 0.
What does it even mean to have a non-real characteristic impedance?
It means the transmission line is lossy. The sum of output power and reflected power will be (at least slightly) less than the input power.
but if hope to achieve a real Z0 this means carefully matching the non-ideal properties of both the conductor and the dielectric,
This isn't practical.
More often you just try to get R and G low enough that they can be ignored for practical purposes.
Minimizing G means choosing a dielectric material appropriate for the frequencies you're using.
Minimizing R means using a larger conductor (wider trace on a PCB or larger diameter coax). But this also tends to decrease the maximum frequency before the transmission line goes multi-mode. End result: at high frequencies, you just have to use small geometries and deal with the loss.
Due to skin effect, R tends to increase in proportion to \$\sqrt{f}\$, and this will typically dominate the frequency-dependent loss of a transmission line, if the dielectric has been chosen appropriately for the operating frequency band.
R and G don't directly correlate to physical quantities,
R is pretty directly tied to the conductivity of the conductor material. Skin effect also plays an important role. This is why you see silver-plated conductors on coax meant for high frequencies.
G relates to the polarization behavior of the dielectric, so it's harder to tie down to a specific physical mechanism. Generally if you're choosing dielectrics for high frequency designs, you'll see the loss specified as a loss tangent, but it will probably only be specified at a single frequency, which naturally won't be the one you're designing for.
Best Answer
Reflections happen everywhere, not just in transmission lines. Transmission line is a model of the physical situation, which is easy to apply to a pair of conductors whose length is comparable to or larger than the wavelength of the signal, and which is regular in cross section.
What determines whether reflections matter is the frequencies in and the physical size of the circuit. If you have unmatched impedances then you do get reflected waves just as you describe, and either you have to deal with them or they are negligible for some reason. Here are two reasons:
For exclusively low-frequency circuits, the reflections reflect repeatedly and settle down on a timescale much faster than the signals change. That is, each double reflection is an extra signal which is merely out of phase with the original signal, but as they get more out of phase their amplitude drops quickly enough that they can be neglected. (Even RF circuits can be built this way, as can be seen from a lot of homebuilt HF amateur radio gear.)
As frequency increases, wavelength decreases, and the physical size of your components becomes relatively larger, and you start having to worry about avoiding impedance “bumps”. This is where you start using microstrip design techniques in printed circuits.
In digital circuits, sharp transitions may have high-frequency components that will reflect but you don't have to worry about this as long as your clock speed is much slower than the length of your traces/wires (there's a conversion via c to make that make sense, of course) because by the time the clock makes its next tick all the signals have settled down to a steady state.
(Note that there are no standing waves here because within the period of a single clock tick the driving signals are steps (high to low or low to high logic levels), not periodic signals.)
As clock speed increases, the settling time available decreases, requiring you to either minimize reflections or minimize signal travel time (so that the settling occurs faster).