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I am aware that there are a lot of answers explaining the fundamentals of a tidally locked "Earth-like" moon around a giant gas planet. For example this one:
Habitable moon of a gas giant: working out the sizes and distances

However, my question is related to a very particular side-effect of the tidal-locking: Imagine this guy standing in the side of the Earth-like moon that is always facing to the giant gas planet, as shown in the picture (I show there 4 positions of the moon around the planet). The drawing is obviously not to scale. It is just to explain the idea graphically:

enter image description here

That face of the moon where the little guy is standing (as you can notice) will almost always be in the twilight and will never have the sun directly perpendicular to the surface. So, my question is:
Is it valid to assume that there will be an ice line (or at least a very very cold zone forever in Winter) in the meridian directly pointing to the planet?
I mean: something like this picture:
enter image description here

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  • $\begingroup$ Wouldn't the shaded area be more of a circle around your guy? I'm not sure how you get a line. $\endgroup$ – userLTK Aug 15 '18 at 4:57
  • $\begingroup$ Perhaps it is difficult to imagine without a simulation (I hope I will be able to upload one later). Consider, since this is a top view of the system, that the light patterns will actually be the same on each meridian of the moon (considering no axial tilt and cero excentricity). So the meridian with less amount of light in all the moon, will be precisely the meridian where this little guy is standing. $\endgroup$ – Carlos Zamora Aug 15 '18 at 17:54
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    $\begingroup$ Altitude also helps the formation of permanent ice. A few well positioned mountain ranges and your setup might become more likely. $\endgroup$ – userLTK Aug 15 '18 at 18:08
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Now I am not really an expert and perhaps you should wait until someone more experienced than me answers your question.

Judging by what you are trying to do seems to point out to the fact that this moon will always pass into the shadow of the gas giant. And thus the near side of this moon seems to have a region bathed in twilight and/or night. But this isn't really true.

Remember this moon orbits around the planet, thus between each 4 states that you showed in the diagram (which I assume is not in scale), there are several states that will eventually bring the little man into sunlight (which you said in your question by "almost always be in the twilight").

enter image description here

The picture above is a more "detailed" version of your diagram (not to scale). The arrows show the location of this "man" and the direction to the planet to show where the near side of the moon is (we can assume the man is at the base of the arrow).

If we consider the moon on the very right as number 1, we can then proceed "prograde" (counter-clockwise) to number all 8 of them. Notice how in number 4 and 6 the near side does catch some sunlight before and after entering/exiting the planet's shadow. Thus that side will receive sunlight. As far as I know those are not conditions to create a line of ice stretching from pole to pole, because during those times the near side will receive heat from the star which will heat the surface like anywhere else, preventing an "eternal winter" state.

In order to do that the little guy must always be in a twilight zone and not "almost always", or else the area will receive heat, even if the time the near side spends in daylight is shorter than when the far side for instance is in daylight (as the near side's daylight is then blocked by the shadow as the planet eclipses the star).

Perhaps what could happen is mild seasonal change.

Also this scenario happens mainly on moons that orbit over the equator and are relatively close to the planet so I assume your moon also follows this model.

Your scenario looks similar to the "equatorial ice cap" scenario, where a planet has ice all over the equator and polar jungles. This has been seen in this question which might help you: Equatorial icecaps and polar jungles a fantasy or reality?

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    $\begingroup$ Our polar caps receive daylight half the year and are still permanently frozen. Your premise of heat = no permafrost ist simply not true. However, on the side facing away from the planet, which receives normal sunlight, there would probably be no endless winter. The other side might be covered in ice, but I doubt it'd go all the way down to the equator; and it would also resemble more of an eye shape rather than this solid line. $\endgroup$ – Otto Abnormalverbraucher Aug 15 '18 at 14:20
  • $\begingroup$ The poles receive light for only half a year. This moon however likely receives a lot more light than Earth's poles, receiving light in every rotation. It would be better if we knew its orbital parameters in that way we can see precisely for how long it receives light $\endgroup$ – Victorbrine Cassini Aug 15 '18 at 14:26
  • $\begingroup$ This moon's planet facing pole receives light somewhere between a third and a quarter of one rotation given the premise it always passes through the planet's shadow. As the numbers don't matter at all, let's just assume this moon takes 18 days to rotate around its planet and the planet takes 360 days around its star: The moon receives 6 days of sunlight per rotation around the planet and thus 120 days of sunlight compared to the 180 days of sunlight Earth's poles receive. It's much less, not more. $\endgroup$ – Otto Abnormalverbraucher Aug 15 '18 at 14:38
  • $\begingroup$ Of course, depending on the size of the planet and the distance of the moon to the planet, those numbers vary, but it's still always less than half a year due to passing the shadow. Half the rotation is in darkness while the other half rotation is in light except for the time it passes the shadow. There are other factors to consider like atmosphere trapping heat but the math considering the light is simple (considering approximation - times of light vary throughout the year due to the usual elliptic orbit both of the planet and the moon but it's negligible). $\endgroup$ – Otto Abnormalverbraucher Aug 15 '18 at 14:45
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    $\begingroup$ Remember, this moon is tidally locked. Otherwise you'd probably be right, if the moon would rotate around itself, this whole calculation wouldn't work. The picture you added even makes it clear: you have 4 positions of darkness, 2 positions of light and 2 positions of twilight ;) $\endgroup$ – Otto Abnormalverbraucher Aug 15 '18 at 14:51

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