Exactly. At one end of the scale, you're not travelling through space (or practically-zero on a relativistic scale), and you're experiencing the full affects of time.
At the other end of the scale, you're travelling through space (or practically-C on a relativistic scale), and you're not experiencing time.
So the whole theory of zipping around space and coming home to find you've barely aged, is just spending more time at a higher "angle" than everyone else.
It is instantaneous from the perspective of the object’s internal clocks.
Traveling at the speed of light results in infinite time dilation. Which means that from the perspective of an outside observer, no time at all is passing inside the spaceship.
What happens to fields (magnetic, gravitational, electrostatic, etc) in that situation? E.g. can a photon 'feel' the surfaces it would ultimately interact with?
One way I like to think of this is the term 'sun-kissed'. From the perspective of the photon, the sun is actually giving you a kiss on a summer day.
'feel' the surfaces it would ultimately interact with
I would like to see the map of the universe at different potential speeds (or thrusts). E.g. you choose a point in space nearby the sun. At thrust zero you only see hot sun everywhere, because there you go anyway. But at greater thrusts the sun turns into a circle and you start to see sections of the “sky” where you could land, given the thrust is constant. Some areas would be still black because of blackholes, orbits and event horizon. I always wanted that simulation but never found it. It would be much more interesting than just looking around via reversed photons flying into your eyes.
That depends on how you define 'conservation of energy'. Consider an arbitrary bounded volume of spacetime. Conservation of energy says that the net flow across the boundary of any such volume is 0. Under this definition moving in space but not time is not a violation, as the same amount of mass enters the volume as exits it. The only odity is that both events happen at the same time
The idea is that the volume we are talking about is a 4 dimensional volume of space-time and is bounded; not a three dimensional volume of space that extends to infinity along time.
Conservation of energy says that it is impossible for energy to enter this volume with that same amount of energy exiting the volume.
Consider what it would mean for this to be violated. For the sake of argument, assume that all particles must move forward in time by a non zero amount at all points along there path. Since the volume is bounded, any particle with an infinite path must eventually have a time coordinate beyond the largest time coordinated in the volume. Therefore, the particle must eventually exit the volume. If you were to work out the geometry more carefully, you could show with relative ease that the particle must exit the volume an equal number of times as it enters. If a particle were to enter the volume without exiting the volume, it would mean that said particle was destroyed within the volume. Similarly, if a particle were to exit the volume without entering, it would have to have been created within the volume. Both of these situations are possible if an interaction occurs within the volume, but the net energy of the particles leaving such an interaction, must be the same as the net energy of the particles entering the interaction.
Put another way, assume that all interactions obey the conservation of energy. If our original volume was V, we can construct a new volume V' from V by carving out sub volumes in which an interaction occurs. Since all such sub volumes obey the conservation of energy (by assumption), the net energy flow into and out of V' must be the same as for V. However, since no interactions occur withing V', all particles entering V' must exit V' an equal number of times, so the net energy flow of V' must be 0. Therefore the net flow of V must also be 0.
That proper time is zero everywhere along a photon's geodesic does not mean that the photon cannot evolve from point to point along it, and we can show this by taking advantage of total coordinate freedom.
We can parametrize (as in make parametric) arbitrary curves through spacetime however we like. Parametric representations of unique curves are generally nonunique.
Some of the infinite possible parametrizations of a chosen curve have useful properties, such as uniquely labelling every point on the curve with some monotonically ordering value and keeping the form of some set of equations reasonably simple.
For timelike geodesics, particularly in the Minkowski space of Special Relativity, proper time (being a Lorentz scalar) is a good option. However that is not true for all geodesics in Minkowski space (as you note, the proper time is everywhere zero on a null geodesic, and so a bad option), much less all curves through general curved spacetimes.
As is noted below the comment directly pointed to by the second link, labelling a timelike geodesic with proper time is choosing one specific affine parametrization on that geodesic, and that this choice is driven by convenience.
One of the neat outcomes of affine parametrization is that we can take a point on an affinely-parameterized null geodesic and look at the derivative with respect to the affine parameter there, and define a momentum k^{\mu} = \dot X^{\mu}. In a Lorentzian spacetime, with curvature, we can compare the momentum at two different points on the null geodesic, giving us the gravitational redshift between those two points of the photon's wavelength equiv. frequency.
