I'm thinking of a little something where a hundred Sun-sized stars would dot the sky (night or day; the Sun just blocks other stars at day after all). These stars are scattered across the sky, but they are all exactly half a light year away from Earth.
So what would the sky look like in this scenario, aside from the fact that there will be new constellations? Will the nights be brighter? Will these stars be visible at day? Any interesting deviations from the typical sky?
Will the nights be brighter? Will these stars be visible at day? Any interesting deviations from the typical sky?
Nope, no big difference.
First of all, take a look at this chart showing how the Sun look like when seen from various bodies in our solar system
68 AU is still 0.1% of 1 light year, so every single star won't stand out that much.
Below another comparison of the apparent size of the Sun.
Just for reference, again from the image, even having 100 sun-like stars at the distance of Eris, you would still get 2% of the light coming from the Sun. And they are much farther and much more feeble.
To quote Zeiss Ikon
Eris is only about 12 light hours away. Your stars, at 1/2 a light year, are 365 times that far, meaning about 100,000 times dimmer than the sun seen from Eris
You would have a sky filled with 101 Venuses
The apparent magnitude of these stars would be just a bit brighter than Venus at its best, less than 10% brighter.
Of course, they would appear in the day, and twilight, and in the darkest night. The nighttime view would be quite spectacular, about similar to looking at the sky over JFK airport during a baggage worker's strike.
I'm surprised no one has given this a more direct quantitative treatment, so here goes.
The perceived brightness at the Earth's surface is related to illuminance, or luminous flux per unit area. Direct sunlight at noon is about 100,000 lux.
Illuminance from a source follows the inverse square law: it is inversely proportional to the square of the distance to the source. A star at the zenith at a distance of 0.5 light years (about 31,620 AU, so 31,620 times further away than the Sun) therefore causes an illuminance 1/316202 ≈ 1/1,000,000,000 = 10-9 times that of the Sun. We can therefore expect about 100,000 / 109 = 0.0001 lux from one such star. Illuminance from multiple sources is additive, so 100 such stars (again, at the zenith) would give us 0.01 lux.
The stars are stated to be randomly distributed around the Earth, so they won't all be at the zenith at the same time. Illuminance is proportional to the cosine of the angle between the surface normal and the line between the surface and the source. The "average" illuminance of a collection of randomly distributed sources around a circle or sphere, relative to the illuminance if they were all at the zenith, is 1/π (this is the integral of the positive part of cos(θ), from θ = -π/2 to π/2, divided by 2π). So the total luminance of 100 randomly distributed stars is about 0.01/π ≈ 0.003 lux.
For comparison, the full moon gives us between 0.05 and 0.1 lux, and starlight gives us 0.0001 lux. Note that human perception of brightness is logarithmic (proportional to the logarithm of the illuminance).
There would therefore be no perceptible difference in brightness during the day, but the difference at night would often be noticeable, especially during a new moon or when the moon is below the horizon.
As for whether the stars would be visible during the day, maybe, just barely: the brightness of a single star at this distance corresponds to an apparent magnitude of -26.74 (Sun) + 5 log100 1,000,000,000 = -4.24. This is comparable to the mean brightness of Venus (-4.14) which is right at the edge of daytime visibility.
